Kinesiology 201 Solutions Kinematics Tony Leyland School of Kinesiology Simon Fraser University 1. a) Vertical ocity = 10 sin20 = 3.42 m/s Horizontal ocity = 10 cos20 = 9.4 m/s B Vertical A-B (start to max height) vf = vi + at 0 = 3.42-9.81t A t = 3.42/9.81 t = 0.35 s Therefore from A-C total time = 0.7s C Vertical C-D (height of release point to contact) vi = -3.42 m/s d = -0.5 m g = -9.81 m/s 2 vf 2 = vi 2 + 2ad vf 2 = (3.42) 2 + (2 x 0.5 x 9.81) vf = (11.696 + 9.81) vf = -4.64 m/s D vf = vi + at -4.64 = -3.42-9.81t t = 0.12 s Therefore: total time (A-D) = 0.7 + 0.12 = 0.82 s Horizontal A-D d = vit + 0.5at 2 as a = 0 d = 9.4 x 0.82 = 7.71 m The long jumper tras 7.71 m horizontally between take-off and landing b) No it is not the optimal angle. The optimal angle would be approximately 40-43 degrees. It would not be exactly 45 o because the projection height is higher than the landing height. Although angle of release is an important factor in the range of a projectile it is secondary to speed of release. If a long jumper wanted to increase his vertical component of take-off ocity (and hence increase the angle of take-off) then he would have to slow up his approach so that he was on the board long enough to generate enough vertical impulse. As this slowing up would reduce his overall speed of release this would be detrimental to the length he or she could jump. 2a The first thing you should have done is draw a rough representation of the movement. This gives you a visual picture of what is happening and will help you determine is you have a realistic answer as you step through the calculations. 1
Forearm Movement Arm Movement Both together This is not plotted, it is just a rough drawing determined from the approximate locations of the joint centres. To obtain the angle of the upper arm: Arctan of (Y elbow -Y shoulde r)/(x elbow -X shoulder ) Then to get angular ocity you subtract the two angles and divide by 2 t As stated you may have omitted this step as it is rather clear from scanning the data that the upperarm segment does not have any angular ocity (hence the angle of the arm is the same for all time frames). Therefore the angular ocity of the forearm about the fixed frame of reference is going to be the same as the angular ocity about the elbow joint. Note: It doesn't matter what the rest of the body is doing, as it is clear that the segment proximal to the forearm is just translating with no rotation. Therefore to answer to this question we need the angle at t=0.01 and t=0.03, subtract them and divide by 2 t. Below is a portion of an Excel file where I entered formulae to calculate the required variables. Below is a sample calculation, if you are still unsure of how these values were obtained refer to question 6, which will show the methodology of getting angle data from co-ordinate data, and the examples of finite differentiation of angles to obtain angular ocities and elerations. Forearm segment angle at t=0.01 tan -1 [(253.7-225)/(267.8-276.4)] = tan -1 (28.7/-8.6) = -73.31 o + 180 o = 106.7 o (1.862 rads) 2
RESULTS Time Shoulder (S) Elbow (E) Wrist (W) θ upperarm θ forearm ω upperarm ω forearm x y x y x y rads rads rads/s rads/s 0.00 251.3 220.1 275.8 225.0 262.3 243.6 0.1974 2.19863 0.01 251.9 220.1 276.4 225.0 267.8 253.7 0.1974 1.86193 0.00-31.392 0.02 252.5 220.1 277.0 225.0 277.0 255.0 0.1974 1.5708 0.00-32.44 0.03 253.1 220.1 277.6 225.0 288.1 253.1 0.1974 1.2132 0.00-28.107 0.04 253.7 220.1 278.2 225.0 294.2 250.4 0.1974 1.00867 Notice than for time 0.00 and 0.01 the forearm is in the 2 nd quadrant (see above diagrams) so we add 180 degrees. It wouldn t make much sense if you failed to do this, as your angular ocity would be through the roof and in the opite direction. At time 0.02 when calculating the forearm angle (θ forearm ) the division returns infinity (division by zero) which would return 90 degrees or π/2 rads (1.57 rads). Angular