Modeling the Jump of a Snowboard

7/25/2019 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.)




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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)




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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)




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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




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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)




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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




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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)+ (




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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




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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:




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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.




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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.



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Modeling the Jump of a Snowboard