Modeling the Jump of a Snowboard
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Transcript of Modeling the Jump of a Snowboard
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Mathematics SL Exploration:
The jump of a snowboard
Name: Ewout Kessels
Student number: 000512-0063
Topic: Projectile motion
Teacher: Eva Watson
School Number: 00512
Date: 11/01/16
Word Count: 1964
[1] http://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/
[2] http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdf
http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdfhttp://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/ -
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Tricks are a fundamental part of snowboarding apart from looking amazing if performed
correctly, it is also a show of extreme skill and practice, as it is impossible for a
snowboarder to innately know how to perform jumps and slides without any knowledge
and experience on the perfect approach. In case of jumps especially, it is crucial to have
insight whether a jump is possible or not, as things could go severely wrong with one
single mistake. It is therefore important for any snowboarder attempting jumps to
approximate how fast they would need to travel to complete the jump safely without any
complications. Of course, jumps vary in height and size, rendering most snowboarders
unable to know the minimum velocity required to complete the jump. The only viable
option is to use the tool of approximation. Of course, more experienced snowboarders
might find relative ease in approximating the velocity they need to successfully land a
jump, yet beginners often struggle with finding a good speed to approach a jump with,
either going to slow or too fast could result in injuries. This investigation is therefore
based around modelling the jump of a snowboard.
To conduct the investigation, a video of a snowboarder jumping has been made
at an indoor skiing park. This is as the jump has been conducted in a controlled
environment to acquire an accurate representation of a common snowboard jump. The
video of the snowboarder jumping has been processed in Logger Pro with the trajectory
of the jump shown below. As reference points, a meter ruler was used to measure the
scale of the video, and the front foot of the snowboarder has been tracked, seen by the
green points on the picture.
(Picture 1: the trajectory of the snowboard jump.)
[1] http://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/
[2] http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdf
http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdfhttp://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/ -
7/25/2019 Modeling the Jump of a Snowboard
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Limitations and assumptions
To address some initial limitations to my investigation, I have firstly chosen to analyze
the trajectory in 2d, not taking into account any anomaly with the depth of the video. It
has been estimated that the camera location was at approximately 4 meters from the
ramp.. Secondly, the air resistance shall be ignored as well, since its effect on the datais close to negligible, thus, the initial velocity will mostly be taken into account of. Lastly,
it should be mentioned that the indoor ski slope lies on a hill. This effectively means that
the video was shot already at an angle while on the slope in order to accurately show
the parabolic movement of the snowboard. This effect has been correct by shifting the
axes to line up with the structure of the building the ski slope is located, seen in picture
1.
Moreover, the movie was shot in 60 frames per second, yet Logger pro has put a frame
limiter of 29.97 frames per second on the movie, making certain data points unstable
while processing the video and its data points. This is most notable on the X velocity
component below.
(graph 1: X velocity over time graph)
[1] http://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/
[2] http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdf
http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdfhttp://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/ -
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Graph 1 shows the instability of the horizontal component of velocity, which should
normally show a linear trend with a constant velocity. This is as, theoretically speaking,
an object moving through air should not increase its velocity in the horizontal direction.
This graph has been created after the axes have been set to fit the natural slope.
Instead, the trend in this graph shows a frequent dip in an overall increasing slope.
Presumably, this effect occurs due to the reduced amount of frames not being able to
process all of the movement seen in the hd video, instead creating frames where the
video does not show any movement (in the future referred to as ghost frames), even
though the time of the video progresses. In this particular trial, the minima seen of every
negative increase of the y axis shows the data point that has been manually entered on
every ghost frame. In an attempt to correct this anomaly, the every ghost frame has
been struck through, with the resulting graph shown below.
(graph 2, X velocity over time graph with ghost frame struck through)
[1] http://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/
[2] http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdf
http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdfhttp://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/ -
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Graph 2 still shows a trend where the horizontal component of velocity increases as the
time in the video progresses, despite the ghost frames having been ignored. Although
still inaccurate, it is believed that this effect is still present due to the snowboarder
shifting the weight of the snowboard closer to their body during the takeoff of the jump,
increasing the density of mass for the whole duration of the jump, which affects the
magnitude of the acceleration due to gravity, thus increasing the x-velocity component.
Calculations
Now ive described the various limitations and assumptions made, the mathematical
model of the jump can be constructed by looking at various components of the jump
time, components of velocity, angle of elevation, initial velocity, range.
While processing the video, Logger pro already calculated some key components
needed in order to construct the model.
