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# Geodesy 1 SURE 452

Ferris

GPA 3.9

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This 78 page Class Notes was uploaded by Mrs. April Bechtelar on Monday October 12, 2015. The Class Notes belongs to SURE 452 at Ferris State University taught by Staff in Fall. Since its upload, it has received 20 views. For similar materials see /class/221622/sure-452-ferris-state-university in Surveying Engineering at Ferris State University.

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Date Created: 10/12/15

Bowring Direct Method Page l of 2 The Direct Geodetic Problem Using the Method by Bowring Some useful angle functions ddang degree oorang 00000000001 radiansang d ddang mins ang 7 degree1000 d 7E minutes oormins 0000000001 1800 seconds mins 7 minutes1000 dmsltang degree oorltang degree mlnutes seconds rem mg 7 degree60 600 36000 mms oorrem reml rem 7 mins 180 0 secs rem1600 r2d 39 mms secs degree 100 10000 The following data refer to a given reference system 1 a 6378160m f b a7af 29825000158005 b 635677471930860 m 2 2 a 7 b second eccentricity squared e 2 P b2 ep2 000673966076 Given data cl 73739155571 7 4355306630 pif lt0711 11 pradians s 54972161m 1 iftlt0711 k11radians7 az 1271027080 112 radiansaz Common equations A 1ep2cos 14 B 1ep2cos 12 A 1001323083 B 1002110004 C 1003364172 Z Crowne Plaza Phoenix mu NDnh am am Phuemx AZ 85004 quot094quot WWWquot 15122733370000 In 611273337518 away72A 73 50057 amenMrs TmusFarzHR Dml Benny s14mm I959 Axuun ALwIzn lM Mkraz anum To 500577 2020194715 quot1 31an ifan nauf as 1 15 55 ame llEM X Y 2 Plug 47 H econ125 x xHl w m A Y WUco 4 sm L AHUm 439 9 pm 5 vx A Buzlowrkl39s lZEZT Ana2mm I 1 F v f o 7 E A May g M AM39 F 151 n J av a 1r IF wig 5141 o Psm P4EF1 Q 251 F zuv we pawsw m Maia1w gun re L AIITESJALI gainAMTES X39X39m 135 2339 Wl v 1 Il JHMJ 5 39 3 4 5547 5 VALUE IX I 1 K caaZDu r ll 4 newin ewpre i 55 F L meFwW Jcmune j g 5 Irwmgmw TZEmer M M g caorwwn n MM Mae01512 ELLIB F aoumlq For A Rer Axon5 m Bsww X SuiM afIr Ell5010 456 Winnr5 M 7amp1 5y I YsLA 32 quotramwzryI 21y 0221 mm W 37 9M Awe a r 570 0 1er 701 zztag f 37 y Imwwsy aw sun7y kHz 71 any1 9 5 a ag ar wig 53932 7 7lya47 19 4 39mf 39 quotulna warm 339 quotJ WW7 Irma41 gtrr7 clarity mma 37 o 1 ynihawr awMy aay dzrxy V 41 M57 noun1m JILILW map21M 057103 if f 5quot 711 19 V s W39f E3quot 5v 9 Laxquot45c mgrvazmg studmi sa 17 JW 7Wrl1j M2 5y Gauchos 521V Myya Ont 50 qu 1 MA10ml mysz galM Myrnar7 397 Transformation Between Cartesian to Geodetic Coordinates Borkowski Method Some useful angle functions ddang degree oorang radiansang d ddang mins ang 7 degree1000 7E minutes oormins seconds mins 7 minutes 1000 dmsltang degree oorltang degree m w rem ang 7 degree60 600 36000 mins oorrem rem1 rem 7 mins 180 I r2d secs reml 600 mins secs degree 100 10000 a and b are the semimajor and emiminor axes of the ellipsoid a 6378137 b 63567523141 e2 0006694355002290respectively 62 and 9P2 are the first and second eccentricities ep2 000673949677548 f 0003352810655118 squared respectively fis the flattening Constants for GRS 80 Given Quantities X 4722390061 Y 744930540133 Z 44875605408 Solution A atan2X Y a fofofofofojoooi Designating X39 by r and 239 by z X r 4517803010902 005k Z Z Z Z 44875605408 7 bz 7 a2 7 b2 E 0980525 50M7au 5 aw 39HEIIPblA EMF IQ LAMquot 1 SaHE ELEMnip AM P W m mm 39c 39 dawn op Guamrm aF HE Ww 541359 F DEMv C p aw N 299M 0r Camrim 7 xn8 t3 k n a u Xu z 9 nu r 331 n uz on 2366 953 V 31me m Ruhrlm eus L X x A a n I T w uxn 8 a R ask u TLumr u if N u Nm kbv SW n u n ang ESXE kn wrunmm ts SWEEP Nu AN mu u mkr wrww KP kn 55in Vi lllmimn x xn u wwm3rm wm xal yn suw Wm yam1m mrs rigyams mm PMZIMET k LAEJ Mr agu 4 Am 54 m m39un mm lj fm wtl J Aux4 lr MHzwan 41 427mm 7m 45 am 2 mm Epym l39o l WW 1 ADD MM 1 247 w1M15A7w a LKIr m1k 15 15 21x sy39fu a 141 galquot rm All12 J is Ska an III 954 firm 6 14 101 l m u A 1191 M7 full 5156 a1bL w 4 DVVaAi Sm 69 m a fan 4 M III42M Fwyop usus ll 3 41 5mm Earme 4w 5 v FMuy 5 may 1 cowWow 0 Pompmag IS 1 cub 4L1 4556011an 9F awn 41177 5 11k dawma39u a ngn Am I 3972 Sunnmi 39aF 5115 39 7 39 494711 Elia7011 Amt4 Hmo u A e PmuELM 5 Btu57 7AM4577LK LIVE 9quot 00M C WEIR7014 F AL lM7E mm ml summon 5 5 h k 4l s M J22 F39 W J57 23375w TE sailmfg 32 sawmoan AXIS a ELLna39t IamWr so Multanx run 1 4baoLzuA Az z lmww39 1 a LJ sy m A39LHL 241124 4429 901mm 1 minian DIIZA HTmJ an 1 pr rm rem 4148 59 v at 1 L Wquot 1 5 er my 251quot by aquotZ a bquot 1T1 A FM ME pl39sToera pleaan a Mr Azqu Ln 7 X 39 f by 905721an J4 Lam mmmiau 1 J1 W a L39Aqu 4 Am mm am who ab 7 FM Ma MAXIMUM DlliEm39wH DUTMTIMJ Shun 12 141 ML EXAMPLE Fran DZ a mum faneDU77F sn rhfnup MAP 7 Enigma an in Panama Puma oxcaznl UIL 429011 3 40 sxwme 9mm 51man 0 55gtng 21 n 9 AtISIAH wannaan o u gx FdZ M5 Dim SWIM rs E 12739 1 F 4 5 Ezm w azr Malw 4125 1 gumnan For 1 HAMHEL AiToFF bik mu g M 4 54321 Z A h JlkwcloauL a 3971 SH IIHIEIIMSL 41 mama gamma hm AP 6 A Ares 21L swswa mewn 5 M J YIQMM AL 01542 50 ALL w42w top L 60 gaging gt4 214wdmLgt339L la 2 WW1 ManL M vii mummy L 65 wads x z 2341 kumXth ammbzl M Awm gymnme QUAWJ IES an 44 myamu SonFA F ARE 1 539 21gtZ mew23L 4 2 u 21 24 u a r M H M 0429370 Transformation Between Cartesian to Geodetic Coordinates Bowring Method Some useful angle functions ddang degree oorang radiansang d ddang mins ang 7 degree1000 d 7E minutes oormins 