Gen. Math. Notes, Vol. 4, No. 2, June 2011, pp. 71-86
ISSN 2219-7184; Copyright © ICSRS Publication, 2011
www.i-csrs.org
Available free online at http://www.geman.in
The Application of the Variational Method
to Optimize a Snowboard Course
Feroz shah Syed1, Xue-Yuan Zhang2, Song Huang3 and Xiao Sun4
1
Department of Mathematics, Beijing Institute of Technology,
Beijing 100081, P.R. China
E-mail: sf_shah@hotmail.com
2
School of Information and Electronics, Beijing Institute of Technology,
Beijing 100081, P.R.China
E-mail: 743546490@qq.com
3
Department of Mathematics, Beijing Institute of Technology,
Beijing 100081, P.R. China
E-mail: william61124@sina.com
4
School of Automation, Beijing Institute of Technology,
Beijing 100081, P.R.China
E-mail: sunxiaoseraph@yahoo.cn
(Received: 16-5-11/ Accepted: 29-6-11)
Abstract
In this paper, by using the variational method, we obtain the optimal design on
the “halfpipe” shape for a skilled player to maximize the production of “vertical
air”. The main idea is that we make the halfpipe shape such that the player can
get maximal kinetic energy. Neglecting the air resistance, friction drag, and so on,
we formulate the variation for the steepest descent curve so that the player gets a
maximum velocity at the bottom of the halfpipe. Then, we consider the bottom
plane of the halfpipe to continue optimization. Having the aid of the obtained
maximum velocity together with the optimizations in the bottom plane, the player
can obtain more kinetic energy to jump up and twist as higher and better as
possible. We also make some numerical simulation to verify the theoretical
Feroz shah Syed et al.
72
analysis. Finally, according to above analysis, we design a better halfpipe for
players to use in practice.
Our conclusion is that, based on our design with the standard sizes, the maximum
vertical air and the time for twist in the air can exceed the known best results
obtained by the world class players.
Keywords: halfpipe, design, variational method, vertical air twist,
optimization.
1
Analysis of the Problem
Snowboard skating is a kind of popular sports in the world nowadays[1,2,3].
Furthermore, since it has become a formal competition of winter Olympics, it
attracts more and more attention. Building a better course to enable players to
perform their ability sufficiently is becoming more and more important.
For this purpose, we need to search for the steepest descent curve which the
player is recommended to slide along showed as Fig. 1.
Fig. 1 the “halfpipe” course
In Fig. 1, we denote the depth of the halfpipe by H , the width of the curved part
of the course in horizontal direction by D and the length of the halfpipe by L .
The incining angle of the whole course is α . Considering the maximal vertical air,
the player is supposed to obtain the maximal kinetic energy when he reaches the
bottom of the course and is about to enter the flat area. Since the kinetic energy
1
ds
is mv 2 , and v = , where ds denotes the arc length and m is the mass of the
2
dt
player (here we assume it is always fixed). Our purpose is to make the shortest
time cost in reaching the bottom so to have the maximum velocity. Thus, the
curved part of the course should be designed by using the nets of the steepest
descent curves ( x, y ( x ), z ( x )) in coordinate.
73
The Application of the Variational Method…
Assume that the width of the flat zone is d , the length of a part of “vertical air” is
y0 . As a result, when a player slides from the edge to the bottom, x changes from
D−d
x = 0 to x1 =
, y changes from y = 0 to y1 = − H + y 0 − tan αz1 and
2
z changes from z = 0 to z = z1 . The values of z depends on the total length along
1
the course. For instance, if the player jumps six times, then z1 = ( L − L0 ) / 6 ,
2
where L0 is considered to be the buffer distance when the player ends the motion.
2
Mathematical Model and the Solutions for the
Jumping Down [7]
Fig. 2 the steepest descent curve
Assume that the initial velocity is v0 , according to the law of conservation in
kinetic energy and potential energy, we have
1 2
1 ds
mv0 − mgy = m( ) 2
2
2 dt
and from which we get
ds = −2 gy + v02 dt
Now, the time cost along the steepest descent curve is
T = ∫ dt = ∫
ds
−2 gy + v02
=∫
x1
0
