EXERCISE BIOMECHANICS KINE 426
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This 18 page Class Notes was uploaded by Pattie Nitzsche on Wednesday October 21, 2015. The Class Notes belongs to KINE 426 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 17 views. For similar materials see /class/226208/kine-426-texas-a-m-university in Kinesiology at Texas A&M University.
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Date Created: 10/21/15
KINE 426 Lab Introduction Tour and Video Group Activity July 6 7 Today we will talk about the components of a modern Exercise Biomechanics laboratory Please see enclosed attachment We will take a tour through the facilities and discuss the history of measurement in biomechanics from serial pictures to lm to VHSbeta to now digital video Digital video is the direction all biomechanics laboratories will continue to head in the next 510 years for movement analysis Visit our Website for an Introduction as well httpexbiomechtamuedu Click on the Introduction LINK A Components of Exercise Biomechanics labs gt Movement analysis lm or Video gt Computers and software gt Force Platform force measurement for walking running jumping gt Electromyography muscle activity gt Cybex Machine torque measurement gt Goniometers joint angles and movement gt Clinical grade anatomy models and full skeleton 1 Use of AMD Athlon 63 bit PC and Mac G4 a Press ON button on the hard drive or more the mouse to activate the computer Make sure the monitor is on b You are now on the Windows XP OS platform desktop 24 GHz 64 bit AMD single processor Asus motherboard w 800 MHZ frontside bus 7 AMD Athlon processor support 1 GB SDRAM 200 GB Serial Caviar ATA Hard Drive 128 MB256 DDR Radeon 9250 Graphics card Sony DVDROM 16X CDROM drive 24X 2quotd DVDRW CDRW drive Here you will be able to locate easily icons for Internet Explorer for browsing the Internet Windows Picture amp Fax Viewer for viewing Digital Video still pictures and Windows Movie Maker for viewing and processingediting digital video segments To open any program or le simply doubleclick on the icon 0 2 We will watch segments of a video produced by PBS in 1991 called The Champion Within It includes interviews with expert researchers in the areas of biomechanics and exercise physiology It includes scientists from USOC training center in Colorado Springs Penn State University University of San Francisco University of Pacific Ball State University and Tufts University We will watch the rst 3 segments USOC Peter Cavanagh Barbara Drinkwater 3 Choose a group of 4 to work with in Group Assignments We will have 6 groups per lab section Group 1 Group 2 Group 3 Group 4 Group 5 Group 6 Exercise Biomechanics Laboratory Equipment and Facilities A Physical Plant Approximately 1880 square feet of space that houses the Exercise Biomechanics computer lab and the Exercise Biomechanics Lab The entire suite including the Exercise Physiology lab is 3000 ftz B Components 1 2 Sony DV Cameras VX2000 2001 IEEE 1394 rewire cables gt capture DV 2 Sony Digital Video Tape Player DHR1000 gt playback and editing 3 2 Sony portable DV playerrecorders DVRs GVD1000 4 Kine o Fluorescent Lighting System 2 banks with non ickering ballasts and full solar spectrum bulbs 5 Monitors gt Two 31 in high resolution analog Toshiba Sony 6 Computer Resources a ONE Apple G4 w DVDR and CDRW 0 60 GB Hard drive amp 733 MHz processor these limits will increase very soon 0 for digital editing of DV lms by students will burn DVDs pending all laboratory experiences and will be disseminated by DVD A DVD lab b Six 64 bit AMD Athlon PCs 0 24 GHz 64 bit AMD single processor 0 Asus motherboard w 800 MHz frontside bus AMD Athlon processor support 0 1 GB SDRAM 0 200 GB Serial Caviar ATA Hard Drive 0 128 MB256 DDR Radeon 9250 Graphics card 0 Sony DVDROM 16X CDROM drive 24X 2quotd DVDRW CDRW drive 0 Windows XP OS 7 Software a Editing Windows Movie Maker PC iMovie2 Final Cut Pro b Production coordinate generation ex Adobe Photoshop c Picture viewing Windows Picture and Fax Viewer d Kinematic Analysis NIH Image J e stick gure generation and kinematic calculations National Instruments LabView and pending Dartish H U MAN 1 DVD CD burning Roxio Easy Media Creator 7 8 Printers a 3 HP Laser 1220 BW Printers b Color Ink Jet Printer HP 1220 Cse 2400 x 1200 dpi resolution 8 MB bu er memory 9 EMG Electromyography Bagnoli 0 electrodes wires 0 AC ampli er 0 AD converter 0 analysis software ex LabView National Instruments 10 Force platform Bertec 0 strain guage technology 0 3D analysis for locomotion walking and running 0 AD converter 0 analysis software ex LabView National Instruments 11 Goniometers mechanical electro 0 joint position and movement 0 AD converter analysis software 12 Anatomy Module 0 full skeleton 0 extremity models 0 large size musculoskeletal charts 13 Cybex