1800 seconds mins 7 minutes1000 dmsltang degree oorltang degree mlnutes seconds rem mg 7 degree60 600 36000 mms oorrem rem1 rem 7 mins 180 lad secs rem1 600 mins secs degree 100 10000 Constants for GRS 80 a 6378137 b 63567523141 e2 000669438002290 Given Quantities X 4722390061 Y 744930540133 Z 44875605408 Solution 2 2 a 7 b ep2 ep2 00067394968 2 b 7 atan2XY W X2Y2 rdmus39tngwra39dn 8 30000000 W 4517803010902 The initial estimate of the parameter B0 is a Z atan 07837191028 B b W B 3394 0 1 K TE quwa DX Jgng5a 1 va wzcg W sas 1 3 LMVJJWJ v mag 2 W54 i VI faIJWIZJU 741112 f sasz D39 S ma A gmqwazx Elli14S 311 1m LzJu mo miss7V7 VFJAWINHWW f r 15710 rrp W 14 mo 5 Warmv WVquotquot quot7 szofazuyf39 o m m 39I s39drn yf H0 goer397 aarls 17w 7 no 0113450 9 x A A 7 prqy 1719 7 1m11 gsmzm quot5 e H L f1 7 I397 99 A A A GlassM Fwmnemt ammm HM Farm 5 z 1 f D E E 1w 1 91 We 445 91 2 972725 Mat2 smmmq lawman21 AWE Ara szpgumcuum lm M w 0 5 gig ca b 94 Big62Ezm b 711 21AA Ilmt A gazef Win 5 oFFwDAII m MA77775 E E a ewD 94 99294617 75724 ms gem M CEHrErL 044 SPHERE Mugs41 FwmusymL M 41 by g Cy Mama 7 53 M 5 2457 Maw szf w Hwy 1w 7 9A1 Fran Me 444 P E Sm anagram msrmi SAX m F P X39PM V 2 snow 3am 2 5Ltsa za 47 SWUWmmw wawxhwp R we M Mttaa xjsc Ly 5 WEDAMT an X7ltan E HMXJDAQ V Vquotgt Y39 SWA 39 Z MKhnq 4amp1 2quot 74m v R 1139 th Me mm pawng vfchZDWATg wr t l1 wannang M ZM XWL A quotE ii 4 39 mwa sm n 1 th av M MA gt7 2 End rbii wl 31 A R Jwx4a q 2 4 a w R wwwrxnxq n D if mensm mum2s 21 c A M x 539 044stk47 F Fet s statenpo ege MVEEIE PIG515M 9 5 EggMam WEDHA 75 V V 39 4739quot g F Earlv76 sng ea M thz C39aa DIv39 TES 4 X AHdent MA 7 MA 0949 Z N39 1Usind IE f Pee15 pay n m A Wyzu X0 2 ELLPram mew2752 diff srrzm zzmm Ma 4 anw J 4P oLUTl y 12y HunMal Mp17 mm5M 5 I xllquot1 SCHWm ream Day B g Harm M rhgan 491m mm 795mm m 3701 Fp laxamp 5y 1 n X Fzm4 Fme 5m 4 WAJsm x1yL 8u7 zAr42e Asu 47 w orz IVWUISPgtW Zthlsaaz EMF 2 eIAxfa So y E up fa 5 2177753 r7125 wait 4 Xy1 lr t l FM Fmsr AManmau r 4 2 wICLJsMa Il WTZTV 95er 71 Hi ml f 55055173925 Ia ewe7f dawnnan mm mmw 414 6010 I an man quotmammaryan an smearac 72 soar76 mpMim VIM YEW7M3 ing Id Emu 77w 5995370115 H 439smlt303 if musrm M714 pmupm By my Mm MM was WZzsx A m vamrm nan MHZ0a 771 504 5y 77 477711 5 imamI Fawn 9125th1 A M11774 971272 Manvar quot l A 4 x 45 5 AMImm Fwdl6 Ad ELM W V X 1 V1 All10h ism 0F I MM7571 quot3901quot Jaw ag 3 my 416 a Z b Wquot l a n 2 w zt z b W1 1 ITEKRTHE fa mf 0F 1 HMM M 5mm 12725 isnm39 m Fzm HOLIDAV mum ANCHORAGE DOWNTOWN 239 West 4m AvenueAnchorage maska 99501723 5907 27978571Fax sumauna w 2 11 1 1 4 Ha 2 9 b FoM 1 0 I39m 4 1 w 104 L3 1 7 Om4 m 4quotquot Hm 5 5 gammy 125 ranen Far rrmrve77 mm39L m cammgr Ta gaze 1 15524 774 HOYT Cum mm mm m w V Hz E W Zltb l 7 4977 War EranIE 4 4 a a 51W MD I yew77 H IMW 22 34 415771 5 g ea2y eww 1 47491 3 my 4 my 5er A VIIIZ4 I bymum mme 5 500V 05 MaiL rm 1 Ar unkr Mayax74 7745 9717 Arm ELLPSpb wA 775 Eva oxr or 1 mum 7 4 k Z 1 at 1391 LL vpgz x y at an g mm12k CmPHHTeTUF ME PDIIJT an M Sun nt aF Embm u lie Wul d 4F FoW lm m 5qu Elamo 425 0min 8V XV Z A Um I DrZAMM MW 5 Alamm T a Ms Empire mm PASSa Miami 5 mw 51 p7 447 mu 5 Dame5w F zw 9 lmWC E u WoAf 6 L 121 wr zz i a fZMbl Wm 11 5 Aru cogenle or 10510 av rr am A m I 4 Flmd k pacewig wim J 133 tamr 49 Us mu 2 we yza mra 111 49534 my 14 r mpllMi 15433 Mic Mia1M 5 Ixr A or miler 45 all I 7 x 4 camp947 5 97 541 lama 945925 2L 739 zquot A km N your M quotwt Hw Y 0 AL 1quot um07 quot Mix 1 o alll5 W Xquot ya Jag My ramp IWfme amt4 77 MW IzaFISw Mgrlap Am 71th Ef iW A m x 3 mo d gnzg rb wquot a a z w 2042 5 b W 5501 ll lfm u aws TAKE 1 ag pawl5 a to m thv0 L L Fc V m 4A4z43blbz bf Aw comm u WM VALUE Av 04 05114 M 6 mi my 39 FYmL 56 441 AEW IEZ ls 255m m 4 Mm VAws rm FHA 1 wa Aw 51554 me mu 1 Imam fyar WMA Hm com5mm 72 29720 4 LAT7w frHa z b vl W 1 L MD a blV N 911 7 lEh lTS Fawn Fw H w W 1 z LYJUL As bu i WAVq Max 76131 or I wdzl lt We 24l w14 H lIZ cp4mzT Welt 1 5 Anuser EM DHHUFE LLIPSIF P 24 Iquot 3E 1 dam If 1 1 Ea it bf 41 quotL 26le Sheraton 35021qu Illlll I Is Ilml Ih rammm 9446 Farr14M VALQEW 4255 99 Ta EMA MES teaFawn FIOIE WQiJF fLLlPQIE 64175949 guia illmay 2ng aFisEgzwnkHL i5gampwgg AL 513m 6431 r 7 iJ Bmvm inarzntifngm Wit75W raj L g fEZLEIAEMR Rama re z W JO M 7 H LpM zi z F LE rm 5mm Mrrj IlLEFSE swm I 39l zlk i 74pwmn EQupZoAI 7 777533 quotif iv A an 4 g39 39 7 ig39 39f39 7 7 777 HT lemmn c f Fina EM J37quot aC Abdwi 51 7M 9 3 0474 04Ia 5W4 yuan D stp h39fsum cmtu J 594quot Spam nw 471 4m ZDBquot3 C39ZDSAZ V 85 M Jaw4k m6 0mm buzz13 and lIE 5amp Fawn 3914 L A47 3539 y D I A49 24 M zr5 43 39 3 E SAD ww F7 mu Bmz mDskz gm 65 m E r any imam y My masmy munmu Jm 21 5F E 01 I 451147 re Iv mm au Hm 00194 LEATI wme I J T7 17mm IIMa 72 wr Doquotmm F39nquot 39Pn quotpm Suc39u39 Pa 1 In le 7539 mu 39 PM a Shus i Mlsl pm I m a mu39 5 12 1 C PM In spy7 in We arr PS AM 4 ma I131quot Lima lswu IVFTE39L rawu 9 Appws eT 445 Duran sz Let 77 I D w u 4 Azstfu llquot 1 n 119 IF 540 I 5 Am DUE17M Am aF ll717 c39mclf r 3 Am Am ck mm Malcan tiIm6 l A 1 12 17 llre x 1 hrqu mum 41 m M couchav 0 Am E uiuMAuzv IS nth 1 my my Ii Pleajchw 7G WHIEAI 1 mm div394 AT 55w m4 uuw 411 KuTME1I M 001417 HIV Iwy 135799le or any Km raw Mn Ara uo7 Vlzaj t 54 5 one1711794 0F 4 fwof Manix ap312 ro ameny a My awepalm 12mme DAFIM 51er 6 Pagan Sum71239 Ma a15m n a e A m39 A39 Du39zecnour a a