Surveying Computation SURE 215
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LINECIRCLE INTERSECTION Surveying Engineering Department Ferris State University INTRODUCTION There are three relationships within coordinate geometry that can exist between a line and a circle They can intersect at one or two points or they will not intersect Ifthere is only one intersection then the line must be tangent to the circle at that point of intersection To nd where the intersection exists one can use either the solution of two equations or by the triangle method To solve the problem in a simple mode one can solve for the equation of a line and the equation of the circle simultaneously Since the equation of a circle is a seconddegree equation the solution is through the quadratic equation The development of the solution by simultaneous equations begins by writing the equation of the line and circle in the following forms Hashimi 1988 y y1 mx x1 gtymx x1y1 R2 x h2 y k2 where x and y are the coordinates of the point of intersection x1 and y1 are the coordinates of a point on the line h and k are the translation between the center of the circle and the origin of the coordinate system in the x and y directions respectively m is the slope ofthe line As a reminder the slope is the cotangent of the azimuth of the line Write the equation of the line and expand the equation by taking the squares and equating the equation to zero x h2 y k2 R2 0 1 x2 2hxh2 y2 2kyk2 R2 0 SURE 215 Surveying Calculations LineCircle Intersection Page 67 Substitute into this equation the relationship developed for the ycoordinate above yielding X2 2hXh2 mX X1y12 2kmX X1y1k2 R2 0 X2 2hX h2 m2 X2 2m2 XX1 2mxy1 m2 K 2ley1 y2 2 2ka2ka1 2ky1 k2 R2 0 Recall that the quadratic equation has the following form aX2 bX c 0 3 Substitute 2 into 3 and factoring results in 1m2 XZ 2hm2X1 my1 mkXh2 k2 y2 mzxf R2 4 2ley1 2km1 2ky1 0 The solution is b i Vbz 4ac X 5 2a where a 1 m2 b 2h m2 X1 my1 mk 6 c h2 k2 y2 mzxf R2 2ley1 2ka1 2ky1 Example If the offset to the center of the circle with a radius of 2 units are de ned a 3 units in the Xdirection and 7 units in the ydirection ie h 3 and k 7 and if a line passing through point 14 has an azimuth of 620 11 40 what are the coordinates of the intersection of this line with the circle Hashimi 1988 Solution Recall that the slope m is de ned by the cotangent 0f the azimuth m cot AZ therefore the parameters a b and c can be found using the relationships shown in 6 a 1 cot2 A2 1 cot2 62 11 40quot 127811 SURE 215 7 Surveying Calculations LineCircle Intersection Page 68 b 2h mZX1 my1 mk b 23 cotZ 62 11 40quot1 cot62 11 40quot4 cot62 11 40quot7 b 972041 chZ kZ yZ II12X12 RZ 2ley12kal 2ky1 c 32 72 42 cotZ 62 1 13940quot1 2cot62 11 40quot14 22 2cot62 11 40quot17 247 c 174423 Solving the equations by using the quadratic equation results in two sets of coordinates For the Xcoordinates the intersections are bJb2 4ac 972041 9720412 4127811174423 X 2a 2 128811 m 972041 9720412 4127811174423 29009 2 128811 Finally substitute these values of X into the equation y mx X1 y1 thus giving the other coordinate value shown as y mX x1 y1 cot62 11 40quotX47044 1 4 59536 cot62 11 40quotX29009 1 4 50025 This approach is best performed using programmable calculatorscomputers An alternative approach is to solve this problem by solving simple triangles The problem is shown in gure 1 O is the center of the circle and points P1 and P2 are the two points of intersection The line passes through point A and naturally P1 and P2 Figure 1 Line cirle intersection problem SURE 215 7 Surveying Calculations LineCircle Intersection Page 69 The distance A0 is found using the Pythagorean Theorem DAD X0 XA2 YO YA2 3 12 7 42 3606 The azimuth between A and O is found to be X X 3 l AZAO tan71 YO YA tan 1 7 4 O A 33 4124quot The azimuth from A to P1 is given as 620 11 40 Also the azimuth from A to P2 is equal to the azimuth from A to P1 Therefore the angle at point A 9 is the difference between the azimuths along the line and the azimuth from A to O 9 62 113940quot 33 4139 24quot 28 3016quot Also from the problem statement we know that Dopl Dopz Radius of the circle 2 Then from the sine law given as D DOPI D sine sinp siny A0 A0 These relationships can be rearranged and relationships for p and y can be developed 39 e 39 28 30 16quot psin 1 51 DAD sin 1 sm 3606 D0 2 120 38 02quot 39 e 39 28 30 16quot ysin71 51 DAD sin 1 sm 3606 D0 2 59 2157quot SURE 215 7 Surveying Calculations LineCircle Intersection Page 70 With p and 7 known 0L and B can be found using the relationship that a triangle contains 180 OL 180 120 38 02quot28 30 16quot OL 30 51 42quot B 180p 59 21 57quot28 30 16quot 5 92 0739 47quot Then compute the distance from A to P1 and P2 by using the relationships from the sine law DAPI D DAPZ DAO A0 5 smoc s1np smB smy from which the distances are found as DAP PAD sin0L sin30 513942quot s1np s1n120 38 02quot DA 2150 D 3606 DAP A0 sinB sin92 073947quot 2 s1ny s1n59 213957quot D 4188 API The X and Y coordinates of the points of intersection can be computed as normal XPl XA DAPI sinAZAPI 1 2150sin62011 40quot 290 YPl YA DAPI cosAZAPI 1 2149cos62011 40quot XA D sinAZAPl 14188sin620113940quot A12 XPIZ YPZ YA DAPZ cosAZAPl 14188cos62 11 40quot E Check by inversing between the points of intersection P1 and P2 and the center of the circle 0 These values should equal the radius of the circle which is known SURE 215 7 Surveying Calculations LineCircle Intersection Page 71 D0 290 32 500 72 200 x D P 470 32 595 72 200 x 0 2 Both values check to the two decimal place level They may not agree perfectly with the known value of the radius because of rounding of the numbers in intermediate steps For another example of how to solve the line circle intersections problem look at the appendix This example was solved using Mathcad The geometry is shown in gure 2 REFERENCE Hashimi S 1988 SUR 221 Surveying Calculations Lecture Notes Ferris State University 90p APPENDIX The following is an example of solving a linecircle intersection problem using Mathcad 130U85 DE1 598699 74 39 17 X Figure 2 Line Intersection example using Mathcad SURE 215 e Surveying Calculations LineCircle Intersection Page 72 LineCircle Intersection Given the following data compute the coordinates of point P Coordinates of the Pl Coordinates of the center of the circle 0 Arc length from the PC to point P Central angle from the PC to point P Coordinates of point 1 on the line and the point on curve PC X1 15946976 Y14801176 XPC 1 5429259 YPCI 4447866 Azimuth from point 1 to the point of intersection A271P12313121 In dddmmss format a121 oorA271P amnlAZ71P 7 a12100 A1P1a12 W amne oora1nn 60 3600 AlP 2315225 in decimal degrees format Forward azimuth ofthe back tangent ofthe curve AZiBT 122o2502ln dddmmss format a121 oorAZ7BT amn AZiBT 7 a12100 mm 7 oor amn 3600 ABT a12W 60 ABT 220418889 in decimal degrees format Radius of curve Rad 1 598699 Central angle of the curve a CA1743917 In dddmmssformaf a121 oorCA amn1CAea12100 100 A1a12 amne ooramni 3600 ooramn Jr 60 A 74655 in decimal degrees format 180 7 rd SOLUTION Tangent Distance T 1Radtanltigt 2rd T 456538 Azimuth from PC to Acho ABT e 90 Center of Curve Acho 13041889 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 z z z z SURE 215 7 Surveying Calculations LineCircle Intersection Page 73 Coordinates of center x0 XPC RadvsmltAZ gt YO YPC Radvcos Acho of circle rd rd XO 58850633103 YO 4059687103 ABT Coordinates of the 1 PI XPI XPCTsmlt gt YPLXPCTVCOSABTgt rrl 3 o3 XPI5133252810 YPI50816851391 num 1X07 X1 den 1Y07 Y1 The azimuth between point 1 on the line and AZ atan Hum the center of the circle is computed using den the arc tangent function as AZ 00833046 AZlO 1AZ7E 0L1A1P7 AZIOEj 7 0L 467494985 D10A X07 X12 Y07 Y12 D10744069 Sin 3 7 rd From the sine law the the B iasm 39D10 5 can be computed as B 11318961 Then the angley is found by subtracting the sum ofthe othertwo angles in the triangle or and B 711807 0L7 pg 7 7 683976341 The azimuth between the center of the circle 0 and the point of intersection P is AZOP1A2107 77 mi 27 180 AZOP 5172726 The coordinates of the point of intersection are now computed as normal XP1XORadsiI1AZOP YP1YO RadcosAZOP XP 53486871 103 YP 4325659103 SURE 215 7 Surveying Calculations LineCircle Intersection Page 74 The central angle between the PC to point P is d 1Acho l 7 7 AZOP 180 d 0245 ang dg ang 140435214 degs I oorang minu 1ang 7 degs 60 Central angle deg min sec ms 1 oorm1nu degs 14 mins 2 secu 1m111u7 mms60 secu 36677 Chord distance between the PC and the point P 014 XPC7 XP2 chi YP2 c 1463774751 AZPPC atanw YPC7 YP AZPPC 05828898 ang IZAZPPCEj TE ang 333971282 degs 1 oor ang minu 1ang 7 degs 60 Azimuth from P to PC deg min sec mms I oorm1nu degs 33 secu I minu7 mins60 mins 23 secu 49661 360 24 Rad 100 11100A A 7 Arc 146745 7E A 7 1 d 180 The arc length from the PC to the point P is m 7 i 39 SURE 215 7 Surveying Calculations LineCircle Intersection Page 75 Example 2 Given X1 5313674 Y1 4200812 X121 4977455 YpI 3951449 Distance from 1 to point P 327387 Forward Azimuth of Back Tangent 1370 54 22 Radius of the circular curve 819524 Central angle A 480 39 53 Compute The coordinates of point P Coordinates of the PC Coordinates of the center of the circle 0 Arc length from the PC to point P Distance and azimuth from the PC to P and Central angle from the PC to point P Solution Solution to parts of the horizontal circle T Rang 819524w 370587 XPC X1I TsinAZPHC 4977455 370587sin317 54 22quot XPC 4729033 YPC Y1I TcosAZPHC 3951449 370587 cos317p543922quot YPC 4226442 SURE 215 7 Surveying Calculations LineCircle Intersection Page 76 Azpw AZPHI 90 137 543922quot 90 47 543922quot X0 XPC Rsin Azm 472903 819524sin47 54 22quot X0 5337159 Y0 YPC RcosAzm 3951449 819524 cos 47 543922quot Y0 4775808 D071 X1 X02 Y1 YO2 5313674 53371592 4200812 47758082 57547539 D3 DEH D3 8195242 5754752 3273872 2DHDO1 2 819524 575475 01 18 1739 03quot X X Azmztar 1 O ztan15313674 5337159 Y Y 4200812 4775808 COSOL 1 O AZM 182 2039 20quot AZH Azo1 oc 182 20 20quot18 17 03 AZH 200 3723quot X1 X0 DH sinAzH 5337159 819524sin200 3723quot X1 5048508 Y1 YO DH cosAzH 4775808 819524 cos 200 373923quot Y1 4008801 Horizontal Curve Calculations COMPUTING MISSING DATA IN A POLYGON Robert Burtch Surveying Engineering Ferris State University TRAVERSE WITH MISSING ELEMENTS A Traverse Shaun TRAVERSE WITH MISSING ELEMENTS 3 forms of missing data problem 1 Missing distance and direction of same line trivial problem Missing 2 distances of 2 adjacent lines or 2 directions of 2 adjacent lines or distance and direction of 2 adjacent lines 2 unknown distances or directions on 2 nonadjacent lines or unknown distance on one line and unknown direction on another nonadjacent line M 8 MISSING DATA ON 2 ADJACENT LINES EXAMPLE quot39MISSING DATA ON 2 0 Compute distance and azimuth of line ADJACENT LINES 41 Do X 7X 0541 l10000071748892 44500007428902 D17522639 XrX m 1 w Y 7Y 500 00 7428 90 A26 275 253924quot Az tan Solve of missing elements in triangle 145 Solve for interior angles 0 Using sine law distances can be computed 0LAZH 7A2 95 25 24quot733 473932quot D445 D145 D144 01 61 3752quot sinOL sinB siny u v u u v u D DH sinB 75 39 sjn39 2554quot LAzH 7A2 4 314 5118 7275 2524 15 Sim Sin78056q4u B 39 2539 54quot DI5 48685 7 180 4MB 180 61 37 52 39 25 54 D DH I a 75226 I 6103752 U y u 1 5111 778 5614 4 5 S111 s1n78 563914 D45 67445 MISSING DATA ON 2 NON ADJACENT LINES x IDEIEIEIEI was 21 222 n angles 0L B and ycan be computed EXAMPLE Compute distance and direction between an B DAB XE 7X92 YB 7Yc2 1 1000007538282 1000007 570222 DAB 63079 AZ tanil XrXA 1 100000753828 AB YB 7 Yc 100000757022 AZAB 47 033907quot 0c AzAc 7 AzAB 145 14 20quotH47 033907quot 98 1 13913quot B AZBA 7 AZBC 227 033907quot7 182 213943 44 4124quot y 360 7 AzCA iAzcg 360 7 325 143920quot7 2 213943quot 37 o73923quot Use sine law to compute distances D 7 15mm 51117 7 63079 DBCJ K si1137 073923quot 103452 D DAC AB JSIHB 51117 63079 Si1144u413924quot sm37 073923quot 73504 5i1198 1 13913quot Check results ANALYTICAL APPROACH Exploit fact that in closed loop traverse that the sum of latitudes and departures is zero Write 2 equations with 2 unknowns Equations solved simultaneously Unknowns in 3 categories 2 unknowns both lengths oftraverse 1 length unknown and 1 direction of another traverse course Direction of 2 lines unknowns CATEGORY 1 0 Write equations D1 sinAz1 D2 sinAz2 AX 0 D1 cosAz1 D2 cosAz2 Ay 0 Rearrange 2nd equation Ay D1 cosAz1 D2 cosAz2 CATEGORY 1 D sinAZ iAysinAzzDcosAzsinAzz Ax0 cosAz2 D sinAz cosAz2 iAysin Az2 7D cosAz sinAz2 Ax cosAz2 0 Aysin Az2 iAxcosAzz D 7 sinAz cosAz2 rcosAz sin Az2 Recognizing SlnX SIHX COSyi COSX slny D 7 AysinAzZ iAxcosAzz 1 sinAzZ 7AA CATEGORY 1 EXAMPLE Given data CATEGORY 1 EXAMPLE Ax D sinAz D3 sinAz3 D5 sinAZS Ax 468 3839sin 36 42 2539723 0039sin193 023956quot967 3039 sin 24672920quot Ax 7109 54739 Ay D cosAz D3 cosAz3 D5 cosA 5 Ay 468 3839c0536a423925quot723 0039c05193 7023956quot967 3039cos 246 283920quot Ay 611635 Difference in azimuths from 2 to 4 A24 Az2 124 41 07quot CATEGORY 1 EXAMPLE The distance D2 is D 7 Aysm AZ rAxcosAz 611 635 sm zzzns os rkm 547 cosZZZquot15 08 sm z 7A2 sm124 41 07 59875 Alternatively can also compute D2 as 7 AysinAZArAxcosAz sin AZ cosAz2 r cosAz sin Az2 6116355in222 15 08 77109547 c05222 153908quot 7 sin222 15 08quot cos97 343901quot7 cos222 15 08 sin97 343901quot 5987539 D 2 CATEGORY 1 EXAMPLE Distance D4 found from CATEGORY 2 One length unknown and direction of another course 39 Rearrange unknown Designate M for sin A DZM AX D 7 7D sjnAz2 Ax 7 59875sin97 343901 77109547 A22 and N for cos A22 s1n z1 D 4 T sinAz4 T sin222 153908quot 1 71930quot DlsmAZ1 D2MAx0 D N A cos AZ1 27y chosAzl D2NAy0 D1 CATEGORY 2 CATEGORY 2 DZM AxZ DZN AyZ 2 sin2 AZl 005Z A21 2 D1 D1 1 DZMAxZ DZNAyZ 1 D5 DgMZ 2DZMAXAXZ D NZ 2DZNAyAyZ D MZ NZDZ2MAX2NAyAxZ AyZ in 0 Form of quadratic equation b i1Ib2 4210 2a where a M2 NZ sin2 AZZ cosZ AZZ b ZMAX ZNAy 2AxsinAzZ 2Ay cosAzZ c sz AyZ D1Z x DZ CATEGORY 2 o 4 can be factored out of quadratic equation leading to this solution 2Vi Jaw2 4U 1 2 V V2 U where UAX2 Ay2 Di and VAXMAyN CATEGORY 2 EXAMPLE CATEGORY 2 EXAMPLE Solution U Ax2 Ay2 iD 71095472 5115352 471980 4320141215 V Ax sinAz2 Ay cosAz2 7109547sin 97 343901 611635cos 97 34 01 718913588 Distance D27V 1m 7718913588 718913588771320141216 59875390r72204839 lgnore second distance CATEGORY 2 EXAMPLE 0 Solve for azimuth of line 4 is D2MAx 5 75987551n97a343901quot77109547 l DA l l 42 153908quot 0 NW or SW quadrant Good drawing 222 15 08 CATEGORY 2 EXAMPLE CATEGORY 2 EXAMPLE SW QUADRANT RESULT NW QUADRANT RESULT CATEGORY 3 CATEGORY 2 ALTERNATIVE 0 Multiply 1st and 2quotd equation in slide 16 by sin A21 and by cos Az1 respectively 7D sinAzl cosAzl 7 DZ cosAzZ sinAzl iAxsinAzl 0 D1 sinAzl cosAzl DZ sinAzZ cosAzl AyCOSAZ1 0 0 After manipulation 7in1lwzvz 7 WZ WzvZ ilivz cosAzz 1v1 7 7inwv2 17WZ 17vZ 0 Square both equation on slide 16 D12 sinZ AZ1 DZ sinAzZ AXZ D1Z cosZ AZ1 DZ cosAzZ AyZ 0 Add equations of D2 sinAz2 Ax2 D2 cosAz2 Ay2 D sin2 A22 2D2sinAz2 Ax Ax2 D cos2 A22 2D2cosAz2Ay Ay2 D 2D2AxsinAz2 AycosAz2 Ax2 Ay2 CATEGORY 3 Rearrange 2 7 2 7 2 7 2 AxsinAz2 AyCOSAZZ w 2D2 0 Divide by 1Ax2Ay2 AxsinAz2 AycosAz2 7 D2 7D isz iAyz 1 1 M Ay2 sz Ay1A 2D2Ax2 Ay1A CATEGORY 3 Let sin9Lcose sz Ayz sz AyzyZ 0 And using trig identity cosu V sinusinv cosucosv then 2 2 2 2 D D Ax A cosAzZ 8 2132AxZ Ayz Z Df D Ax2 Ay2 2D2Ax2 my A22 9 cosl z 7 z 7 3 7 z AZZ cos 1 D 2 DDZAXZAA ZAy cos 1 AXZAy Z 2 y Ay D2 cosAz2 Ay All 2 005 1 D1 CATEGORY 3 CATEGORY 3 ALTERNATIVE Recall 71 A W e O Ax2 y2 COSAZZ AyJz AAy2 AX J y Ax 0 yields where J7 sz Ay2 DfD 2AxD2 CATEGORY 3 ALTERNATIVE Ay K i 1M2 Ay2 AXZKZ AX2 Ay2 AX sz AyZ D D1Z 2AxD1 cosAz1 whe re K CATEGORY 3 EXAMPLE Line Azimuth Distance 1 36 42 25 46838 2 A22 59875 3 193 02 56 72300 4 A24 71980 5 346 28 20 96730 CATEGORY 3 EXAMPLE I Ax2 Ay2 7D D 71095472 6116352 459875 719 802 2 7109547719 80 73 450335 1 Ayi1AX2 Ay2 AX22 AX2 Ay2 AX 1211646279 4 42424042622 COS 35244956 A24cos Solution set of A22 97 34 002 and A21 222 15 076 TRAVERSIN G AND TRAVERSE ADJUSTMENTS Surveying Engineering Department Ferris State University Introduction Control surveying is that part of surveying in which high precision instruments and techniques are employed to locate points for subsequent surveying operations Because it is used as a base for further work it needs to be performed with more care and greater accuracy Control is used for many purposes Photogrammetry and topographic mapping require control as does construction layout Control provides the basic framework of coordinates required for many other surveying applications The basis for evaluation of errors and corrections is based on a very simple relationship T R C WhereT the true value R the recorded eld value C the correction to be applied to the eld measurement What we see is that the correction term is always added to the recorded value The correction term could be either positive or negative It should be recognized that the error E has the same magnitude as the correction but with opposite sign Thus E C The error can be expressed as ER7T This formula states that the error is equal to the difference between the recorded value and the true value The true value is always subtracted from the measured field recorded value Since the true value is never really known it is usually approximated by some value such as the mean Kinds of Traverse There are a number of different kinds of traverses employed in surveying They include de ectionangle traverse interior angle traverse exterior angle traverse and a traverse where all the angles are measured to the right as one would experience with a directional theodolitel Each has its own geometry from which the angular error of closure can be computed 1 Conceptually one could also have all angles measured to the left although there are no instruments I know that measures angles just to the left SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 2 De ection Angle Traverse The de ectionangle traverse gure 1 is primarily used in route surveys such a highways One can see that the measured angle is that formed from the prolonged backsight line from the last station to the line in the direction of next point The direction left or right from the prolonged line is a necessary part of the angle The general form from the geometry states that 140 103900 R 0L1 49 303900 R 0L2 73 203900quotL 0L2 054 50 203900 R AZMN 1040423930quot M as 52 253900 L U Figure 1 De ection angle traverse from Anderson and Mikhail 1998 Az1iocK 0cL Az2 360 0 i1 i1 where Azl the forward azimuth at the origin Azz the azimuth at the closing and OLRYL the de ection angles to the right and left respectively The equation expresses the geometric relationship that should theoretically exist Thus this represents the true value 0 In the example in gure 1 Anderson and Mikhail 1998 the angular error of closure is found using the following relationships Zak 0c1 0c3 0c4 140 103900quot 49 303900quot 50 2000quot 240 0039 00quot SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 3 Z XL 0L2 15 73 203900quot 52 2500quot 125 4500quot The angular error is AZsrR 20 ZOCL AZMN 360 Ez 350 303900quot 240 003900quot 125 453900quot 104 42 30quot 360 0 0239 3039 Note that in a de ectionangle traverse that the angles to the right are considered as positive angles while those to the left are treated as negative angles The correction to each angle cpa is found by dividing the total angular error by the number of angles recognizing that C E E o y u cpa A w 30quot n 5 In this formula n is the total number of angles measured The corrected angles are computed by adding the correction per angle cpa to each of the measured angles 11 140 10 00 30 1400 09 30 0c 730 20 00 30 730 20 30 13 490 30 00 30 490 29 30 0L4 500 20 00 30 500 19 30 0L5 520 25 00 30 520 25 30 114 12 30 Check to make sure that the corrections have been applied correctly 350 303900quot 114 123930quot 104 42 30quot 360 0 The corrections have been applied correctly Note that in this example the angles to the right decreased numerically while the angles to the left increased Interior Angle Traverse The interior angle traverse gure 2 is probably the most common kind of traverse encountered in surveying Here all of the angles within the polygon are observed The general form of the geometric relationship of the angles is SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 4 20 i n 2180 0 From the example in gure 2 Anderson and Mikhail 1998 the angular error of closure is any difference found in the previous equation that deviates from 0 x1 66 40 30quot x2 131 35 00quot x3 97 353900quot x 64 00 30quot x 227 26 30quot x6 132 453930quot 4 5 Figure 2 Interior angle traverse from Anderson and Mikhail 1998 ocl oc2 oc3 oc4 oc5 oc6 6 2180 E4 720 0339 00quot 720 339 00quot The correction per angle cpa is determined next 3039 00quot 6 30quot cpa The corrected angles are found by adding the cpa to each observation SURE 215 e Surveying Calculations Traversing and Traverse Adjustment Page 5 011 66 403930quot 30quot 66 4000quot 012 131 35 00quot 30quot 131 3430quot 013 97 35 00quot 30quot 97 3430quot 014 64 003930quot 30quot 64 0000quot 015 227 26 30quot 30quot227 2600quot 016 132 453930quot 30quot132 4500quot Z ori 720 0039 00quot A special case of gure 2 is the case where the exterior angles have been measured Here geometric angular relationship is shown as Z aim n 2180 0 Note that the interior and exterior angle traverses are all closedloop traverses Angles t0 the Right 290 013910quot 13 210 013932quot 11 140 003930quot 12 u 000000 YZ 25 90 013913quot OLA Figure 3 Traverse where angles were measured to the light from Anderson and Mikhail 1998 SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 6 With directional instruments angles are measured to the right An example is shown in gure 3 Anderson and Mikhail 1998 The general form for the angular geometry is shown as Az1ioci Azz n 1180 0 11 From the example the