1 + ( y '( x)) 2 + ( z '( x)) 2
dx ≡ T ( y , z )
−2 gy ( x) + v02
(1)
Notice the fact that v0 = 0 implies the player first starts to go down. Set the
74
Feroz shah Syed et al.
function
1 + ( y '( x))2 + ( z '( x))2
F ( y, y ', z ') =
−2 gy ( x) + v02
(2)
For any smooth function v satisfying v(0) = 0, v( x1 ) = 0 , the time along any
admissible curve is
T ( y + tv, z + τ v ) ，where t , τ are any parameters.
Denoting φ (t ,τ ) = T ( y + tv, z + τ v ) − T ( y , z ) , since T ( y , z ) is the time cost along
the steepest descent curve, then we have
T ( y + tv , z + τ v ) ≥ T ( y , z ) ，and so φ (t ,τ ) ≥ 0, ∀ t ,τ namely, φ (0,0) = 0 achieves
the minimum of φ (t ,τ ) .
According to the extremum principle, we have
∂φ
∂φ
(0,0) =
(0,0) = 0
∂t
∂τ
It is easy to get
x1
∂φ
(0,0) = ∫ ( Fy ( y , y ' , z ' )v + Fy ' ( y , y ' , z ' )v ' ) dx = 0
0
∂t
(3)
and
x1
∂φ
(4)
(0,0) = ∫ Fz ' ( y , y ' , z ' )v 'dx = 0
0
∂τ
Notice that v(0) = v( x1 ) = 0 , using the integration by parts, from (3) and (4) we get
∫
x1
( Fy ( y, y' , z' ) −
0
dFy ' ( y, y' , z' )
dx
)vdx = 0, ∀ v
(5)
and
−∫
x1
0
d
Fz ' ( y, y ' , z ' )vdx = 0, ∀ v
dx
(6)
From the fact, that v is arbitrary, by the variational principle, from (5) and (6) we
get
Fy ( y, y ' , z ' ) −
dFy ' ( y , y ' , z ' )
and
d
Fz ' ( y, y' , z' ) = 0
dx
dx
=0
(7)
(8)
75
The Application of the Variational Method…
By the expression of F ( y, y ' , z ' ) in (2), we can get from (7)
1 + ( y ')2 + ( z ')2
( y ')2
−
= C1
−2 gy + v02
(−2 gy + v02 )(1 + ( y ')2 + ( z ')2 )
or
1 + ( z ') 2
= C1
(−2 gy + v02 )(1 + ( y ')2 + ( z ') 2 )
From (8) we get
z'
(−2 gy + v02 )(1 + ( y ') 2 + ( z ') 2 )
= C2
(9)
(10)
Note that C1 ≠ 0 , from (9) and (10) we get
z'
= k0
1 + ( z' ) 2
Solving this equation we get
z ' = k1 , z = k1 x + k 2
Since z (0) = 0, z ( x1 ) = z1 , we get k 2 = 0 and
z = k1 x, k1 = z1 / x1
Substituting ‘z’ into (9), we get
( −2 gy + v02 )(1 + ( y ') 2 + ( k1 ) 2 ) = C3
(11)
By setting y ' = 1 + k12 cot t , from above equation we get
(1 + (k1 ) 2 )(1 + cot 2 t )( −2 gy + v02 ) = C
namely,
y=−
v02
C
2
sin
t
+
2 g (1 + ( k1 ) 2 )
2g
Now, we attempt to get x = x(t ) . In fact,
x = ∫ dx = ∫
dy
2C sin 2 t
2C
t 1
=∫
dt =
( − sin 2t ) + C 4
2 3/ 2
2 3/ 2
y'
2 4
2 g (1 + ( k1 ) )
2 g (1 + ( k1 ) )
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Feroz shah Syed et al.
Notice that when x = 0 , y = 0 so that C4 = 0 . Therefore, we obtain the steepest
descent curve:
C
(12)
x=
(2t − sin 2t )
4 g (1 + (k1 ) 2 ) 3 / 2
v02
v02
C
C
2
sin
t
+
=
(cos
2
t
−
1)
+
2 g (1 + (k1 ) 2 )
2 g 4 g (1 + (k1 ) 2 )
2g
z
z = k1 x, k1 = 1
(14)
x1
Denote
C
(15)
R=
, θ = 2t
4 g (1 + (k1 ) 2 )
y=−
(13)
Therefore, the above curve can be rewritten as
x=
R
1 + (k1 ) 2
(θ − sin θ )
y = R(cos θ − 1) +
v02
2g
z1
x1
z = k1 x, k1 =
(16)
(17)
(18)
where the variable θ and the constant R can be determined by the initial point
coordinate ( x0 , y0 , z0 ) = (0, 0, 0) and the end point coordinate ( x1 , y1 , z1 ) .
When the initial velocity v0 = 0 , we obtain the steepest descent curve:
x=
R
1 + (k1 ) 2
(θ − sin θ )
(19)
y = R (cos θ − 1)
(20)
z1
x1
(21)
z = k1 x, k1 =
As a result, when the initial velocity v0 ≠ 0 , curves in x direction and z direction do
v
not change, while y curve has a positive direction displacement − 0 g (pointing
2
to the ground). Physically, this means the bottom of the snowboard course is
inclining downwards. This fits exactly to the existing design of the snowboard
77
The Application of the Variational Method…
course. The declining angle is given by
v2
tan α = 0
2 gz1
(22)
When v0 = 0, z1 = 0 , we get the steepest descent curve on a plane:
x = R (θ − sin θ )
y = R (cos θ − 1)
(23)
(24)
Now we design the course according to the data measured.
First of all, as it is depicted, we might as well take the depth of the course H = 6 m
and the width of the course D = 6 m according to the standard condition of
international snowboard halfpipe competitions [2, 3].
By observing the athletes in the videos [1], we know that the coordinates of the
falling point on z axis is 10 meters. So the coordinates of the falling point B is (6,6,10).
Consider the coordinates of falling point as the initial condition and solve the
steepest descent curve function with it to obtain the fomula below:

1 34
1
(t − sin 2t )
 x= 2
C1 9
2


34 sin 2 t
y=−
, C1 = 0.688, t ∈ [0,1.5]

9 C12


5 34 1
1
z =
(t − sin 2t )
2
3 9 C1
2

(25)
or
 x = 4.016t − 2.053sin 2t

y = −7.980sin 2 t
, t ∈ [0,1.5]

 z = 6.843t − 3.422sin 2t

(26)
78
Feroz shah Syed et al.
Depict them in Matlab：
Fig. 3 the steepest descent curve observed in different angle
We gained the steepest descent curve and create the curvature using parallel
translation along the z axis by Matlab.
Fig. 4 the 3D figure of the curved part
79
The Application of the Variational Method…
3
The Consideration of the Bottom
In this section, we consider the minimum width of the bottom of the course and
we conclude that the velocity almost does not change while the players glide on it.
Denote the angle between the snowboard and the ground of the course by θ , β be
the angle between the snowboard and the horizontal line and α is the inclining
angle of the whole course respectively.
Snowboard
θ
Ground
Fig. 5 angle between the snowboard and the ground
F ' denotes the projection of the force acted on the horizontal direction and ‘ a ' ’
indicates the acceleration created by F ' . We denote the time cost by t when the
player is on the flat area of the course.
Fig. 6 the analysis of the stress
In order to obtain the state of the playing on the flat bottom, we need the analysis
of the stress. Due to the fact, that sliding is on the snow ground, we consider the
following hypothesis:
1.
2.
3.
Neglect the frictional force on the snowboard caused by the ground, only
consider the support force.
Neglect the air resistance.
The player is well skilled.
Fig. 1 shows the bearing force that the snowboard exerted, while the player is
sliding on the plane. The snowboard is not always parallel to plane. In order to
change the direction of the motion of the player, the angle between the snowboard
with ground and the angle between the snowboard with the horizontal direction
needs to be adjusted by the player. Therefore, the supporting force that the
snowboard actually exerted is not vertical to the incined plane upwards, as shown
80
Feroz shah Syed et al.
F in Fig. 2.
In Fig. 3, we give the orthogonal decomposition of F ' which is the component of
F along inclined plane and F '' is vertical to the inclined plane. F '' does not affect
the motion on the ground and only F ' does.
v0
d
β
Fy’
F’
Fx’
mgsinα
vt x
Area D：flat
area
vt y
vt
Fig. 7 analysis of motion on the flat area
According to the practical experiment, the results can be concluded as below:
1.
2.
when θ = 0 , ∀ β satisfies F ' = 0 .
π
when β = 0 , θ = , we have F ' = mg sin α .
2
Due to the negligence of the fluid feature of the snow, F is only related to the
mass of the player, θ , β and α .
1.
2.
3.
F ' is directly propotional to the the mass of the player.
F ' is correlated positively with θ and α and correleted oppositely with
β.
when θ = 0 , there is no force acted on the snowboard.
when θ =
4.
π
, the player is in the state of braking.
2
when β = 0 , the player has the most tendency to slide sideways, when
β=
π
2
, the player has the most tendency to slide directly along the course.
Therefore, in order to make calculation easier, we get exprimental formula below.
As the purpose of the design is to enable player to perform best by increasing the
“vertical air”, so we observed and measured the parameters to obtain the average
datas of skilled athelets as below: θ = 45 , β = 20 .
In this paper, we suppose that α = 1 7 because the inclining angle α is
usually between 16 and 18 [4, 5],
81
The Application of the Variational Method…
Formula derivation
Solving the parameter t and l , we have
a ' = g sin θ cos β sin α
1
d = v0t cos β + a ' t 2 cos β
2
l = v0 t sin β + ( g sin α − a ' cos β )t
(27)
(28)
(29)
2
From above equations, we get
−2v0 cos β + 2 v02 cos β + dg sin θ sin 2β sin α
t=
g sin θ sin 2β sin α
l = v0t sin β +
(30)
1
g sin α (1 − a sin θ cos 2 β )t 2
2
(31)
d
D
l
Fig.8 the track of the motion on the flat area
By calculation, we can get the fly out velocity vt as
vty = v0 sin β + ( g sin α − a 'cos β )t
(32)
vtx = v0 cos β + a ' t sin β
vt = vtx2 + vty2
(33)
(34)
From the above analysis, we have come to know that when α = 17 , β > 20 , the
magnitude and the direction of vt seems not to be changed, therefore, we consider
that the direction is not changed [10,13, 14].
Hence, we see that:
1.
Passing the plane does not affect the velocity almost. Hence, in order to
save the expenses , the width of the plane should be as narrower as we
82
Feroz shah Syed et al.
can.(in ideal case)
In fact, the flat area plays an important role for the players to keep their
balance and to prepare for the next jump. For a person, the total time cost
in adjusting balance and controlling the direction for the next jumping is at
least 0.2s . Therefore, the minimal width of the plane is d min = 2v0 x t ≈ 5m .
2.
4 Mathematical Model and the Solutions for the Jumping
up
Here, we study the process of jumping up. Our purpose is to find the maximum
height of the jumping of the player.
Let v0 denote the initial velocity of the parabola, and γ denote the angle as the Fig.
3. Notice that parabola passes through the origin (0, 0) , we can get the parabola
g
equation as y = − 2
x 2 + x tan γ and the edge line of the curved part of the
2v0 cos 2 γ
halfpipe can be given by y = − x tan γ
Further, we can get the coordinates of point A as
2v02 cos 2 γ
−2v02 cos 2 γ
A(
(tan α + tan γ ),
tan α (tan α + tan γ ))
g
g
To find the maximal height of the jumping (vertical air), we suppose the
coordinates of that point is ( x0 , y0 ) . Then we have
y0 = −
g
x02 + x0 tan γ
2
2v cos γ
2
0
Therefore, we can see that the distance from the highest point to the edge (denoted
by d ) is expressed as
y + tan α x
v02
d=
=
(sin γ + tan γ ) 2
2
1 + tan α 2 g
(35)
Assume that the velocity at the end of the flat area is vm . By orthogonal
decomposition of v0 as shown in Fig. 9, we observe that