EXPZEJTEp ram 27 JarLpr EmuMW 5F 1 Spark Flurle 471 l5 4mm Mamm do 45 4A JET4 M417 4 Eda 44 Jim1 7w 0 ME LEI4739 12szdJ Aim4 A ups an h M Am 4 ups 94 k JLW d mw rm x w39 L 5 27 Mire Ta FWD ll 44mg chub 2 M 5am enrzu n WE mm Jig 51401 M74774 4 M u I m u b 5 u Js39no FNj lf71 my ma A Herrera A 1 4157sz 4 41 3 A M A W 11m 4 8AM Eat779 M 3 WW M Ill1 41 yLTILy Ml 17147 Eggmay 5V 1 A 117 Sin Vm u A1 swv39mu MERIDIAN ELLIPSE Surveying Engineering Department Ferris State University The X and z coordinates of a point on the meridian ellipse can be calculated given the reduced geodetic or geocentric latitudes Using the reduced latitude B see Figure l the formula are derived in the following manner A 0 7X4 P2 Figure 1 Meridian ellipse using the reduced latitude From geometry one can see the triangle OPle that the application of the Pythagorean theorem results in 0122lez2 a2 1 The equation of the ellipse can be shown to be 2 2 a 2b 21 Since X CF and z P2P substitute these relationships into 2 results in 1 3 SURE 452 7 Geodesy l Meridian Ellipse Page 27 which can be rewritten in terms of the semimajor axis by multiplying both sides of the equation with a2 2 2 a 2 0132 2 PzP bl 61 3a Combining 1 with 3a results in Z 012 13sz a2 OPZ 2 PZPIY 4 Solve for P2P Z PZPIY 13sz 2 2 2 b2 2 BF 703211 b PZP 303213 5 From Figure 1 one can see that PzPl a sin Therefore b b zPZP PZPl asin3 a a 01 6 and lt7 Example IfB 430 37 25 and using the GRS 80 ellipsoid the meridional coordinates are found from N acosB 6378137mcos43 373925quot 461705437l6m SURE 452 7 Geodesy l Meridian Ellipse Page 28 2 b sin 5 63567523142msin 43 373925quot 43856372730m Intuitively one should be able to look at equations 6 and 7 to see if they look correct Let s take an example Ifthe latitude is 00 then we see that X a and z 0 which makes sense In a similar vein one can look at the example and say that it looks correc in its presentation Since the latitude is less tan 450 and agtb then we know that X must be larger than 2 Above 45 2 will begin to become larger than X A Z Figure 2 Meridian ellipse using the geodetic latitude The determination of the X and z coordinates using the geodetic latitude is shown neXt Looking at Figure 2 the slope of the tangent line dx dz is given as dy cos tan 900 cot 8 a 0 SM Rewrite the equation of ellipse 2 into the following form bzx2 azz2 czle 9 Rearranging Page 29 SURE 452 Geodesy l Meridian Ellipse 2 2 2 2 2 2 2 2 a b b x a b b x 22 gt 2 10 2 2 a 61 Taking the derivative with respect to X will give 2 2 b2 2xb2xcs 11 dx 2 a2b2b2x2 a a z s1n0 Z a One can rearrange 11 into bzxsin 0 azz cos 0 Square both sides yields b4x2 sin2 0 61422 cos2 0 0 12 Multiply 9 by b2 sin2 0 will result in the next equation bquotx2 sin2 0 albzz2 sin2 0 61 sin2 0 0 13 Adding 12 and 13 then multiplying by 1 results in 61422 cos2 0 czle sin2 0 czle sin2 0 0 14 Solve for z z2 61 cos2 0 czle sin2 0 azb4 sin2 0 61 sin2 0 b4 sin2 0 2 612612 cos2 0 b2 sin2 0 a2 cos2 0 b2 sin2 0 z 2 b s1n0 15 a2 cos2 0 b2 sin2 0 The X coordinate can be found using a similar method of reduction yielding SURE 452 Geodesy l Meridian Ellipse Page 30 2 a cos0 16 612 cos2 0 02 sin2 0 But recall 2 b2 2 a gtb2 azl e2 17 e 2 0 Substitute 17 into the denominator of 15 and 16 gives 1 612 cos2 0 02 sin2 0A 612 cos2 0 a2 sin2 0 ale2 sin2 0 020082 0 sin2 0 62 sin2 0 a 1 62 sin2 0 Upon substituting this relationship back onto the denominator of 15 and 16 yields acosq 18 19 Example If p 430 43 11 and again using the GRS 80 ellipse the meridional coordinates X and z are found by X acosp l e2 sin2 p 6378137mcos43 43 1 1quot 1 00066943800229 sin 43 43391 1quot 46170542002m SURE 452 7 Geodesy l Meridian Ellipse Page 31 al e2sinp l e2 sin2 p 6378137m1 00066943800229sin 43 43391 1quot 1 00066943800229 sin 43 43391 1quot 43856374523m You will notice that these two values are very close to the meridional coordinates found using the reduced latitude This is not by change The reduced latitude and for that the geocentric latitude that follows were computed based on a given geodetic latitude The discrepancy is due to rounding off the latitude to the nearest second of arc Otherwise they would be exactly the same zll A X X Figure 3 Meridian ellipse using the geocentric latitude The determination of the X and z meridional coordinates using the geocentric latitude is developed as follows From Figure 3 one can write x r cost 20 z r siny 21 where r the geocentric radius and V the geocentric latitude SURE 452 7 Geodesy l Meridian Ellipse Page 32 Substitute 20 and 21 into 2 results in r2 cos2 l r2 sin2 l 2 a 13 2 1 0 r2b2 coszla2 Sinzll1 Solving for r czle 22 2 ab b2 cos2 l a2 sin2 l 23 Substitute 17 which was written as b2 6121 ez into 23 2 612612 1 ez 6141 e2 lcos2 l a2 sin2 lI 612162 cos2 l 62 cos2 l sin2 l 24 Substituting 24 into 20 and 21 results in the equations to compute the meridional coordinates using the geocentric latitude 25 z 26 Example Given V 430 31 39 and the GRS 80 ellipse the meridional coordinates are SURE 452 7 Geodesy l Meridian Ellipse Page 33 a1 e2 cosw X quotl e2 cos2 V 6378138m1 00066943800229 0084303139quot J1 00066943800229 0084303139quot 46170557337m 1 a1 e2 sinw Z 1 1 e2 cos2 V 6378138m1 00066943800229 0084303139quot J1 00066943800229 cos43 313939quot 43856358487m Some useful relationships between the three different latitudes can now be developed For example using the geometry in Figure 3 write