angular error of closure is found as AzYZ 0L10L2 03 0 0 AzTU 5 1180 E4 250 003900quot 789 59 20quot 319 593945quot 720 0039 25quot The correction per angle is 25quot c a 5quot p 5 The corrected angles are found by adding the cpa to each measurement ocl 210 013932quot 5quot 210 013937quot oz 140 003930quot5quot140 0035quot 13 290 013910quot 5quot 290 013915quot oz 90 013913quot5quot 90 01 18quot 0c 59 543955quot5quot 59 5500quot 20ci 789 5939 45quot Check using the general relationship for the angles to the right 250 003900quot 789 59 45quot 319 59 45quot 720 0 x Steps in Traverse Computations The process of computing and adjusting a traverse can be broken down into 6 basic steps 1 Balanceadjust the angles 2 Compute the azimuthsbearings of the survey lines Note that the directions of the lines can be computed using the raw eld angles and these preliminary directions can then be adjusted for the angular misclosure 3 Compute the latitudes and departures SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 7 4 V39 LatsgtcosocAY DepsgtsinocAX s distance between two points or azimuthbearing of the line AX difference in the Xcoordinate between the two ends of the line AY difference in the Ycoordinate between the two ends of the line where The latitudes and departures are also referred to by the northing or casting value Determine the closure error in departure and latitude eD and eL respectively Expressed in this form the closures are error terms recall that E R 7T nil er DX 214339th Xn X1 i1 nil 6L DY ZDepiJH Yn Y1 i1 In other words the error in departurelatitude is equal to the sums of the departureslatitudes minus the difference in the coordinates between the last and first point If the traverse is a closed loop traverse where the beginning and ending points are the same in other words Xn X1 and Yu Y1 then the closure errors are simply the sums of the departures and latitudes n71 eD DX 2mm 3971 nil 6L DY ZDepiJH i1 Compute the linear error of closure accuracy ratio and the direction of the closing line The linear error of closure EC is the square root of the sum of the latitude and departure closure errors squared 2 2 Ec 1 eD eL The accuracy ratio is de ned as the ratio of the linear error of closure to the perimeter distance P Accuracy 13 Finally the direction of the closing line is found by the arctangent of the error in departure to the error in latitude SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 8 Be tan 1 6 D eL 6 Adjust the traverse and compute the adjusted distances directions and coordinates The methods of adjusting the traverse will be discussed after the example is introduced The last step is to normally compute the area of the traverse assuming that the traverse is a closedloop traverse gt1 Example A traverse was run as shown in gure 4 The data are also given in the following table 34 120 D 483406 08 405 D 384926 52 175 D 369173 Note that the angle codes are as follows R angle to the right L angle to the left D de ection angle with a positive being to the right and a negative to the left 6703412 0quot a Azimuth 256 493924 8quot 15912920 6quot Figure 4 Example for traverse adjustment SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustinth Page 9 Because this problem uses mixed angle types the rst part of the problem is to compute the azimuth of the lines based on the measured quantities and then nd the angular error by comparing the known azimuth from point 1 t0 the azimuth mark to the computed valued based on the measured angles The azimuths 0f the lines are then found as follows Azimuth 1 gt Az Mark 104 05 050 180 180 00 000 Back Azimuth 284 05 050 41 67 34 120 Azimuth 1 2 216 30 530 zz 256 49 248 40 18 318 180 180 00 000 Azimuth 2 3 139 41 282 180 180 00 000 43 259 29 206 Azimuth 3 4 60 12 076 z4 64 08 405 360 360 00 000 Azimuth 4 5 356 03 271 45 64 52 175 Azimuth 5 1 291 11 096 180 180 00 000 Ato Az Mark 356 53 287 360 360 00 000 Azimuth 1 7 Az Mark 104 04 382 The angular error of closure is the difference between the calculated azimuth of the closing line to the known or true azimuth E4 Az 1 gt Az Markcm Az 1 gt Az MarkTme 104 04 383quot 104 05050quot 267quot The correction per angle is found using E I u cpa A 445quot n 6 The adjusted azimuths become SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 10 12 076 134 12 210 03 271 178 03 449 In this particular example the correction accumulates when being added to the preliminary azimuth The rst line receives the computed cpa The second preliminary azimuth must be adjusted not only by the cpa but must also accommodate the adjustment that was placed on the rst line In other words the correction is 2cpa The next line receives a 3cpa correction This continues along the entire traverse each new azimuth receiving he same correction as the previous line plus the cpa for the angle measured at the new station The latitudes and departures are shown as follows in the Excel program The error in closure for the latitude and departure are 0075 ii1 n71 eD Dx ZLat 391 1 n71 eL DY ZDepiH 0176 i1 SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 11 If the error in closure is expressed as eD or eL the corresponding corrections can be expressed as ch and clL respectively where the value of ch and clL are equal to the corresponding error in closure with opposite sign In this example ch 0075 while clL 0176 The linear error of closure is computed as E 00752 01762 019139 The azimuth ofthe closing line becomes2 5 tan71 l 203 0450quot 0176 To compute the accuracy ratio or relative error of closure of the traverse divide the linear error of closure by the perimeter distance 0191 1 Accuracy 2109684 11045 x l 11000 Traverse Adjustment Compass RuleBowditch Method There are several different types of traverse adjustment only a couple of which will be discussed here The first method is called the Compass Rule or Bowditch Method named after Nathaniel Bowditch3 see gure 5 who devised this method of adjustment 2 The arctangent function requires some thought in its application In order to determine the proper quadrant one must evaluate the sign of the numerator and denominator If both the numerator and denominator are positive then the azimuth is the in the northeast quadrant and 3c is the value returned from the calculator If the signs are both negative 3 will still be positive and the proper quadrant is in the southwest Therefore add the angle returned from the calculator to 180 In this example the calculator will show an angular value of 23 04 50 Add 180 to place it in the proper quadrant and the correct azimuth becomes 203 04 50 If the numerator is positive and the denominator is negative 3 will be a negative angle In this particular case the proper quadrant is the southeast Therefore add the angle to 180 and the result should fall between 90 and 180 Finally when the numerator is negative and the denominator is positive then the azimuth of the line is in the northwest quadrant The angle displayed in the calculator should be a negative value Add this angle to 360 to obtain the proper azimuth The value should fall between 270 and 360 3 Nathaniel Bowditch March 26 1773 March 16 1838 was a selftaught mathematician whose formal education ended at the age of 10 During his teens he studied at night and taught himself Latin and Greek as well as mathematics science and astronomy Taking the navigation tables of John Hamilton Moore Bowditch made several thousand corrections to the calculations eventually publishing The New American Practical Navigator that is still being published today else 12 for sea naviga on purposes This method of adjustme 4 akes eertain assumptions about the data Fist it assunes a ors in distanee are 39 to e square root of the lengths of e tr verse sides Seeond it assumes that the pmbable e or in these distanees are equal to the probable error 0 the direetion multiplied by the distanee The eorreetions to the latitudes and depaitures ean be written as where chop con39ections for latitudes and departures respeetively L engthoftheline perimeter distanee around the Jl m Figure 5 Nathaniel traverse and Bawd39nch LelD closure correction in latitude and departure respectively for the traverse From the example For line 12 4 me Dr David Gibsun uf the University uf Flunda Buwddtch vvas an early Amenean navigatur and authur uf a fzmuus bunk un nangatiun Amerieai praetieal Nangatur lle develuped a methud fur starting map puint by plutting arimuths measured by eumpass aid distanees by dead reckumngquot time x velueity when cluslng39 untu areliable map feature the end ufthe last plutted euurse vmuld miss the Amypulnt une quarter ufthe my alung the mute shuuld be muved a quarter ufthe errur uf elusure Thls methud was later applied tu eumpass surveys and beeame lmuvvn in surveying as the Cumpass Rule traverse adiust eaeli latitude and dep arture separately dep enth of Line Error latdep Ttav Perimeter Tu eurreet a cheekthat the sum uf euueetiuns equals the tutal errur uf latitude ur depature wth uppusite S31 EllI IZOutllne dun SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 13 cL amp 483406 0040 2109684 cD j483406 0017 2109684 Line 23 cL 446622 0037 2109684 cD 446622 0016 2109684 etc Using the spreadsheet As a check sum the latitudes and departures and their sum should add up to zero for a closed loop traverse Once the latitudes and departures have been adjusted the surveyor may want to determine the coordinates of each point along the traverse Given the Xcoordinate of a point at one end of the line the Xcoordinate of the point of the other end is found by adding the departure from the first to second point to the Xcoordinate of the rst point In a similar fashion the latitude is applied to the Ycoordinate This is shown as SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 14 Xn1 Xn Depnn1 Y 1 Y Lat nn1 Assuming that the coordinates for point 1 are known to be X1 5460445 and Y1 6238012 the coordinates of the remaining traverse points are Make sure to compute the coordinates of point 1 again from the last traverse station point 5 in this case For a closed loop traverse this coordinate must agree with the coordinates of the rst point This provides a check on the calculations Ifthe traverse is a closed traverse but not a closed loop traverse the nal coordinates must agree with the known coordinates of that point Since the latitudes and departures are adjusted the corresponding distances and directions do not coincide with these adjusted values Therefore the adjusted distances and directions need to be determined The distance is found using the Pythagorean theorem Drum Xn1 Xn 2 Ymr Yn 2 2 2 I Depnn1 Latnn1 The adjusted directions are computed using the arctangent function Xquot1 Xn tan1Deprn1 Y Y L Lat n1 71 ann1 tan nn1 j For the example the adjusted distances and azimuths for the lines are found as follows Line 12 SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 15 D12 517281339 5460445392 5849543 6238012392 J 483364 5172813 5460445 216 3139 018quot 5849543 623801239 OCH tan 1 Line 23 DH 546173739 5172813392 5508988 5849543 2 J 446604 1 5461737 5172813 OCH ta 5508988 5849543 139 4139 204quot etc The following table shows the results from the distance and azimuth calculations for all points within the traverse rave rse Transit Rule A second traverse adjustment method is called the Transit Rule This method is preferred when the angular measurements are more precise than the distance measurements There is also some evidence to suggest that this method may be better than the Compass Rule even when the distance and angle measurements have comparable precisions when the traverse legs run parallel to the axes used in the traverse coordinate system The corrections to the latitudes and departures for the traverse are computed as SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 16 clL 2M SL c CI Ds D ZlDePI D where cL cD corrections to the latitudes and departures 0f the line clL ch closure corrections for the latitudes and departures for the traverse ZlLatl ZlDepl sum of the absolute values of the latitudes and departures and sL sD latitude and departure length of the line absolute values Using the example the adjusted latitudes and departures are shown for the Compass Rule as follows First the sums of the absolute values of the latitudes and departures are found and then the corrections to the latitudes and departures ZlLatl 388509 340592 211453 384017 133455 1458026 leepl 287632 288924 369320 26419 344194 1316501 Line l2 cL 388509 0047 1458026 cD 287649 0016 1316501 Line 23 cL 340592 004l 1458026 cD 288908 0016 1316501 etc The adjusted values for all lines are presented in the following spreadsheet SURE 215 7 and Traverse As with the Compass Rule the next step is to compute the coordinates and adjusted distances and directions of the traverse lines This is shown in the following table raverse Again make sure that the coordinates of point 1 are recalculated instead of just putting the known value into the solution Crandall Method The third method of traverse adjustment is based on the method developed by Charles L Crandall and referred to as the Crandall Method Crandall based his development on least squares and assumed that after the angular error is accounted for that the random SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 18 errors remaining were attributed to the distances This approach implies that the angle measurements are much more accurate than the distance measurements To compute the corrections to the latitudes and departures Crandall identi ed two intermediate calculations A and B shown as follows quot Lati rDepi quot Dep2 e e A D 100si L 100s Dep2 Lat2 Lati rDepi 2 100s ilroos 1 100g quot Lati rDepi quot Lat2 e e B L 100si L 100s Dep2 Lat2 Lati rDepi 2 H 1008 H 1008 100i where eL eD total errors in latitude and departure Lat Dep latitude and departure length for a given line and s distance between the two points The corrections applied to a particular line c Z is found from the following relationships cll Lat1A DeplB cZZ LatZA DepZB c Lat3A Dep3B cl LatnA DepnB With these corrections to the lines the corrections to the latitudes and departures are given in this general form c c Lat L L 100s Dep c L L 100s Substituting the value for c Z into these equations yields the nal form of the correction to the latitudes and departures SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 19 Lat2 Lat r Dep cL 100s 100s 2 CD A Lat Dep B Dep 100s 100s Using the same example as before the adjusted latitudes and departures using the Crandall Method are presented in the next table The coordinates of the traverse points and the adjusted distances and azimuths are computed as before and are shown in the next table A comparison of the results of the three different adjustments is given in the next Excel spreadsheet While there are some differences they are all less than 004 It is evident SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 20 that any of the adjustment techniques will have similar results although one cannot state that this level of agreement will occur with all traverses Area Calculations Once the traverse has been adjusted the area is normally calculated assuming that the traverse is a closed polygon The area of a polygon can be computed using the basic geometry for computing the area of a trapezoid gure 6 The area is simply the average height times the width h1 h2 W Figure 6 Area ofa trapezoid h1h2 Area r w SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 21 The double area is then written as 2A h1 h2 w There are two basic methods of computing the area of a polygon Double Meridian Distance DMD or by coordinates Both methods will give identical results Referring to gure 5 the area by Double Meridian Distance is based on the area of a series of trapezoids Draw a meridian through the left most point in the traverse and then draw lines perpendicular to that meridian line to each of the other points in the traverses The width of the trapezoid is designated as the distance along the meridian the latitude of the line while the heights are the perpendicular distances from the meridian to the traverse point departures of the line Thus from gure 5 the double area of traverse l 234 5 is found as follows 2A 2A 2A33394439 2A44395 2A56639 2A11396639 2A11392239 2233 2 JQeILartgml 7 2 Latitude Figure 7 Area by Double Meridian Distance from Mof tt and Bouchard 1992 In other words the area of the traverse shown as shaded in the gure is the total area of the polygon 52 234 minus the area of polygon 52 2l6 The computation of the DMD is given in 3 basic rules 5 Drawing the line through the western most point is done for convenience so that areas come out as positive quantities In theory the meridian line can be placed anywhere SURE 215 7 Surveyng Calculations Traversing and Traverse Adjustment Page 22 1 The DMD of the rst line is equal to the departure of that line 2 The DMD of each subsequent line is equal to the DMD of the preceding line plus the departure of the preceding line plus the departure of the line in question 3 The DMD of the last line should equal the departure of the last line with opposite s1gn In other words the double area is found for the traverse in gure 5 using 2A DMDH Lat DMDH DepH Dep 1 4 LatH DMDH Dep 1 DepH 4 LatH DMDH Depl2 DepH LatH DMDH DepH DepH 4 LatH DMDH DepH DepH 4 LatH Using the results from the Compass Rule adjustment the area by DMD of the example traverse is 2A 287632 388469 1117361154 287632 4 287632 288924 340555 975145187 286340 288924 369320 211489 786536051 371904 369320 4 26418 384049 2745205295 714806 4 26418 4 344194 133486 459450803 2A 6083698489 A 30418482 sq ft 698 ac6 The area of a polygon can also be written in terms of coordinates The area can be written as 2AX1Yz YnX2 Y3 Y139 Xn41Yn Yn42Xn Y1 Yn41 or 2A Y1X2 XnY2 X3 X139 Yn41Xn Xn42Yn X1 Xn41 Using the values from the Compass Rule adjustment the area is calculated in the following manner 6 1 acre 43564 square feet THREE POINT RESECTION PROBLEM Surveying Engineering Department Ferris State University INTRODUCTION The threepoint resection problem in surveying involves occupying an unknown point and observing angles only to three known points Today with the advent of total stationsEDMs the problem is greatly simpli ed Ifthe unknown point P lies on a circle de ned by the three known control points then the solution is indeterminate or not uniquely possible There are theoretically an in nite number of solutions for the observed angles Ifthe geometry is close to this then the solution is weak In addition there is no solution to this problem when all the points lie on a straight or nearly straight line There are a number of approaches to solving the resection problem KAESTNERBURKHARDT METHOD Figure 1 Three point resection problem using the KaestnerBurkhardt method In the KaestnerBurkhardt approach Blachut et al 1979 Faig 1972 Kissam 1981 Ziemann 1974 also referred to as the PothonotSnellius method Allan et al 1968 the coordinates of points A B and C are known and the angles 0L and 5 measured at point P Inversing between the control points we can compute a b AzAc and Ach using the following relationships SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 176 AZAC tanl a XC XA2 YC YA2 Yc YA X X Ach tan 1 b XC XB2 YC YB2 Yc YB Compute y Y AZCA AZCB AZAC AZBC Compute the auxiliary angles p and 9 First recognize that the sum of the interior angles is equal to 3600 the sum of interior angles of a polygon must equal n 7 2180 ocB9y360 Rearrange e180 0c5y61 From the sine rule compute the distance 5 39 b 39 9 as1n and S sln sm 0L smB Combining these relationships yields 39 b 39 0L sm sm cotk sin 9 a sinB where M is an auxiliary angle with an uncertainty of i1800 We then have sin 1 sin 9 COt7M 01 sin sin9 cot 1 sin sin9 cot7t1 SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 177 cot Since cot M and using trigonometric theorems one can write l l ZCOSE e51n3 e cot450 cotk l cot cot 45 1 1 251n3 9cosa 9 But recognizing that cot 450 1 and tan 9 tan 9cot45 7 tan 51 cot450 11 Therefore 9 tan 1 tan 51 cot450 1 52 Then i 5r 52 9 51 52 Recall that 52 has an uncertainty of i1800 due to the uncertainty in 11 Next using the sine rule compute the distances 01 and 02 a sin 71 a sin180 06 a sin0 1 sin 1 sin 1 sin 1 sin 72 sin180 5 9 sinB 9 2 b b b srnB srnB srnB If M was picked in the right quadrant then y is in the right quadrant and 01 and c are positive If they turn out to be negative 52 I and 9 have to be changed by 180 As a check recall that 0L B y I 9 360quot The next step is to compute the azimuths to point P SURE 215 7 Surveying Calculations Three Point Resection Problem Page 178 AZAP AZAC l AZBP AZ 9 BC Finally compute the coordinates of point P X1 XA csinAzAPXB czsinAzBP Y1 YA clcosAzA1YB c2 cosAzBP An example prepared using Mathcad is presented as follows Three Point Resection Problem KaestnerBurkhardt Method ddang degree oorang radiar1sang d ddang mins ang e degree1000 TE minutes oormins d 1800 seconds mins 7 minutes 1000 minutes seconds degree 600 36000 dmsang degree oorang rem ang e degree60 mins oorrem reml rem 7 mins secs rem1600 ms secs m degree 100 10000 TB 180 trad tdeg 180 TE Given SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 179 XA 100000 YA 530000 XB 310000 YB 500000 XC 220000 YC 630000 at 1 1093045 393 1 1150520 Solution Find the coordinates of point P using the KaestnerBurkhardt Method Begin by computing the azimuths and distances between the known points AZAC 7am2yc 7 YA xC 7 gt901 1 rhslt4i24cfclt t4eiggtl 5044599 AZ atan2YC 7 YBXC 7 XB Az 7060554 AZBC A2 27 a XC 7 XA2 YC 7 YA2 b XC 7 KB2 YC 7 YB2 The angle at point C is computed as are the auxiliary angles 7 AZAC 7 AZBCtdeg 360 51 1807 lmdw ddB 7 dms51 2515163 I k0 1053482162 A tdegatmT dInSOL 4330291 Note that M has an uncertainty of 180 degrees 527atmtanradiansdmssl ll meg tanradiansdms45 dms52 04214 13 51 52 SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 180 9 3 51 7 52 Compute the distances between the point P and control points A and B a sinradiar1s