ocity of the forearm at t = 0.02 = -32.44 rads/sec or (-1859 degrees/sec) b) If the upper arm is not rotating and the forearm is rotating clockwise due to concentric muscular activity, then the triceps must be the agonist. Refer to the rough diagrams and the picture itself to see that clockwise rotation is elbow extension. Note: If you had a itive value for part a) then you should have reported that the biceps would be the agonist (elbow flexion). I want to give marks for part b) based on whether you can take the answer and interpret the movement it indicates. Don t secondguess yourself and start putting in answers that don t correspond to each other. Another type of response that I would ept for full marks is to say that the answer you got (itive angular ocity) indicates elbow flexion but you consider this to be an error but have no time to check the answer. Then you could say you believe the true motion to be elbow extension. c) This part could be answered by two methods. The linear ocity of the wrist is due to two factors: its rotation about the elbow, plus the linear ocity of the elbow. Much like the ocity of the bottom of a spinning ball relative to the ground is due to the ocity of the ball's centre of gravity plus the tangential ocity due to the spin. Therefore, the linear ocity of the wrist can be obtained by calculating the linear ocity of the elbow with respect to the external frame of reference and then adding this to the linear ocity of the wrist with respect to the elbow. The linear ocity of the wrist with respect to the elbow is given by: v t = rω Note that radians must be used in this equation. 3
Therefore this is 0.3 x -32.437 = 9.7311 m/s The negative sign can be dropped as we know exactly the direction of this ocity. It is at 90 degrees (tangential) to the alignment of the forearm and as the segment is aligned vertical 90 o (1.5708 rads) vt will be horizontal. Boy am I good to you or what! The ocity of the elbow is easy to calculate as there is no vertical change in direction. Therefore, zero vertical ocity. Horizontal elbow ocity = (277.6-276.4)/0.02 = 60cm/s = 0.6 m/s Velocity of wrist = 9.7311 + 0.6 = 10.3311 m/s at 0 o to the right horizontal. You may ask why you would use this equation when you could simply differentiate the data for the wrist. The answer is that you may want to calculate the linear ocity of the segment s centre of gravity (we often do in fact) and the problem with this is that you do not have co-ordinate data for the centre of gravity. In this case you must calculate the linear ocity of the proximal joint and then use the equation v CG = rω (where r is the distance from the proximal joint centre to the centre of gravity, and ω is the segment angular ocity). This gives you the ocity of the centre of gravity with respect to the proximal joint (then add the ocity of the proximal joint to this value). I mentioned a second method whereby you could have simply differentiated the data for the wrist. This is an eptable solution to the problem and the slight difference in values obtained is due to the data having been made up and therefore probably encompassing a high frequency component that would not normally be present in human movement. Horizontal ocity of wrist Velocity 0.02 = (X 0.03 -X 0.01 )/2 t Velocity 0.02 = (288.1-267.8)/0.02 = 10.15m/s Vertical ocity of wrist Velocity 0.02 = (Y 0.03 -Y 0.01 )/2 t Velocity 0.02 = (253.1-253.7)/0.02 = -0.3m/s Resultant (pythagorus)= 10.2 m/s Angle = -1.69 o (4th quadrant) Note: There is a 1.7% difference in magnitude with the different methods. If t was good enough for the frequency of movement there should be no difference. As stated, I made up the data so that is why it appears to have some high frequency component. d) Acceleration of shoulder. There is no vertical movement of shoulder at all throughout movement. Therefore zero vertical eleration. Horizontal data Velocity 0.01 = (252.5-251.3)/0.02 = 60 cm/s = 0.6 m/s Velocity 0.03 = (253.7-252.5)/0.02 = 60 cm/s = 0.6 m/s Therefore: Acceleration 0.02 = (0.6-0.6)/0.02 = 0 m/s 2. Therefore total eleration of the shoulder joint equals zero. B 4