tanlI Z x Substituting the ratios and 18 into this last relationship will yield tany tan l eztan0 a From this basic equation one can write the following basic relationships between the latitudes tanlll eztan l eztan0 27 tan l eztan0 28 tan a 1 e tan 3 29 SURE 452 7 Geodesy 1 Men39dian Ellipse MERIDIONAL ELLIPSE Robert Burtch Surveying Engineering Ferris State University Meridian ellipse using the reduced latitude where aandb are the semi major and semiminor axes ofthe ellipse and 3 is the reduced latitude Meridian ellipse using the reduced latitude 0 Example using GRS 80 and B 43 37 23 Find the X and z meridional coordinates X 6378137m cos 43 373925quot 46170543716m z 63567523141m sin 43 373925quot 43856372730m Page 34 SURE 452 7 Geodesy 1 Men39dian Ellipse Page 35 Meridian ellipse using the geodetic latitude 0 Begin by writing equation of ellipse O Differentiate with respect to x and z the meridional coordinates Finds slope of ellipse at a Meridian ellipse using the geodetic latitude 0 Meridiona coordinates found by Meridian ellipse using the geodetic latitude 0 Example p 43 43 1 1quot Find the meridional coordinates 7 6378137 c0543 4311quot 1 r 0 00669438002 sin2 43 433911quot 17 e2 sin2 6 4617054 ZOOZm 31 62 Sini 63781371 r 0 00669438002 sin 43 433911quot 39 2 9 v u 1 7 e2 sin2 9 17 000669438002 sin 43 4311 4385637 4523m SURE 452 7 Geodesy 1 Men39dian Ellipse Meridian ellipse using the geocentric latitude O From the g lrg r costI Z z r sinll 0 Substitute into equation of ellipse and PX 4quot X solve for r Meridian ellipse using the geocentric latitude r a l e2 wll EZCOSZIl 0 Substitute into previous equations yields the meridional coordinates 1 x a l eZgtZcosl wlI QZCOSZll z 11 eZ cos2 1 al eZ Zsin V w Meridian ellipse using the geocentric latitude OExampej l 43 3139 39quot Find the meridional coordinates 7 307624 COSV 7 6378138m140066943800229 cos43quot31 39 lie cos2 v J17 00066943800229 cos43quot31 39 46170557337m allre mi 7 6378138m r00066943800229jx c0543 31 39 z I17 2 cosz w 1170 0066943800229 cos43 31 39 4385635 8487m Page 36 A n mu Aum3mnsmw I 1 u 5x8 Q3 um 1x 4m 5 u 38 a as g 533 owwmwmeqh CLXFSV ea envwkw 3x3 nurCQv u Q an k umh kuNmSQI QMQA PMV V S Amtd NE QMQ3N WQ mi tuam Qi nom A Hg zw sw any 9 5mg Sim 47 25545 m 24Itsud u41 dipa a 7 5quot a 94422475 4 4 ain39L39swm s L4 H 4 4 ED thaw Z W 4 f 3quot 5amp4 6450 4 1 3 I Him 2 SM staked Ts 1 44 47m gt anmur Si 5A 47 wage a w Fm quotampAd Am coxr 39Ez 1 As MAIM4 O39ZEkazl 03mm 5 541 9 427 y MIMAwJA DEI liVM Far olm lw ld nave If 5y 74m wx amp 1 By rum IZEarML of yuansxlt r my g x1suao m 4 4x1 u nub x1 I N amauw 171 aZ 127 IZe Sfft fV FB 39a 1 3w 2 4 law via a m on A I 1 unauuf I ME I flawsaway MM w cndJas wumx 5 Ma sumo rz MillMAM cs coAIIDBL A A mk a EFEIz IF AIME t 667101 sux JM 5 DEMHAIZDIFK 9 on Fan 1 at 905043 267qu 54439 17 541 110 WU gt11 EV nil114 EECI ch a 41011ng mm Qwuzvznwt 51 AP Wm 0 4 47 5130 W145 lack in S 3 o 8 quot 4quot 5 quotW 9247 anuspent Pasn rIf Niam M 44w I a an of Mal Mega of u Hard 5 3 I3 I H on A U g xn a j awvlm z w pm 1139an raggapew Eavrkrwely quotquot quot9 39FMH rr 41m l y l nmprl o 5 Ms Marat 3 Mr I 10qu HM no ltpltnlt 2L mum uAI Moan IM 91 If 4mm 1 or nggmy f i A IZEWEL mEzy i u 57 quot 3 u 957Nuzo 1m 593720411 360 OBTrZACTIiVA 3wquot J h A 434 a s 425 cn a Z 39L M F 6 392 SW pgm sgt RANK4 1 A A IF A vb90 Mai A1 Ky ADDAM ae smTumAz 3amp0 Adamm 79 mtg4m IT a man Pm at 155 Sir1 1 m474 9 53 1 amp m 1gm4 sw yshampuI S B 1 we willEV 1 m 7 1 m 51 61 5 T 03937 EL 9mm Aunt175 mam 9 gm 41 amp l an M41 0 4 tL MIEIZIF faAuVM39J A IL1 on ArM Sbyr a wxWaan L a 0 Wilt ne AA I 161041175 VALIE lt an gt 541 Accomm To WIMm 4 51017421 9 MarHm Lawn 45mm u mam 5n lErlw u M4 AA 92 ma mnoquot w 51 AA 3117 aux0 7 1 t ialfz i 1 4 gun 7 Ix y M 545 Am m an Max M gt4 51 M75 yhf I 4037 w i w 0951 my a 5 MM 7 7 0imw 1y 1 M 1y C m m3 a 5547 156 Sn slur1m N24 i 3 42 quot quot491 C A SAM Eli 17 Orv I c z 47 s 1mm a f siwlt1 4504 er W ZJo zsww WW1 44 wwle m wwm 39mw 55sz W fJJn1cha 0 11 ma le The Inverse Geodetic Problem Using the Gauss MidLatitude Method Some useful angle functions ddang degree 7 oorang 00000000001 radiansang d 7 ddang mins 7 ang 7 degree1000 1 minutes 7 oormins 0000000001 1800 seconds 7 mins 7 minutes1000 dmsltang degree 6 OOmang minutes seconds degree rem 7 ang 7 degree60 600 36000 mms 7 oorrem reml 7 rem 7 mins 180 0 secs 7 rem1600 r2d mins secs 7 degree 100 10000 The following data refer to a given reference system a 6378160m b a 7 af 397 298257222028 b 635677523702048 In first eccentricity squared e2 2 e2 0006694380024537 Given datai A 73739155571 AA 4355306630 p if Alt 071 1 1 ifAAlt 041 9 1 pradians A Al 1radians7 A B 737570912874 AB 44252481672 pif Blt0711 1 ifABlt0711 2 pradians B 12 1radians7 B Inverse Problem The mean latitude 4 1 t 2 m T mr2d 73780342859 Gauss MidLatitude Method Inverse Page 2 of 3 Problem The difference in longitude and latitude AA 7 712 7 711 AAer 049837603 A 7 2 7 1 A r2d 7029821434 Wm 7 J1 7 62Sin m2 Wm 7 099874163 a Nm 7 W Nm 7 638619622720554 m m 1 1 Am 7 N Am 7 000000016 m m a 1 7 e2 Mm Mm 7 635943964734370 m 3 Wm B 7 L B 7 0000000161 111 Mm 111 In F 7 i 39 2 F 7 0 03188828 7 1251 mcos m 7 7 51 Q AA 7 AA n m FAA3 dmsAAr2d 7 7058597381 M COS K 2 J J 4N i A 7 A A 2d7 0 29821401 p7 M pr 77 T J Ax 51 AA 7 AA 2 Al r2d 049837446 p AA p 7 J cos X1 7 AlpM X1 7 4389019860931 m Am X2 73309940217128 m si 