at trad sinradians 1 c 2V Smradians The azimuths between the control points A and B are now determined AZAP AZAC trad AZBP AZBC 7 etrad X1 XA C139SlI1AZAP YP YA clcos AZAP Check XP XB c2sinAzBP Yp YB c2 cos AZBP Allan et A1 1968 present a slightly different approach called the PothonotSnellius method Recall that the distance from C to P was designated as s and was expressed asind S bsinG sinoc sinB method is basically that already presented above The second method is described as follows Write the ratio of p to 9 by a constant K as From this there are two methods of solving this problem The rst cosS K sine sinS p sinScosQ cosSsinqp sinScosp sinp sinp sinp sinp where S 360 0LBY This relationship is based on the fact that the sum of the interior angles in polygon ACBPA must equal 360 Thus one can write from this basic relationship refer to gure 1 G 360 0L 77 p S p S represents the known angles Manipulation of this last relationship yields SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 181 sin S cos p KcosS sinScotp sm p From which K cos S cot p sm S Solve for p and then compute 01 and the azimuth to determine the coordinates of point P Alternatively use lineline intersection to nd the coordinates of the unknown point Another modi cation of the KaestnerBurkhardt Method is that reported by the United States Coast and Geodetic Survey USCampGS now the National Geodetic Survey NGS Hodgson 1957 Reynolds 1934 Figure 2 identi es three cases of the three point resection problem This is a modi cation of the USCampGS method presented in Kissam 1981 and with a slight modi cation in Anderson and Mikhail 1998 The solution can be broken down into a few steps given here without derivation Figure 2 Three scenarios for the threepoint resection problem SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 182 Compute gh360 0Bij if the problem is the same as that indicated in gure 2a and b For the con guration depicted in gure 2c gh ij0ltB b Then de ne a V asinoc l cot 450 0bsm B asmoc bsinB where amalgam asmoc 0 Further tang h cot450 0tangh d Then gghgh and hghgh 2 2 e Finally i180 goc and j180 hB Now that all of the angles are known the lengths of the different legs of the triangles can be found using the sine law From the previous example we can see that this follows the Case 2 situation shown in gure 2 For this example we will renumber the points so that they coincide with the gure for Case 2 Thus from the original example point C is now designated as point B and the original B coordinate is now C Therefore the coordinates are XA 100000 YA 530000 XB 220000 YB 630000 Xc 310000 Yc 500000 06 109 30 45quot 31150 05 20quot It was already shown that the azimuths are SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 183 COLLINS METHOD The Collins or Bessel s method Blachut et a1 1979 Faig 1972 Klinkenberg 1955 Zeimann 1974 is different in that the problem is broken down into two intersections A circle is drawn through two control points and the occupied point as A B and P in gure 3 The line from P to C is extended until it intersects the circle at a point labeled H This point is called the Collins Auxiliary Point Figure 3 Three point resection problem using the Collins method e 9e 8C Figure 4 Geometry of circle showing that an angle on the circle subtending a base line is equal SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 184 From the geometry of a circle shown in gure 4 one can state that the angle formed at a point on the circumference of a circle subtending a base line on the circle is the same anywhere on the circle provided that it is always on the same side of the base line This property is exploited in the Collins Method The solution involves five distinct steps 1 Compute the coordinate of the Collins Auxiliary Point H by intersection from both control points A and B 2 Compute the azimuth AzHc which will also yield the azimuth between C and P since AZHC AZCP 3 Compute the azimuth of the lines AP and BP AzAP AzCP X AZBP AZCP 4 The J39 can be r J by 39 quot from A and C and also from B and C If desired the solution can be performed using the auxiliary angles p and V V39 Then using the sine law DAC sin0 p sinoc D AP DEC sinB v DBP smB This gives X1 XA DAP sinAzAP XB DBP cosAzBP Y1 YA DAP cos AzAP YB DBO cosAzBP Following is a MathCAD program that solves the same problem as presented earlier but this time using the Collins method Page 125 Three Point Resection Problem Collins Method the same fuheuohs as de ned m the KaesmerrBurkhardt MathcAD program x x YA sznnnn E mnnn YE Sunnnn E 22nnnn YE 63nnnn e mums B nsnszn Solutmh e Fmdthe coordmates of pomt P usmg the Collms Method Begm by lookmg at the man 1 e d ge ABH Angles ar backsxght station and foreseght 1e ttenh em 12 MB em 12 Me 2 7 Xe XA Ye YA ehnzwryerxg zmnemxne MED 12m ddu esxgnated by the vanable a thh subsehpt showmg g Q180 E b180 r d emshm 64544 emshm 7n2915 em 212132n34 new tee e 92 n7424 emshm e 4436M SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 186 AZAHZ AZAB trad1807 ddB DAH 7 jsjn1807 dd0Ltrad DBH 7 jsjn1807 ddBtrad XH 7 XA DAHsinAzAH YH 7 YA DAH cos AZAH Az 7 atan2YH 7 YC XH 7 KC AzCH 7 ifAz gt 0AzAz 27 AZ 7 atan2YA 7 YC XA 7 KC AZCA 7 ifAz gt 0AzAz 27 aACP 3 AZCA AZCH 71807 dda aACPtdeg AzAP AZCA 7 71 CDtrad DAC 3 XC XA2 t YC YA2 From the sine law unfilsmm smdd0Ltrad XP 7 XA DAPsinAzAP Yp YA DApcos AZAP dmsAzAHtdeg 7 16302284 DAH 7 284758555 DBH 7 273605413 dmsAzCHtdeg 7 18539552 dmsAzCAtdeg 7 23011399 drnsaACptdeg 7 4431447 1136st DAC 7 156204994 For a check compute the coordinates from point B by solving for the elements in triangle SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 187 CASSINI METHOD The Cassini approach Blachut et a1 1979 Faig 1972 Klinkenberg 1955 Ziemann 1974 to the solution of the threepoint resection problem is a geometric approach It breaks the problem down to an intersection of two circles where one of the intersection points is the unknown point P while the other is one of the three control points This is depicted in figure 5 The solution is shown as follows Figure 5 Three point resection problem as proposed by Cassini Compute the coordinates of the auxiliary points H1 and H2 First the azimuths between A and H1 and B and H1 are determined AzAHl AZAC 90 AzBHl Ach 90 From triangle ACHl the distance from A to H1 can be computed D tanoc AC AHl D DAC Xc XA Yc YA AHl tan 1 sm AZAC tan 1 cos AZAC tan 1 SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 188 Since the angle at A is 90 sm AZAHl cosAzZ1c cosAzAHl sm AZAC XHI XA DAHI sin A2 XA YC YAcotoc AHl YHI YA DAHI cos AzAHl YA XC XAcotoc The coordinates for H are computed in like fashion XCXB YCYB DBC DBH2 tanB sm Asz tanB cos Asz tanB sm AZBHZ cos AZBC cos AZBHZ sm Asz XHZ XB DBHZ sin Azbhz XB YC YB cotB Y H 2 YB DBHZ cos AZBHZ YB XC XB cotB An alternative approach to coming up with the formulas for XH and YH can also be presented This approach breaks the solution of the Cassini Method down to 5 equations From the equation of the intersections of two lines we can write XC XB YC YBtan Ach This can also be written as XC XB YC YA tan Ach YA YB tan Asz Xc XA Yc YA tan AZAC Solving these last two equations can be done by subtracting the last equation from the preceding equation resulting in SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 189 XC XB YC YA tan Ach YA YB tan Ach X XAYCYAtanAZAC C XA XB YC YA Xtan Ach tan AZAC XA XB tan Ach Rearranging yields XA XB YA YB tan Azbc YCY tanAzBC tanAzAC A Using the form of this last equation one can write express the Ycoordinate of the Cassini auxiliary point H1 as Yc YA tan AZCHl Xc XA tan AZ tan AZ CHl AHl YY Hl A Yc YA ta nAZCA Xc XA and tanAzAHl tanAzCA 1 then the Ycoordinate for H1 becomes after multiplication by tan AZCA YHl YA Xc XAta n AZ AZCA CHl The Xcoordinate can also be developed in a similar fashion yielding XH XA Ye YA tan AZCHl AZCA But AZCHI AzCA 90 oc Then XHI XA D sin AZAHI XA Yc YAcotoc AHl YHI YA DAHI cos AzAHl YA Xc XAcot 0c SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 190 The coordinates for H can be developed in a similar fashion and they are given above Next compute the azimuth between the two auxiliary points H1 and H2 1XH2 XH1 AZHH tan l 2 LYHz YH1 J As before one can write the equation of intersection containing the unknown point P as XHl Xc Yc YHl tan AZCP Y1 Y tanAzCP tanAzHlP H1 or H1H2 YC tan Ach YHI tan Ach xc le tan AZCP tan AZHIP Y YH1 tanAzCP YH1 tanAz P YC tan Ach YHI tan AZHIP xc le tan AZCP tan AZHIP But n YH1Yc xc le Y1 f Thus 1 tan AZCP tan AZHIP where n tan AZHIP Nn1n The Xcoordinate of the unknown point can be expressed in a similar form as SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 191 nXCXH1YC YHI XP The same problem used in the previous methods follows showing the application of the Cassini method to solving the resection problem Three Point Resection Problem Cassini Method See the same functions as de ned in the KaestnerBurkhardt MathCAD program Given XA 100000 YA 530000 XB 310000 YB 500000 XC 220000 YC 630000 at 1 1093045 393 1 1150520 Solution Find the coordinates of point P using the Cassini Method XH1 XA YC 7 YAcotdd0Ltrad XH1 64563588 YH1 YA XA 7 XCcotddottrad YH1 572523694 KHZ XB YB 7 YccotddBtrad KHZ 37086571 YH2 YB XC 7 XBcotddBtrad YH2 5421378 AzHlH2 7 atan2YH2 7 YH1XH2 7 XHI dmsAzH1H2tdeg 9539552 11 tanAzH1H2 1 Nn n 1 nYH1 YCXC7XH1 I Y P N SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 192 1 nXC E XH1 YC 7 YHI X1 N TIENSTRA METHOD Figure 6 Basic geometry outlining the principles of the Tienstra Method The Tienstra method see Bannister et a1 1984 is also referred to as the Barycentric method An easy to understand proof is given in Allan et a1 1968 Figure 6 shows a triangle formed from the known control points Line CD divides the angle at C into two components X and V Line AB is also divided into two components m and n The angle 9 is formed by the intersection of the line CD with the line AB From gure 6 one can also see that line CE is perpendicular to line AB Thus D cotzA AE CE cotzB EB CE D cote DE D SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 193 Then m DAD DAE DDE cot AA cot 9 cot AB cot 9 DCE n DDE DDE DEB DCE which upon further manipulation yields mcot A cot9 n cot AB cot 9 mcotzB mcot9ncotzA ncot9 or mncot9n cotzA mcotzB Since lines AF and BG are perpendicular to line CF one can write D D cotx CF gt AF CF AF cotx cote DF gt AF DDF DAF cot 9 DGD 2 D DGD DBG BG cot 9 D D cotw CG gt DBG CG DBG cotw From these relationships equate D AF DCF DDF DCF COtX cotx cote DDF cote and equating the distance DBG DGD DCG Z DGD cote cote cotw DCF cotw From figure 6 we can also write SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 194 D cotx DCD DCF DDF F0TDDF cotx DCF DDF DDF cote 1J DCF DDF cotx cot9 DDF cot 9 Also we have D t DCDD D DG VD CG DG cote DG cotw D 1 DG cote J DCG DDG cotw cote DDG cote From above one can see that the distance from C to D can be expressed as cotx D D 1 CD DF cot e But from gure 6 we can write the following two relationships cos 9 DDF DDF AD m DDF m cosG D cosG DG DDG ncosG n Substitute these values for DDF and DDG into the relationships derived above This is shown as SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 195 cotw cot DDGDDGcot9 1 DCDDDFcotGC 1 t t ncosG co 1 mcosG ll cote cote Equating the two values for DCD yields noose cotwcot9 mcose cotX cote cote cote ncotwncot9mcotW mcot9 mncot9mcotX n cotw The threepoint resection problem is shown in figure 7 Point P is the occupied point and points A B and C are the control points that are observed The measured angles are 0L 5 and y The other angles are numbered in a clockwise manner from point A Recall that from the intersection problem the coordinates of a point such as point C can be computed as Figure 7 Three point resection problem using the Tienstra Method SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 196 YA YBXA cotBXB cotOL XC cot0LcotB XC cotoccot5XA cotBXB cot0LYA YB where 0c is the angle at A and B is the angle at B Using this basic relationship the X coordinate at point P can be computed as follows Adding these three equations yields X1 cot1 cot2 cot3 cot4 cot5 cot6XAcot3 cot6 XB cot2 cot 5 Xccot4 cotl This is usually represented as XPL1 L2 L3 LIXA LZXB LSXC where L1 cot 3 cot 6 Lzcot2cot5 L3cot4cot1 The Xcoordinate is computed as LIXA LZXB LSXC L1 L2 L3 X P X cot2cot3XA cot3XB cot2YA YB X1 cot4cot5XBcot5XCcot4YB YC X1 cot1cot6chot1XA cot6YC YA In a similar fashion the Ycoordinate can be written from the intersection problem Y XE XAYA cotBYB cotOL C cot 0L cotB which can be shown after the same manipulation performed on the Xcoordinate as LIYA LZYB L3YC L1 L2 L3 Y P SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 197 From gure 7 the line BP was extended until it intersected the line AC at a point labeled Q This divides the line into two parts m and n Recall that the angle ZCPQ 180O 0L and ZAPQ 1800 y Recall that we wrote earlier mncot9mcotX ncoty Using the geometry from figure7 this becomes mncot9mcot4 ncot3 Recall that earlier we wrote the relationship m ncot 9 m cot AA n cot AB which can be written as considering the geometry in gure 7 mncot9ncot6 mcot1 Equating these last two formulas yields the following formula m cot 4 cot1 n cot3 cot6 Using m ncot9 m cotx n cotw and m ncot9 m cotzA n cotzB again write mncot9 m cot0Ln coty mncot9ncotzC mcotzA Equating these last two equations gives m cot AA cot 0L n cot AC cot y Using this formula equate it with m cot4cot1n cot3cot6 giving us the next equation cot1cot4 n cot A cot0c cot3cot6 m cotzC coty Ilt1 1 A1 cotzA cotoc where lt cot AB cotB Z 1 A3 cotzC coty SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 198 In a similar fashion one can easily show that L3 K3 L2 K2 Therefore h h i w 1 K2 K3 from which L1 KIW L2 KZW L3 KSW and uuuuwnmKQ Thus n K 1 nuunmm L2 K2 nuu nmm L3 K3 L1L2L3 K1K2K3 Substituting these relationships back into the equations for Xp and Yp which were expressed in terms of L1 L2 and L3 that were presented earlier yields the nal form for computing the coordinates using the Tienstra method X KxAKmmXC P Kmm Knmnmn Kmm Y P SURE 215 7 Surveying Calculations Three Point Resection Problem Page 199 An example using MathCAD follows Three Point Resection Problem Tienstra Method See the same functions as defined in the KaestnerBurkhardt MathCAD program This MathCAD example is the same example used in the other methods There is a slight difference in that the triangle is lettered in a clockwise manner and 01 is the clockwise angle from line PB to line PC 3 is the clockwise angle from line PC to line PA and y is the clockwise angle from line PA to line PB See the following figure B a C Given XA 2 100000 YA 2 530000 0c 2 1150520 XB 2 220000 YB 2 630000 3 2 1352355 XC 2 310000 YC 2 500000 7 2 1093045 Solution Find the coordinates of point P using the Tienstra Method AZAB 2 501140 AZBA 2 AZAB 180 AZ CB 2 3251817 AZBC 2 AZCB 180 AZAC 2 980748 AZCA 2 AZAC 180 A 2 dd AZAC dd AZAB dms A 475608 B ddAzBA ddAzBC dms B 845323 C ddAzCB ddAzCA dms C 471029 Place the angles into radians Ar 2 Atrad Br 2 Btrad Cr 2 Ctrad oar 2 radians 0c Br 2 radians yr 2 radians y SURE 215 7 Surveyng Calculations Three Point Resection Problem Page 200 Solve for the constants used in the Tienstra Method K1 7 cotAr 7 cotocr 1 K1 072959 K2 7 cotBr 7 cot5r 1 K2 090626 K3 7 cotCr 7 cotyr 1 K3 078052 The solution is KlXA KZXB K3XC X1 K1 K2 K3 KlYA KZYB K3YC YP K1 K2 K3 REFERENCES Allan A Hollwey J and Maynes J 1968 Practical Field Surveying and Computations American ElseVier Publishing Co Inc New York Anderson J and Mikhail E 1998 Surveying Theory and Practice 7Lh edtion WCBMcGrawHill New York Bannister A Raymond S and Baker R 1984 Surveying 63911 edition Longman Scienti c amp Technical Essex England Blachut T Chrzanowski A and Saastamoinen J 1979 Urban Surveying and Mapping SpringerVrlag New York Faig W 1972 Advanced Surveying I Preliminary Copy Department of Surveying Engineering Lecture Notes No 26 University of New Brunswick Fredericton NB Canada 225 p Hodgson C 1957 Manual of Second and Third Order Triangulation and Traverse USCampGS Special Publication No 145 Reprinted 1957 US Government Printing Of ce Washington DC Kissam P 1981 Surveying for Civil Engineers 2quotd edition McGrawHill New York Klinkenberg H 1955 Coordinate Systems and the Three Point Problem The Canadian Surveyor XII8508518 MISSING DATA IN A POLYGON Surveying Engineering Department Ferris State University COMPUTING MISSING DATA USING THE GEOMETRIC APPROACH The conventional approach of solving the missing data problem is shown in gure 1 On the left is the traverse as it exists in the eld But two data items are missing Depending on the situation a combination of two items given from s1 sz on or 0c are unknown Then one simply draws the sides of the traverse which are known Then the unknown lines are grouped together Thus the traverse consists of an open traverse with line numbers 12 14 and 15 drawn in a group The coordinates of points A and C are computed by processing this open traverse Then the distance and directions between A and C can be computed and the triangle composed of AB C is computed using coordinate geometry There are three basic forms in which the missing data problem can be formulated An example from Hashimi 1988 is shown for each case A Traverse Station Olt Azimuth of line 3 Distance of line By B Line number E D39 S DE a DE D Figure 1 Traverse showing missing elements 1 Missing the distance and direction of the same line This problem is very simple to solve Determine the X and Y coordinates of each of the points in the traverse This usually means starting at one known point and going both clockwise and counterclockwise until the unknown line is reached Once the coordinates are found inverse between them to nd the distance and direction of the line SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 2 2 Missing two distances of two adjacent lines or two directions of two adjacent lines or a distance and a direction of two adjacent lines For example look at gure 2 The data that is known is given in the accompanying table 8073839 92 413952quot Figure 2 Example problem for computing missing data on two adjacent lines Compute the distance and azimuth of line 41 DH JX1 X4 2 Y1 Y4 2 J100000 1748892 50000 428902 DH 75226 X1 X4 1 100000 174889 Y Y 50000 42890 1 AzzH tan 1 4 A24 1 275 2539 24quot Then solve for the missing elements within triangle 145 For example if we know that the azimuth between points 1 and 5 Az15 was 33 47 32 and AZ45 3140 51 SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 3 18 then the problem is one of nding the distances between points 1 and 5 and between 4 and 5 Solve for the interior angles of the triangle by OL AZH AZH 95 253924quot 33 473932quot CL 61 3739 52quot B AzH AZH 314 51 18quot 275 253924quot 5 39 2539 54quot y 180 0LB 180 61 3739 52quot39 2539 54quot y 78 56 14quot Then using the sine law the distances can be easily computed D D sin 0L sinB sin y D 75226 D15 H sinB sin39 25 54quot s1ny s1n 78 563914quot DH 48685 D 475 175 174 D DH H sinoc amps1n61 37 52quot s1ny s1n78 56 14quot DH 67445 If the problem was de ned in terms of the distances with the unknowns being the directions a similar approach would be used Here the cosine law would be utilized to determine the unknown angles 0L and B The relationship of these angles with the azimuths of the lines between 1 and 5 and between 4 and 5 are shown above E The last case involves 2 unknown distances or directions on two nonadjacent lines or an unknown distance on one line and an unknown direction on another non adjacent line The conventional approach is one of reconstruction of the problem For example Hashimi 1988 gives the following shown in gure 3 and the accompanying table SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 4 STA Distance Azimuth 5 Figure 3 Example of two unknown distances on two non adjacent lines Starting with assumed coordinates of 100000 and 100000 for point 1 compute the coordinates of the lines which have both distance and direction known For example the traverse would be shown by leg l2 followed with leg 34 with point 3 of this leg coinciding with point 2 in the rst line followed with the last leg 45 Then the new con guration looks like that shown in the gure 4 With the coordinates now known for points A and B compute the distance and direction between the two points DAB JXB XA 2 YB YC2 J100000 538282 100000 570222 DAB 63079 AZ 2 tan XB XA mil 100000 53828 AB YB YC 100000 57022 AzAB 47 0339 07quot With the azimuth between points A and B now know the angles 0L 5 and y can now be computed SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 5 1276 54 X 1000 00 Y1000 00 1085 21 i 823 03 5 Figure 4 Construction of traverse by combining known sides into a group 0c AZAC AZAB 145 143920quot 47 033907quot 98 1113quot 5 AZBA Ach 227 033907quot 182 21 43quot 44 4139 24quot y 360 AZCA AZCB 360 325 143920quot 2 213943quot 37 0739 23quot Now that the angles are known use the sine law to compute the corresponding distances 63079 Sin9801113u103452 D DBC AB sinoc s1ny s1n37 07 23quot AC AB s39 63079 sin44 41 24quot73504 s1n37 07 23quot Finally check to see that the results are correct by computing the latitudes and departures for the whole traverse SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 6 42 17 35774 27654 14 20 60386 41909 60253 11 32 25280 103452 21 43 103364 As one can see the values work COMPUTING MISSING DATA USING THE ANALYTICAL APPROACH While the method of rearranging the traverse is commonly used to explain how missing data from a polygon can be computed there are alternatives Root 1970 suggests another approach by exploiting the condition that in a closed loop traverse the sum of the latitudes and departures must both equal zero From this two equations can be written with two unknowns These equations are solved simultaneously The unknowns can be classi ed into three different categories Category 1 is the situation where the two unknowns are both lengths of