machine in 267A 0 force and torque measurement at different movement speeds 0 AD converter analysis software Schematics for 276 Read storage closet cabinets 95in Exercise Ph siolo Teachin Lab cybex O O O O D O IE Computer Lab DV Computer Module Dual G4 DVD treadmill AMD Athlton 64 bit PCs O ECG I I I I I I I I I r I I I I I ANATOMY MODULE DV V I I I I I I I I I I Biomechanics E E cabinets MonitorT ape Playercomputer modules Lighting system Exercise Biomechanics Lab A IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIII Force platform 5 V KINE 426 Lab Notes Vectors and Torque Vectors Vectors represent the actual progression of some measurable physical process Vectors have two characteristics that de ne them magnitude or size and direction A measurement with magnitude and no direction is called a scalar 50 mph due east is a vector 50 mph is a scalar Is weight a vector Yes because weight is a force that always acts straight down Vectors are graphically represented by a line segment with an arrowhead at one end The length of this line segment represents the vector magnitude and the direction of the arrow represents the vector s direction t 1 Up The starting end of a vector representation is called the tail and the arrowhead end is called the tip This concept is used when graphically representing operations on a vector In performing operations on a vector it can be moved anywhere for the purpose of performing these operations as long as it remains the same length and points in the same direction This is used when showing how several vectors interact with or affect each other Vector Resultant This is a vector that results from the combination addition of two or more vectors One method of adding vectors involves placing them together using the tip to tail method where the tail of one vector is placed next to the tip of another with the resultant vector going from the tail of the rst to the tip of the last Vector Resultant Example 1 Show the graphic representation of the resultant of the following three illustrated vectors named v1 vz and V3 V2 V1 gt V1 T V3 The resultant vector show in the diagram is the single e ect that results from the three vectors acting together When adding two vectors together with the tip to tail method the resultant would actually form the third side of a triangle The Law of Cosines could be used to nd the magnitude of the resultant and the Law of Sines would be used to nd the direction Vector Resultant Example 2 Given two vectors Vector 1 v1 which is a force of 5 ON acting horizontally and Vector 2 v2 which is a force of 100N acing at an angle 600 above horizontal final the magnitude and direction of the resultant of these two vectors acting together When drawn individually the vectors would be represented as x shown on the left Their relative 0 sizes are due to their stated values Since v has a value that s twice as big as v1 its graphic representation is drawn twice as long as that ofvl As shown on the left when the two vectors are put together they form two sides of a triangle with the resultant being the third side We can label the sides of the triangle as A B and C and label the appropriate angles or B and y We use the values of the vectors as the values of the corresponding sides of the triangle v1A50NandvzB100N y 1800 600 1200 We can now use the Law of Cosines to nd the magnitude of the resultant C2 A2 B2 2ABcos y 50 N2 100 N2 7 250 N100 Ncos 120 2500 N2 10000 N2 7 250 N100 N5 Be careful not to drop the negative sign from the cosine of the angle This leaves us with the nal calculation for magnitude C V2500 N2 10000 N2 7 250 N100 N5 13229 N To nd the direction of the resultant we need to nd the angle that it makes relative to horizontal This would be the value of angle B To do this we use the Law of Sines sin BB sin yC sin B Bsin yC 100 Nsin 120013229 N 655 This isn t the value of the angle B To nd the value of B you would calculate the inverse sine also called the arc sine or arcsin of 655 The inverse sine reproesented as sin39l yields an angle For example sin 300 5 Therefore sin391 5 30 So 5 sin39l 655 40890 The nal answer would be that the resultant C is a force of 13229 N acting at an angle of 40890 above horizontal Vector Component Just as two or more vectors can be combined to nd a resultant a single vector can be broken into components Vector components are vectors themselves Vector components always originate at the same point as the original vector Each component indicates the effect of the original vector in the direction of the components For example suppose you have a cart being pulled by a rope indicated by the vector labeled V in the picture and the rope is acting at some angle above horizontal Some of the force pulling the wagon would be acting horizontally represented by the vector labeled vh and some of the force would