e) Vertical ocity = 3sin80 = 2.954m/s Horizontal ocity = 3cos80 = 0.521 m/s Vertical A-B (start to max height where v f = 0 m/s) vf 2 = vi 2 + 2ad 0 2 = 2.954 2 + (2 x [-9.81] x d) d = 8.728/19.62 d = 0.445 m Height jumped = 44.5 cm We will need time to calculate the range vf = vi + at 0 = 2.954-9.81t t = 2.954/9.81 t = 0.301 s Therefore from A-C total time = 0.602 s A C Horizontal A-D d = vit + 0.5at 2 as a = 0 (ignore air resistance) d = 0.521 x 0.602 = 0.314 m The player tras 31.4 cm horizontally during the jump shot. 3. Components Horizontal ocity = 20 cos 40 = 15.321 m/s Vertical ocity = 20 sin 40 = 12.856 m/s t = -12.856 ± (12.856 2 + (2 x -9.81 x -1.5)) The vertical displacement is -1.5 m -9.81 t = -12.856 ± (165.277 + 29.43) t = -12.856 ± 194.707-9.81-9.81 t = -12.856-13.954-9.81 The itive result from the square root is not an option. t = -26.81/-9.81 = 2.733 seconds Range = s = V horizontal t S = 15.321 x 2.733 = 41.9 m 4. a) Vibration of markers on the subject and the camera(s). b) Estimate of the frequency of the movement to get correct sampling frequency (i.e. avoid alaising error) will sibly be slight erroneous. c) When smoothing data prior to differentiation we must be careful to filter out most of the noise and leave as much of the signal as sible. However, no filter is good enough to filter out all of the noise and leave the entire signal. d) Parallax errors where applicable. e) 60 Hz alternating current imes a small 60 Hz component in the digitised data. f) When differentiating, we magnify % error. g) Extreme care is needed when locating the joint centre of rotation relative to the bony landmarks where markers are placed. 5
Note: Other assumptions we make in the analysis of human movement are necessary and we can do nothing about them. Therefore they are not the answer to this question as I asked for "errors we must ount for". For example, any error due to assuming segment is a rigid structure (bones will bend and length of segment may change slightly) is inherent in this type of analysis and nothing can be done about it. Similarly, once we have located the best location we are ging to assume is the joit centre of rotation we cannot ount for any slight variation during the movement (i.e. the joint centre of rotation is not really fixed as our joints are not true hinge joints.) 5. Basically the answer is to reduce errors. Sometimes we actually filter the signal before we collect any digital data from the signal (low-pass and high pass filters are common in some data collection protocols). I did not necessarily discuss this in lecture so it is not necessary to include this type of filtering in your answer. Your answer could have included sources of error that have been previously described (question 4). We need to filter because error is often (usually) unavoidably included in the data we obtain when collecting analogue data into a digital format. Sources of error in biomechanics may be vibration of the camera or joint markers, 60 Hz interference from electrical sources, slight errors by assuming segments are rigid and have centres of rotation that are constant, etc. Differentiation and Noise Signal vs Noise The differential of the line between these markers is much larger than the difference in their location. Dark stars = true location of markers Light stars = location of markers due to noise Your answer should have discussed the problems with differentiating data that is unfiltered. The two slides above from the lecture on analogue-to-digital data collection explain how the errors are magnified if we differentiate the data before filtering (smoothing). In the slide on the left we see that a slight variation in the marker location (due