Kinesiology 201 Solutions
Kinematics
Tony Leyland
School of Kinesiology
Simon Fraser University
1. a)
B
Vertical velocity = 10 sin20 = 3.42 m/s
Horizontal velocity = 10 cos20 = 9.4 m/s
Vertical A-B (start to max height)
vf = vi + at
A
0 = 3.42 -9.81t
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/s2
2
2
vf = vi + 2ad
vf2= (3.42) 2 + (2 x 0.5 x 9.81)
vf = √(11.696 + 9.81)
vf = -4.64 m/s
vf = vi + at
-4.64 = -3.42 - 9.81t
Therefore: total time (A-D) = 0.7 + 0.12 = 0.82 s
Horizontal A-D
d = vit + 0.5at2
as a = 0
D
t = 0.12 s
d = 9.4 x 0.82 = 7.71 m
The long jumper travels 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 45o 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 velocity (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 (Yelbow-Yshoulder)/(Xelbow-Xshoulder)
Then to get angular velocity 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 velocity (hence the angle of the arm is
the same for all time frames).
Therefore the angular velocity of the forearm about the fixed frame of reference is going
to be the same as the angular velocity 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 velocities and accelerations.
Forearm segment angle at t=0.01
tan-1[(253.7-225)/(267.8-276.4)] = tan-1 (28.7/-8.6) = -73.31o + 180o = 106.7o (1.862 rads)
2
RESULTS
Time Shoulder (S) Elbow (E)
0.00
0.01
0.02
0.03
0.04
x
251.3
251.9
252.5
253.1
253.7
y
220.1
220.1
220.1
220.1
220.1
x
275.8
276.4
277.0
277.6
278.2
y
225.0
225.0
225.0
225.0
225.0
Wrist (W)
x
262.3
267.8
277.0
288.1
294.2
y
243.6
253.7
255.0
253.1
250.4
θupperarm θforearm ωupperarm ωforearm
rads
rads
rads/s
0.1974 2.19863
0.1974 1.86193
0.00
0.1974 1.5708
0.00
0.1974 1.2132
0.00
0.1974 1.00867
rads/s
-31.392
-32.44
-28.107
Notice than for time 0.00 and 0.01 the forearm is in the 2nd quadrant (see above
diagrams) so we add 180 degrees. It wouldn’t make much sense if you failed to do this,
as your angular velocity would be through the roof and in the opposite direction.
At time 0.02 when calculating the forearm angle (θforearm) the division returns infinity
(division by zero) which would return 90 degrees or π/2 rads (1.57 rads).
Angular velocity 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 positive 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 accept for full marks is to say that the answer you
got (positive angular velocity) 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 velocity of the wrist is due to two factors: its rotation about the elbow, plus the
linear velocity of the elbow. Much like the velocity of the bottom of a spinning ball
relative to the ground is due to the velocity of the ball's centre of gravity plus the
tangential velocity due to the spin.
Therefore, the linear velocity of the wrist can be obtained by calculating the linear
velocity of the elbow with respect to the external frame of reference and then adding this
to the linear velocity of the wrist with respect to the elbow.
The linear velocity of the wrist with respect to the elbow is given by: vt = 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 velocity. It is
at 90 degrees (tangential) to the alignment of the forearm and as the segment is aligned
vertical 90o (1.5708 rads) vt will be horizontal. Boy am I good to you or what!
The velocity of the elbow is easy to calculate as there is no vertical change in direction.
Therefore, zero vertical velocity.
Horizontal elbow velocity = (277.6-276.4)/0.02 = 60cm/s = 0.6 m/s
Velocity of wrist = 9.7311 + 0.6 = 10.3311 m/s at 0o 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 velocity 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 velocity of the proximal joint and then use the equation vCG = rω (where r is the
distance from the proximal joint centre to the centre of gravity, and ω is the segment
angular velocity). This gives you the velocity of the centre of gravity with respect to the
proximal joint (then add the velocity 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 acceptable 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 velocity of wrist
Velocity0.02= (X0.03-X0.01)/2∆t
Velocity0.02= (288.1-267.8)/0.02 = 10.15m/s
Vertical velocity of wrist
Velocity0.02= (Y0.03-Y0.01)/2∆t
Velocity0.02= (253.1-253.7)/0.02 = -0.3m/s
Resultant (pythagorus)= 10.2 m/s
Angle = -1.69o (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 acceleration.