5497199248763 m Fet s statenpo ege MVEEIE PIG515M 9 5 EggMam WEDHA 75 V V 39 4739quot g F Earlv76 sng ea M thz C39aa DIv39 TES 4 X AHdent MA 7 MA 0949 Z N39 1Usind IE f Pee15 pay n m A Wyzu X0 2 ELLPram mew2752 diff srrzm zzmm Ma 4 anw J 4P oLUTl y 12y HunMal Mp17 mm5M 5 I xllquot1 SCHWm ream Day B g Harm M rhgan 491m mm 795mm m 3701 Fp laxamp 5y 1 n X Fzm4 Fme 5m 4 WAJsm x1yL 8u7 zAr42e Asu 47 w orz IVWUISPgtW Zthlsaaz EMF 2 eIAxfa So y E up fa 5 2177753 r7125 wait 4 Xy1 lr t l FM Fmsr AManmau r 4 2 wICLJsMa Il WTZTV 95er 71 Hi ml IMRECTANDINVERSEGEODEHCPROBLEM Surveying Engineering Department Ferris State University The direct and inverse geodetic problems are one of the most basic questions posed to the geodesistsurveyor The direct problem can be simply stated as given the latitude and longitude of a beginning point and a distance and azimuth to the second point compute the latitude and longitude of that second point along with the back azimuth from point 2 to l The inverse problem can be formulated as given the latitude and longitude of two points compute the distance between them along with the forward and reverse azimuths The problem is complex because the earth is not a plane or even a sphere Thus to solve these problems will require assumptions that limit the accuracy of the results Bow ling Formulas Direct Problem Bowring developed a formulation for the direct problem using a conformal projection of the ellipsoid on the sphere and it is called a Gaussian projection of a second kind The simplicity of the system lies in that the ellipsoidal geodesic is projected to its corresponding line on a sphere thereby allowing the formulation using spherical trigonometry Both the direct and inverse solutions are noniterative These formulas are valid for lines up to 150 km Without derivation the common equations used by Bowring are A 1 e392 cos4 p1 B 1 e392 cos2 p1 C 0e AOVZ 7H 2 Then the formulas for the direct problem are presented as follows SURE 452 7 Geodesy 1 Direct and Inverse Geodetic Problem Page 42 sB2 aC k2 XI itanl A tano sinoc12 J B cosp1 tano s1np1 cosoc12 1 1 1 D Esm s1n039 cosoc12 Xs1np1 s1noc12 tanw 3 2 4 p2p12D B Ee Ds1n 2p1 BD a t 1 B sinoc12 all 12 coso tano tanp1 B cosocu Inverse Problem The inverse problem begins by computing a series of constant values A 3e 2 2 D ltP1 4B2 Aps1n2p1 EAQJ E E sin Dcos w F isinwBcosp1 cosD sinp1 sin D tanG3 sin E2 F2 tan H sin p1 Bcos p1 tan Dtan w SURE 452 7 Geodesy 1 Direct and Inverse Geodetic Problem Page 43 Then the inverse geodetic values azimuth and distance are found using the following relationships oclG H OLZGHi180 aCO39 s B2 Gauss Mid Latitude Formula This approach was rst published in English in 1861 This method is based on spherical approximations of the earth This method is based on using spherical approximations of the earth Thus azimuths and distances would be the same on the ellipsoid as they would be on a sphere This assumption is not true Thus this method is best used for distances less than 40 km at latitudes less than 80 Direct Method Without derivation the formulas for using the Gauss MidLatitude approach are M ssin0c12 A Nm cosqpm s cos0c12 A mp Mm c0sA7VA where tan tan MM 3 or AOL Msinqpm sec M 2 12 sinqpm secA2 P sec3 pm sec3 Because of the nature of these equations they have to be solved iteratively The following steps are used in this solution SURE 452 7 Geodesy 1 Direct and Inverse Geodetic Problem Page 44 N E MA Solve for the approximate change in latitude Np by using the measured azimuth instead of cc12 A and using the radius of curvature in the meridian at point 1 M1 instead of the mean radius of curvature Mm Compute the rst approximation of the latitude of the second point P2 P1AP Determine the rst approximation of the change in longitude and the longitude of the second point x2xlm Find the rst approximation of the change in azimuth Using these approximations update the values M N pm and the other values listed in the rst four steps Inverse Problem The Gauss MidLatitude formulae for the inverse problem can be developed into direct computation requiring no iteration It can be presented as follows without derivation where S1 2Nm 1 51 MN J 1 X1 Aoc Wm x 7 2 SS 121 0c12 Aoci 180 s1 X12 X Then SURE 452 7 Geodesy 1 Direct and Inverse Geodetic Problem Page 45 X1 s1 sinoc12 NmAk39cosqpm A M X2 s1 cosoc12 706 MmAp cos7 sinA W P W 2 MA A06 Aksinqpm sec AN M SinA l A pFAk3 2 1 F sin cos2 12 Pm Pm This approach has an accuracy of about 1 ppm for lines up to 100 km in length PUISSANT FORMULAS The Puissant formulas are not generally used for lines greater than 100 km long The formulation of this method is based on de ning spheres passing through the rst point Direct Problem Without derivation the difference in the latitude can be presented as where Ap scosocuB s2 sin2 0612C hs2 sin2 06le EXP2 D SURE 452 7 Geodesy 1 Direct and Inverse Geodetic Problem Page 46 l B M1 C tampl 2M1N1 D 3e2 sinp1 c0sp1 21 e2 sin2 p1 E1mmfg 6N12 h scosoc12 M 1 The term 5p can be found either from summing the first three terms in the formula used to compute Ap or from the relationship sin2 0612 tanqp1 sin2 1120 3tan2 p1 5p ic0s06 SZ M 2MN 1 1 1 6 MINI2 The latitude is then found P2 P1 AP The difference in longitude is found from the follwing equation S s1nM s1nA12s1n 0612 sec p2 Then the longitude of the second point is found by adding this change in longitude to the