to vibration for example) can result in large differences in the slope of the line joining them. The slide on the rights shows the result of differentiating unfiltered displacement data twice to get the eleration of the toe marker during walking. Also included on this graph is the eleration data obtained from differentiating filtered data. 6. There are two parts to these types of questions. One pure is to see if you can do the necessary mathematics (which is basically some geometry/trigonometry and finite differentiation), but just as important is the issue of whether you can see the movement (the gait cycle) which is probably more important. 6
a) To calculate the right knee angle at frame 14 you will need the shank and thigh angles at frame 14. Remember the format of the equation to obtain angle. θ = 43 tan y x 1 3 4 y x 3 4 Thigh angle (q 21 ) at frame 14 tan -1 [(92.9-53.8)/(160.5-168.4)] = tan -1 (39.1/-7.9) = tan -1 (-4.949) = -78.6 o + 180 o = 101.4 o By looking at the co-ordinates and/or plotting out the rough alignment of the segment you can determine what quadrant the segment is in. This segment is in the second quadrant as the femoral condyle is in advance (greater x value) of the greater trochanter. Shank angle (q 43 ) at frame 14 (use tibial condyle not femoral) tan -1 [(50.9-25.8)/(165.1-134.2)] = tan -1 (25.1/30.9) = tan -1 (0.812) = 39.1 o (1 st quadrant) Knee angle (relative) = q 21 - q 43 (+ve for flexion, -ve for extension) Knee angle = 101.4-39.1 = 62.3 o This is quite a high le of flexion (see opite) In relation to the gait cycle, frame 14 is 4 frames after toe off. This indicates the knee is being flexed prior to swinging forward towards heel strike. The knee angle vs. time graph slide opite (from lecture), shows that maximum knee angle occurs early in the swing phase of walking and that for the subjects used that peak angle was indeed close to the 63 o I have just calculated. Angle (degrees) 80 60 40 20 0 Support Phase KNEE 62.3 o Swing Phase b) To calculate shank angular ocity at frame 24 we need to find the shank angle at frame 23 and 25 and then differentiate using the first central differences method. Shank angle (q 43 ) at frame 23 tan -1 [(53.8-14.0)/(194.9-188.9)] = tan -1 (39.8/6) = tan -1 (6.633) = 81.4 o (1 st quadrant) Shank angle (q 43 ) at frame 25 tan -1 [(53.1-13.0)/(200-201.4)]=tan -1 (40.1/-1.4)= tan -1 (28.64) = -88 o +180 o = 92 o (2 nd quad) Frame 24 is just after the mid-point of the swing phase in terms of time. It certainly makes sense that the shank could be swinging through the vertical around this frame number. Shank anglular ocity (w s ) at frame 24 (finite differentiation) ω s = (θ frame25 - θ frame23 )/2 t = (92-81.4)/0.0333 = 317 o /sec or 5.54 rads/sec If you look at the slide above (used in answer to par a) you can see that the slope of the knee angle vs. time graph is steep during middle of the swing phase (frame 24 is mid- 7
swing) indicating a high knee angular ocity. As this is walking the thigh will not be rotating very quickly at this point so the knee angular ocity will bear a close relationship to the shank angular ocity. Hence our high value here makes sense. c) Calculating ankle angular eleration at frame 7 requires the same procedures used above (lots of them in fact) so I will not repeat them here. The steps and answers are listed below. If you got this question correct you can rest assured you can manipulate co-ordinates and perform finite differentiation well! Shank angle (absolute) = θ 43 Foot angle (absolute) = θ 65 Ankle angle (relative) = θ 43 - θ 65 + 90 o (+ve for plantarflexion, -ve for dorsiflexion) Step 1: Calculate shank angles at frames 5, 7 and 9. (46.7 o, 40.7 o, 36.1 o ) Step 2: Calculate foot angles at frames 5, 7 and 9. (128.8 o, 117.6 o, 112.0 o ) Step 3: Calculate ankle angle at frames 5, 7 and 9. (7.9 o, 13.1 o, 14.1 o ) Step 4: Calculate ankle angular ocity at frames 6 and 