Horizontal data
Velocity0.01= (252.5-251.3)/0.02 = 60 cm/s = 0.6 m/s
Velocity0.03= (253.7-252.5)/0.02 = 60 cm/s = 0.6 m/s
Therefore: Acceleration0.02 = (0.6-0.6)/0.02 = 0 m/s2.
Therefore total acceleration of the shoulder joint equals zero.
B
4
e)
Vertical velocity = 3sin80 = 2.954m/s
Horizontal velocity = 3cos80 = 0.521 m/s
Vertical A-B (start to max height where vf = 0 m/s)
vf2 = vi2 + 2ad
02 = 2.9542 + (2 x [-9.81] x d)
d = 8.728/19.62
d = 0.445 m
A
Height jumped = 44.5 cm
C
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
Horizontal A-D
d = vit + 0.5at2
as a = 0 (ignore air resistance)
d = 0.521 x 0.602 = 0.314 m
The player travels 31.4 cm horizontally during the jump shot.
3.
Components
Horizontal velocity
Vertical velocity
= 20 cos 40
= 20 sin 40
= 15.321 m/s
= 12.856 m/s
t = -12.856 ±√(12.8562 + (2 x -9.81 x -1.5)) The vertical displacement is -1.5 m
-9.81
t = -12.856 ±√ (165.277 + 29.43)
-9.81
t = -12.856 - 13.954
-9.81
t = -26.81/-9.81
=
t = -12.856 ±√194.707
-9.81
The positive result from the square root is not an option.
2.733 seconds
Range = s = Vhorizontalt
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 possibly 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 possible. 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 imposes 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 account 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 account 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 acceleration of the toe marker during walking. Also
included on this graph is the acceleration data obtained from differentiating filtered data.
6. There are two parts to these types of questions. One purpose 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
−1




y −y
x −x
3
3



4
4
Thigh angle (θ21) at frame 14
tan-1[(92.9-53.8)/(160.5-168.4)] = tan-1(39.1/-7.9) = tan-1(-4.949) = -78.6o + 180o = 101.4o
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 (θ43) 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) = 39.1o (1st quadrant)
Knee angle (relative) = θ21 - θ43 (+ve for flexion, -ve for extension)
62.3o
o
Support Phase
Swing Phase
80
Angle (degrees)
Knee angle = 101.4 - 39.1 = 62.3
This is quite a high level of flexion (see
opposite)
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 opposite (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 63o I have
just calculated.
60
KNEE
40
20
0
b) To calculate shank angular velocity 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 (θ43) at frame 23
tan-1[(53.8-14.0)/(194.9-188.9)] = tan-1(39.8/6) = tan-1(6.633) = 81.4o (1st quadrant)
Shank angle (θ43) at frame 25
tan-1[(53.1-13.0)/(200-201.4)]=tan-1(40.1/-1.4)= tan-1(28.64) = -88o+180 o = 92o (2nd 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 velocity ( ωs) at frame 24 (finite differentiation)
ωs = (θframe25 - θframe23)/2∆t = (92-81.4)/0.0333 = 317o/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 velocity. As this is walking the thigh will not be
rotating very quickly at this point so the knee angular velocity will bear a close
relationship to the shank angular velocity. Hence our high value here makes sense.
c) Calculating ankle angular acceleration 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) = θ43
Foot angle (absolute) = θ65
Ankle angle (relative) = θ43 - θ65 + 90o (+ve for plantarflexion, -ve for dorsiflexion)
Step 1: Calculate shank angles at frames 5, 7 and 9. (46.7o, 40.7o, 36.1o)
Step 2: Calculate foot angles at frames 5, 7 and 9. (128.8o, 117.6o, 112.0o)
Step 3: Calculate ankle angle at frames 5, 7 and 9. (7.9o, 13.1o, 14.1o)
Step 4: Calculate ankle angular velocity at frames 6 and 8. (156os-1, 30os-1)
This can be done using first central differences method across the frames in step 3.