be acting vertically indicated by the vector labeled vv If we found the value of these vector components we would know how much force the original vector was producing in the horizontal and vertical directions In representing the relative size of component vectors imagine that the original vector labeled V in the A picture is the diagonal of a rectangle V In terms of their relative sizes the Vy components of this vector would be like the sides of the rectangle that start at the origin of the vector The components are labeled vx and vy in the picture Vector Component Example 1 Final the horizontal and vertical components of a 100N vector that is acting at an angle that is 600 above horizontal This problem is illustrated to the left with the original vector labeled v and the horizontal and vertical components labeled vh and vV respectively To find the value of the vector components we can treat each of the vectors as part of a right triangle Using the right triangle functions of sine and cosine we can then solve the problem 60 Vh Remember that for the purpose of performing calculations a vector can be moved as long as it stays the same length and points in the same direction Therefore we can temporarily move the vertical component vv to the opposite side of the rectangle This gives us a right triangle with v as the hypotenuse and vh and vV as the remaining two sides Set vh A vv B and V C To maintain the conventional method of illustrating triangles this would make the 600 angle 3 V VV Remember the basic formulas for sine and cosine sin 0 opposite sidehypotenuse cos 0 adjacent sidehypotenuse 60 Vh C Substituting 5 for 0 we get sin 5 BC OR B sin 3C sin 600100 N 866 N 60 cos B NC A OR A cos BC cos 600100 N 50 N Therefore the values of the vector components are Avh50NandBvv866N It s important to remember that the vector components actually act in the locations shown in the original illustration The vertical component was moved just for the purpose of solving the problem It also wasn t necessary to change the names of the vectors to A B and C This was done to relate the vector diagram to the previous examples used for right triangles Torgu Creates a tendency to turn or rotate around a speci c point Symbols T or F Basic Formula T perpendicular forcedistance to point of rotation OR T FJd Units English inlb or ftlb Metric 7 Nm or Ncm The force must be perpendicular to the surface on which it s acting in order to be able to calculate torque F 1000 N Torque Example 1 Assume a force of 1 m j JOOON is acting at one end ofa beam that is 1 m long Also assume theforce is perpendicular to the beam How much torque is this force producing at the opposite end of the beam T FJd 1000 Nl m 1000 Nm If the force is not perpendicular then some perpendicular relationship must be established One way is to nd the component of the force that is perpendicular to the surface Torque Example 2 Assume that a F 10 0 N beam 1 m long has a 1000 N force FL acting at a 3 00 angle to one end How much torque is thisforce producing at the opposite end I 1 m First you have to nd the value of the component of the force that is perpendicular to the beam Using the rules for nding vector components we have 30 F1 Fsin 30 1000 N5 500 N T F1d 500 N1 m 500 N m What if you have a vertical force like a weight that is acting on a structure that isn t horizontal In this case you can nd the horizontal or perpendicular distance to the point of rotation and multiply that by the weight in order to calculate torque Torque Example 3 Assume a weight of 000 N is hangingfrom a beam that is 1 m long anal is mounted at an angle of 600 below horizontal How much torque does this weight produce on the opposite end of the beam Since weight acts vertically and is perpendicular to our horizontal reference line the horizontal or perpendicular distance d1 from the weight to the end ofthe beam can be used to calculate the torque Remember you just need to establish a perpendicular relationship between distance and force wt1000N d1 dcos 60 1 m5 5 m T wtdL 1000 N5 m 500 N m So in general when you have a weight acting on a structure that isn t horizontal like a weight held in the hand T wtdcos 9 where d is the actual distance from the weight to the point of rotation and 9 is the angle of the structure relative to horizontal It can be above or below horizontal It won t make any difference in the calculation of torque KINE 426 Lab Notes 2 Kinematics amp Kinetics Change Represented by the Greek letter delta A Change Example 1 A runner travels from the 10 yd line to the 50 yd line of a football field What is the distance traveled from the starting point to the finishing point When this is measured as a straight line from one point to the next it is called displacement position 1 or p1 10 yd position 2 or p2 50 yd The distance traveled could be called