initial longitude x2 xlm The back azimuth is found by A06 Msin pm sec A A AQ Ac 12 s1npm sec AJ sm3 pm sec3 3394 0 1 K TE quwa DX Jgng5a 1 va wzcg W sas 1 3 LMVJJWJ v mag 2 W54 i VI faIJWIZJU 741112 f sasz D39 S ma A gmqwazx Elli14S 311 1m LzJu mo miss7V7 VFJAWINHWW f r 15710 rrp W 14 mo 5 Warmv WVquotquot quot7 szofazuyf39 o m m 39I s39drn yf H0 goer397 aarls 17w 7 no 0113450 9 x A A 7 prqy 1719 7 1m11 gsmzm quot5 e H L f1 7 I397 99 A A A GlassM Fwmnemt ammm HM Farm 5 z 1 f D E E 1w 1 91 We 445 91 2 972725 Mat2 smmmq lawman21 AWE Ara szpgumcuum lm M w 0 5 gig ca b 94 Big62Ezm b 711 21AA Ilmt A gazef Win 5 oFFwDAII m MA77775 E E a ewD 94 99294617 75724 ms gem M CEHrErL 044 SPHERE Mugs41 FwmusymL M 41 by g Cy Mama 7 53 M 5 2457 Maw szf w Hwy 1w 7 9A1 Fran Me 444 P E Sm anagram msrmi SAX m F P X39PM V 2 snow 3am 2 5Ltsa za 47 SWUWmmw wawxhwp R we M Mttaa xjsc Ly 5 WEDAMT an X7ltan E HMXJDAQ V Vquotgt Y39 SWA 39 Z MKhnq 4amp1 2quot 74m v R 1139 th Me mm pawng vfchZDWATg wr t l1 wannang M ZM XWL A quotE ii 4 39 mwa sm n 1 th av M MA gt7 2 End rbii wl 31 A R Jwx4a q 2 4 a w R wwwrxnxq n D if mensm mum2s 21 c A M x 539 044stk47 F Puissant Method Direct Geodetic ro m Page 1 of 3 ble Puissant Method Direct Problem Some useful angle functions ddang degree oorang 00000000001 radiansang d ddang mins ang 7 degree1000 d 7 minutes oormins 00000000001 1800 seconds mins 7 minutes 1000 dmsltang degree oomang minutes seconds 600 36000 degree rem ang 7 degree60 mins oorrem rem1 rem 7 mins 1800 I r2d secs rem1 600 mins secs degree 100 10000 Constants a 3 6378160 b 63567747193 e2 00066945418 e2 is the first eccentricity squared Given data 1 73739155571 7 4355306630 pif lt0711 1ifvlt0711 11 pradiaIlS X1 3 1radians S 2 54972161 112 radians1271027080 Solution M1 a 7 62 M 7 6359277 924313 M1 is the radius Of 2 3 1 39 curvature in the meridian 1 7 e2sin 1 for the first latitude while N1 is the radius of curvature in the prime a vertical N1 N1 638614243900121 Puissant Method Direct Geodetic Page 2 of 3 Problem 1 B Z B 0000000157 M1 t 1 3 an C 950002 X 10 15 2M1N1 e SIeZISinwlycoswl D7 0004869 21 fezsiw i 2 1 3081101 B 1138621 x 10 14 e T h S39COSX12 M1 h 70005223 ewowBeszisniaercehs2ltsmltaegtgt2E Aq seese12e e 52s1ne122c 7h52sin0c122E e 51813 112 dms 1 A r2d if 01 mp gt 0 dms 1 A r2dlt71gt if m Act lt o The latitude ofthe second point is the 39 I p3 if 2 lt0e11 12 pradians a N2 is the radius of curvature N2 6386250478147 2 in the prime vertical for the 1 7 eZsin 2 second latitude AA ism 12l1i1sm 122H N2 c002 c002 N2 HISTORY OF GEODESY Robert Burtch Surveying Engineering Ferris State University EARLY IDEAS OF EARTH I Earth stood still I World center of omerwise we would feel universe OUFSEIVES mOV39ng I Looking up one saw the C Must be at moon and stars revolving Otherwise we would fall ab ear off I Each society thought I Immense stretching itself as the center of into unknown oceans Unlverse and Ian s called ompalos syndrome Greek for navel FLAT DISK EARTH IHomer 9th Century BC flat earth supporting the sky over hemisphere IThales of Miletus earth disk resting on water IAnaximenes earth held fixed over air of infinite thickness which could not escape upward due to the large extent of the disk C Y LINDRI CAL EAR TH IAnaXimander I rst to hold that the earth was held at the center of a celestial sphere Being in the center earth needed no support because it was equidistant to the sphere at all poinls No preference between directions to fall toward one end and no up and down I Earth cylinder 31 ratio of diameter to height Top location of inhabited world Axis oblique to path of sun SPHERICAL EARTH 393 ARISTOTLE 4d CENT BC ERATUSTHENES 2 757195 BC my a 3 quotmm m a man hm mmnzlnamm Myquot umuumv mum md km a he m m min a mg m w mg the tummy 0F M ERATDSTHENES ERA T OS THENES ERA T 0S THENES GHow was distance measured sDifficulty in GErrors lCamel caravan traveled 100 stadiaday and took converting stadia I Sun could not be about 50 days to travel distance into meters directly overhead at I Probably determined by Egyptian cadastral maps therefore one time Of measuremlent made by bematists clerks who measured estimate is the quot Off by abOUt 22 distances by walking evenly and counting their circumference is too lAlexandria and steps Syene not on same large by 16 and meridian OCalculated dIstance 5000 stadIa therefore another 05 too circumference of earth 250000 stadia small 39 un39Elconvers39Oquot pro em POSIDONI US amp ALMAMUYlI WILLEBRORD SNELL US a ggidonius 13550 Caliph aMamum 9th centUry C 1580 1626 S One of first to use Melirslugeg SIZe 0f triangulation enaeasurgd length I Measured 33 triangles along using wooden nearly meridional arc between rods over a Alkmaar and Bergen op Zoom degree of latitude in Netherlands with 1 angular 56 23 Arabic precision and astronomic miles near latitudes at ends Bagdad 0 about 01 different from current average a Used star Canopus to determine size of 3 Found star grazed horizon at Rhodes but came up 14 of zodiacal sign over Alexandria 148 of circle 1 Reduced size to 180000 stadia which was used by Columbus in voyage zek z Baa zek z an v w 32513 OBLA TE SPHEROID OPendulum clock provided practical evidence lJean Richer experienced trouble adjusting his clock on his expedition to Cayenne Guiana in South America STwo expeditions commissioned to determine size OPeru now Equador ACADEIlIIE ROYALE DES SCIENCES GFound 