Team 9819
1
Research of the Snowboard Course
Summary
In order to design a snowboard course, we discuss the radius of the transition and the width
of the flat-bottom, which influences the “vertical air”, the twist of the player and the time the
player need to adjust their posture.
As to maximize the “vertical air”, assume that the transition is a semicircle. Suppose that the
“vertical air” is determined by the radius of the transition and the obliquity of the snowboard
course. Furthermore, we analyze the forces on a player, including the snow-friction, the air
resistance and the gravity. To apply the energy conservation law, we get the speed of the player at
any time which is related to the radius of the transition and the obliquity of the snowboard course.
Since the “vertical air” is determined by the speed of the player, we utilize Mat lab to calculate the
maximum “vertical air”, the result is that the radius of the transition is 7.0m, the obliquity of the
snowboard course is 0.3 rad, and the “vertical air” is 3.1m. Through the stability analysis, we
conclude that our model is reliable.
To maximize the twist in the air, two factors must be considered, including the friction and
the “vertical air”. Since these two factors are all related to the radius of the transition, we design a
transition which is a combination of two circles with different radius. Through calculating the
friction and the speed of the player which are related to the two different radiuses, the friction
work and the kinetic energy to twist can be calculated.
To develop a “practical” course, we consider the size of the flat-bottom. The width of it
influences the kinetic energy loss and the time the player needed to adjust their posture on the
flat-bottom.
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Introduction
The shape of a snowboard course is generally known as a “half-pipe” (shown in figure 1). In
appearance, it resembles a cross section of a swimming-pool, and in its most basic form, it consists
of two concave ramps (or quarterpipes). Originally half-pipes were simply half sections of a large
pipe. Since the 1980s, half-pipes have had extended flat ground (the flat bottom) added between
the quarterpipes; The flat ground gives the athlete time to regain balance after landing and more
time to prepare for the next trick..
In order to enable the player to perform their tricks sufficiently, the design of a snowboard
course is significantly important.The requiirements include the maximum vertical distance above
the edge of the halfpipe (“vertical air”), and the player’s maximum twist in the air. To satisfy these
kinds of requirements, many factors should be considered, such as the obliquity of a snowboard
course and the radius of the transition
1.1Transitions and snowboard course design
The character of a snowboard course depends on four qualities: most importantly, the
transition radius and the height of the vert, the obliquity and less so, the amount of flat-bottom.
The flatbottom, while valued for recovery time, serves no purpose if it is longer than it needs to
be.
1.2Objectives
Our goal is to design a snowboard course to maximize the vacated height or the twist.
2. Scheme 1
2.1 Assumption
(1)The own skill of the player won’t be thought over.
(2)The surface of the snowboard course is smooth.
2.2Model
Suppose that the maximum vertical distance above the edge of the half-pipe is related to the
following factors, which are the radius of the transition, the height of the vert,the obliquity of a
snowboard course, the friction, the air resistance and the initial speed of the player.
2.2.1 Definition and symbol
Table 1 Symbols
v
The speed of the player
v0
The initial speed of the player entering the
transition
0
v1
The angle that the player enters the transition
The speed perpendicular to the slide(shown in
Figure 1)
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
The friction coefficient
N
The supportive force
3
The radius of the transition
r
m
The mass of a human
g
The acceleration of gravity
Figure 1: the snowboard course
The friction
According to the formula
f friction   N
(1) ,
The fiction is determined by the friction coefficient and the supportive force N . First, we will
calculate the value of N .
Define  as the obliquity of the snowboard course, Figure 2 shows the side face of a snowboard
course.
Figure 2: The side-face of a snowboard course
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Depict the half-pipe in Figure 3,
Figure 3: The cross-section of a half-pipe
The supportive force N contains two parts, the centripetal force and the component force of the
mv 2
gravity. The centripetal force is r , and the component force of the gravity is mg cos  cos  ,
we could get
N
mv
r
2
 mg cos  cos 
(2)
where
 is the included angle with the perpendicular line
so
f friction   ( mvr  mg cos  cos  )
2
(3)
Air Resistance
The formula
f airresis tan ce 
Where
A is the player’s sectional area
Cw
is the drag coefficient
p is the density of the air
So the total resistance
1
ACw pv 2
2
(4)
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f  f friction  f airresis t a cen   ( mvr  mg c o s
2
5
1
c o s ) ACw pv 2
2
)
(5)
In order to obtain the maximum height, the kinetic energy of the player should be figured out.
The velocity can be decomposed into the horizontal velocity, which has no effect on the “vertical
air” and the vertical velocity. The resistance work is concerned with the vertical velocity, which
reduces the kinetic energy of the player, thus causing the reduction of the height:
The resistance work is:


0
0
W f   dW f ( )   [  (mv( ) 2  mgr cos  cos  ) 
1
ACw pv( ) 2 r ]d 
2
Where
v is the vertical velocity
According to the law of conservation of energy, when the angle is
(6)
 (shown in Figure3), we get
1 2
1
mv0  mg (r  r cos  )  mv( , r ) 2 
2
2


0
[  (mv( , r ) 2  mgr cos  0 cos( , r ) 
1
ACw pv( , r ) 2 r ]d 
2
through equation 7, we could know the relationship between v and
(7)
,
as v  v( )
When the angle is

, the equation is
2

1 2
1

1
mv0  mgr  mv( )2   2 [  (mv( )2  mgr cos  cos  )  ACw pv( ) 2 r ]d
0
2
2
2
2
In equation 8,
1

mv( ) 2 is
2
2
(8)
the player’s kinetic energy when leaving the course, so when 1 mv(  )2
2
2
reaches the maximum, the player could reach the maximum height.
3
In order to get ideal height, assume that v0  10m / s , m=50kg,  =0.2, p=1.293 kg / m ,
A=0.5, Cw
 1 , and we use Mat lab to get r and  . Since the equation is too complicated, we
first remain  unchanged. And get the following Table 2
Table 2 The relationship between Vertical air and  , r
when  =0.1

r
Vertical air
0.1
1
3.070
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0.1
2
3.073
0.1
3
3.074
0.1
4
3.075
0.1
5
3.077
0.1
6
3.079
0.1
7
3.081
0.1
8
3.081
0.1
9
3.080
Then we remain r unchanged
Table 3 The relationship between Vertical air and  , r
when r =1

r
Height
0.1
1
3.070
0.2
1
3.071
0.3
1
3.072
0.4
1
3.072
0.5
1
3.071
We find that when  changes, the Vertical air won’t change a lot, so we focus on the
parameter r .
From Table2, when  =0.3, the vertical air is the maximum, assume that  is 0.3, and we get
r changed, and we can obtain the following Table 4.
Table 4 The relationship between Vertical air and  , r
when  =0.3

r
Height
0.3
1
3.072
0.3
2
3.073
0.3
3
3.074
0.3
4
3.075
0.3
5
3.077
0.3
6
3.079
0.3
7
3.081
0.3
8
3.081
0.3
9
3.080
Through this method we could find the best solution, and  =0.3, r =7 is the best answer.
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3 Scheme 2
If the twist is produced, the friction is needed, for the friction produces a torque. Apparently,
the more the friction is, the more a torque is produced. At the coping of the half-pipe, the
supportive force N equals to the centripetal force. As we know, the less the radius of the
transition is, the more the centripetal force is. On the other hand, the twist is also influenced by the
“vertical air”. As we try to balance these two factors, we design a transition which is a
combination of two circles with different radius. Through descriptive geometry, depict the
following Figure 4,
Figure 4: The descriptive geometry
And we could figure out the transition GEH as shown in the following Figure 5
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Figure 5: the transition depicted by the descriptive geometry
As is shown in the figure, the player’s route is GEH, the radius of the ellipse could be divided into
two parts.
1  CAE ,  2  EFG . When the player is located between E and H, the radius
AH= r1 , while the player is located between E and G, the radius EF= r2 . Through geometrical
relation, r1
 3r2 , while 1   2 