8. (156 o s -1, 30 o s -1 ) This can be done using first central differences method across the frames in step 3. Step 5: Calculate ankle angular eleration at frame 7. (-3780 o s -2 ) Your answer should be approximately -66 radians per second 2 (-3780 o s -2 ). Note that this eleration is in the direction of dorsiflexion. Frame 7 is shortly before toe-off (toe-off at frame 10). Toe-off is when the foot leaves contact with the ground so it is reasonable that the high plantar flexion eleration required to propel the body forward has taken place before this point in time. So the foot is still plantar-flexing (itive ocities at frame 6 and 8) but the magnitude of these ocities are reducing prior to having to start dorsiflexing to make sure the toe clears the ground. The ankle does not actually move into a dorsiflexed ition until mid-swing phase (frame 16) prior to ground contact. Note: Peak plantar-flexion ocity occurs at frame 4 and peak plantar flexion eleration occurs at frame 2. If you look at the linear eleration time graph for the 100m sprint for question 9 below, you will see that peak eleration takes place very early in the movement (first few 10 ths of a second). This is typical for human movement (think of that muscle force-ocity curve!). On the next page are the angle, angular ocity and angular eleration for the four segments and three joints of the lower body. This data is calculated from the Kinematic data given on page 19 of this booklet. Not all the frames are included as I still use this data for assignments. However, there are plenty of answers there if you want to choose a segment and/or joint and calculate these variables as was done in question 6 above. Note that rounding differences (e.g. writing 4.5378 as 4.5 versus say 4.54) will result in quite large differences once you start differentiating. Do not worry as long as the magnitude is similar. Important: Winter uses the following equation to calculate ankle angle (from page 18). Ankle angle (relative) = θ shank - θ foot + 90 o (+ve for plantarflexion, -ve for dorsiflexion) Hamill and Knutzen on the other hand use the following equation Ankle angle (relative) = θ foot - θ anklet - 90 o (-ve for plantarflexion, +ve for dorsiflexion) 8
You should be able to determine which equation was used in the answers overleaf. 7 a) Linear (radial eleration) b) ar = v t 2 /r = 5 2 /15 = 25/15 = 1.67 m/s 2 Towards the centre of the curve (centripetal). c) Sliding. Radial (centripetal) force is perpendicular to the tangential ocity the bike has (therefore across the tires). 8. Relative angle: Angle at a joint formed between the longitudinal axes of the adjacent body segments. Useful in assessing extreme movement ranges like wrist flexion and/or extension and ulnar and/or radial deviation when using hand tools. Absolute angle: Angular orientation of a body segment with respect to a fixed line of reference (usually the right horizontal). Useful in assessing required muscle torques to counteract moments due to weight of held objects (e.g. trunk ition in lifting task). 9. Tables and graphs for question 9 are on the page after the Kinematic data listed below. Frame (#F) TIME [s] Thigh Segment Shank Segment Foot Segment -1 0.000 1.31-0.364 16.40 1.12-2.44-10.3 2.810-3.58-29.9 0 0.017 1.30-0.091 22.29 1.08-2.61-13.4 2.75-4.08-40.0 1 0.033 1.31 0.379 22.20 1.03-2.89-16.9 2.67-4.92-53.0 2 0.050 1.32 0.649 19.66 0.985-3.18-14.0 2.59-5.85-51.0 3 0.067 1.33 1.03 25.82 0.929-3.35-6.40 2.48-6.62-32.7 4 0.083 1.35 1.51 25.55 0.873-3.39-0.890 2.37-6.94-4.55 5 0.100 1.38 1.89 20.31 0.816-3.38 6.73 2.25-6.77 32.1 6 0.117 1.41 2.19 18.27 0.760-3.17 16.2 2.14-5.87 64.6 7 0.133 1.45 2.49 18.05 0.710-2.84 22.6 2.05-4.62 88.3 8 0.150 1.50 2.79 6.608 0.666-2.41 38.6 1.99-2.93 103 9 0.167 1.55 2.72 2.381 0.630-1.56 49.7 1.95-1.17 110 10 0.183 1.59 2.87 9.039 0.614-0.755 49.8 1.95 0.753 108 11 0.200 1.64 3.02-4.118 0.605 0.105 58.4 1.98 2.44 87.3 12 0.217 1.69 2.73-17.39 0.617 1.19 55.2 2.03 3.66 78.3 13 0.233 1.73 2.44-13.82 0.645 1.95 51.1 2.10 5.05 57.8 9