Frames Time (s) X (m) Y (m) X Velocity (m/s) Y Velocity (m/s)
1 0 0 0 2.228 0.926
2 0.0334 0.0757 0.0315 2.444 0.924
3 0.0667 0.1326 0.0492 3.167 1.142
4 0.1001 0.2730 0.1034 4.208 1.493
5 0.1335 0.4439 0.1614 4.437 1.511
6 0.1668 0.5841 0.2069 4.138 1.408
7 0.2002 0.7289 0.2610 3.567 1.131
8 0.2336 0.8160 0.2813 0.901
9 0.2669 0.9080 0.3102 3.973 0.935
10 0.3003 1.0750 0.3465 4.944 0.86
11 0.3337 1.2580 0.3825 5.346 0.385
12 0.3670 1.4450 0.3792 5.189 -0.276
13 0.4004 1.6320 0.3498 4.284 -0.583
14 0.4338 1.7280 0.3307 -0.635
15 0.4671 1.8150 0.3205 4.328 -0.968
16 0.5005 2.0100 0.2691 5.355 -1.385
17 0.5339 2.2090 0.2134 5.405 -1.426
18 0.5672 2.3740 0.1800 5.134 -1.551
19 0.6006 2.5730 0.1025 4.279 -1.485
[1] http://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/
[2] http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdf
http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdfhttp://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/ -
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20 0.634 2.664 0.07474 -1.326
21 0.6673 2.742 0.03852 4.27 -1.83
22 0.7007 2.941 -0.04337 5.371 -2.513
23 0.7341 3.14 -0.134 5.513 -3.093
24 0.7674 3.312 -0.2459 5.52 -3.749
25 0.8008 3.501 -0.4017 5.569 -4.268
(Table 1: Values for: time, X (m), Y (m), X-velocity, Y-Velocity)
From the data it is derived that the jump lasted for about 0.8 seconds, and that 3.5
meters have been traversed in this duration. Using frame by frame analysis, logger pro
has processed this data and calculated the for both the X and Y vectors. Asv
reference, an example calculation from the first frame to the second frame for will bevx
given./t Sx
= vx
(0.0757)/(0.0334)= 0.0746 m/0.0334 s
.0334 0 s0 3 1
0.0757 x 30 = 2.271 ms-1
The result shows a systematic error of 0.200ms -1due to the many decimals that
normally exceed the numbers shown in the table.
Now, using the data from the table above, the first thing that should be calculated is the
angle of elevation, as many other components rely on it. Using a simple trigonometry,
data for the for the elevation in X and Y can be used to find the angle of elevation. We
can assume that the angle of elevation stays consistent for the first quadrant of the
jump, seen by graph 3. The following trigonometric equation will be used to find the
angle of elevation between frames: 1-2, 1-3, 1-4, 1-5, 1-6, in order to find a consistent
angle:
an( )t i =x
y
rctan( )i=a xy
frames 1-2 1-3 1-4 1-5 1-6
i() 22.61 20.35 20.74 19.98 19.50
(Table 2: Angle of elevations comparing the first frames to the 5 following frames)
[1] http://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/
[2] http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdf
http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdfhttp://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/ -
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(Graph 3: Y(m) over X(m) graph showcasing the trajectory of the snowboard jump)
The values shown in Table 2 are seen to decrease for every frame the the first frame is
compared to, hence it is seen that the trajectory immediately begins to move in a
parabolic motion. Thus, the most accurate angle should lie in between the first and and
the second frame, as it shows the initial stages of the jump.
Now, having calculated an accurate angle of elevation, the equation for the initial
velocity of the jump can be modelled with two equations for each initial velocity of X, vxi
and initial velocity of Y, .vyi
for the vertical component ofsin(22.61)vyi=vi
vy
for the horizontal component ofcos(22.61)vxi=vi
vx
[1] http://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/
[2] http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdf
http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdfhttp://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/ -
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To find initial velocity, these two equations can be rearranged to:
vi
=
vxi
cos(22.61)
and
vi
=
vyi
sin(22.61)
Using the first frame values to find of and respectively, the initial velocity can bevyi vxi
found. Results are shown in the following table:
usingvi vyi usingvi
vxi
2.4085 ms
-1 2.4134 ms
-1
The two results given for initial velocity differ by a few decimal points, and it has been
found out this is due to various factors. Mainly, the instability of the X-velocity
component which was addressed previously. Nevertheless, both values are close
enough to call the difference negligible. One main concern with these values, however,
is that the snowboarder traversed 3.5 meters in 0.8 second, meaning that the initial
velocity should have been . This is mostly the result when the calculation.5ms vi > 3 1
was done with values of a ghost frame.
This means the initial velocity should have beencalculated using frame values in between ghost frames, but since the snowboard is
already in air by that time, the effect of gravity now comes into play for the vy
component. the gravitational constant on Earth equals to, g=9.81 ms-2, meaning that
every second the vertical component of velocity changes, , by 9.81 m/s. Whenvyi
integrated into our previous model of initial velocity, the equation for changes to:vyi
, where g=9.81sin(22.61) tvy=vi
g
Rearranged to:
9.81)(t)vi
= vy
sin(22.61)+ (
[1] http://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/
[2] http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdf
http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdfhttp://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/ -
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Now, using both the new equation for the vertical component and the previous equation
for the horizontal component of velocity we get the answers, we can calculate the initial
velocity by looking at the fourth frame :
usingvi vx usingvi
vy average
4.55 ms-1 4.86 ms-1 4.70 ms-1
These values are a lot more accurate and realistic, rendering the average of these two
as an accurate representation of the initial velocity.