the traverse The following two equations can be written D1 sinAz1 D2 sinAz2 AX 0 1 D1 cosAz1 D2 cosAz2 Ay 0 where D is the distance between two points Azi is the azimuth of the line 1 and 2 indicates the lines with the unknown parameters and AX and Ay are the sums of the departures and latitudes of the known lengths respectively In the rst scenario the lengths are unknown This means that D1 and D2 are the unknown values Rearrange the second equation in 1 so that Ay D1 cosAz1 cosAz2 D 2 SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 7 This is then inserted into the rst equation in 1 to determine the distance of the unknown leg D1 AysinAz2 D1 cosAz1 sinAz2 D1 sinAzl AX0 cosAz2 D1 sinAz1 cosAz2 AysinAz2 D1 cosAz1 sinAz2 AxcosAz2 0 AysinAz2 AxcosAz2 2 1 sinAz1 cosAz2 cosAz1 sinAz2 With D1 known solve for D2 using either 1a or 1b Stoughton 1975 simpli es this equation by recognizing the trig identity sinX y sin X cosy cos X siny Therefore 2 becomes AysinAz2 AxcosAz2 sinAz2 Az1 3 Stoughton presents this example Given the following data Line Azimuth Distance 1 36 42 25 46838 2 97 34 01 D2 3 193 02 56 72300 4 222 15 08 D4 5 346 28 20 96730 The sums of the eastings AX and northings Ay or the departures and latitudes for the known links are computed as AX D1 sinAz1 D3 sinAz3 D5 sinAz5 AX 46838 sin36 42 25quot72300 sin 193 02 56quot96730 sin 246 28 20quot AX 109547 Ay D1 cosAz1 D3 cosAz3 D5 cosAz5 Ay 46838 cos36 42 25quot72300 cos193 02 56quot96730 cos246 28 20quot Ay 611635 The difference in the azimuths is Az4 Azz 124 41 07 Then the distance D is computed using 3 SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 8 AysinAzA AxcosAz4 611635 sin222 15 08quot 109547 cos222 15 08quot sinAz4 Az2 sin124 413907quot 59875 D 2 Alternatively using 2 we can nd the same value using the relationship AysinAz4 AxcosAz4 sinAz4 cosAz2 cosAz4 sinAz2 61 1635sin 222 15 08quot 109547cos 222 15 08quot sin222 15 08quotcos97 34 01quot cos222 15 08quotsin97 34 01quot 59875 D 2 In the calculations the distance is shown in a negative quantity Since a distance cannot be negative use the absolute value of this value Finally rearranging 1a gives the distance to line 4 D2 sinAzZ AX 59875sin97 34 01quot 109547 sinAz4 sin222 15 08quot 71980quot D 4 The next category covers the problem where one length is unknown and the direction of another traverse course is unknown Lets assume that Azl and D2 are unknown Designating M for the sin AZ and N for the cos Azz then 1a and 1b can be rewritten D1 sinAz1 D2MAX 0 D1 cosAz1 DZN Ay 0 Rearrange into the following form sinAz1 D2MAX 4a D1 cosAz1 D213 Ay 4b Square each equation SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 9 D M A 2 sinZAz1 5a D1 D N A 2 0052 All 5b D1 Add 5a and 5b D M A 2 D N A 2 sin2 AZ1 cos2 AZ1 2 X 2 t y D1 D1 1 DZMAXZ D2NAy2 1 Di D M2 2DZMAXAX2 D N2 2D2NAyAy2 D M2 N2D22MAX2NAyAXZ Ay2 Df 0 6 6 is in the form of a quadratic equation and the solution is in the general forms as bi1b2 4ac X 7 2a where a M2 N2 sin2 AZ2 cos2 AZ2 b 2MAX ZNAy 2AxsinAz2 2AycosAz2 cAX2 Ay2 D12 X D2 With Dz known substitute into 4a or 4b and solve for Azl As we can see there are two values for D2 Ifany of the values are negative these can be ignored Ifboth values are positive than an accurate sketch of the area is used to identify the proper values for D2 and Azl Looking at a M2 N2 we see that a 1 Also recognize that a 4 can be factored out of the radical This leads to the solution of the quadratic equation shown in the following form using the notation of Stoughton 1975 SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 10 2ViI4V2 4U 1 2 8 Vi V2 U where UAX2 Ay2 D and VAXMAyN Example from Stoughton 1975 Given the following traverse data compute D2 and AZ4 42 25 46838 34 01 71980 28 20 96730 The solution is as follows U M Ay2 D 1095472 6116352 719802 1320141216 V AxsinAzz AycosAz2 109547sin97 34 01quot6ll635cos97 34 01quot 18913588 Substitute these values into 8 yields D2 vnv2 U 18913588i 189135882 1320141216 59875 or 22048 The last value is not possible therefore the solution is D 59875 Substitute this value into 4a or 4b to solve for the azimuth of line 4 D MA 59875 39 97 34 Olquot 109547 Az4 sin 1 2 X sin 1 Sm 71980 4 42 15 08quot This result shows that the azimuth is in either the northwest 0r southwest quadrant A good drawing of the traverse would show that it needs to be in the southwest quadrant therefore the azimuth is 2220 15 08 This should be verified by computing the latitudes and departures for the traverse For example the azimuth of 2220 15 08 results in a closed traverse as shown in the accompanying spreadsheet SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 11 Ifthe angle 3170 44 513 were used the traverse data would be as shown in the next spreadsheet Stoughton 1975 presents an alternative method of computing the azimuth of the missing line His derivation is as follows Multiply 1a by sin Azl and 1b by cos Azl D1 sinAz1 cosAz1 D2 cosAz2 sinAz1 AxsinAz1 0 DlsinAz1 cosAz1 D2 sinAz2 cosAz1 AycosAz1 0 Add the equations together and rearrange D2 sinAz2 cosAz1 cosAz2 sinAz1 AxsinAz1 AycosAz1 SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 12 Divide this relationship by D cos Azl yields AX sin AZ1 AycosAz1 sinAz tanAz cosAz 9 2 1 2 D2 COSAZ1 De ne V tan AZ1 and AxsinAz1 AycosAz1 D2 cosAz1 Then 9 becomes sinAz2 VcosAz2 W 10 sinAz2 WVcosAz2 Recall the trig identity sinoc 11 cos2 1 which results in 10 taking on the form il cos2 AZ2 WVcosAz2 Square both sides and rearrange l cos2 AZ2 W2 2WVcosAz2 V2 cos2 AZ2 v2 1cos2 Az2 2chosAz2 w2 1 0 In this form use the quadratic equation to nd the cos Azz 2WV r 1 4w2v2 41 v2 w2 1 2 1 v2 Factor out W 4 yields the formula for computing the cosine of the azimuth WVi1 W2V2 w2 W2v2 1 v2 1v1 inxv2 1 W2 lvz cosAz2 cosAz2 SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 13 Stoughton points out that there are four solution for the azimuth Generally two of the solutions can be rejected outright To check the results of the remaining two one should have a goodquality sketch from which the comparison can be made Using both acceptable azimuths into equation 1 can help identify which one is correct for the particular problem The next category pertains to the situation where the direction of the two lines are unknown ie Azl and A22 are unknown Beginning with 1a and 1b rearrange into the following form D1 sinAz1 D2 sinAz2 AX D1 cosAz1 D2 cosAz2 Ay Square both sides Di sin2 AZ1 D2 sinAz2 AX2 Di cos2 AZ1 D2 cosAz2 Ay2 Add these two equations yields Di D2 sinAz2 AX2 D2 cosAz2 Ay2 Di sin2 AZ2 2 D2 sinAz2 AX AXZ Di cos2 AZ2 2 D2 cosAz2 Ay Ay2 Di 2D2 AxsinAz2 AycosAz2 AX2 Ayi Rearranging D2 D2 A 2 A 2 AxsinAz2 AycosAz2 2 DiVide both sides by JAXiAyi AycosAz2 Di Di AX2 Ay2 AX sinAz2 AXZ Ay2 AXZ Ay2 2 D2 AXZ Ay2 AX Ay sin9W 39 COS9W SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 14 D2 D AX2 Ay2 sinAz2 sin9cosAz2 cos9 1 y 11 2D2 AXZ Ayz 2 Using the trigonometric identity cosu V sinusinV cosucosv 11 becomes D2 D2 A 2 A 2 005AZ2 2 D2 AXZ Ayz 2 D2 D2 A 2 A 2 A22 9 cos 1 12 2 D2 AXZ Ayz 2 But recall that 9 cos 1 Ay AXZ Ay2 which when substituted back into 12 yields 2 2 3 2 A22 cos 1 D1 DZA ZAXA ZAy cos 1 A ZAyA 2 13 2D2 X y X y The main problem with this algorithm is that there are two possible solutions to the problem The nal determination of the azimuth of the second line will be dependent upon a good drawing of the area in questions With this azimuth now known the azimuth of the rst line can be found using 1b expressed in the following form AZI 2 0081 D2 cosAz2 Ay D1 Stoughton 1975 presents a different form of this solution that of a quadratic equation Here the azimuth of the second line is computed using without derivation AyJi1 AXZ Ay2 AX2J2 COSAZZ AXZ Ay2 where SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 15 J AXZ Ay2 Df D ZAXD2 Stoughton then shows that the azimuth of the rst line can be computed using the same formula except for the value J which is updated and referred to here as K Thus Ay K i sz Ay2 AXZ K2 cosAz1 AXZ Ay2 AX where K AXZ Ay2 D Df ZAXD1 Using the example presented in Stoughton 1975 the data are given as 42 25 46838 59875 71980 28 20 96730 Then using the formulas above adjusted for the unknowns in this problem compute J and the azimuth from 23 J M Ay2 D D 1095472 6116352 598752 719802 2 AX D4 2 10954771980 3460336 SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 16 AyJiiAXZ Ay2 AX2J2 AX2 Ay2 AX 611635 3460336iJ 1095472 6116352 1095472 3460336 109547 6116352 AZ4 cos 1 2 1 S 109547 c051111646279 r 12424042622 35244956 As Stoughton points out there are four possible answers to this quadratic equation Recall that with the ambiguous case that the angle could be either the value determined directly from the computation or its complement The possible solutions are 1370 44 524 or 2220 15 076 when the radical is added and 1170 26 214 or 2420 33 386 when the radical is subtracted Then solve for the azimuth between points 2 and 3 K AX2 Ay2 D D 2A D2 l726506 2 cos 1 Ax2 Ay2 35244956 AX As before there are four solutions to this quadratic equation They are 117052 312 or 2420 07 288 when the radical is added and 970 34 002 or 2620 25 598 when the radical is subtracted AyKrle2 Ay2 AX2K2 110559916i3503264140 AZ COS A good drawing of the traverse would show what two angles t the situation the best Absent a drawing there will be two solutions These are shown in the following two spreadsheets Traverse calculation using the solution set of AZ 970 34 002 and A21 2220 15 076 is shown as SURE 215 7 Surveying Calculations Missing Data in a Polygon Page 17 Traverse calculation using the solution set of A22 2420 07 288 and Azl 1170 26 214 is shown as The other solution possibilities do not yield acceptable results when inserted in the traverse calculations program CONCLUSION As it has been shown computing missing data in a polygon requires different solutions depending on what is unknown The solution can be found using a geometric reconstruction of the problem or by analytical methods The former may be more intuitive but does require more computations I LINELINE INTERSECTION Robert Burtoh Surveying Engineering Department I 1 EQUATION OF A LINE 0 Slope rise mtan6cotoc run SURE 2157 LinErLinE imersecnun EQUATION OF A LINE Ratios of difference in coordinates between 2 points on straight line are equal General equation YTYI XTxi Y2 TY1 X2 Txl Rearrange aX bY c 0 7 Where ayl iyz b Xzixl and CX1 YZ Y1TY1 Xszl SURE 2157 LinErLinE imersecnun a EQUATION OF A LINE EXAMPLE 0 Write equation of line passing through A43 and B105 0 Solution a YA RYE 3 5 8 bxB ixA 10746 CXAYB TYATYAXB TXA 45773773107450 0 Equation of line 4X 3y 25 0 SURE 2157 LinErLinE imersecnun EQUATION OF A LINE EXAMPLE 0 Alternative solution y3x 4 2 y3x 4 53 10 4 8 6 6y188x 32 2 8x6y500 4x3y250 SURE 2157 LinErLinE imersecnun 5 GEOMETRY OF A STRAIGHT LINE Slope yinteroept cb xinteroept ca writercept rcb Xrintercept rda SURE 2157 LInErLInE imersecnun 6 GEOMETRY OF A STRAIGHT LINE 0 Azimuth 0Ltan39l tan39l72 YZ TY1 a 0 From example my i 6 Lula 0L infri mquot 075 36 52 12 8 SURE 2157 LInErLInE imersecnun 7 GEOMETRY OF A STRAIGHT LINE Expand relationship by looking at slope and coordinates of one point y y1cotocx X1 Rearrange cotek X y y1 COtX X1 0 Then aXbyc 0 where a cot0Lm cx1cototy1 mx1 y1 SURE 2157 LInErLInE imersecnun E GEOMETRY OF A STRAIGHT LINE 0 From example 131 0 Equation of lineci4e731 53 3 3 3 3 4X 25 0 3 y 3 4X3y250 SURE 2157 LlnErLlnE imersecnun SOLVING LINEAR EQUATIONS 0 Equation for 2 lines 4x5y12 0 12x30y200 0 First multiply first equation by 3 and add it to the second 712x715y7360 12x30y200 15y7160 y15 SURE 2157 LlnErLlnE imersecnun IEI SOLVING LINEAR EQUATIONS Solve for x 4x5 1615120 4x1120 12X52 52 13 X 12 3 12x30y 20 0 12 1 30 0 20 0 75232200 0 Check results 0 0 check SURE 2157 LlnErLlnE imersecnun ll SOLVING LINEAR EQUATIONS 0 Alternatively isolate one variable like x in first equation 4X5y120 4X75y712 X7Ayi3 0 Substitute into the second equation 127y7330y200 715y73630y200 15y16 y15 SURE 2157 LlnErLlnE imersecnun 12 LINELINE INTERSECTION SIMULTANEOUS SOLUTION 3 Methods In first Substitute yinto 2quotd Simultaneous solution ofthe equations of equai39onv39SPIat equat39onv SOIVB for X 2 intersecting lines 9 var39ab39e I39ke y Using the triangle solution a2xb2y02 0 Base angle method a x t c 7 a1xblycl0 a2xb2 b 2 0 I azb xia bzxibzc b c2 0 7amp1er 01 y azbi aibzx bzci iblc2 b1 xb2 i bi 2 azbraibz SURE 2157 LlnErLlnE Imersecnun IS SURE 2157 LlnErLlnE Imersecnun IA SIMULTANEOUS SOLUTION SIMULTANEOUS SOLUTION 0 If slope is given Yy1m1X X1 Solve for x and y Pm y mzx2 yi yyzmzXXz mz mi mixi mzxz 7y W Then a m 7 miimz b1 y77mlxmlxliyl c IIIX y m1X X1YI SURE 2157 LlnErLlnE Imersecnun 15 SURE 2157 LlnErLlnE Imersecnun IE EXAMPLE Given SOLUTION TO EXAM PLE x1 125452 x2 147734 x 1200194 2 178872 m109103909 m264797612 mm mzxz y1y2 mt m2 7 52r7 64797612X1477347 200194178872 0910390947 647976 142104 Find the point of intersection 7 1 x xmy 09103909142103r125452200194 215354 SURE 215 7 L712er Intevsectiun SURE 24511121112 Intevsectiun TRIANGLE SOLUTION TRIANGLE SOLUTION B A Solve for missing elements of triangle 4 180 401 D11Jltxrx1gt2ltvarv1gt2 Az tan391 XETXA AB YE iYA SURE 24511121112 Intevsectiun TRAVE RSING AND TRAVERSE ADJUSTMENT Robert Burtch Surveying Engineering Department COMPUTING TRUE VALUE True value is recorded value plus the correction TRC Always added to the reading Error EzR T Always subtracted from the reading SURE 215 Traversing and Traverse Adjustments TRAVERSES Open v closed traverses Open loop v closed loop traverses Types of angles measured Deflection angles Interior angles of traverse Exterior angles of traverse Angles to the right DEFLECTION ANGLE TRAVERSE 140 103900quotR on gt N a 49 303900quotR 0L3 73 203900quotL 0 4 0 4 50 203900quotR OOlOEoOLl KAZMN 104 423930quot 0L3 52 253900 L N Az1 20 oaL Az2 360 0 11 11 1 SURE 215 Traversing and Traverse Adjustments 201R 011 013 014 2 140 103900quot 49 303900quot 50 2000quot 2 240 0039 00quot 20L 2 012 015 73 203900quot 52 2500quot 125 453900quot AzsR 201R 20LL AzMN 360 E4 350 303900quot 240 003900quot 125 453900quot 104 42 30quot 360 0 0239 3039 EA 00 02 30quot c a 30quot p n 5 0L1 140 10 00 30 140 09 30 0L2 73 20 00 30 73 20 30 0L3 49 30 00 30 49 29 30 0L4 50 20 00 30 50 19 30 0L5 52 25 00 30 52 25 30 114 12 30 350 3039 00quot 114 123930quot 104 42 30quot 360 0 1 SURE 215 Traversing and Traverse Adjustments INTERIOR ANGLE TRAVERSE 0Ll 66 40 30 0L2 131 35 00 0L3 97 35 00 0LA 64 00 30 0L5 227 26 30quot 0L 132 453930 11OLZ 0L3 0L4 0L5 0L 7672180 Ez 720 0339 00quot7 720 339 00quot The correctionper angle cpa is next determined if 3039 00 6 730quot The corrected angles are found byadding the cpato each observation 66 403930quot7 30quot 66 4000quot 131 353900quot730quot 131 3430quot cc 97 3539 00quot7 30quot 97 3430quot 04 64 003930quot730quot 64 0000quot 227 263930 730 227 2600 05 132 453930quot730quot132 4900quot 2a 720 0000quot SURE 215 Traversing and Traverse Adjustments EXTERIOR ANGLE TRAVERSE Special case of interior angle traverse aim n2180 0 i1 Note that interior and exterior angle traverses are closed loop traverses ANGLES TO THE RIGHT 290 013910 03 210 0132 on 140 0030quot on Az120ci Az2 n 1180 0 S 90M or SURE 215 Traversing and Traverse Adjustments From the example the angular error of closure is found as AZYZ oc1oc2 x3 x4 x5 AzTU 5 1180 EZ 250 00 00quot789 59 20quot 319 593945quot 720 0039 25quot The correction per angle is c a 5quot p 5 The corrected angles are found by adding the cpa to each measurement ocl 210 013932quot5quot210 0137quot a2 140 003930quot 5quot140 003935quot a3 290 013910quot5quot290 013915quot a4 90 013913quot5quot 90 01 18quot d5 59 543955quot 5quot 59 5900 Zai 789 5939 45quot Check using the general relationship for the angles to the right 250 003900quot 789 59 45quot 319 593945quot 720 0 SURE 215 Traversing and Traverse Adjustments STEPS IN TRAVERSE ADJUSTMENT 1 Balanceadjust angles 2 Compute azimuthsbearings of lines 3 Compute latitudes and departures LatscoaXAY DepssinocAX 4 Determine closure error nil 6D dX ZLath T Xn TX1 i1 nil 6L DY ZDGPMH T Yn T Y1 i1 STEPS IN TRAVERSE ADJUSTMENT 5 Compute linear error of closure accuracy ratio direction of closing line EC 1jcg of Accuracy 13 e Betan1 D 6L 6 Adjust traverse and compute adjusted distances directions and coordinates 7 Compute area of traverse SURE 215 Traversing and Traverse Adjustments EXAMPLE 256 4939ZA 8quot 259 29 20 6quot SURE 215 Traversing and Traverse Adjustments Azimuth 1 gt A Mark 104 05 050 0 180 180 00 000 Back Azimuth 284 05 050 11 67 34 120 Azimuth 1 2 216 30 530 12 256 49 248 40 18 318 180 180 00 000 Azimuth 2 3 139 41 282 180 180 00 000 13 259 29 206 Azimuth 3 4 60 12 076 4 64 08 405 360 360 00 000 Azimuth 4 5 356 03 271 15 64 52 175 Azimuth 5 1 291 11 096 180 180 00 000 A to Az Mark 356 53 287 360 360 00 000 Azimuth 1 7 Az Mark 104 04 382 The angular error of closure is E1szl aAszka QvlaaAszkmg 2 104 0439383quot 104 09050 2 267 The correction per angle is found using E4 267quot 6 cpa 445quot SURE 215 Traversing and Traverse Adjustments ADJUSTED AZIMUTHS LATITUDES AND DEPARTURES rave rs e SURE 215 Traversing and Traverse Adjustments Error in closure n71 eD dX 1me 7007539 il n71 eL DY ZDepmH 7017639 il Linear error of closure E 1K7 0075Z 7 0176Z 019 Azimuth of the closing line 3tanquot 039075 203 04 50quot 7017639 Accuracy ratio 019139 1 Accuracy 2109684 11045 1 11000 COMPASS RULE ADJUSTMENT cL LAM cD LAch where cL cD corrections for latitudes and departures respectively length of the line P perimeter distance around the traverse and clL ch closure correction in latitude and departure respectively for the traverse SURE 215 Traversing and Traverse Adjustments From the example For line 12 CL 483406 004039 210968439 CD 483406 0017 210968439 Line 23 CL 446622 003739 210968439 2109684 CD j44662239 001639 SURE 215 Traversing and Traverse Adjustments DETERMINE Xn Depniltnlgt COORDINATES Y 1 YR Lat n7n1 COMPUTE ADJUSTED DISTANCES AND DIRECTIONS Dn7n1 erl Xn 2 Yn1 Yn 2 I 2 2 Depn7nl Latn7n1 X X De 11 11 n1 n tan 1 p lt 1 Yn ocn n tan 391 1 I L Lat n1 n n1 J SURE 215 Traversing and Traverse Adjustments FROM EXAMPLE Line 12 DH 5172813 5460445 Z 5849543 6238012 Z 483364 x172 tan 1 5172813 546044539 2 216 31 018quot 5849543 6238012 Line 23 DH 5461737 517281339Z 5508988 584954339Z 446604 Xer etc tan15461737 5172813 139 41 204quot 5508988 5849543 Traverse Adjustment Program Adjusted Adjusted Azimuth Sta X Y Distance Deg Min Sec 1 7885572 7097635 483363 216 31 18 2 7597940 6709166 446604 139 41 204 3 7886864 6368611 425588 60 12 97 4 8256185 6580100 384957 356 3 534 5 8229766 6964149 369172 291 11 506 7885572 7097635 SURE 215 Traversing and Traverse Adjustments TRANSIT cL clL S RULE EM L cl cD D SD XIDGPI Where cL cD corrections to the latitudes and departures of the line clL ch closure corrections for the latitudes and departures for the traverse ZlLatl ZlDepl sum of the absolute values ofthe latitudes and departures and sL sD latitude and departure length of the line absolute values FROM EXAMPLE Line 12 38850939 0047 145802639 007539 287649 0016 1316501 Line 23 340592 0041 1458026 007539 1316501 288908 0016 j H J saw H J SURE 215 Traversing and Traverse Adjustments raverse rave rse SURE 215 Traversing and Traverse Adjustments CRANDALL METHOD Lati Depi Dep2 e e A D 100si J L 100s EnlDep2 EniLat2 1 IL21tiDepi 11 100s 11 100s 11 100si Lati Depi Lat2 C e B L 100si J D1005 j EnlDep2 EniLat2 RILatyDepi 11 100s 11 100s 11 100si j CORRECTION APPLIED TO PARTICULAR LINE cg1 LatlA DeplB cg2 LatZA DepZB CE Lat3A Dep3B CI LatnA DepnB SURE 215 Traversing and Traverse Adjustments CORRECTIONS TO LATITUDES General form of correction CL 02 m Lat2 DZCZ AND DEPARTURES Substituting into previous equation Lat Lat Dep 1005 1005 CL Dep 2 A Lat Dep B Dep C 103 D 100s 100s Traverse Adjustment Program Poirit Distance 1 483 406 2 446 622 3 425 557 439 384 926 5 369 173 Surn 2109 684 Departure e287 6492 288 9084 369 3052 726 43238 7344 207 70 075009 FROM EXAMPLE Correction to Adjusted Latitude 1 2 3 Departure Latitude Departure Latitude 7388 5091 3122412 1 711648 2 311811 0 048 0 064 7287 602 7388 445 7340 5924 2 597346 1 868874 72 203205 70 022 0 025 288 887 7340 567 2114531 1 05068 3 20489 1 835024 0 050 0 029 369 356 211482 384 0174 3 831109 0 018151 70 2637 70 004 0 058 726 436 384 075 133 4551 0 482437 3 209293 71 2443 0 002 70 001 7344 204 133 454 70175764 11 08398 10 01286 0 435631 0 075 0176 0 000 0 000 SURE 215 Traversing and Traverse Adjustments Traverse Adjustment Program Adjusted Adjusted Azimuth Point Departure Latitude X Y Distance Deg Min Sec 1 5460445 6238012 287602 388445 483326 216 30 575 2 5172843 5849567 288 887 340 567 446 589 139 41 371 3 5461730 5509000 369356 211482 425615 60 12 21 4 5831086 5720483 26436 384075 84984 356 3 449 5 5804649 6104558 344204 133454 17 291 11 319 1 5460445 6238012 h1 h 2 Area W Double area 2A h1h2w SURE 215 Traversing and Traverse Adjustments DOUBLE MERIDIAN DISTANCE damage L altitude 6l 3 Iquot 3 M 4i 5 2A 2A22 3339 2A33 4439 2A44395 2A56639 2A11l6639 2A113922 D M D 1 DMD of first line equal departure of that line 2 DMD of each subsequent line equals the DMD of preceding line departure of previous line departure of line in question 3 DMD of last line should equal departure of last line with opposite sign SURE 215 Traversing and Traverse Adjustments 2O DMD 2A DMDH gtxlt Lat DMDH DepH Dep671 Lat DMDS1 Dep i1 Dep172 Latli2 DMDI2 Depli2 DepH Latzi3 DMD3 Depzi3 Dep34 Lat374 DMD34 Dep374 Dep475 Lat45 FROM EXAMPLE 2A 287632 388469 1117361154 287632 7 287632 288924 340555 975145187 286340 288924 369320 211489 786536051 371904 369320 7 26418 384049 714806 7 26418 7 344194 133486 2A A 2745205295 459450803 6083698489 30418482 sq ft 698 ac SURE 215 Traversing and Traverse Adjustments 3 8026044 7271556 3Point Resection Problem pquot 8 2839 210quot p39 8quot 1539 468quot Given points and coordinates of points 1 2 and 3 given in black Identify point P from the angles 7861772 measured from P to points 1 2 and 3 