2
,we could know 1  1 (r1 , r2 ) ,
 2   2 (r1 , r2 )
At the coping of the snowboard course, the centripetal force is
f friction 
 mv(1 , 2 )2
r2
(9)
where
 is the friction coefficient
v is the speed of the player
We could work out the kinetic energy between EC and EG separately.
When the player is located in EC,
The friction work
1
W f (1 )   dW f (1 ) ( )
0
1
  [  mv( 1 ) 2  mgr1 cos  cos  1 
0
where
1
ACw pv( 1 ) 2 r1 ]d  1
2
(10)
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 is the obliquity of the snowboard course]
A is the player’s sectional area
Cw is the drag coefficient
p is the density of the air
And get
1 2
mv0  mg (r1  r1 cos 1 ) 
2

1
1
mv(1 ) 2   [  mv( 1 ) 2  mgr1 cos  cos  1  ACw pv( 1 ) 2 r1 ]d  1
0
2
2
(11)
Through equation (11), we could get to know the relationship between v and 1 ,
as v  v(1 )
When the player is located in EG
The friction work
2
W f (2 )   dW f (1 ) ( ) 
0

2
0
[  mv( 2 ) 2  mgr2 cos  cos  2 
1
ACw pv( 2 ) 2 r2 ]d  2
2
(12)
And
1
mv(1 ,  2 )
2
 mg (r2  r2 cos  2 ) 

2
0
1
mv(1 ,  2 ) 2 
2
[  mv(1 ,  2 ) 2  mgr2 cos  cos  2 
1
ACw pv( 2 ) 2 r2 ]d  2
2
(13)
Thus the player’s kinetic energy at the coping is v(1 , 2 ) , which is determined
by 1 (r1 , r2 ), 2 (r1 , r2 ) , that is to say, it is determined by
r1 , r2 .
Furthermore, we could calculate the friction work to twist, so
W friction   f friction
（14）
where
 is the coefficient
Thus the total work to twist
1
Wtwist   f friction  mv(1 , 2 ) 2
2
4 Scheme 3
（15）
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The size of the flat-bottom should be considerd to make the half-pipe pracitical. The width of
it influenced the energy loss and the time the players need to adjust their posture. The energy loss
will cause the decrease of the “vertical air” and fewer twists. Therefore, the width should be short
and meet the requirement of adjusting the player’s posture at the same time.
Apparently, assume that time t is needed for the players to adjust their posture and the angle
between the velocity V and the lengthway is  .
f friction is the friction which is backward to the
velocity V. Figure 6 is as below,
Figure 6: The snowboard course
The friction
f friction   mg sin 
(16)
Acceleration of gravity component is
a   g sin 
So the size of the flat-bottom is
s  vx t 
1
 g sin  t 2
2
(17)
Strengths and weaknesses
5.1Strengths
1. A useful model is set to get the “vertical air” maximized.
2. The model can be applied in many ways, such as other different kinds of slide.
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5.2Weaknesses
1. Many resistances haven’t be considered, such as push-snow resistance.
Conclusions
We make a model to solve the maximum “Vertical Air”, and discuss the shape of the
snowboard course that determines the twist of the people. Our model could be used in many ways,
and we also admit that the model contains a lot of weaknesses.
Reference
[1] BaoHeng Wang. Thought for vertical air and sliding in the snowboard course. The newspaper
of ShenYang Institute ,Vol24, No.2
[2] Zhimin Bai, Hailiang Yang. The research of the factors that influence the friction coefficient.
The discovery and the research in the laboratory, Vol.25,No.11
[3]Hongguang Yan, PingLiu, Feng Guo. The analysis of the speed of leaving the snowboard
course. The newspaper of ShenYang Institute. Vol 28, No.3
[4]Naipeng Yan . The main factors that influence the sliding resistance when skiing. China
Winter Sports. No.5
[5] http://en.wikipedia.org/wiki/Halfpipe