the change in position or Ap and would be calculated with the following formula Appzp150yd10yd40yd The value represented by the symbol with the lowest subscript in this case p1 is called the initial or starting value The value represented by the symbol with the highest subscript in this case p2 is called the nal value Relative Change This involves comparing the change in one unit of measure with the change in another unit of measure OR Velocity The relative change of some distance over a unit of time such as meters per second or miles per hour Symbol v In calculations velocity v distancetime interval change in positionchange in time v ApAt Units distance units time units ftsec msec mileshr etc Velocity Example 1 If the runner in the example above takes 54 sec to cover the 40 yd distance what is the velocity during this time Appzp150yd10yd40yds At t2 t1 54 sec 0 sec 54 sec Wheneveryou are givenjust the valuefor an interval of time distance etc you assume that the value of the interval is the final value and that 0 is the initial value v ApAt 40 yds54 sec 741 ydssec Velocity Example 2 A cyclist travelling at a constant velocity goes 400 m in 20 sec What distance was covered in 15 sec during this time interval From our original data v ApAt 400 m20 sec 20 msec We39re lookingfor the value opr when v 20 msec and At 15 sec Since v ApAt and we 39re looking for Ap we can say that Ap vAt Ap 20 msec15 sec 300 m Acceleration The relative change of velocity compared to a change in time Symbol a In calculations acceleration a AvAt ApAtAt Units distance unitstime unitstime units OR distance unitstime units2 ftsecsec OR ftsecz msecsec OR msecz etc Be cautious when expressing values in calculations The units are as important as the numerical values For example let39s say that the answer to an acceleration problem is 10 msec2 but it is written as 10 msec The answer would be incorrect because 10 msec would be implying that the answer is a velocity That is the units should actually define the physical process that is being calculated as acceleration Acceleration Example 1 A runner from a dead start that is v1 0 msec is travelling at a velocity of 16 msec after 8 sec What is this runner39s acceleration during this time interval a AvAt 16 msec8 sec 2 msecsec OR 2 msec2 This actually implies that during this period of time for every second of time traveled the velocity increases by 2 msec 1 sec after the start the velocity is 2 msec 2 sec after the start the velocity is 4 msec etc Force Symbol F In calculations F mass X acceleration F ma Units English Force pounds mass slugs acceleration ftsec2 Metric Force Newtons symbol N mass kg acceleration msec2 In order to correctly calculate force all of the values used must be in the listed units Force Example 1 What force is required to accelerate a 5 kg object at a rate of 20 msecz F ma 5 kg20 msecz 100 N Gravity Imposes an effective acceleration on any object it in uences This acceleration acts towards the center of the attracting body usually the Earth Symbol g ag is also sometimes used Value English Units g 32 ftsec2 at sea level Metric Units g 981 msec2 orjust 98 msecz at sea level The value of g decreases as you move away from the center of the attracting body That is there is less gravitational effect at high altitude than there is at sea level That s why the value for g is expressed as the value at sea level Weight A special case of force in which the acceleration is equal to the acceleration due to gravity g Symbol W In calculations W mg Units The units are the same as the units for force English pounds Metric Newtons Weight Example 1 At sea level what is the weight of an object that has a mass of 10 kg W mg 10 kg98l msecz 981 N Always make sure that the units are consistent between English and metric values when doing calculations of this type That is don 39t try to calculate weight by using a mass expressed in slugs English units with g expressed in msecz Metric units Because the value of g decreases as altitude increases an object weighs less at the top of a mountain than it does at sea level Mass however is always the same Force of Impact A result of mass and negative acceleration Negative acceleration or deceleration occurs when something slows down Force of Impact Example 1 A gymnast weighs 539N After dismounting the horizontal bar he impacts a landing mat with an initial velocity of 10 msec As he lands he bends his knees to absorb landing shock From the moment of first contact with the mat until all downward movement stops the time is 25 sec Calculate the force of the impact of the gymnast with the landing mat F ma Since in this case acceleration will have a negative value we could say that F ma or F is equal to the absolute value of mass times acceleration W mg Therefore m Wg 539 N981 msecz 55 kg a AvAt V2 V1tz t10 msec 10 msec25 