6749484 The lines from P to the points are shown in red This method uses the approach presented by the US Coast and Geodetic Survey 8002734 6354664 31quot1 1 1252ka SaLm ML lt 2555 Mgr799 DamEr Azxw m an 23 I 1723 I I X3 vhf Vs V1quot 47 3M n1 1wquot I 2739 169 quot 1 L Ourwe Azmmi an z munI L I axz x 1v vquot r4la zzaz Egg XL x 74W39H1VIgt a4o zl 07 quot 01721ch Hamm FM 3 as Jagma N 407 293911593 71 A X 23 4 VIAJ 3 7 227 r 157 x 0 5 0 c DAL AAz3 Az1 Z39 quotquot sum impw zw39mquot nary52quot 3 c ma 1 203945Jquot ll Munu P I 25m MSW 3 70 2339245quot A2 A2 3 50 05 24039 b D W SWAP Inf 3544 5n wins 2 quot1091an a 7nnsHEEYs Lw sii mmn it QWT PT F X D3 55 A2 Yr y 1 1 M A Milk4 EESuATJ m 50 Wm 5V P 5 AP39Z Dz A2 A x J L M DHATES a linw P XrIXL DLp 55 A117 Yp 1er D1 may 625574 E 3a39zda quot 557459 153722142quot Example 3Point Resection Problem ddang degree oorang radiansang d ddang mins ang 7 degree1000 d 7E minutes oorm1115 1800 seconds mins 7 miI1utes1000 dmsltang degree oorltang minutes seconds degree rem ang 7 degree60 600 36000 mms oorrem reml rem 7 mins trad 1 secs rem1600 180 d mms secs e ree td 180 g 100 10000 eg 7 Given the following values X1 8002734 Y1 6354664 X2 7861772 Y2 6749484 0L is the angle between points x3 8025044 Y3 7271555 1 and 2 designated by the USCampGS as pquot while B is the angle a 82821 5 315468 between points 2 and 3 designated by the USCampGS as p39 Solution Distances and azimuths between the control points D 397 X X 2 Y Y 2 D 7 547 30655 237 3 23 2 23 A223 atan2Y3 4 Y2X3 4 22 dmsA223tdeg 17275899 D 397 X X 2 Y Y 2 D 741922919 127 2 12 1 12 Az12 atan2Y2 4 Y1X2 4 X1 27 dmsA212tdeg 340210715 D31 X3 7 X12 Y3 7 Y12 a 3 D23 b 3 D31 0 3 D12 Az31 atan2Y3 7 Y1X3 7 X1 7 A A231 7 7 Az12 A7 radia11s0L7 radiansB 2 Z atan2bsinradians 1 csinradians s 7 atan2tanZ gtans X S S y 3 S 7 S In triangle P13 Ap133 7E y radians A231 AZ31 Y b D 3P sin radians 61395in1AP13 The coordinates of point P XP X3 D3PsinAZ3P YP Y3 D3PcosAz3P Checking the results AP12 7t 7 radians0L X D2P 3 00139sm1APIZ 0 sin radians AZZP A212 7 7E X D31 91718826 dmsAz31tdeg 181272271 dmsAtdeg 2106156 dmsStdeg 211039 dmsZtdeg 7 2401483 dmsStdeg 050152 dmsXtdeg 301190 dmsytdeg 120487 dmsAP13tdeg 17023245 dmsAZ3Ptdeg 18006340 D3P 106538438 dmsAP12tdeg 16830200 D2P 5670159 dmsA22Ptdeg 163222620 Robert Burtch SURE 215 Surveying Computations LINECIRCLE INTERSECTION Robert Burtch Surveying Engineering Ferris State University LINECIRCLE INTERSECTION V Write equation of line and circle yeyi mXXIeymXXIyi R2 xh2yk2 V Expand equation of circle xh2 yk2 18 0 x22hxh2 y22kyk2R2 0 LINECIRCLE INTERSECTION V Substitute into equation of line written in terms ofy x2 thh2 mXxy2 2kmXxy k2 R2 0 x2 thh2 rer2 Zmzxx 2mxy quotBx2 zmxm y2 zkmx2kmx2kyk2R2 0 V Has form of quadratic equation 1m2x2 2hm2x1 my mkx h2 k2 y2 m2x2 R2 zmxy 2kmx zky 0 LINECIRCLE INTERSECTION V Solution of quadratic equation x7 biIb2 4ac 7 2a there a 1 m2 b 2hm2x my mk c h2 k2 y2 m2x2 R2 2mxy 2ka 2ky SURE 215 Surveying Computations Robert Burtch LINECIRCLE INTERSECTION LINECIRCLE INTERSECTION EXAMPLE EXAMPLE V R 2 h 3 k 7 XA yA ofpoint on line 313 181 sz 62 40 14 AZ 62 11 40 37 V Solution recall m is slope de ned by b zlhmzxx myxmk b 72347cot2 62 11 40 17cot 62H1 40 4cot62quot1140 7 cotangent of azimuth b 7972041 ch2 k2 y2 m xi 7R2 72mxy2mx72ky c 32 72 42 cot2 62 1 1 40 1 7 Zcot62 11 40quot147 22 2cot 62 1 1 40 17 7 247 c 717 4423 LINECIRCLE INTERSECTION LINECIRCLE INTERSECTION EXAMPLE EXAMPLE V Solve using quadratic equation 7 b lb2 4 7 79 72041 J79 720412 74127811174423 7 A 7044 2a 2 128811 79 720417 J79 720412 741 27811 174423 2 9009 2 128811 V Solve for y cot 62 11 40 47044 71 4 7 59536 ymxixlyl 7 7 cot 62 11 40 29009 71 4 7 50025 Robert Burtch SURE 215 Surveying Computations LINECIRCLE INTERSECTION LINECIRCLE INTERSECTION EXAMPLE TRIANGLE APPROACH EXAMPLE TRIANGLE APPROACH V Angle at point A is V Distance and Azimuth difference in azimuths V Given from A to center of 7 R 2 h 3 k 7 XA yA ofpoint on line 14 7 Az 62 11 40 a circle found by along the line and DAO or A2Y07YA2 fromAtoO 2 2 I31 7 4I e 52 11 40 7 33 4124 505 28 3016quot LINECIRCLE INTERSECTION LINECIRCLE INTERSECTION EXAMPLE TRIANGLE APPROACH EXAMPLE TRIANGLE APPROACH V From this V Using sine law p and y are found as a 180 7 1 20 383902quot28 3016quot P SM sine DAD 5111 5111283016 3606 a 30051 42 DO 2 B 180 r 59 213957quot28 30 16quot 3 92 0739 47quot 120 38 02quot 39 e 39 28D303916quot Y sin I 5m DA0 sin I sm 3606 DOF 2 59 2157 Robert Burtch SURE 215 Surveying Computations LINECIRCLE INTERSECTION EXAMPLE TRIANGLE APPROACH V Using the sine law again the distances from point A to points 1 and 2 are D 3606 DAP A0 smot sm30 51 42 smq sm120 383902quot 2150 DAP D 3606 DAP A sin fjstTWWW 2 any 511159 2157 D 4188 AP LINECIRCLE INTERSECTION EXAMPLE TRIANGLE APPROACH V The coordinates of the points of intersection XF XA DA sinAZAP 1 21505in620113940 y Y YA DAF cosAZAF 1 2149c05620113940 m X XA DAPZ sinAZAF 1 41885in620113940 sz YA DAB cosAZAF 1 4188c05620113940 g LINECIRCLE INTERSECTION EXAMPLE TRIANGLE APPROACH V As a check compute distances 2 2 from the center Of Don 290 3 500 7 200 circle to the pOthS DOE 470 32 595 72 200 l of intersection 1 and 2 7 Should be equal to radius of circle EXAM P L E 2 X1 5313 674 Y1 42111 812 Km 4977455 Yn 3951449 Distance 17m 1 to point P 327 387 D Radius ofthe Circular curve 8 5 Caml angle A48 39 53quot Forwa d Azimuth ofBack Tangmt137 54 22 19 24 Robert Burtch SURE 215 Surveying Computations EXAMPLE 2 V Solving part of horizontal circle TRtan 81952412 w 2 2 37058739 XPC XP1 TsinAzMC 497745539370587sin317 54 22 XPC 4729033 no YP TcosAzMC 395144939370587cos317p543922quot ch 422644239 EXAMPLE 2 V AzpcO AZFH 790 137 54 22quot790 47 543922quot x0 XPC RsinAzPCO 472903 819 524sin47 543922quot x0 5337159 YO YPC RcosAchO 395144939 81952400547 54 22quot YO 477580839 EXAMPLE 2 Du i XrXu2 r r x u2 5313674753371592 42UU812747758U8Y 575475 Law s rDh W ZDume 28195210575475 18quot17 U3 eusoc PE I 3105B7 PT x Yi Yu AZD 182 2020 X dm 531367475337159 AZD Lan 4200 81274775 808 AZD P AZD Hut 182 ZO ZO 18 17 03 AZD P 200 37 23 EXAMPLE 2 V The coordinates of the point of intersection between the circle and line is XP XO DH sin AZH 5337159 819524 sin 200 3723quot XP 5048508 YP YO D0 cosAzH 4775808 819524 cosZOO 373923quot YP 400880139 3Point Resection Problem Given points and coordinates of points 1 2 and 3 given in black Identify point P from the angles measured from P to points 1 2 and 3 Construct a circle through points P 1 and 3 as shown in red Extent line P2 until it intersects the circumference of the circle This point is labeled 0 Draw lines from points 1 and 3 to 2 This construction is shown in green 4908975 oc 40 3539 221quot 5 9 1839324 4905726 7221493 5297154 7050825 3Waqu Easy Thou SaLW u V rmcg 539 2mm rm z a 0 xg xl val quot r 437 45 xx1 a I A u39 YPV 925334 04mm 3 zzmm From 12 I DA ffxx XJ NYm AY J 1 421mm 5 1quot xL x a y n J a w Emur5 g llama p5 34 1 I 3911 0 e X X 9 Y 5 via1554 quot Ag 1quot LEquotXxvsl 4391 g 072 MIgmrz Aw 75 manm z3 5 A12 I1st a7395 3quot 41quotAeu A2 32 5 zataquot gur e A2 33 n3935naquot Ham 5 Jammy at 4 mzcai 435 130 211774 40 373922J 5 3le 437175 quotquot0 gl z39 MflI 4316 mac raai zm lia o yoLyh Ml munLE orgl wan SA MW 061 24 s39quot 3457 gm 15m 0 pg shhm 1525134 w A quot AR quot5 7 quottquotn3945 431 A 3 elrt39u39m quot X X3039Du35 hAl 595412437 Yr Y Wu M A 7144po 4514 zw39dv39aezquot r1424 0 W X quot Y 40 quotL 120315 7 sz 20 Jlao lt44 o2 194quot 4 m 5 Pa w 772mm Hz 4P1quot quot xny ar 1quot9 p39 quot391r51 9l 6339ar395quot Pu S 3 5 4 p 4 15 4f0 I 6 39 I14quot 39 3m 2413933953 EmmaTH at gun 7 x9 x o m w 42 Yr YMHrWllzw T mm 432 15511 r 39o393 7quot 4 wt 4 33 m 739a py EX xyap Manx 19234 r A e A I M za oza X61 D MS39h Raquot v YF Ysr u mlz 413 57 4 Example 3Point Resection Problem ddang degree oorang radiansang d ddang mins ang 7 degree1000 d 7E minutes oorm1115 1800 seconds mins 7 miI1utes1000 dmsltang degree oorltang degree m seconds rem ang 7 degree60 600 36000 mms oorrem reml rem 7 mins trad secs rem1600 180 d mms secs e ree td 180 g 100 10000 eg 7 Given the following values X1 5297154 Y1 7050325 X2 4905726 Y2 7221493 0L is the angle between points X3 4908975 Y3 7653629 1 and 2 while B is the angle between points 2 and 3 11 4035221 5 918324 Solution Distances and azimuths between the control points D23 A223 X3 4 X22 Y3 4 Y22 atan2Y3 Y2 X3 7 X2 X2 7X12 Y27 Y12 atan2Y2 4 Y1X2 4 X1 27 X3 7X12 Y3 4 Y12 D23 43714807 dmsA223tdeg 0253303 D12 42701691 dmsA212tdeg 293332831 D31 72118558 A231 atan2Y3 7 Y1X3 7 X1 7 dmsAZ31tdeg 147260719 The lnterionr Angles to the triangle formed with the control A123 3 A212 7 A223 A231 3 A223 t 75 AZ31 A312 A231 7t 7 A212 From the geometry of a circle A130 radians0L AOPI radians 0L A310 radianSB A3P0 radians Then A301 3 75 A130 t A310 ln triangle 031 using the Sine Law D D013 31 31119130 SmA301 D1 D03 3 W39S WAMO 3 301 Then computing the coordinates of point 0 on the circle AZ30 3 AZ31 A130 A201 A230 1 7 A301 X0 X3 D03sinAz30 Y0 3 Y3 D03COSAZ30 Then A202 atan2Y2 7 YOX2 7 X0 275 AZOP A202 D02 3 X2 7 X02 AP01 3 AZOP AZ01 2 Y02 dmsA123tdeg 11307553 dmsA231tdeg 3259258 dmsA312tdeg 3352389 A2P1 radians0L A3P2 radians5 dmsA301tdeg 13006055 D01 61344720 DO3 715251339 dmsAZ30tdeg 10650451 dmsAzOltdeg 15644396 X0 505494372 Y0 761443094 dmsA202tdeg 20047392 D02 42031672 dmsAP01tdeg 4402596 LINELINE INTERSECTION Surveying Engineering Department Ferris State University Introduction A common problem in surveying is the solution of a problem using coordinate geometry This is the basis of computing latitude and departure and the inverse relationships of distance and azimuth that were discussed in the last section Now we shall move our attention to other forms of coordinate geometry mainly how to determine the location of a point of intersection between two lines a line and a circle and two circles Equation of a line The slope of a line m can be given by the following relationship found in analytical geometry gure 1 Y 0 Figure 1 Basic relationship for equation of a line mtan9 where 9 is the angle measured from the XaXis In surveying we often use the azimuth or bearing designated as 0L Then tan 9 cot900 9 cot 0L Thus from the surveying perspective the slope is written as rise m cot X run SURE 215 7 Surveying Calculations LineLine Intersection Page 2 From basic trigonometry one can also easily see that 1 run 1 tan X cot 0L r157 rlse run To write the equation of the line one needs two items either the coordinates of two points on the line X1 Y1 and X2 Y2 or the coordinates of one point on the line and the slope of the line We should easily recognize that the ratios of differences in coordinates between all points on a straight line are equal The differences are always taken from the same point Lets look at a very simple line whose azimuth is 45 We know that if X 1 unit then y 1 unit if X 2 units then y 2 units if X 150 units then y 150 units etc Lets identify three points on the line y1 1 y 2 and y3 3 units respectively The X coordinates are then X1 1 X2 2 and X3 3 units respectively Then 373 371 312X3X1 372 371 21 1 X2X1 This will be valid for all lines Again for simplicity lets look at a line whose run is twice the rise X 2y The appr0Ximate azimuth ofthis line is 630 26 06 If y1 1 then X1 2 if y 2 then X2 4 if y3 3 then X3 6 etc Then using the same relationship as above 2i 2 X2 X1 Therefore we can write this general equation given two points on the line with the third point being unknown Rearranging yY1X2 X1XX1yZ Y1 YX2 X1Y1X2 X1XYZ Y1X1YZ Y1 XY1 yZYX2 X1X1yZ Y1Y1X2 X10 In a general form this can be represented as SURE 215 7 Surveying Calculations LineLine Intersection Page 3 aX bY c 0 where a y1 7 y b X2 7 X1 and C X1072 YI yl X2 7 X1 Example Write the equation of a line that passes through point A 4 3 and B 10 5 Solution a 73 5 78 b 10 4 6 c 45 3 310 4 50 The equation of the line is 8X 6y 50 0 which reduces to 4X 3y 25 0 Alternatively the equation of the line can be formed using the coordinate ratios y3X 4 y3X 4 53 10 4 8 6 6y188X732 gt 78X6y500 74X3y250 Figure 2 shows the geometry of a straight line The yintercept is given as 7cb while the Xintercept is de ned as wa The slope of the line is found by SURE 215 7 Surveying Calculations LineLine Intersection Page 4 yintercept cb xintercept ca Figure 2 Geometry of a straight line Y2 Y1 X2 X1 crlxs The azimuth is determined using b OL tan 1 0r OL tan 1 Y2 Y1 339 From the previous example we can compute the slope and azimuth OL tan 1 i tan 1 075 36 52 12quot 8 We can expand our relationships by looking at the slope and the coordinates of one point X1 yl Then we have yY1 COtOCXX1 Rearranging cotocxy y1 cotocgtlltx1 0 SURE 215 7 Surveying Calculations LineLine Intersection Page 5 0r aXbyc0 where acot Xm b1 cchotOCy1mX17y1 Continuing our example we can de ne the coefficients for the equation of the line are computed as a A b 1 c i 43 2 3 3 3 3 which becomes after inserting into the equation of the line j g 0 y 3 or 4X 3y 25 0 Solving a Linear Equation Given the equation for two lines 4X5y120 12X30y200 solve for the variables X and y To do this lets rst multiply the rst equation by 3 and add this to the second equation 4X5y120 3 12X 15y 360 12X 15y 360 12X30y200 15y 160 16 y 15 Substitute the value for y into either of the two equations and solve for X SURE 215 7 Surveying Calculations LineLine Intersection Page 6 4x5 1512 0 4x163120 12x 52 2 E 12 3 Finally check the equations to see if the equalities are correct 12X30y200 12 1 30 1615 20 0 5232200 00 check An alternative approach to solving the linear equations is to take one of the equations and isolate one of the variables such as X 4X5y120 4X 5y 12 X y 3 Substitute this value for X into the second equation 12 y 330y200 15y 3630y200 15yl6 16 y d 5 Next solve for X by substituting the value of y into either equation and check the results EXample Find the coordinates of the center of a circle passing through the following three points Solution Recall from analytical geometry that the perpendicular bisectors to lines chords connecting points on the circle all will pass through the center of the circle SURE 215 7 Surveying Calculations LineLine Intersection Compute the slope of line 12 Page 7 m use y2 y1 run 7 2 3 X2 X1 3 1 The slope of a perpendicular line from the center of the circle is the negative reciprocal of the slope of the chord Inf A Compute the midpoint of line 12 by nding the average values of X and y i2 2 x 72 y 45 2 With the slope and coordinates of a point on the line write an equation of a line aX by c 0 where acm m b1 ccot0cgtlltx1 y1 mX1 y1 2 45 22 26 3X 2650 5 y 5 2X5y 2650 Hu For line 23 the slope and corresponding slope of the perpendicular bisector are 4 7 3 4 3 The midpoint coordinates are SURE 215 7 Surveying Calculations LineLine Intersection Page 8 Xz z 5 2 2 The coef cients for the equation of the line become a A b1 4 20 165 35 Ah 55 A 4 A Finally the equation of the line is shown to be Axy353 0 4X3y350 2 Solve 1 and 2 simultaneously by rst multiplying 1 by 2 and then adding that to equation 2 22X5y 2650 gt 4X10y 530 4X10y 530 4X3y350 13y 495 0 y 380769 13 Substitute y into either one of the equations yields 2X5 2650 13 26X 2475 3445 0 26X 97 X 373077 As a check compute the radius of the circle to the 3 points SURE 215 7 Surveying Calculations LineLine Intersection Page9 r1 373077 12 380769 22 V 327488 r2 373077 32 380769 72 V 327488 r1 373077 72 380769 42 327488 Line Line Intersection The determination of the intersection point of two straight lines can be accomplished in three different ways 1 Simultaneous solution of the equations of 2 intersecting lines 2 Using the triangle solution 3 Base angle method Simultaneous Solution of Two Equations The point of intersection can be computed by solving two equations simultaneously Take the rst equation of a straight line and isolate one of the variables such as y a1Xcr b a1Xb1ycl0 gt 1 Substitute this value into the second equation and solve for the unknown variable X a2Xb2yc2 0 a2Xb2 M cz0 b1 azle a1b2X bzc1 blc2 0 azbr a1b2X bzcr blcl b c b c X 2 1 1 2 albl arbz Ifthe equation of the lines are expressed in terms of slope 37 31 m1XX1 yyZ m2XX2 one can solve for X and y in a similar fashion as shown above Alternatively one can substitute the following equalities into the equations for X and y shown above SURE 215 7 Surveying Calculations LineLine Intersection Page 10 a m b1 cmX y X In1X1 Y1 m2X2 YZ m2 m1 In1X1 m2X2 Y1 372 In1 1112 m1Xm1X1 Y1 y 1 m1XX1Y1 Example Given the coordinates of two points on two different lines and the slope of the lines compute the point of intersection X1 125452 X2 147734 y1 200194 yz 178872 m1 09103909 m2 64797612 Solution In1X1 m2X2 Y1 372 In1 m2 09103909125452 64797612147734 200194 178872 09103909 64797612 142104 ym1XX1Y1 09103909142103 125452 200194 215354 Solution Using Triangle Solution Ifthe coordinates of two points are given and the observed angles to the unknown point from the two points along with a direction are known then the coordinates of the unknown point can be easily computed SURE 215 7 Surveying Calculations LineLine Intersection Page 11 A a 6 Figure 3 LineLine Intersection Using the Triangle Solution The distance between points A and B is found using the Pythagorean theorem DAB VXB XA2 YB YA2 The azimuth from A to B is found using the arctangent function XE XA YB YA The angle at point C is found using the fundamental property that the sum of the angles of a triangle must equal 180 Thus AZAB tan 1 4C 180 0cB The sine law then can be used to compute the missing parts ofthe triangle The sine law states that the ratio formed by the sine of an angle to the distance of the line opposite the angle must be equal Therefore sinoc sinB sinzC DEC DAC D AB from which we can write SURE 215 7 Surveying Calculations LineLine Intersection Page 12 D D Sin AC Bsin AC J D DBC 5m 1 AB smz C Next compute the latitudes and departures as normal and apply these values to the known coordinates A ready check is available by computing the coordinates from both points Base Angle Method The Base Angle Method allows the surveyor to compute the coordinates of the unknown point C without having to compute the directions or lengths of the sides as was done in the triangle method Begin by writing the equation for computing the Xcoordinate of the unknown point from the Xcoordinate of one of the known points like point A gure 3 XC XA DAC sinAzAC Using the sine law write an expression for the distance between points A and C D D D AC AB gt DAC AB JsinB 1 sinB sinzC sin AC Therefore the Xcoordinate for C can be expressed as DAB 5m 3 5m AzAC X X C A sin AC Next using two basic trigonometric identities sinzC sin1180 0BJsinocB 2 sinAzAC sinAzAB OCSin AZAB cosoccos AZAB sinoc 3 substitute these identities into the equation for Xc yields DAB sinB X X C A sinocB sin AZAB cos 1 cos AZAB sin X SURE 215 7 Surveying Calculations LineLine Intersection Page 13 We can also write the sine and cosine of the azimuth in terms of the difference in coordinates to the distance shown as follows sin AZAB XBD XA 4 AB cos AZAB YBD YA 5 AB Substitute these relationships into the equation for Xc X X IDABsinB XE XAcosoc I YB YAsinoc C A I sinocB DAB I DAB which yields XcXAfi E KXB XAkmag Y n smocB or Xe XA XB XAsinBcosocYB YAsinBcosoc sin0 B sin0 B This relationship can be developed without the use of the sine of the sum of angles From trigonometry we have the identity cosoc cot ocsinoc 6 Substituting this identity into our equation for Xc gives us the following equation XB XAcotocsin ocsinB I YB YA sin ocsinB sinocB I sinocB XCXA Again from our list of trigonometric identities we can write the following equation sinocB cotoccotB smocsmB 01 SURE 215 7 Surveying Calculations LineLine Intersection Page 14 1 sin ocsinB 7 cotoc cotB sinoc B Substitute this equality into the equation for Xc Xc XAXBXACOHX YB YA cotoccotB cotoccotB XAcotocXAcotBXBcotoc XAcotocYB YA cotoccotB XC This equation reduces to YB YAXA cotBXB cotoc Xc cotoccotB Using a very similar approach similar equations can be written for Yc Begin with the basic relationship Yc YA DAc cosAzAc First use the relationship in 1 into this equation DAB schosAzAc Y Y c A sinzc We have the following trigonometric identity cos AZAc cosAzAB 06 cosAzAB cosoc sinAzAB sinoc Substitute this relationship and that shown in equation 2 into the relationship for Yc DAB sinB Y Y c Asinoc5 cos AZAB cos 1 sin AZAB sin X The relationships that were described in equations 4 and 5 are now substituted into the previous equation SURE 215 7 Surveying Calculations LineLine Intersection Page 15 LDABsinB YB YAcosoc XB XAsinoc I c A sm 06 D D Y AB AB This reduces to the following form sinB Y Y c Asinoc5 YB YA cosoc XB XAsinoc 01 YeYAltYB YAgt EES ltXB gtltAgt m Next using the trigonometric identity shown in 6 in this last equation we can begin to derive a relationship for the Ycoordinate 0f the unknown point without the sum of angles as we did for the Xcoordinate Yc becomes YB YAcotocsinocsinB XB XAsinocsinB sin0 sin0 YCYA Substituting the identity shown in 7 results in YB YACOtX XB XA Yc Y cotoccotB cotoccotB A YA cotocYA cotBYB cotoc YA cotoc XB XA cotoccotB This reduces to the nal form XA XBYA cotBYB cotoc Yc cot X cotB It is very important to look at the sign of the angles 06 and B in applying the base angle method Standing at point A and looking towards point B if the azimuths put the intersection to the right then 06 and B are positive If they are to the left then 06 and B are negative Example Refer to the data given in the gure SURE 215 7 Surveying Calculations LineLine Intersection Page16 Bx 174409 Y 170159 A 1 0 6703515 1 6 8201535 X 89503 Y 14265139 1 1 1 l bc Figure 4 Example of a lineline intersection problem Solution Example of LineLine Intersection Problem ddang degree 7 oorang radiansang d 7 ddang mins 7 ang 7 degree 1000 11 minutes 7 oor mins d 1800 seconds 7 mins 7 minutes1000 drnsang degree 7 oorang degree rum mes seconds rem 7 ang 7 degree60 600 36000 m1ns 7 oorrem reml 7 