Step 5: Calculate ankle angular acceleration at frame 7. (-3780os-2)
Your answer should be approximately -66 radians per second2 (-3780os-2). Note
that this acceleration 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 acceleration
required to propel the body forward has taken place before this point in time. So the
foot is still plantar-flexing (positive velocities at frame 6 and 8) but the magnitude of
these velocities 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 position until
mid-swing phase (frame 16) prior to ground contact.
Note: Peak plantar-flexion velocity occurs at frame 4 and peak plantar flexion
acceleration occurs at frame 2. If you look at the linear acceleration time graph for the
100m sprint for question 9 below, you will see that peak acceleration takes place very
early in the movement (first few 10ths of a second). This is typical for human movement
(think of that muscle force-velocity curve!).
On the next page are the angle, angular velocity and angular acceleration 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) = θshank - θfoot + 90o (+ve for plantarflexion, -ve for dorsiflexion)
Hamill and Knutzen on the other hand use the following equation
Ankle angle (relative) = θfoot - θanklet - 90o (-ve for plantarflexion, +ve for dorsiflexion)
8
You should be able to determine which equation was used in the answers overleaf.
7
8.
a) Linear (radial acceleration)
b) ar = vt2/r = 52/15 = 25/15 = 1.67 m/s2
Towards the centre of the curve (centripetal).
c) Sliding. Radial (centripetal) force is perpendicular to the tangential velocity the
bike has (therefore across the tires).
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 position in lifting task).
9. Tables and graphs for question 9 are on the page after the Kinematic data listed
below.
Thigh Segment
Frame
(#F) TIME [s]
-1
0.000
0
0.017
1
0.033
2
0.050
3
0.067
4
0.083
5
0.100
6
0.117
7
0.133
8
0.150
9
0.167
10
0.183
11
0.200
12
0.217
13
0.233
pos
[rad]
1.31
1.30
1.31
1.32
1.33
1.35
1.38
1.41
1.45
1.50
1.55
1.59
1.64
1.69
1.73
vel
acc
[rad/s] [rad/s2]
-0.364
16.40
-0.091
22.29
0.379
22.20
0.649
19.66
1.03
25.82
1.51
25.55
1.89
20.31
2.19
18.27
2.49
18.05
2.79
6.608
2.72
2.381
2.87
9.039
3.02
-4.118
2.73
-17.39
2.44
-13.82
Shank Segment
pos
[rad]
1.12
1.08
1.03
0.985
0.929
0.873
0.816
0.760
0.710
0.666
0.630
0.614
0.605
0.617
0.645
vel
acc
[rad/s] [rad/s2]
-2.44
-10.3
-2.61
-13.4
-2.89
-16.9
-3.18
-14.0
-3.35
-6.40
-3.39 -0.890
-3.38
6.73
-3.17
16.2
-2.84
22.6
-2.41
38.6
-1.56
49.7
-0.755
49.8
0.105
58.4
1.19
55.2
1.95
51.1
Foot Segment
pos
[rad]
2.810
2.75
2.67
2.59
2.48
2.37
2.25
2.14
2.05
1.99
1.95
1.95
1.98
2.03
2.10
vel
acc
[rad/s] [rad/s2]
-3.58
-29.9
-4.08
-40.0
-4.92
-53.0
-5.85
-51.0
-6.62
-32.7
-6.94
-4.55
-6.77
32.1
-5.87
64.6
-4.62
88.3
-2.93
103
-1.17
110
0.753
108
2.44
87.3
3.66
78.3
5.05
57.8
9
Toe Segment
#F
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
Knee Joint
Ankle Joint
Metatarsal Joint
pos
vel
acc
pos
vel
acc
pos
vel
acc
pos
vel
acc
[rad] [rad/s] [rad/s2] [rad] [rad/s] [rad/s2] [rad] [rad/s] [rad/s2] [rad] [rad/s] [rad/s2]
3.36 -0.217 13.0 0.189 2.08
26.7 -0.117 1.14
19.6 -0.552 -3.37
-43.0
3.36 0.000 -20.0 0.224 2.52
35.7 -0.098 1.47
26.6 -0.608 -4.08
-20.0
3.36 -0.883 -59.9 0.273 3.27
39.1 -0.068 2.03