sec 0 sec 40 msec2 The negative value of acceleration is due to the fact that the gymnast slows down or decelerates after hitting the mat It is still a form of acceleration since it involves a change in velocity relative to a change in a time interval F ma 55 kg40 msec2 2200 N 2200 N KINE 426 Lab Notes Load Stress and Strain Load Stress and Strain Load An applied force or weight Units English pounds Metric Newtons Stress A load divided by the area of the structure supporting it Units English pound per square inch lbin2 or psi Metric 7 Newtons per square centimeter Ncmz Stress Example 1 A square post that is 2 on each side supports a weight of 100 lb What is the stress on the post Using the formula stress loadarea we rst have to determine the area bearing the load In this case we have a post with a square crosssectional area area 2 in2 in 4 in2 stress loadarea 100 1b4 inz 25 lbin2 There are three types of stress 1 Compressive Stress or Compression This is the stress seen in pressure Bones and brocartilage are designed to withstand compressive stress 2 Tensile Stress or Tension Stress seen when pulling something apart Ligaments and tendons are designed to withstand tensile stress 3 Shear Stress A stress that tends to separate across parallel planes Shear stress is most often seen injoints when opposing surfaces slide across each other Shear Stress Compressive Tensile Stress Stress v One effect of shear stress in a joint can be seen in the knee When bending the knee while the body s weight is supported as in sqatting down or walking up or down stairs the knee will experience noticeable shear stress especially if the lower leg bends forward as shown in the illustration to the left This is the way the knee naturally moves when walking downstairs shear stress If the lower leg stays roughly perpendicular to the ground as in the next picture the shear stress is greatly reduced This is a primary reason for trying to keep the knees from moving forward when performing squats Strain This is deformation or a change in shape caused by the application of a load There are two ways to express strain 1 As a Ratio strain change in length original length In this case strain would be expressed as a unit of length over a unit of length English in in Metric cm cm 2 As a Percent strain change in lengthoriginal length 100 Strain Example I 39 A segment 0ftend0n with a length of cm has a load applied to it and it stretches an additional 2 mm Calculate the strain Load We need to be consistent in the units of measure so we need to convert 2 mm to cm There are 10 mm in a cm so 2mm 02 cm strain change in length original length strain 02 cml cm 02 cmcm expressed as a ratio OR strain 02 cml cm100 2 strain Stress Strain Curve A graph of the relationship between stress and strain in a speci c material Even though the shape of stressstrain curves will be different from one material to the next all of them have certain common characteristics elastic deformation 7 This is the type of deformation that occurs when stress is first applied During elastic deformation if the stress is removed the material will return to its original shape The stressstrain plastic deform ation elastic limit OR 2 yield point curve is a straight line during this phase 91 elastic limit OR yield point 7 After this point an increase in stress will permanently change the material s shape The shape of the stressstrain curve is no longer a straight line after this point plastic deformation 7 This starts at the elastic limit In plastic deformation the shape of the material is permanently changed It also may take decreasing amounts of stress to yield a given amount of strain or deformation failure 7 The material tears or separates failure elastic deformation V strain cmcm inin OR The basic stressstrain process can be seen in the body For example someone may experience a slight lateral stress or turning of the ankle If the load is mild enough there may be a sensation of the stress in the ankle but no damage is done The ligaments have probably experienced strain in the range of elastic deformation and they will return to their original shape A slightly greater stress may create the initial stages of damage Besides the visible signs of pain and swelling there is also some slight permanent deformation of the ligaments The strain has entered the range of plastic deformation As those that have experienced repeated ankle sprains can attest after the initial injury it often takes less stress to injure the ankle in succeeding incidents This is characteristic of the range of plastic deformation Increased or repeated stress to the area may result in failure tearing of the ligaments or at least enough permanent deformation that the joint is excessively unstable and must be repaired surgically
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