rem 7 mins 1800 todeg secs 7 rem1600 mins secs degree 100 10000 Given 01de 673515 5de 821535 0L radianslt0 dms 393 radianslt dms 0L 117962 XA 3 89503 XB174409 5 14357 YA 3 142651 YB170159 SURE 215 7 Surveying Calculations LineLine Intersection Solution AZAB atan2YB 7 YAXB 7 XA AngC7E70L B From the sine law DABN DACsmmi 1 AZAC AzAB at The Coordinates of the point of intersection XC XA DACsinAzAC YC YA DAGcos AZAC As a check compute the coordinates at point C from point B AZBC3 AZAB TE 7 393 DAB 1 DEC S 1a XC XB DBC 5111AZBC YC YB DBC cos AZBC Page 17 dmsAzABtodeg 720255 DAB 89250876 dmsAng Ctodeg 30091 DAC 176062867 dmsAzACtodeg 139381 s 203155 an dmsAzBCtodeg 169472 DBC 164260197 3 QO ESZE check 00 check SURE 215 7 Surveying Calculations LineLine Intersection Page 18 Alternatively using the base angle method xC xA 7 XAcos0L YB 7 YAsjnoc YC YA UYB 7 YAcos0L 7 XB 7 XAsin0L m 7 033523 E C 78390 These values agree to those calculated above Ambiguous Case A problem occurs when computing the intersection of two lines when the length of two lines and an angle opposite one of them is given This is referred to as the ambiguous case since there could be more than 1 intersection Actually four situations are possible In gure 5 an acute angle 06 measured from the xaxis sets off point C with respect to A The distances a and b are given In 5a we can see that only one intersection occurs and because the arc is tangent to the xaxis a right triangle is formed From trigonometry one can see that if a b sin 1 then only one triangle exists If bsin X gt a as shown in 5b then no intersection can exist If bsin 1 lt athen two intersections exist Finally if a gt b then 5c is formed and if a 2 b then 5d occurs Figure 5 Four possible situations for the ambiguous solution SURE 215 7 Surveying Calculations LineLine Intersection Page 19 An easy way to determine this is by evaluating the sine law Thus a b sin 06 sinB If lsin gt1 then no triangle can exist case b On the other hand when sinB1we have a right triangle as indicated in case a Finally when lsin lt 1 two solutions are possible Example A gtlt5822690 Y4879169 Figure 6 Example for the ambiguous case 0c AzAc AZAB 229 56 15quot 169 12 26quot 62 4339 49quot From the sine law a b sin 1 sinB from which we can write SinB smoc b sm62 4349 893714 a 802308 09901 Therefore since sinBlt 1 there are two solutions one with B 81 03 30quot and another using the complement B 98 03 30quot SURE 215 7 Surveying Calculations LineLine Intersection Page 20 Example 1 Line Line Intersection Problem Find the coordinates of the point of intersection of two lines given the azimuth from point 1 to point 3 being 3340 48 47 the azimuth between point 2 and point 3 being 308 39 58 and the coordinates ofpoint 1 and point 2 being X1 5447330 X2 5752796 Y1 4080822 Y2 4377 864 Solution The geometry for this problem is depicted in gure 7 The solution is shown in the following MathCAD program Figure 7 Example 1 ddlanq dcqrcc e oorlonq radianslonq d eddlanql min e long 7 dcqrcc 4000 d 7 minutes e oor mins 1800 seconds elmins 7 minutes 1000 dmslanq dcqrcc e oorlanq minutes seconds rem e long 7 degree l 50 degree 500 35000 min lt oor rem rem e ram 7 mins 1800 secs e rem 500 7 mins secs dcqrcc 100 10000 todcq The given date are X 1 5447330 Y 4080822 A2133 radians3344847 X 2 5752796 Y2 4377864 Az 23 radiunsI 3083958 SURE 215 7 Surveying Calculations LineLine Intersection Solution 01 A2137 AZ23 A212 u1onZY27Y1X27 X 1 D123 XZXZYZYZ B3 AZ1ZZTAZ3 D 23 i sin 5 sin 01 X 3 X 2 D 233951T1AZ 2 Y3 Y2 D 23C05AZ 2 As 0 check compute the coordinates of point 3 from point 1 7 z n 7 01 5 D 13 sin7 D12 sina X33X1 31351110542139 Y3Y1 D 13cosAz1 Example 2 Line Line Intersection Page 21 dms011odeq 260849 dmsAz121odeq 454804 D 12 426079 dms 5 1odeq 705917 D23 9141358 dms71odeq 825154 D 1 9593917 22 50 check check Problem Find the coordinates of the point of intersection of two lines given the azimuth from point 1 to point 2 being 2290 56 15 and distance of 893714 the azimuth between point 1 and point 3 being 1670 12 26 and a distance of 802308 and the coordinates ofpoint 1 being X1 5822690 Y1 4879169 Solution The geometry for this problem is depicted in gure 7 The solution is shown in the accompanying MathCAD program SURE 215 7 Surveying Calculations Figure 8 Example 2 LineLine Intersection 1 Page 22 ddang degree 7 oorang mins 7 ang 7 degree1000 minutes 7 oorrnins seconds 7 mins 7 minutes 1000 radiansang d e ddang 1 d 1800 drnsang degree 7 oorang degree minutes seconds fame Eng 7 degree60 600 36000 m1ns e oorrem rem1 e rern 7 mins secs 7 rem1600 1800 todeg m1ns secs TE degree 100 10000 The given data are X1 3 5822690 Y1 4879169 D12 893714 Az12 2295615 Az123 radiansAz12 D23 3 802308 A213 3 1671226 A213 radians Az13 SURE 215 7 Surveying Calculations LineLine Intersection Page 23 Solution 1 0 A212 A213 drns0ctodeg 624349 B 39 asin Sin 0 D r D23 12 dmsBtodeg 815630 7 3 7 0 t B dmsytodeg 351941 D23 D13 3 7 D13 5219485 A223 A212 7 7 7 dmsA223todeg 851556 X2 x1 4 D125mAZ12 x2 5138692 Y2 Y1 D12cos A A212 Y2 4303954 X3 X2 D23sinAz23 98 6 Y3 Y2 D23 cos A223 2 2 D23 quot X3 X2 3 Y2 D23 8023080 check Solution 2 point 3 is indicated as 339 in the figure B1 3 7B drnsB1todeg 980330 7137E0L B1 drns71todeg 191241 D 23 D133m3m71 D13 2970151 AZ2312212 7E Y1 dmsA223todeg 690856 X3 3 X2 D23 sinA223 Y3 Y2 D23 cos A223 33 D23 X3 7 X22 Y3 7 Y22 D23 8023080 check Example 2 Line Line Intersection Problem Find the coordinates of the point of intersection of two lines given the distance from point 1 to point 3 of 1201764 to the right and the distance between point 2 and point 3 being 527808 to the left and the coordinates of points 1 and 2 being SURE 215 7 Surveying Calculations LineLine Intersection Page 24 X1 5822690 Y1 4879169 X2 5605521 Y2 4077758 Solution The geometry for this problem is depicted in gure 8 The solution is shown in the accompanying MathCAD program 3 Figure 9 Example 3 ddang degree 7 oorang radiansang d 7 ddang mins 7 ang 7 degree 1000 d 1 minutes 7 oorrnins 1800 seconds 7 mins 7 rninutes1000 dInsang degree e oomang minutes 5 econds degree rem lt7 ang 7 degree60 600 36000 rmns 7 oorrem reml 7 rem 7 mins secs 7 reml600 1800 todeg mrns secs degree 100 10000 The given data are X1 5909403 Y1 497479 X2 5605521 Y2 4077752 D13 1201764 D23 52780 AREA PARTITION IN G Surveying Engineering Department Ferris State University INTRODUCTION The division of a parcel into two or more separate polygons is a common surveying application For a rectangular gure the process is pretty straightforward Unfortunately the process becomes a little more complex for irregular gures This paper will look at the problem from a number of view points First the problem of dividing a quadrilateral will be discussed Then a gure with more than four sides will be presented and nally how to divide a polygons with a curved side will be shown Division is usually performed by Anderson and Mikhail 1998 Mof tt and Bossler 1998 I Dividing the parcel with a line whose direction is known such as parallel to another line within the traverse I Dividing the parcel with a line from a given starting points on the perimeter of the polygon SUBDIVIDING A QUADRILATERAL Stoughton 1986 presents an excellent overview of subdividing a quadrilateral He has identi ed six different cases on how the subdivision can be computed Figure 1 shows the general polygon that will be used to develop the theory Stoughton presented in his paper The original parcel is de ned by points A B C and D whereas the ends of the dividing line are represented by points E and F B E Parcel A Parcel B Figure 1 General form of quadrilateral used in developing area partitioning theory SURE 215 7 Surveyng Calculations Area Partitioning Page 110 Case I is the simplest case Here lines BC and AD are parallel to each other and the dividing line EF is parallel to one of the sides such as AB The perpendicular distance between lines BC and AD h is found as hDAB sine where 0 is the acute angle at point A The area A of the resulting parallelogram that forms parcel A is found by ADFAh DFADAB sin 0 Solving for the unknown distance to the dividing line DFA A D DAB 1n0 FA Example Divide the area for the quadrilateral shown in gure 2 into two equal areas with a dividing line parallel to line AB 20848539 210934 1750533923 57897 21263539 153186 100000 111550 B 100000 A 50000 Figure 2 Example quadrilateral with lines AD parallel to BC and dividing line parallel to line AB Solution The following table shows that the total area of polygon ABCD is 680490 square feet The desired area for parcel A is therefore 340245 square feet SURE 215 7 Surveyng Calculations Area Partitioning Page 111 The angle 9 is found as the difference in azimuth between lines AD and AB 9 47 30 25quot 0 00 00quot 47 30 25quot The distance DAF is found using the relationship developed by Stoughton 340245sq 749695 61550 sin 47 303925quot AF Inserting this value for the distance DBE and recognizing that DEF DAB and AZEF AZAB 1800 we can see from the results in the following table that the correct distance was found The slight discrepancy is due to round off While the approach by Stoughton is very simple the calculations can also be done directly using coordinates To develop the formula for computing the distance to the dividing line write four equations for determining the coordinates of the two ends of the line XE XB DBE sinAzBE XB Dsin0c YE YBDBE sinAzBE YB DcosOL XF XADAF sinAzAF XADsin0L YF YADAF sinAzAF YA DcosOL SURE 215 7 Surveyng Calculations Area Partitioning Page 112 In these equations D is the distance to the dividing line We simplified the equations taking into account that DAF DBE D In addition since the line BC is parallel to line AD then AZBE AZAF CL A fifth equation can also be developed and this is the double area formula for parcel A 2AXAYB YFXBYE YAXEYF YB XFYA YE Substitute the value for the coordinates into the area equation 2A XA YB YA Dcos 0LXB YB Dcos oc YB XE Dsin 0LYA Dcos OL YB XA Dsin 0LYA YB Dcos 0L 2KXB XAmmaYA yggnaD Solving for DAF recall that DAF D and AZFA CL A D XB XA cosocYA YB sinoc AF Using the data in the example 34d245 000000 100000cos7 3025950000 111550kn197 30259 74969 DAF This corresponds to the same value as shown in the previous examplel Case II is very similar to Case I in that line BC is parallel to line AD but here the dividing line is fixed on one of the parallel lines such as point F along line AD The result is a trapezoid The area of a trapezoid is the average length times the height The height is the perpendicular distance between the two parallel lines It can be computed as h D AB sin 9 The area of parcel A is then defined as DMD A BEh 2 Solving for the unknown distance DBE gives us 1 Note that the value for the distance DAF does come out as a negative number in this calculation Computing area in a clockwise manner as done in this example results in a negative quantity To alleviate this problem the area could be entered as a negative value SURE 215 7 Surveyng Calculations Area Partitioning Page 113 D D BE AF h Ifthis relationship results in a negative number then this means that parcel A is a triangle The unknown in this case is the altitude h which can be computed using the relationship 2A D h AF The distance of the dividing line of the triangle DEF can be found from 2A D DAFsmG EF Using the previous example divide the polygon ABCD into equal areas where point F is xed at the midpoint along line AD To solve this problem rst nd the coordinates of point F by averaging the coordinates of the two ends of the line Thus XF 1563175 YF 1015930 The perpendicular distance between the two parallel lines is h DAB sin e 61550 sin 47 303925quot 453845 The unknown distance in parcel A is next to be calculated 2340245 ft DBE2A AF Sq 763775 h 453845 735614 The coordinates of point E can now be computed XE XB DBE sin AzBE 100000 735614 sin 47 30 25quot 154241 YE YB DBE cos AzBE 111550 735614 cos 47 30 25quot 161241 Using the distance formula and the azimuth relationship DEF and OLEF can be computed SURE 215 7 Surveyng Calculations Area Partitioning Page 114 DEF J156317539 154241 2 10159339 161241 59684 0 tan11563175 154241 101593 161241 1780 0039 23quot Inserting these values into the following spreadsheet shows that the calculations are correct Again the slight discrepancy is caused by roundoff errors As it was shown in Case I the coordinates of the unknown point E can be found directly using coordinates of the quadrilateral Write two equations one for the double area and one for the azimuth of line BF Thus we have 2AXAYB YFXB YE YA XEYF YB XFYA YE X X an AZBE Y YB In the azimuth equation rewrite it in terms of YE Y XE XBcotAzBE YB E Substitute this into the double area equation and solve for XE The result is 2A XAYB YFYA XF XBXBXB XFcotAZBE YB XF XB X XB XFcotAzBE YF YB E The results of this approach are shown in the following Mathcad program Note that the area is negative because the area computation is clockwise Also the variable COL is the cotangent of the azimuth between B and E SURE 215 7 Surveying Calculations Area Partitioning Page 115 Area Partitioning of a Quadrilateral Using Coordinates ddltanggt degree lt oormg radiansang 7 d lt ddltang mins ang 7 degree1000 TE minutes oormins d39 180 0 seconds mins 7 minutes1000 minutes seconds 600 36000 degree The given quantities are A 7340245 XA 100000 YA 50000 AZBC 473025 1 X 100000 Y 111550 at B B tanradians AZBC XD 212635 YD 153186 Ca 09161082 The midpoint along line AD is found using the average coordinates of the ends ofthe line X7XAXD Y7YAYD F39 2 F39 2 Solving for the coordinates ofthe unknown point E 7 2A 7 XAYB 7 YF 7 YAXF 7 KB XBXB 7 XFc0L YBXF 7 KB T KB 7 XFc0L YF 7 YB XE1542415 YEXE7 XBc0L YB YE 1612410 Because the quadrilateral is such a simple gure the approach used by Stoughton is easier to solve unless the solution is already encoded in a program The remaining cases presented by Stoughton 1986 will only be discussed using the geometric method he uses in the paper Later on a generalized approach using coordinates for polygons of any size will be presented Case 111 approaches the problem a little differently Here lines BC and AD are not parallel The dividing line though is parallel to line AB Stoughton 1986 identi es four solutions based on the location of the acute angles These solutions are SURE 215 7 Surveyng Calculations Area Partitioning Page 116 Case IIIa AA and AB are acute Case IIIb AA and 4E are acute Case 1110 AB and AF are acute and Case IIId 4E and AF are acute 993 Case IIIa is depicted in gure 3 The distance DAB is made up of three segments namely u V w where V is equal to the distance between E and F DEF The acute angles are 91 and 92 located at A and B respectively The distance h is the perpendicular distance between line AB and the dividing line EF Using trigonometry these angels are shown as B The distance from A to B can be represented as DAB hcot91 DEF hcot92 The area of the trapezoid ABEF is h AEDAB DEF hcot 91 cot92 hz AB This can be shown in the following form SURE 215 7 Surveyng Calculations Area Partitioning Page 117 t9 t9 thDABhl0 The solution is performed using the quadratic equation bib2 4ac h 2a cot 91 cot 92 where a 2 b DAB 0 A Example The problem is to nd the location of the dividing line EF within polygon ABCD shown in gure 4 such that the area of both new parcels are equal From analysis ofthe data the desired area for parcel A is 316212234 square feet 95 1539 30quot 147127v C 169 43 24quot Parcel B 23773 Figure 4 Example using Case IIIa from Stoughton 1986 Solution The acute angles are computed rst 91 80 42 15quot 0 00 00quot 80 4215quot 92 180 00 00quot 95 15 30quot 84 4430quot SURE 215 7 Surveyng Calculations Area Partitioning Page 118 Then a cot 80 42 15quot cot 84 44 30quot 2 0127858 b 6155039 c 316212234ft2 The offset distance h is found using the quadratic equation h 61550 iJ 61550 2 40127858316212234 2 20127858 422916 or 584787 The obvious correct answer is 584787 The distances DAF and DBE are found to be h 584787 259257 Slne1 Sln80 423915quot AF h 584787 258726 Slne2 Sln84 44 30quot BE The distance of the dividing line becomes DEF 61550 584787 cot 80 42 15quot 584787 cot 84 44 30quot 46596 As shown in the following spreadsheet used to compute the area of parcel A the answer passes the check In Case IIIb the acute angles are located at points A and E figure 5 The dividing line EF is parallel to line AB The angles 91 and 92 can be defined as SURE 215 7 Surveyng Calculations Area Partitioning Page 119 Figure 5 Geometry for subdivision of a quadrilateral using Case IIIb from Stoughton 1986 tanG1 3 gt uhcot91 u tanG2 3 gt whcot92 w Again the area of the trapezoid ABEF is h AEDAB DEF cot 92 cot 91 hi DABh 2 Therefore cote2 cot91 h2D hA0 AB Solve the equation using the quadratic equation Example Given the data shown in gure 6 nd the location of the dividing line that divides the area in half where the dividing line is parallel to line AB Solution The total area of the quadrilateral is 6484211502 square feet Half of this area is 32421058 square feet SURE 215 7 Surveyng Calculations Area Partitioning Page 120 C 178 2939 54quot 66290 5 9 D A 267 1539 25quot 12079539 Figure 6 Example area partitioning using Case IIIb Then 91 87 15 25quot 20 353920quot 66 4039 05quot 92 79 38 45quot 20 35 20quot59 0339 25quot a cot 59 033925quot cot 66 40 05quot 2 08408980 b 5771539 c 32421058 sq ft The offset distance of the dividing line from line AB is h 5771539 Jr 577152 408408980 32421058 2 08408980 522038 and 73855339 The obvious correct answer is 522038 The distances DBE and DAF become SURE 215 7 Surveyng Calculations Area Partitioning Page 121 BE 522038 260866 s1n59 03 25quot AF 522038 256853 s1n66 50 05quot The line segments u and w are also found using basic trigonometric relationships u z zmm tan 66 40 05quot W z znggg tan 59 03 25quot The length of dividing line EF becomes DEF 5771539 225170 312966 66494639 As a check compute the area of quadrilateral ABEF The results are shown in the next table and it can be seen that the dividing line has been correctly located While Stoughton uses the acute angle at E one can also use the angle at point B The acute angle at B opposite the line segment w can be written by the following trigonometric relationship gure 5 tanG4 gtw htan94 Since 94 93 7 90 one can use 93 directly using the trigonometric identity tan 9 90 cot 9 Therefore the line segment w becomes SURE 215 7 Surveyng Calculations Area Partitioning Page 122 w h cot 03 Then the area function can be represented as cot01 cot03 hi D hA0 AB From the example a can be found to be 2 cot 66 40 05quot cot 120 56 35quot 2 08408980 which is identical to the value using the acute angle at E Therefore the solution will be the same Stoughton 1986 identi es Case 1110 where lines BC and AD are not parallel but the dividing line EF is parallel to line AB The acute angles 01 and 02 are located at points B and F figure 7 The line segments u and w are found as before Figure 7 Geometry for Case 1110 after Stoughton 1986 u h cot 01 whcot02 Substitute these values into the area of the quadrilateral SURE 215 7 Surveyng Calculations Area Partitioning Page 123 h AEDABDEF cote1 cot92 h2D hA0 2 AB Example Given the data in gure 8 find the location of the dividing line EF that is parallel to line AB such that both of the new parcels have the same area 32901539 40quot 68744 1830 4239 11quot 87962 Figure 8 Example for Stoughton39s Case IIIc Solution The total area of the quadrilateral is 89153935 square feet The desired area for the two parcels is 44576961 square feet The acute angles are 91 329 153940quot 292 053935quot 37 1039 05quot e2 149 153940quot 95 403925quot53 353915quot azwmmw b 6874439 c 445769677 sq ft SURE 215 7 Surveyng Calculations Area Partitioning Page 124 h 68744 r J68744392 4 02906887 445769677 sq ft 2 02906887 529771 0r 289464 The obvious correct answer is 529771 The distances of the dividing line and the offsets from line AB to line EF are DEF DAB hcot 92 hcot91 68744 529771 cot 37 103905quot cot 53 35391 9954439 BE 65829339 s1n 53 353915quot FA 876379 s1n 37 103905quot A check of the results are shown in the following table The slight discrepancy is due to round off errors Stoughton identi es Case IIId by the situation where the acute angles occur at the dividing line namely points E and F gure 9 The distance DEF is recognizing that V DAB DEF uDABw The line segments are SURE 215 7 Surveying Calculations Area Partitioning Page 125 To Point C To Point D Figure 9 Geometry for the area partitioning using Case IIId from Stoughton 1986 tanG1 2 gt uhcot91 u tanG2 3 gt w hcot92 w Therefore DEF DAB hcot91 hcot92 The area of the quadrilateral is A 2 EF hDAB DAB hcot91 hcot92 from which cot91cot92 h2D hA0 AB Example Using gure 9 the results of the area computation of quadrilateral ABCD is given in the following table DiVide the area in half with the dividing line oriented parallel to line AB SURE 215 7 Surveyng Calculations Area Partitioning Page 126 Solution See the following Mathcad program for the solution to this example Note that 091 represents the cotangent of Si Solution to Area Partitioning Problem Using Case llld as Presented by Stoughton ddang degree oorang radiansltang d ddltang mins ang 7 degree1000 7E minutes oormins d39 180 0 seconds mins 7 minutes1000 minutes seconds degree 600 36000 The given values for this problem are 91 543327 Ce 1 1 tan radians 9 1 92 402245 C92 1 DAB 58265 tanradians92 A 773168236 The Solution is cel c92 a 3 b DAB c 3 A 2 2 71241676050958684227 ah bh c solveh gt 62434500948755793002 h 624345 SURE 215 7 Surveyng Calculations Area Partitioning Page 127 u hcel u 444396 w hc92 w 734144 DEFTDABMW DEF176119 D a AF V sinradians91 DAF 766352 DB L E sinradians92 DBE 963729 Check the results They are given for parcel ABEF in the next table A rather interesting geometric approach to solving the Case 111 area partitioning problem was presented by Stoughton 1986 This involved projecting the lines BC and AD to a point of intersection point P in gures 10 and 11 There