1 meridional arc in Lapland was longer than 1 meridional arc in lClock lost 2 12 minutesday and pendulum had to in 1735 led by Peru be shortened Charles Marie de la ONewton s concept I Period of clock depends on length of pendulum CONdamine COFFECt and value of gravity farther from center of Lapland in 1736 by Beginning of gravity slower the clock Pierre L M de ellipsoidal era in Maupertuis geodesy DETERMINING LONGI T UDE DETERMINING LONGITUDE Galileo discovered celestial timekeeper eEarly explorers used sNot knowing precise dead reckoning I Never agreed with one another because estimates of distance not know where they and direction varied were accurately hourbyhour and enough daY39to39day lCaused death and loss of ships and goods longitude was costly I Shipwrecks common because sailors did Jupiter had 4 satellites and their passages g in front and behind the plane could be predicted for some months in advance Because Jupiter is so far away astronomic parallax no problem motion would appear the same everywhere If observed at 2 points and if local time were determined when observations were made one could subtract to obtain the difference amp at a rate of 4 mindegree one would have the longitude Still not possible at sea DETERMININGLWGITUDE um Himan mum 2 mm mm New w me Mg mquot WVle And My delermme mm mmm a 0F amvenmmq mam 2nd 2 m a mm m John Harmon ramW CARL FRIEDRICH JOHANN CARL FRIEDRICH GA USS 177771355 GA USS nmmma Wu aumm yum M u A Vathmd HELMER T omxmda mzheqmdz MWW Wme mm W My iwmmmm WWW Wigwamqu 391mm undmwhednemnz w quotawndxmmm mm 3 cum 4 5500mm SUR mm mm m 0F rem inHumequot mm uvmvi 2nd w M New And mum m mamas w my cum 4 GEUDETIC SURVEY w aFFebmzvy m m7 qzvu the my x mm MW quta be um mm with ma 5mm m M be damned Khuxhnd 2nd m4 md thexafznzhau Q mm Yhama JeKaxan ma m uent mpr Saueth m 0F mm m m ma rm mm FERDINAND HAXSLER mu m Erwm iq v cum 4 GEUDETIC SURVEY anhmq mm dawn m Mar Mm T a m thmaem We um mm mm Swarm mm m n Cami smogrzcsmzm x n seam M 2mmquot 0f uned m ms 25 an in muggy u n Wzvaf mm mm w mum Wm ms WM SPUINIK Sm Beqzn mun magma m hunzh 0F SVulmklaan 0 95 mm 0F Puma whom 0F We GEOMETRIC RELATIONSHIPS ON THE SPHERE Surveying Engineering Department Ferris State University Arc Distance on a Sphere The arc length on a surface of a sphere can be easily computed by multiplying the angle measured at the center of the sphere by the distance Here the angle is measured in radians Hence E Ry where g the arc length between two points on the surface of the sphere R the radius of the sphere and y the angle measured at the center of the sphere subtending the two points on the surface On the surface of the earth considering the earth as a sphere there are three different types of arc measurements see gure 1 They are the length of arc on a meridian length of arc on a parallel and a length of arc on any great circle NP Meridian B XA A XE 95A pB 4 6 Meridian Equator SP Figure 1 Length of an arc on a meridian The length of the arc on a meridian is depicted in figure 1 Using the basic relationship for computing an arc length on the sphere the arc length from point A to the Equator can be shown to be SmRpA SURE 452 7 Geodesy 1 Geometric 39 39 on the Sphere Page 14 where sm the arc length along the meridian from the Equator to point A R the radius of the earth spherical radius and pA the latitude of point A The arc length can also be computed from the nearest pole in this case the North Pole using the colatitude X Thus S m RXA Here s m is used to designate the arc distance from A to the North Pole The arc distance between points A pA 1A and B pB XB where both points have the same longitude 1A XB is found in a similar fashion squotm R 8p where s m is being used to designate the arc length between two points and 5p pA pB Meridian O Meridian k B 6 A XE lt4 0 wB Q 5 A Equator Figure 2 Length of an arc along a parallel The computation of the arc length along a parallel can be shown to be similar to that shown for the arc length along a meridian There are two main differences First the difference in longitude 57 is used as the angle Second the radius of the sphere measured along the rotational aXis of the earth changes At the Equator the arc distance between the two meridians arc distance between D and E shown in figure 2 is found as SURE 452 7 Geodesy 1 Geometric 39 39 on the Sphere Page 15 SER57M where S is the arc distance along the Equator R is the radius of the earth and 57 is the change in longitude 57 IXD 7 E Looking at the triangle OO A one can see that this is a right triangle Therefore the radius r from O to A can be found using either the colatitude or latitude as rRsinX rRcosp Since the arc distance between A and B can be shown to be equal to r67 the arc distance along the parallel sp can be presented as sp Rcosp57t It should be evident that this formula is also valid at the Equator If p 00 then cos p l and the arc distance becomes sE R 57 The length of an arc for any great circle on a spherical earth can be computed using spherical trigonometry The angular distance 2 between A and B is the unknown The corresponding arc length on the surface is s Points A and B lie on meridional arcs which have a lengths of XA and XB which are the colatitudes of points A and B respectively figure 3 SURE 452 7 Geodesy 1 Geometric 39 39 on the Sphere Page 16 NP Meridian Meridian S B A Z Q WA B 5 A Equator SP Figure 3 Arc