36.1 -0.688 -4.03
6.88
3.33
-2.00
-62.0 0.332 3.83
33.6 -0.031 2.67
37.0 -0.743 -3.85
11.1
3.30
-2.95
-78.7 0.401 4.39
32.2 0.021 3.26
26.3 -0.817 -3.66
45.9
3.23
-4.62
-91.2 0.479 4.90
26.4 0.078 3.55
3.66 -0.865 -2.32
86.6
3.14
-5.99
-103 0.564 5.27
13.6 0.139 3.39
-25.4 -0.894 -0.776
135
3.03
-8.04
-91.8 0.654 5.35
2.05 0.191 2.70
-48.3 -0.891 2.17
156
2.87
-9.05
4.54 0.742 5.34
-4.60 0.229 1.77
-65.6 -0.822 4.44
83.7
2.73
-7.89
99.8 0.832 5.20
-32.0 0.250 0.516 -64.8 -0.743 4.96
3.51
2.61
-5.72
138
0.916 4.27
-47.3 0.246 -0.385 -60.7 -0.656 4.55
-28.1
2.54
-3.27
139
0.975 3.62
-40.8 0.237 -1.51
-58.5 -0.591 4.02
-30.5
2.50
-1.10
145
1.04 2.91
-62.5 0.196 -2.34
-28.8 -0.522 3.54
-57.4
2.50
1.55
108
1.07 1.54
-72.6 0.159 -2.47
-23.0 -0.473 2.11
-29.5
2.55
2.50
72.7
1.09 0.491 -64.9 0.114 -3.10
-6.74 -0.452 2.55
-14.9
pos = position (angle)
vel = angular velocity
acc = angular acceleration
10
Displacement - Time
9. Obviously the table for this
graph opposite was given in the
question so I will not duplicate it
here.
110.000
100.000
90.000
80.000
Below is the graph and table of
velocity-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
velocity 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 velocity for that time period
and you assume that to occur at
0.25 seconds).
You were told
velocity at time 0.0 was zero
otherwise
you
couldn't
have
calculated it from the displacementtime table.
Displ. (m)
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
Time (s)
Vel. (m/s)
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
Accel. (m/s2)
5.000
4.000
3.000
2.000
1.000
0.000
0
2
4
6
-1.000
8
10
12
0
0.25
0.75
1.25
1.75
2.25
2.75
3.25
3.75
4.25
4.75
5.25
5.75
6.25
6.75
7.25
7.75
8.25
8.75
9.25
9.75
10.25
Velocity
(m/s)
0.000
1.715
4.606
6.648
8.160
9.292
10.111
10.675
11.053
11.287
11.414
11.465
11.476
11.476
11.467
11.476
11.476
11.476
11.476
11.476
11.467
11.339
Time (s)
11
The table here is for the acceleration-time graph above. Notice that you can calculate
acceleration at time 0.125 seconds because I gave you the information about zero
velocity at time zero.
Now F = ma so if you multiply the acceleration by the mass (70 kg) you can obtain a very
crude force-velocity graph. The table is below right and the graph below that.
Time (s)
0.125
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
7
7.5
8
8.5
9
9.5
10
10.5
Accn (m/s2)
6.858
5.783
4.084
3.025
2.264
1.639
1.127
0.757
0.468
0.252
0.102
0.022
0.000
-0.018
0.018
0.000
0.000
0.000
0.000
-0.018
-0.256
Force - Velocity
600.000
500.000
Force (N)
400.000
300.000
200.000
100.000
0.000
0.000
2.000
4.000
6.000
8.000 10.000 12.000 14.000
-100.000
Velocity (m/s)
Velocity
(m/s)
Force
(N)
0
1.715
4.606
6.648
8.16
9.292
10.111
10.675
11.053
11.287
11.414
11.465
11.476
11.476
11.467
11.476
11.476
11.476
11.476
11.476
11.467
11.339
0
404.787
285.86
211.738
158.484
114.702
78.858
52.999
32.772
17.666
7.169
1.536
0
-1.28
1.28
0
0
0
0
-1.28
-17.922
I say crude as the times do not
actually match up (see tables) and
because this is a force-velocity graph
for whole body motion, not a forcevelocity 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 impossible
to calculate. Still the force-velocity
curve still has the basic shape we
saw in the muscle mechanics lecture.
12