Next, we can use the initial velocity and angle of elevation to model equations for
maximum height, the distance the snowboard has travelled.
Distance
To get a more accurate idea of how far the snowboard travels in its parabolic trajectory,
we can remodel the equations previously used to find initial velocity to incorporate
distance travelled.
It is known in kinematics that if velocity is integrated, the distance travelled can be
found.[1]So if we integrate the previous equation correlating and with withvy vx vi
respect to time, we see that:
sin(22.61) tdt sin(22.61)t gt
vi
g =vi
2
1 2 =Sy
cos(22.61) dt v cos(22.61)t
vi
= i
=Sx
Sx shows the distance throughout the snowboard flight on the x-axis, while Syshows the
altitude of the snowboard at any given time. To adjust this equation to our current model
we get:
4.70)cos(22.61)t ( =Sx
and
(4.70)(22.61)t (9.81)tvi
21 2 =Sy
Maximum height
[1] http://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/
[2] http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdf
http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdfhttp://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/ -
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Following the equation for the distance in y direction, it is known that the maximum
height the snowboard achieved is when the vertical velocity is equal to 0. The maximum
point of the vertex is yet unknown due to the inconsistency of the frame rate, this is
seen in graph 3 below. Yet if we set the derivative of the equation for initial velocity in
the y axis to 0 in respect to time, we can find when the slope of the parabola is parallel
to the x-axis[2], meaning we have found the maximum height.
4.70)sin(22.61) tdt
dy= ( g
= 0
or rearranged
t=g
(4.70)sin(22.61)
(Picture 2: Y(m) over time (s) graph)
As a last step, the equation for time taken can be incorporated in the equation for
distance for the y component, adjusting it to altitude in order to find maximum height
changing the model to:
[1] http://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/
[2] http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdf
http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdfhttp://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/ -
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4.70)sin(22.61)( ) (9.81)( )ymax= ( 9.81(4.70)sin(22.61)
21
9.81
(4.70)sin(22.61) 2
This equation can be simplified to:
ymax = 2(9.81)
((4.70)sin(22.61))2
Alternatively, one could directly use the model for the time taken to reach maximum
height and use the t value and substitute this directly into the equation for altitude, Sy.
Now, the theoretical value for maximum height reached would be approximately 0.166m
after 0.17 seconds, though this clearly not the case, as seen in table 1, where the
maximum height reached exceeded around 0.3825m at the 11th frame before passing
the vertex at the 12th frame. Then how is this possible? Well, if the snowboarder in thevideo were to be an inanimate object, the maximum height reached would indeed be
0.166m after 0.17 seconds, though this is obviously not the case. Instead, it is seen
during the video, that the snowboarder pulls up his feet as a reflex to brace himself for
the land, seen in the following two frames. This causes the trajectory to change, as the
mass of the snowboarder outweighs the mass of the snowboard, resulting in the
snowboard getting pulled towards the rider. This effect renders a large part of the
graphs processed through Logger Pro illegitimate, as they do not represent a complete
parabolic trajectory.
[1] http://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/
[2] http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdf
http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdfhttp://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/ -
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(frames 6 and 9: showcasing the snowboarding retracting his feet, causing the trajectory to shift as the
density of mass becomes higher)
To conclude, the model created for the movement of the snowboard throughout the air
is overall accurate for theoretical cases only. It is due to the various factors such as the
limiting frames per second of the video, and the movement of the snowboarder in the air
itself which render the model unable to completely and flawlessly represent the full
movement. However, it should be stated that the trend seen on the graphs does show a
close representation of a parabola. For further investigation is might be useful to see
how the mass of the snowboarder exactly affects the trajectory, and how an even more
accurate model could be created to fit certain jumps. Perhaps using the mass to find
how the potential energy of the snowboarder varies throughout the jump could help find
a more accurate idea of how the snowboarder shifted his mass by bringing the
snowboard closer to his body.
Bibliography:
[1] "War Maths - Projectile Motion." IB Maths Resources from British International
School Phuket. N.p., 10 Nov. 2013. Web. 10 Jan. 2016.
[2] "13.2 Modeling Projectile Motion." 13.2 (2011): n. pag. Delmar.edu. Del Mar College.
Web. 10 Jan. 2016.
[1] http://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/
[2] http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdf
http://dmc122011.delmar.edu/math/pjohnson/Webpage/calculusIII/notes/13.2.pdfhttp://ibmathsresources.com/2013/11/10/war-maths-projectile-motion/