are two possible scenarios for this geometric construction The rst is when the dividing line lies between the line AB and the point of intersection as depicted in gure 10 The other is when the line AB lies between the dividing line and the point of intersection gure 11 BI DEl F B 93 153926quot 140256 C gt P 14 26 15quot 62548 194 523943quot 1570339 156 100000 50000 Figure 10 General solution to the Case 111 area partitioning problem where the dividing line lies between line AB and point P SURE 215 7 Surveyng Calculations Area Partitioning Page 128 Lets look at the rst con guration where the dividing line is between line AB and point P The height or altitude of triangle ABP is designated as h and de ned by the line B P The coordinates of point P can be determined using a lineline intersection algorithm Then using the sine law the angle at A can be represented as sinzA AP De ning AX XP XA and AY YP YA then sin AZDA 180 sinAzDA and cos AZDA 1800 cos AZDA From this one can write A AX DAPsinAzAD gt sinAzAD X DAP AY AY DAPcosAzAD gt cosAzAD D The height of triangle ABP becomes h DAP sin AA DAP Sin AZAD AZAB DAPsinAzAD cosAzAB cosAzAD sinAzAB DAP sinAzDA cos AZAB cos AzDA sin AZAB Substitute the Sines and cosines of line AD h DAPAX 015 AzAB AY sin AZAB AP AP AX cos AZAB AY sinAzAB The area of triangle ABP is AABP DABh The area of triangle EFP can be de ned as A A ABP ABEF A EFP Using similar triangles one can write the following ratios SURE 215 7 Surveying Calculations Area Partitioning Page 129 DAB DAP DBP h where h is the height or triangle EFP shown by the distance E P in gure 10 We know that the area will increase by the square of the ratio of the lengths In other words AEFP k2 A ABP Since the two areas are known the unknown distances to triangle EFP can be computed DEF kDAB DFP kDAP D kD EP BP Then DEE DBP DEP DAF DAP DFP Example Given the data shown in gure 10 where the area of the quadrilateral ABCD is 5377974441 square feet find the location of the dividing line EF such that the areas of the two new parcels are the same Solution The desired area of quadrilateral ABEF is 2688987221 square feet The solution to the problem is given in the following Mathcad program Solution to Area Partitioning Problem Using the General form of Case Iquot as Presented by Stoughton ddang degree e floorang mins e ang 7 degree1000 radiansang d e ddang 7 d mmutes e floormms 1800 seconds e mins 7 minutes1000 minutes seconds 600 36000 degree SURE 215 7 Surveyng Calculations Area Partitioning Page 130 The given values for this problem are AZAB 142615 OLAB radians AZAB AZAD 760753 GLAD radians AZAD Az 931526 BC OLBC rad1a11sAzBC D AB 62548 X 1000 00 assumed DBcz 140256 A DCD15703 YA 50000 assumed DD A 156144 XB XA f DAB39SIH AB Y 3 Y D cos CL A ABEF 2688987221 B A AB AB The solution is as follows The angles at the vertices of triangle A B P are designated by Ai First find the coordintes of point P through lineline intersection AA01AD701AB AA10768 AB OLAB 7t 7 OLBC AB 17659 AP7E7 AA 4 AB AP 02989 DAB DBPWS1HAA DBP18700955 DAB DAP WS1HAB DAP 20837675 XP XB DBPsin0LBC XP 30230214 YP YB DBPcos0LBC YP 9994712 Next perform the steps for the area partitioning AX XP 7 XA AX 20230214 AY YpeYA AY4994712 h AXcos0LAB 7 AY39SIH AB h 18346048 A w 1 D h ABP39 2 AB A ABP 5737542912 AEFP 3 AABEF AABP AEFP 73048555691 SURE 215 7 Surveyng Calculations Area Partitioning Page 131 k g AEFP AABP k 07289 DEF k D AB DEF 4559293 DFPkDAP DFP15189146 DEP kDBP DEF 13631633 DBE DBP DEP DBE 5069322 DAF DAP DFP DAF 5648529 As a check compute the area of the quadrilateral ABEF The results are shown in the following table The slight discrepancy is due to roundoff errors Division of Polygon Using an Initial Guess of the Dividing Line A common approach to partitioning an area of a polygon with more than four sides is by establishing an initial guess of where the dividing line may be located This position is then adjusted to meet the desired area Mof tt and Bossler 1998 Anderson and Mikhail 1998 Lets rst take a look at the rst case where the line is of a known direction Figure 11 shows an example polygon with 6 sides The problem is to divide the main polygon into two polygons of equal area using a line that is parallel to line AF This is depicted as line H1 in the gure The solution can be broken down to four main steps 1 First compute the area of the entire polygon 2 Draw a line from one of the nodes on the polygon In this example we will construct a line BG that is parallel to line AF Compute the area of the new polygon ABGFA SURE 215 7 Surveying Calculations Area Partitioning Page 132 C A Figure 11 Example area partitioning problem using a line parallel to another line 3 Chances are that this new polygon will not be equal to half the total area of the original polygon Assume that the area is too small Then we will need to add the trapezoid BHIG to the second polygon to arrive at the correct area To do this we will need to perform a series of calculations to determine the unknowns within the trapezoid a Compute the distances DBG and DGF along with the angles 9 and p b Find the distance DHI using the expression DHI DBG Xcot9Xcotp 0 Compute the area using the next relationship A X DEG DEG X cot e cot p l BHIG E X DEGcot9 cotpX2 d The solution to the distance offset of line HI to BG is calculated using the quadratic equation X bi1lb2 4ac 2a where SURE 215 7 Surveyng Calculations Area Partitioning Page 133 a cot 9 cot p 2 b DBG C ABHIG 4 Finally determine the distances DBH and DGI X D BH sine X DGI smp 5 Check the results by computing the area of the two new polygons Example Hashimi 1988 The following table presents the information pertaining to a 6 sided polygon shown in gure 1 34 2006 190 1997 14408 287 728 00 00 The problem is to divide the area into two equal areas using a line that is parallel to line AF in the original polygon The solution is shown as follows First we will compute the distance between B and G using a lineline intersection Then the distance between G and F will be computed by inversing between the coordinates The following values will be used X1 XB 1085 X2 XE 6214 y1 yB 301539 y2 yE 6441 XF 6023 yF 1970 m1 cot710 53 m2 cot1820 2739 SURE 215 7 Surveyng Calculations Area Partitioning Page 134 In1X1 m2X2 Y1 372 G In1 m2 0327172 1085 233717776214 30156441 0327172 23371777 6169 yG m1xG X1 y1 0327172 6169 1085 3015 5388 DEG KG XE2yG YB quot6169 1085Z 5388 3015Z 7632 DGF XFXG2 YF YG2 quot6023 61692 1970 53882 3422 AZBG AZAF e AZBG Ach 71 5339 5 38 66 1539 p AZGB AZGF 251 5339 182 27 69 26 Next compute the area of the new polygon ABGFA The results are shown in the following table 27 3422 146 3418 14362 00 00 Evaluate how close the new polygon is to the desired area Total Area 539870 sq ft Half Area 269935 sq ft SURE 215 7 Surveyng Calculations Area Partitioning Page 135 AABGF 223 720 ABG1H sq ft sq ft gt Area to be added to the polygon ABGF Since the area of the initial division does not diVide the original polygon in half we need to add polygon BGIH to arrive at the correct area To do this we need to compute the offset distance X using the quadratic equation where cot 9 cotp cot66 1539 cot 69 26 a T f DEG 7632 c ABIG 46215 sq ft 00323995 X 7632 76322 400323995 46215 607 200323995 With this data compute the distances BH GI CH FI and HI using the following relationships BH X 6 397 66339 sme Sln66 1539 X 607 2648 GI sinp sin69 2639 DCH DEC DEH 6186 663 5523 DH DFG DGI 342239 648 407039 DHI DBA Xcot9Xcotp 3204 607cot66 15 607cot69 26 7593 Next check the results SURE 215 7 Surveyng Calculations Area Partitioning Page 136 The slight discrepancy between the results here and the true half area are due to roundoff errors in the calculations Carrying values out to more signi cant gures would result in a much closer value to the true half area desired This result is well within the signi cant gures for the example The next example is to divide the polygon using a de ned point on the edge of the polygon For example gure 12 shows the same polygon except that this time the division will be made at point G on the line FA G is located midway between F and A although it could be any distance Again the example is to divide the area into two equal areas The solution is as follows Hashimi 1988 1 Compute the area of the polygon ABCDEF 2 Compute the azimuth and length from the division point to another point on the opposite side of the polygon In this case we compute these values for line CG 3 Compute the area of the quadrilateral ABCG A Figure 12 Example area partitioning from a point located on one sided of a polygon SURE 215 7 Surveyng Calculations Area Partitioning Page 137 4 Since C is most probably not the desired division point the area of the triangle CHC will de ne the area that needs to be added to the quadrilateral to arrive at the correct area of the new polygon 5 Compute the distance from C to H using the relationship 1 A E DGC DCH smoc 6 Check the results Example Using the same values as in the previous example we will divide the original polygon into two equalarea polygons from point G The distance GA can be computed as half the distance between F and A The azimuth and distance for line CG is found by adding up the latitudes and departures of lines GA AB and BC Since this is a closed gure the latitude and departure of line C must have the same magnitude with opposite sign Thus 3490 8186 D tanAzCG Lei gt AZCG tan 1 a CG 156 548 DCG JDeng Lat G J34902 81862 88985 The area of the initial polygon is shown as follows Since the area of the polygon is less than the desired area we compute the area of the triangle CGH Required Area 269935 sq ft AABCG 182 990 sq ft ACGH 86945 sq ft Then using the formula for the area of a triangle where the area is onehalf the base times the height The height is found using the trigonometric relationship DCH sin 600 29 Thus we have SURE 215 7 Surveyng Calculations Area Partitioning Page 138 86945 8898 DCH sin 60 29 from which the distance from C to H is found 2 86945 D 8898s1n 60 29 2246 CH Check by computing the area of the polygon ABCHG Again the slight discrepancy of about 30 square feet is due to roundoff errors Direct Division of Polygon Using Coordinates The problem with the preceding approaches where an initial guess is made of where the dividing line occurs is that the problem changes with different polygon figures Danial 1984 presents an approach using coordinates where the dividing line is computed directly Looking at figure 13 we can write the area using coordinates as follows Amer4 AP39KPKPAP39A An 121l AP39lPlPZP39Z AP39ZPZPKP39K l l Y3 Y4 X3 X4E Y4 Y1 X4 X1 gonna x ltYY3ltx3 xz The double area becomes after manipulation 2A Y2 X1 X2 Y2 Y3 X2 X3 Y3 Y4 X3 X4 Y4 X4 X1 If the computation is done using horizontal trapezoids off the YaXis then the equation takes on the following form SURE 215 7 Surveyng Calculations Area Partitioning Page 139 Figure 13 Geometry for computing the area of a polygon 2A X1 X2Y1 Y2X2 X3Y2 Y3X3 X4Y3 Y4X4 X1Y4 Y1 As we know if the area results in a negative value the negative sign is neglected because area cannot be less than zero The reason for the negative sign exists is because of the order in which points are selected But in area partitioning it is necessary to retain the sign to ensure that the correct results are found Danial 1984 If the polygon area is computed clockwise then the area must be represented as a negative value For an nsided polygon the double area can be computed using the general formula 2AiY1 Y2X1 X2Y2 Y3X2 X339 Yn71 YnX Yn Y1Xn X1 X nil Figure 14 from Danial 1984 shows a situation where the area partitioning is performed using a line with a given direction Points P1 Pi PM PM Q and T are points on the original polygon whose coordinates are known The dividing line PHPM has a known direction but the coordinates of the ends of the line are unknown To solve this problem we need to solve four equations simultaneously to determine the values of the four unknowns Xn1 Yn1 X and Y The four equations entail three azimuth equations and one area equation SURE 215 7 Surveying Calculations Area Partitioning Page 140 Figure 14 Area partitioning by a line with a given direction The three azimuth equations are refer to gure 4 X 7 X an Xniz anl anz X n n71 tan 1H Y nil X tan 0Ln 1 quot 1 1 Ynil where 111 is the azimuth of line PHPM 1411 is the azimuth of line PMPH and 041 is the azimuth of line PnPl The area equation is written as 2A2Uanz Yn1Xn2 Xn1 anl YnXnil Xn Yn Y1Xn X1 where 2U is the area of the remaining part of the partitioned area It is shown as SURE 215 7 Surveyng Calculations Area Partitioning Page 141 ni3 2U 2Yi Yi1Xi Xil 11 Rewrite the azimuth equations into the following form Xn2 Y anz tan 0Lan nil Xn Xn71 Yn anl tan Ina X1 Xn Y1 Yntan 0Lquot Adding these three equations eliminates two unknowns Xn1 and X The result is X1 Xn2 tan 112 tan 1111er1 tan 1H tan 0Ln Yn Y1 tan 0Ln Yn2 tan 1H Solving for one of the unknowns like Yn gives Y 2 X1 Xr2 Y1 tan 0Ln Yni2 tan 1H tan 1H tann1 Yn1 n tan 0L 1 tan 0Ln Designate p as a constant consisting of the known values in the equation for Y p X1 Y1 tan 0Ln Xr2 Yni2 tan Xnizl Then the Ycoordinate for point n becomes p tan 112 tann1Y nil tan 1H tan 0Ln The third azimuth equation can be rewritten as Xn X1 Y1 tan 0Ln Yn tan 0Ln Substitute the value for Yu into this equation tanOL Xn X1 Y1 tan 0Ln n p tan 112 tan 1H YH tan 1H tan 0Ln The rst azimuth equation can be rearranged into the next formula SURE 215 7 Surveyng Calculations Area Partitioning Page 142 Xquot1 X 2 Y 1 tan 112 Yn2 tan 112 I So far we can compute three of the unknowns Yn Xquot and Xml as a function of the other unknown Yn1 Rewrite the double area equation by multiplying the terms on the right hand side of the equation 2A 2Uanzxnrz Yn71Xn72 Yn2Xnil Yan71 Yn1Xn Y1Xn YnX1 Y1X1 Next substitute the values for Y X and Xn1 Lets look at each substitution individually namely the 411 5th 6m 7m and 81h terms on the right hand side of the double area equation 2 anzxnrl anzxnrz anz tan 0Lan anz tan Oererrl Y X 13an2 anz tan 0Lan P tan 0Lan tan 0Lan tan Ina Xanz anz tan 0Lan lanl quot quot71 tan 1H tan 0Lquot tan 1H tan 1H tan 112 inl tan 1H tan 0Lquot Y X 2 X1 Y1 tan 0Ln ptan Gen Y tan 112 tan OLlHtan LnYl1 quot71 quot tan 0Lquot1 tan 0Ln quot71 tan 1H tan 0Lquot Y tan 0L tan 0L tan 0L Y tan 0L Y YIXn Y1X1 Y12 tan 0Lquot p 1 n n72 n71 1 n n71 tan 1H tan 0Lquot tan 1H tan 0L Ifwe substitute these values into the double area one arrives at the next equation aYl1 an1 c 0 p X1 tan 112 tan 1H XIYn1 Y X tan 1H tan 0Lquot tan 1H tan 0Lquot n 1 SURE 215 7 Surveyng Calculations Area Partitioning Page 143 where the coefficients are de ned as Danial 19842 tan OCH tan 06 1 tan In tan 06H a tan 111 tan 06quot tan 1 tan 1 b n72 n71 tan OCH tan 06quot X Y tanoc Ytan0c 7X cp H H H 1 n 1 thanocnYniztanocn2 2U 2A tan OCH tan 06quot The solution for Yn1 is found using the quadratic equation Y bi b2 4ac nil 2a Using the example from the previous section dealing with the estimated partitioning using line HI which is parallel to line AF the results using the method presented by Danial is given in the following Mathcad program Designate H as PH and I as PM Solution to Area Partitioning Problem Using the Method Presented by Danial ddang degree 7 00rang radiansltang d e ddang ruins 7 mg 7 degree1000 1 minutes 7 oorrnirrs d39 180 0 seconds 7 ruins 7 minutes 1000 dmsang degree 7 00rang minutes 5 econds degree 36000 rem 7 mg 7 degree60 ruins e oorrern reml e rem 7 ruins 180 secs 7 rem1600 todeg TE ruins secs degree 100 10000 2 Note that Danial gives the formula for the coefficient 0 as follows 1 7Y2tanoc Y2 tanoc 2U 2A r n n72 n72 tan OCH tan orquot Instead the formula should be presented as given in the text of this paper and shown above pXn2 7Yn2 tan 112 Y1 tan 0Ln 7X1 tan 1H tan 0Ln le tan 0Ln inz tan 1H 2 U 2 A SURE 215 7 Surveyng Calculations Area Partitioning Page 144 The given values for this problem are XA 710000 YA 7 5000 AZFE 7 227 KB 7 8915 YB 7 8015 anz radians AZFE XC 7 9522 YC 14171 AZIH 7 25153 KB 16024 YD 13438 am radiansAzIH XE 16214 YE 11441 AZCB 18538 XF 7 16023 YF 6970 an 7 radians AZCB A 7 269935 xn2 7 XF Yn2 7 YF X1 XB Y1 YB X1 XB Y1 Y The solution is as follows Note that the subscript n1 is shorthand for n1 and the subscript n2 is shorthand for n2 U 7 UYB YAXB 7 XA YA YFXA 7 U 7431082925 p 3 X1 T Yl39tan n T X112 T Yn239tan n2 p 77601932 7 tmlanzl tanlamlltanlan tanlanzll tanxn1 7 tanx a 7005711 tanxn2 7 tanxn1 tanxn1 7 tanxn V Plan Ynz39tan0 n2 t Y139tan0 n X1 C 167 7 617 b 21 b 15491977 7 letanxn YHZZtanxn2 2U 7 2A 0 716401298624 The Solve function is used to arrive at the solution Two possible answers are possible The logical answer is assigned to the value for Yn1 shown here as Ym 2611267655117494897 Next the remaining unknown coordinates are computed as follows 2 1103593007021940253 aX bX c solvex 7 an x 2 4 Ynltanocn2 7 Ynztanotn2 xnl 16196966 Xn X1 Yl39tanlo n WM an 39p 0512 tf nl lnlll39Ynl tanotnl 7 tan p 7 tanan2 7 tanxn1Yn1 Xn 8980208 Y tanan1 7 tanan Yn 7 8674776 SURE 215 7 Surveyng Calculations Area Partitioning Page 145 Check the results As a rst check the azimuth of the dividing line should be equal to the line AF Hence X X AZIH tan71 H 1 tan4 89801 16197 YH YI 86747 110359 251 52 59quot The azimuth checks with the azimuth from F to A Next check the area to ensure that it is correct The results are presented in the following spreadsheet Note that the third column is the difference Xi Xm the fourth column is the sum Yi YM and the fifth column is the product of the difference and the sum From the results it is obvious that this method can be used to easily partition an area using a line with a given direction The difference between the exact value and the computed value is only about 34 square feet This difference is attributed to roundoff problems n2 R l A Figure 15 Example polygon showing the dividing line used in the development by Easa 1989 SURE 215 7 Surveyng Calculations Area Partitioning Page 146 Easa 1989 presents a very similar direct approach to area partitioning using coordinates Using the azimuth relationship refer to gure 15 the Ycoordinates for the ends of the dividing line can be written in the following form YA Y1XA X1cot 11m YB Yn XB Xncotoc nn1 Following the notation from Easa designate S as the cotangent 0f the azimuth We can then rewrite these equations as YA Y1 XA X1SA YB Yn XB XnSB where SA cot 11m SE cot 0L nn1 Also recall that the area of polygon A 1 2 n B can be written as 1 1 A E XE Yn YA XA YB Y1X1YA XnYB 2 Xi1Yi XiYi1 i1 These last three equations form the basic design formulas for area partitioning as identi ed by Easa 1989 There are four unknowns which means that a fourth equation is required to solve the problem This last equation is based on the nature of the dividing line In essence Easa also uses three azimuth equations and an area equation for the solution to the areapartitioning problem Lets look at the same situation already presented by Danial where the dividing line is xed in direction Following after Easa designate the cotangent 0f the azimuth of the dividing line with P direction from A to B Ifthe dividing line is parallel to another line in the polygon de ned by the end points u and v then the cotangent 0f the azimuth of that dividing line can be defined as Ifthe dividing line is perpendicular to the line uv then the relationship is written as XV Xu YV Yu P SURE 215 7 Surveyng Calculations Area Partitioning Page 147 Write the equation for the cotangent of the azimuth of the dividing line in terms of the four unknowns Substitute the values for YA and Y3 Yn XB Xn SB Y1 XA X1SA X P A Rearrange the equation to in a form for solving XA XBP XAP Yn XBSB anB Y1 XASA XlsA XAPXASA XBPXBSB Yn anB Y1XISA A X PSB Yn Y1X1SA XnSB P SA P SA Designating k YnY1X1SAXnSB and 2 P SA P S k3 B P S the equation for XA can be rewritten as XA k2 k3XB Ifthe dividing line is along the meridian or nearly so then P co and k2 0 and k3 1 Then the unknowns can be easily solved by simultaneous equations using the area formula the two azimuth relationships YA and Y3 and the formula above for XA Substitute the two equations for Y A and YB into the area equation SURE 215 7 Surveyng Calculations Area Partitioning Page 148 1 AEXB Yn Y1 XA X1SA XA Yn XB XnSB Y1X1Y1 n71 X1XA X1SA XnYn Xn XE XnSB 2XilYi XiYi1 11 nil Designate k1 24XH1Yi XiYH1 2A Rearrange the area formula i1 XBYn XBY1 XB XA X1SA XAYn XA XE XnSB XAY1 XAX1SA XBXnSB X1Y1 XIZSA XnYn Xf SB k1 0 SB SAXAXB XA XBYn Y1 X1SA XnSB X1Y1 stA XnYn X sB k10 k4 X1Y1 stA XnYn X sB k1 k5 YN Y1XISA anB k6 SB SA Then we have the following formula k4 1lt5XA XB1lt6XAXB 0 Substitute the value for XA into the equation k k5 k2 k3XB XB k6k2 k3XBXB 0 1lt3k6X123k3k5 k5 k2k6XB k4 k2k5 0 This is the quadratic form which can be succinctly written as where ak3k6 b k3k5 k5 k2k6 ck4 k2k5 SURE 215 7 Surveyng Calculations Area Partitioning Page 149 Solve for XB using the quadratic equation Then back substitute into the equations for XA YA and YB The same example used before for dividing a polygon with a line parallel to a line on the edge of the polygon is given in the following Mathcad program Solution to Area Partitioning Problem Using the Method Presented by Easa ddang degree e oorang mins e ang 7 degree1000 minutes e oormins seconds e mins 7 minutes1000 minutes seconds radiansang d e ddang d 1800 dmsang degree e oorang degree 600 36000 remlt7 ang 7 degree60 mins e oorrem reml e rem 7 mins 180 secs e rem1600 todeg 7 3 mins secs degree 100 10000 The given values for this problem are XA 10000 YA 5000 AZFA 25153 1 X 8915 Y 8015 P B B tanradians AZFA XC 9522 YC14171 AZFE 227 XD 16024 YD 13438 S 1 XE 16214 YE 11441 A 397 tanradians AZFE XF 16023 YF 6970 AZBC 538 A 269935 SB 7 1 V tanradians AZBC Xn XB Yn YB X1 XF Y1 YF SURE 215 7 Surveying Calculations Area Partitioning Solution First compute the constants k39s k1XAYF 7 XFYA KBYA 7 XAYB 7 2A k2 k3 Yn Y1 XI39SA Xn39SB 1gt7sA P7sB 1gt7sA 2 2 k43X139Y1 X139SA Xn39Yn Xn 39SBtk1 k5Yn7 Y1 XlSA 7 XIISB k6 a2 