length of any great circle Then from the cosine formula cos 2 cos XA cos XE sin XA sin XE cos 5 or expressed in terms of the latitude cos 2 sin DA sin DB cos DA cos DB cos 5 and the arc length AB is calculated according to s RZ There is a problem with this approach though Snyder1 points out that it does not work for values close to 0 In other words it is an illconditioned system Instead he suggests using the following form sin sin 2 005 l A 005 l B sin 2 2 l Snyder John Map I 39 quot 7 A Working Manual US Government Printing Office Washington DC SURE 452 7 Geodesy 1 Geometric 39 39 on the Sphere Page 17 This formula is exact and very accurate in practice for values of z from 0 to nearly 180 Then the arc distance s is computed as before Area of the Surface From basic math we know that the area of a circle is A rcR2 5 D2 4 From calculus the general formula for the surface area of a surface of revolution is S J2 7 p ds where S is the surface area p is the distance from the aXis of revolution to the surface element and ds is the arc length Figure 4 shows the geometry of the circle Assume that it is revolved about the XaXis Given the coordinates of a point on the circle the radius can be found using the Pythagorean theorem 3 ll ds d Figure 4 Geometry for area of a surface SURE 452 7 Geodesy 1 Geometric 39 39 on the Sphere Page 18 r2 X2 y2 The surface area for a differentially small spot on the sphere can be represented as dS 2 T y ds The differentially small surface distance ds is defined as The X and y coordinates of a point on the sphere can be computed using simple trigonometry Xrcos9 and yrsin9 Differentiating dX rsin9d9 and dy rcosGdG It is also clear that ds r d9 Substituting the values for dy and ds into the general equation for the surface area we have dS 27Ersin 9r9 27Er2 sinGdG It is fairly eVident that if we integrate this equation as 9 varies from 0 to 7 that the upper semicircle surface area is Sr2nr2sin9d9 0 2717r2 cos 91 47Er2 The general formula for the computation of the surface area of a zone as shown in figure 5 is SURE 452 7 Geodesy 1 Geometric 39 39 on the Sphere Page 19 Figure 5 Surface area ofa zone S 2 TE R h but h can be written h R sin 0 therefore S 2 TER2 sin 0 Ifwe use the latitude p instead of the angle 0 the general equation for the surface area becomes S 2 TER2 sin p Defining the zone by p1 and p the surface area of the zone between these two latitudes is the difference in their surface areas S 27ER2 sinqp1 27ER2 sinqp2 27ER2sinq1 sinp2 Example Find the surface area for a portion of the earth defined by 300 N 500 N 700 W and 1200 W Assume a mean radius ofthe earth of 6370000 In Solution The surface area of the zone is defined by the latitude as S 27ER2 lsinqp1 sinp2 27c63700002 lsin 50 sin 30 SURE 452 7 Geodesy 1 Geometric 39 39 on the Sphere Page 20 This is the total area between 30 and 50 north latitudes around the whole sphere The desired area is only 50 wide 120 W 7 70 W Therefore s 50 J2 rc63700002 sin 50 sin 30 360 94206 XlO12 m2 9420640 km2 Slide 1 H i GEOMETRY ON THE SPHERE Robert Burtoh Surveying Engineering Department I Ferris State University 1 Slide 2 ARC DISTANCE ON SURFACE General relationship involves the radius ofthe sphere R and the angle A measured at the a center 7 A B SE R5y Tangent SURE 452 7 Geodesy l Geometiic Relationships on the Sphere Page 21 Slide 3 Slide 4 Slide 5 LENGTHS OF ARCS ON EARTH S SURFACE 3 kinds of arc measurements Length of are on a meridian Length of are on a parallel Length of are on any great circle LENGTH OF ARC ON MERIDIAN Arc length from Equator to point A is NP A M1 sm R m XE all WE Arc length from Equator North Pole IS 1 s m R XA LENGTH OF ARC ON MERIDIAN Arc length on the meridian between points A and B Use the difference in latitude 5qgtqgtA ech squotm R 5p SURE 452 7 Geodesy 1 Geometric Relationships on the Sphere Page 22 Slide 6 LENGTH OF ARC ON A PARALLEL Triangle OO A is a right triangle Radius r computed using either the CO latitude or latitude rRsinX rRcosltp 514 7M Slide 7 LENGTH OF ARC ON A PARALLEL Arc length is equal to r6 which is after substitution for r sp R COSp 8 At Equator p 0 and cos p 1 thus sER5A Slide 8 LENGTH OF ARC ON ANY GREAT CIRCLE Use spherical trigonometry Angulardistance 2 unknown Corresponding arC length is s Can use either latitude or CO latitude SURE 452 7 Geodesy l Geometiic Relationships on the Sphere Page 23 Slide 9 Slide 10 Slide 11 LENGTH OF ARC ON ANY GREAT CIRCLE From cosine formula cosz cos XA cos XB sin XA sin XB cos 5 or cos 2 sin A sin B cos A cos B cos 5 Arc length computed as s LENGTH OF ARC ON ANY GREAT CIRCLE Previous formulas forz are ill conditioned Do not work for values close to 0 Alternative form Exact and very accurate sinsi112 B A cos in 005 1 11le LA 2 AREA OF SURFACE Area of circle AMU D2 4 Surface area of surface of revolution SJ2npds SURE 452 7 Geodesy l Geometiic Relationships on the Sphere Page 24 Slide 12 Slide 13 Slide 14 AREA OF SURFACE Surface from differentially small area d8 2 it y ds Differentially small surface distance ds ldxz dy2 AREA OF SURFACE Coordinates x y of point on sphere Xrc059 and yrsin9 Differentiating 1X 2 rsin6d6 and dy rcosGdG Also ds I de AREA OF SURFACE Substitute dy and ds Integrating from 0 to 1 into general equation for surface area 7 2 dS2nltrsinGgtrG 8 J2Tcr smede 215r2 sinede 211r2 coseE 411r2

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