#F Toe Segment Knee Joint Ankle Joint Metatarsal Joint -1 3.36-0.217 13.0 0.189 2.08 26.7-0.117 1.14 19.6-0.552-3.37-43.0 0 3.36 0.000-20.0 0.224 2.52 35.7-0.098 1.47 26.6-0.608-4.08-20.0 1 3.36-0.883-59.9 0.273 3.27 39.1-0.068 2.03 36.1-0.688-4.03 6.88 2 3.33-2.00-62.0 0.332 3.83 33.6-0.031 2.67 37.0-0.743-3.85 11.1 3 3.30-2.95-78.7 0.401 4.39 32.2 0.021 3.26 26.3-0.817-3.66 45.9 4 3.23-4.62-91.2 0.479 4.90 26.4 0.078 3.55 3.66-0.865-2.32 86.6 5 3.14-5.99-103 0.564 5.27 13.6 0.139 3.39-25.4-0.894-0.776 135 6 3.03-8.04-91.8 0.654 5.35 2.05 0.191 2.70-48.3-0.891 2.17 156 7 2.87-9.05 4.54 0.742 5.34-4.60 0.229 1.77-65.6-0.822 4.44 83.7 8 2.73-7.89 99.8 0.832 5.20-32.0 0.250 0.516-64.8-0.743 4.96 3.51 9 2.61-5.72 138 0.916 4.27-47.3 0.246-0.385-60.7-0.656 4.55-28.1 10 2.54-3.27 139 0.975 3.62-40.8 0.237-1.51-58.5-0.591 4.02-30.5 11 2.50-1.10 145 1.04 2.91-62.5 0.196-2.34-28.8-0.522 3.54-57.4 12 2.50 1.55 108 1.07 1.54-72.6 0.159-2.47-23.0-0.473 2.11-29.5 13 2.55 2.50 72.7 1.09 0.491-64.9 0.114-3.10-6.74-0.452 2.55-14.9 = ition (angle) = angular ocity = angular eleration 10
Accel. (m/s2) Vel. (m/s) Displ. (m) Displacement - Time 110.000 100.000 90.000 80.000 70.000 60.000 50.000 40.000 30.000 20.000 10.000 0.000 0 1 2 3 4 5 6 7 8 9 10 Time (s) Velocity - Time 14.000 12.000 10.000 8.000 6.000 4.000 2.000 0.000 0 1 2 3 4 5 6 7 8 9 10 11 Time (s) Acceleration - Time 8.000 7.000 6.000 5.000 4.000 3.000 2.000 1.000 0.000 0 2 4 6 8 10 12-1.000 Time (s) 9. Obviously the table for this graph opite was given in the question so I will not duplicate it here. Below is the graph and table of ocity-time. Notice that the times are not the same as the other two tables. This is because differentiating over one time frame means you estimate the average ocity of that time frame to occur at the mid-point of the frame (i.e. differentiating displacement from 0.0 to 0.5 seconds gives you the average ocity for that time period and you assume that to occur at 0.25 seconds). You were told ocity at time 0.0 was zero otherwise you couldn't have calculated it from the displacementtime table. Time (s) Velocity (m/s) 0 0.000 0.25 1.715 0.75 4.606 1.25 6.648 1.75 8.160 2.25 9.292 2.75 10.111 3.25 10.675 3.75 11.053 4.25 11.287 4.75 11.414 5.25 11.465 5.75 11.476 6.25 11.476 6.75 11.467 7.25 11.476 7.75 11.476 8.25 11.476 8.75 11.476 9.25 11.476 9.75 11.467 10.25 11.339 11
The table here is for the eleration-time graph above. Notice that you can calculate eleration at time 0.125 seconds because I gave you the information about zero ocity at time zero. Now F = ma so if you multiply the eleration by the mass (70 kg) you can obtain a very crude force-ocity graph. The table is below right and the graph below that. Force (N) Time (s) Accn (m/s 2 ) 0.125 6.858 0.5 5.783 1 4.084 1.5 3.025 2 2.264 2.5 1.639 3 1.127 3.5 0.757 4 0.468 4.5 0.252 5 0.102 5.5 0.022 6 0.000 6.5-0.018 7 0.018 7.5 0.000 8 0.000 8.5 0.000 9 0.000 9.5-0.018 10-0.256 10.5 600.000 500.000 400.000 300.000 200.000 100.000-100.000 Force - Velocity 0.000 0.000 2.000 4.000 6.000 8.000 10.000 12.000 14.000 Velocity (m/s) Velocity (m/s) Force (N) 0 0 1.715 404.787 4.606 285.86 6.648 211.738 8.16 158.484 9.292 114.702 10.111 78.858 10.675 52.999 11.053 32.772 11.287 17.666 11.414 7.169 11.465 1.536 11.476 0 11.476-1.28 11.467 1.28 11.476 0 11.476 0 11.476 0 11.476 0 11.476-1.28 11.467-17.922 11.339 I say crude as the times do not actually match up (see tables) and because this is a force-ocity graph for whole body motion, not a forceocity curve for isolated skeletal muscle. The force estimate is also incorrect as you would have to push air out of the way, deal with friction, etc.so the true force is imsible to calculate. Still the force-ocity curve still has the basic shape we saw in the muscle mechanics lecture. 12