b SB SA a k3k6 b 7 k3k5 k5 k2k6 c k4 k2k5 X csolvex gt 89800778884360475364 34114405649423651734 Page 150 k1 799977000 k2 123816 k3 042 k4 75255991368 k5 2853293 k6 71325 Designate the X and Y coordinate values in terms of Eastings and Northings to avoid the confusion of using the values for already assigned to the traverse Thus EA XA in the text of this paper EA XI Also E B XH The same can be said forthe Northings EB EA 898008 3 k2 k3EB NA 7 Y1EA 7 X1SA NB Yn BB 7 XnSB The results are XH EB XI 3 EA Y1 NA YH NB XH 7 89801 XI 7 161969 The other common problem that has already been identi ed is the division of an area from a de ned point on the boundary of the polygon Again Danial 1984 presents a direct method to solve this problem using coordinates Since one of the end points of the dividing line is known only two equations need to be solved simultaneously This involves an azimuth equation and the area equation The azimuth equation can be written as refer to gure 16 SURE 215 7 Surveying Calculations Area Partitioning Page 151 Figure 16 Example for area partitioning after the developments by Danial 1984 Xn anl Yn Yn71tan Inn The area equation was presented earlier as 2A 2UYH Yn XH Xn Yn Y1Xn X1 nil where 2U Z Yi YH1Xi XH1 11 Rearrange the double area formula 2A 2U XnY1 Yn1 YnX1 Xn1Yn1Xn1 Y1X1 Substitute the value for Xn from the azimuth equation 2A 2U XH Yn Yn71tan earl Y1 Yquot1 Yquot X1 XH YHXH YIXI 2U Y1Xn4 anlxnrl Y1Yn anl tan Ina Yn1Yn anl tan Ina Yn X1 Xn71 Yn1Xnil Y1X1 2U YnYn71 tan Ina Y1 tan Ina X1 Xn71Y1X1 Xn71 Yn1Y1 anl tan Ina Solve for Yquot SURE 215 7 Surveyng Calculations Area Partitioning Page 152 YnX1 Xn71Y1 Yn71tan Ina 2U 2A Y1X1 Xn71 Yn71Y1 Yn71tan an71 2U 2A Y1X1 Xn71 Yn71 Y1 Yn71tan an71 I X1 Xn71 Y1 Yn71tan an71 l Y n Following is the same example that was used by the estimated division method As one can see with the results of the area computations given at the end of the Mathcad program the correct area is accurately determined Solution to Area Partitioning Problem Using the Method Presented by Danial ddang degree e oorang mins e ang 7 degree1000 radiansang d e ddang 71 minutes e oormins d39 180 0 seconds e mins 7 minutes1000 minutes seconds dmsang degree e oorang degree 60 0 3600 0 remlt7 ang 7 degree60 mins e oorrem reml e rem 7 mins 180 secs e rem1600 todeg 7 3 mins secs degree 100 10000 The given values for this problem are XA 10000 YA 5000 XB 8915 YB 8015 XC 9522 YC14171 AZCD 9626 XD 16024 YD 13438 am radiansAzCD XEZ 16214 YE11441 XF 16023 YF 6970 A 3 7269935 The area is negative because the determination of the area is clockwise in this problem SURE 215 7 Surveying Calculations Area Partitioning Page 153 The solution is as follows Note that the subscript n1 is shorthand for n1 The point to be held fixed G is the point midway between points F and A XF XA YF YA KG 2 YG 2 XG 130115 YG 59850 1 UEYG YAXG7 XA YA YBXA 7 KB YB YCXB7 XC U 168678503 X1 XG Y1 YG an XC Yul YC Y 2U7 2A 7 YlXl 7 Xn17 YulY1 7 Yn1tanocn1 n X1 X111 Y1 Ynl39tananl Xn an Yn 7 Yn139tananl XH Xn YH Yn XH117534 YH 139194 Check the results to ensure that the area is correct The results are given in the following table As we have seen before Easa 1989 uses the same method of partitioning a polygon using coordinates Again the approach is a little different Recall that the azimuth equation was written in the next form YB Yn XB Xn SB SURE 215 7 Surveyng Calculations Area Partitioning Page 154 where the subscript B indicates the unknown end of the dividing line point A is xed on this line SB is the azimuth of the line on the polygon where point B will be located see gure 15 Also recall that the area equation was given as 1 1 A E XE Yn YA XA YB Y1X1YA XnYB 2 Xi1Yi XiYi1 11 Further the Easa de ned the constant k1 as nil k1 Xi1Yi XiYi1 2A i1 Thus the area equation can be written as xB Yn YA xA YB Y1X1YA anB k1 0 Substitute the value for YB from the azimuth equation into the area equation and expand the equation XBYn XBYA XAYn XB XnSB XAY1 X1YA XnYn XB XnSB lk1 0 YA Yn XA XnSBXB Yn XnSB Y1XA Yn XnSB Xn X1YA k1 Then solve for X3 k1 XAY1 X1YA XA XnYn XnSB X B YA Yn XA XnSB Next solve for the Ycoordinate using the azimuth equation If the xed point is B and A is the unknown on the dividing line the solution uses the following form k1 XBYn XnYB XB X1Y1 X1SA X A YB Y1XB X1SA and YA Y1 XA X1SA where S A is the cotangent of the azimuth of the line on the original polygon where A will be placed Using the same example that was presented previously the results are depicted in the following Mathcad program The results for XH and YH are the same as found using the approach by Danial SURE 215 7 Surveyng Calculations Area Partitioning Page 155 Solution to Area Partitioning Problem Using the Method Presented by Easa ddang degree e oorang radiansltang d e ddltang mins e ang 7 degree1000 n d m1r1utes e oorm111s 1800 seconds e mins 7 minutes1000 minutes seconds dmsang degree e oorar1g degree 600 36000 remlt7 ang 7 degree60 mins e oorrem reml e rem 7 mins 180 secs e rem1600 todeg 71 mins secs degree 100 10000 The given values for this problem are XA 10000 YA 5000 KB 8915 YB 8015 XC 9522 YC14171 AZCD 9626 XD 16024 YD 13438 0cm radiansAzCD XE16214 YE11441 SB 1 XF 16023 YF 6970 tan 111 A 269935 The solution is as follows The point to be held fixed G is the point midway between points F and A X 7XFXA Y7YFYA Gquot 2 Gquot 2 XG130115 YG 59850 Note that the lowercase variables x and y with subscripts A and B represent the ends ofthe dividing line presented in the paper XAXG yAYG XnXC Y YC X1XA Y1YA SURE 215 7 Surveyng Calculations Area Partitioning Page 156 k1XBYA 7 XAYB XCYB 7 XBYC 7 2A k1 7139577635 k1 XA39Y1 t X139YA t XA Xn39Yn Xn39SB XB YA Yn XA ansB YB Yn t XB X1139SB XH 3 XB YH 3 YB XH 7 117534 YH 7 139194 Easa 1989 also presents a solution to the problem of placing the dividing line through an interior point of the polygon Since there are an in nite number of dividing lines that can pass through this point for example point Q in gure 15 the edge of the polygon where the dividing line will fall A and B in gure 5 needs to be identi ed We can write the relationship between Q and the end points A and B as YA YB YQ YB X A XB XQ XB Recall that earlier the Ycoordinate of the ends of the dividing line were presented as YA Y1XA X1SA YB Y XB X 3 Substitute these two formulas into the previous equation Y1XASA 7xlsA Yn XBSB anB YQ Yn 7X33B anB XA 7XB XQ 7XB XQSA 7vQ Yn 7anBXA YQ Y1XISA XQSBXB sB XAXAXB Y1 XISA Yn anBXQ 0 Following Easa let SURE 215 7 Surveyng Calculations Area Partitioning Page 157 XQSA YQ Yn anB k7 k8 YQ Y1XISA XQSB and k9 Y1 Yn anB XISAXQ k5XQ Remember that k6 SB SA Then k7XA k8XB 1lt6XAXB k9 0 But earlier we saw that k 1lt5XA XBk6XAXB 0 Substitute the value for the product XAXB into the previous equation This results in k7XA k8XB k4 k5XA k5XB k9 0 Solving for XA yields k4 k9 k5 k8 XB k7 k5 X A Substitute this back into the equation before k4 k5 XA XB 1lt6XAXB 0 which leads to k Lks k4 k9k5 k5 k8XB Lk X LkGXBk4 k9k6 k5 k8X123 I I 5 B I k k5 k7 k5 0 7 k6 k5 k8 X123 k5 k7 k8 k6 k4 k9 XB k4k7 k5k90 Partitioning Pie Shaped Polygons Up until now the area partitioning problem discussed the method of dividing a polygon consisting of a boundary of straight lines For various reasons there are numerous situations where one or more of the boundaries consist of a curve One set of derivations for pieshaped lots was developed by Danial 1990 SURE 215 7 Surveying Calculations Area Partitioning Page 158 Ll Figure 17 Area computation on a pie shaped polygon Figure 17 shows the geometry ofa pieshaped parcel The distance S from A to the PC is found as S DorPc D07A R R 1 2 From the principles of horizontal curve geometry the arc lengths become L1 R1A L2 RZA where A is in radians We also know from curve geometry that the angle at the PC from the chord t0 the center of the circle is 90 Since AB is parallel to the long chord then the angle chiAiB 90 The area of are l PCPTO is found as l RfA PchTro 2 A while the area of the second arc AB O is l RZA AiBio 2 2 A where A is again in radians The area of the lot de ned by APCPTB is the difference in the two areas SURE 215 7 Surveyng Calculations Area Partitioning Page 159 The problem of dividing the pieshaped lot can be broken down into two different forms just as we have seen before These are partitioning by a random line through a point and partitioning by a line parallel to one of the radial sides Lets look at the former situation rst Case 1 Danial 1990 involves dividing parcel PCPTAB into two polygons The arc length from the PC to C gure 18 is L11 while the arc length from A to D is L21 These lengths are known Solving for the area of polygon PCCD A and CPTB D commences as follows L1 PC PT 1 A vg D B l r 62 v39 i 1 l 31 r V 0 Figure 18 Area partitioning of a pie shaped gure through a point The central angle between the PC and C 51 and between A and D 52 are found from 52 from which ACOD 52 51 Using the law of cosines DCD R12 R 2R1R2 cos 4m Then from the law of Sines SURE 215 7 Surveyng Calculations Area Partitioning Page 160 I Z 0csin1suJR2 CD The angle at D then becomes AODC 1800 XACOD The area of parcel PCCD A is found by computing the area of the sector PCOC and adding the area of triangle COD and subtracting the area of sector OAD A PcrchrA APcrorc Aorch AOAD Substituting the formulas for the different areas this becomes 2 2 l l l APcrchrA ERIZSI ER1R2 Sln ACOB ER26 where the angles 51 and 52 are in radians The area of parcel CPTB D can also be determined in a similar fashion AchTrBrD AorchT Aorch A07B7D l l l ER12 A61ER1R2 smzcora ERA62 Danial shows that the angles between the chords and the sides can be calculated from the following relationships For parcel PCCD A o 5 PcrArD 90 o 5 A71ch 90 For parcel CPTB D A 90 CiPTiB A Sl 2 Jr 2 A 90 PTiBiD SURE 215 7 Surveyng Calculations Area Partitioning Page 161 A Sl A 90 PTrch 0 A62 4H 90 oc ES1 The second situation for Case 1 is where the dividing line is parallel to one of the radial lines Line CD is located at a distance w from the line PTB gure 19 Figure 19 Area partitioning of pie shaped lot by a line parallel to one of the radial lines 51 sin 1 1 R1 Then 52 sin 1 1 R2 Because line CD is parallel to PTB then X 51 and B 52 The arc length L11 is found using R5 L11 1 1 while the arc length L21 becomes LR5 where 51 and 52 are in radians Recognizing that ZCDO 180 Oc 52 51 then the cosine law can be used to compute the distance DCD SURE 215 7 Surveyng Calculations Area Partitioning Page 162 DCD R12 R 2R1R2 cos ADOC This distance can also be computed as DCD 1 R12 w2 quotR w2 The area of parcel PCCD A is found to be APcrchrA APcrcro Achro AA7D70 l l l ER12 A61ER1R2 SlnzDoc ERA62 where A 51 and 52 are in radians The area of parcel CPTB D becomes AchTrBrD AchTro AcrorD AD7B70 1 1 1 31112 S111 ZDOC ER62 Danial 1990 identifies Case 2 as being the division here one side is not a radial line gure 10 Line AB within polygon ABCD is not a radial line The given data are R1 R2 L1 and L2 From basic curve geometry the central angle to the two curves can be shown as in radians Using the cosine law the distance from A to B is S1 1 R12R 2R1R2 0055 The angle at B between A and the center of the arc is computed from the sine law R X sin 1 Zsin 5 R1 4m 180 oc6 The angle at A then becomes The other angles in polygon AQBCD are then SURE 215 7 Surveyng Calculations Area Partitioning Page 163 o A AADC 90 o A ADCB 90 ACBA 90 Aoc o A ABAD90 ocA1 The area of polygon ABCD is then de ned as A A A ABCD SEOC TEAO ASAOD l 2 l l 2 ER1A1 ER1R2 sme ERZA2 The subscript S indicates the area of a sector while the subscript T delineates the area of a triangle A1 and A2 are in radians The second case presented by Danial 1990 involves constructing a dividing line that is parallel to a nonradial side of the pieshaped parcel gure 20 The known quantities are the two radii R1 and R2 the two are lengths L1 and L2 the distance along the non radial side of the polygon 1 DAB and the offset distance from the nonradial side to the dividing line w The angle at A from B to the center of the curve 0 is 4m 180 oc1 61 From this 3 0c1 51 There are two possible solutions to this general problem of dividing a pieshaped polygon Figure 20 depicts the situation where A1 is smaller than A2 This case will be discussed rst Draw a line from A to a point G on line EF or its extension that is perpendicular to that line The distance of line AG w Line AH is the tangent to the second or lower curve The angle between this tangent and line AG is de ned as Designating 5 as the central angle subtending the are from AF then the angle from the chord AF t0 the line AG is 5 AFAG 15B3 Then the offset distance w can be de ned trigonometrically as SURE 215 7 Surveyng Calculations Area Partitioning Page 164 w DAF cos 1 2R2 sincos 1 Substitute the value for 13 gives w 2R2 sincos l Solve for 5 by rst rearranging this formula w 5 5 E sm3cos Using the trigonometric function of a difference in two angles and rearranging gives cos B sincos sin 55in2 From the trigonometric function of half angles the equation becomes w 1 0055 1cos 2 1 0055 cosB smB 2R2 2 2 2 R cosBsin5sinB sinBcos Z Reducing yields Recognize the double angle formula gives R sinBsin5 B Z From this we can see that W 5Bsm 1 smB R2 where 5 and B are in radians SURE 215 7 Surveyng Calculations Area Partitioning Page 165 With 5 known 13 can now be computed The chord distance between A and F is computed as D R5 AF 2 where 5 is in radians The angle at F from A to G is 4m 180 90 1390 4 The complement at F from A to E becomes 1800 7 900 13 900 13 The angle at F from A to O is 4m 180 90 B1 gt6 90 B 6 From these relationships the angle at F can be represented as A ZAFOA 180 B 5 OFE AFE Using the law of Sines for triangle OEF the angle 062 is found using R 062 sm 1R 5m ZOFE J 52 180 ocz 4 ocZ B5 The angle 52 in triangle OEF is 180 0cz 180 B5 OFE The angle at the center of the are from B to E designated as 53 is 83 5 51 52 ES1 Boc2 Recall that B 061 51 Substitute this into the equation for 53 gives us 63 0c1 0c2 With the central angle between B and E known the arc length is found by multiplying 53 in radians by the radius SURE 215 7 Surveyng Calculations Area Partitioning Page 166 Danial 1990 gives the angles for the two parcels For polygon ABEF the following angles are de ned 4m 90 4 0 6 ZABE 90 72O1 0 6 ZBEF 90 3OC2 4m 90 2 For polygon CDFE the angles are 0 A 5 ZECD 90 12 3 A 5 4CDF90 2 2 0 A 5 ADFE90 12 3Ocz A 6 AFEC900 22 oc262 The area for both polygons are shown as follows A A A A A ABEF SEOE TAOE TEOF SAOF R253 IRIRZ sinES1 IRIRZ sinfS2 R125 1 2 39 2 39 2 2 where the subscript S indicated the area of a sector and the subscript T identi es the area ofa triangle For parcel CDFE the area is A A A SEOC TEOF SFOD A CDFE R12A153 Rle Sinsz R A2 5 2 2 2 Case 2 also can take on a different con guration where A1 is greater than A2 This is depicted in gure 21 Using the same approach as shown in the situation where A1 lt A2 the angle 5 can be shown to be SURE 215 7 Surveyng Calculations Area Partitioning Page 167 W 5 Bsm 1R 2sm5 Danial also shows that the interior angles of polygon ABEF can be shown as ABEF 90 1 A o 53 BEF 90 0 2 4m 90 1 For polygon CDFE the angles are the same as presented before The angle at F is AEFO 180 B 6 The area of both polygons can be computed as before R1263 RIRZ sin51 RlesinSZ R126 AABEFZ 2 2 2 2 RfA1 63 RlesinSZ R A2 6 2 2 2 A CDFE There is a second solution to this problem as Danial points out Draw a line parallel to the nonradial side gures 20 and 21 at a distance w1 This offset distance is shown to be w1 R1 sin 061 When A1 is smaller than A2 the distance from this new line to the dividing line wz is w2 w R1 sin 061 and the distance of the dividing line becomes and SURE 215 7 Surveyng Calculations Area Partitioning Page 168 8 51 52 63 When A1 is greater than A then w2 WR1 sinoc1 63 oc1 0c2 6431 62 63 One method of dividing a foursided area where one side is a circular curve was presented by Easa 1992 In gure 20 OJ is a radial line running parallel with AD It is desired to diVide parcel ADFE in such a manner that the side EF is parallel to AD From the geometry the angle 1 is computed as a 11 90 Figure 20 Geometry of an area partitioning problem with one side being a curved line SURE 215 7 Surveyng Calculations Area Partitioning Page 169 where M is a constant that is equal to 1 for convex circular curves and 71 for concave circular curves The distance X between OJ and AD is found from trigonometry as X R sin oc The angle 3 is found using the simple relationship B AADC 900 Finally the distance EF is found from DEF DAD IRCOSXI R2 X w2 wtanB where w is the offset distance given between EF and AD In the previous discussion the offset distance was given Much more frequently one needs to divide the parcel using a specified area Easa 1992 presents an iterative approach for four different situations Case 1 where EF is parallel to AD Case 2 where EF is not parallel to AD Case 3 where the line dividing the parcel is f1Xed at F and Case 4 where the dividing line is f1Xed at E CASE 1 From basic geometry we know that the area of the figure OAE using the arc distance AEm is A91cR2 2 where p is the central angle for the arc The area of the same figure using the chord distance AEchord is R2 A sm 2 P The area of the segment between the arc and the chord is the difference between these two areas R2 p7 A sm 5 2 180 p D 25in 1 AB The central angle is SURE 215 7 Surveyng Calculations Area Partitioning Page 170 The chord distance AEchord is computed as DAEEM DEF DAD wtan52 w2 From these developments the area A of the gure ADFE is computed using the following relationship A DAD DEFXA Seg Solving for w yields 2A xASeg W DAD The length of the dividing line DEF is then computed using the formula above To solve this problem an initial guess for w is inserted into the equation for the DEF Then using the last equation for w a better estimate of w is computed This process is iterated until the computed value of w from 11 does not deviate by more than some criteria ie 0001 from the value of w used in computing DEF Easa 1992 gives the steps as follows Compute 06 X and Estimate a value for w Compute the distance of the dividing line DEF Compute the chord distance AEchord Compute p and Aseg Compute a new value for w Compare it to the current estimate of w Repeat steps 36 until the difference between the current estimate of w and the computed value is within the tolerance established for the calculation QMr39eP Nt CASE 2 Figure 21 shows the geometry when the dividing line EF is not parallel to AD A line AD is constructed parallel to the dividing line EF The angle between AD and AD is referred to as the offset angle and is designated as 5 For the figure AEFD use the same approach as presented for Case 1 Angle BAD is equal to 900 A2 What remains to be done is the computation of the area ADD Using the sine law the distances DDD and DAD are computed as follows DAD sm5 DD39 smzADc D DAD smzADC AD39 sm 4AM SURE 215 7 Surveying Calculations Area Partitioning Page 171 Figure 21 Geometry of area partitioning for Case 2 The area is computed as AADD EDADDAD sin5 The total area used in the calculations is the area from AEED plus the area ADD if 5 lies to the left of AD minus if on the right The process is iterated as before CASE 3 In Case 3 the dividing line is xed through point F distance DF is given It is also assumed that the other end of the dividing line lies on the arc ABArc The lines AD and OJ are constructed to be parallel to the dividing line EF The direction of the dividing line 5 is unknown From gure 3 one can see that ABAD ABAD 5 A A 5 ADC ADC Using the same relationships presented earlier dealing with 06 X and B one can develop the following 0c x LEAD 90 XRsinoc BzzAD39C 900 The offset distance w is SURE 215 7 Surveying Calculations Area Partitioning Page 172 W DDD39 DDF Sin AAD39C Figure 22 Geometry for area partitioning for Case 3 The length of the dividing line is found using a modified version of formula to compute DEF presented earlier DEF DAD 7 Rcosoc7 R2 W2 wtanB The area of the segment between the chord and the arc is found using the formula for computing the area of the segment between the arc and the chord Aseg The area relationship is W A 3DAD39 DEF XASeg AADD The direction of the dividing line ZEFC is the same as ADC from which one can state AEFC AADC 5 Easa 1992 points out that the unknown is 5 and this must be solved for iteratively by using the area relationship Seg f6 A DAD DEF XA AADO 0 To solve this problem two values for 5 must be selected 51 and 52 such that the corresponding f51 and f52 from the relationships for f5 have opposite signs Then SURE 215 7 Surveyng Calculations Area Partitioning Page 173 A new value for 51 is then found using the test if f53 has the o posite sign of f51 then 51 53 This process is continued until 52 51 I S 81 and f53 S 82 where 81 and 82 are the tolerances established for this problem Easa 1992 identi es the steps necessary for the solution of the Case 3 problem Compute 06 X and DDD and DAD Compute AADDx Calculate w DEF The chord distance AEchord is computed using DAD instead of DAD Determine p Determine the area Aseg Calculate f5 pW QP39HeP Nf Steps 19 are performed rst with 5 51 and then with 5 52 The equation for 53 is then applied and the appropriate changes are made to either 51 or 52 and the process is repeated until the solution converges CASE 4 This case is very simple because the point is xed on the curve Because of that the distances DEF and DDF can be easily calculated REFERENCES Anderson J and E Mikhail 1998 Surveying Theory and Practice 7Lh edition McGrawHill New York 1167 p Danial N 1984 Area Cutoff by Coordinates Survey Review 27212269281 Danial N 1990 Partitioning of Pie Shaped Lots Surveying and Land Information Systems 5014349 Easa S 1989 General Direct Method for Land Subdivision Journal of Surveying Engineering 1154402411 Easa S 1992 Partitioning of a FourSided Area with One Circular Side Surve 39ng and Land Information Systems 5228691