v m2 y + 2 g ∆ h ,
 v0x =

 v0 y = vmx ,

vmy
 ta n γ = v 0 y =
2

v0x
vmy + 2 g ∆ h

(36)
83
The Application of the Variational Method…
where ∆h denotes the difference of the vertical height between the two points, that
is; the starting point of the curved part of the course at entrance and the exiting out
of the ending point of sliding on the course.
vmx , vmy are the components of orthogonal decomposition of vm on the plane, as
shown in Fig. 9. Substituting above γ into the expression of d , we obtain
d=
vtx 2 + vty2 tan 2 α
2g
+
vty tan α
g
vty 2 + 2 g ∆h + tan 2 α∆h
(37)
Fig. 9 analysis of the track in the air
Since vmx , vmy and α are constant, only ∆h changes and ∆h ≤ 0 . Therefore,
When ∆h = 0 , we have d max =
(vmx + vmy tan α )2
2g
which is the maximal height of
the jumping that the player can enjoy.
Here, we consider the time of the player’s twist [6] (twisting time of the player in
the air).
From above calculation, we have known that β =20 , v0 = 12.976m / s . Thus, we
have
vtx = v0 cos β = 11.760 m / s , vty = v0 sin β = 5.484 m / s , when taking ∆h = 0 , we get
d max =
(vtx + vty tan α ) 2
2g
= 9.356m and we get the coordinates of the point C as
2v0 cos 2 γ
2v 2 cos 2 γ tan γ
( t anα +t anγ ) , - 0
( t anα +t anγ ) ) . Denoting by t1 , the
g
g
passing time from O to O ' , and t2 the passing time from O ' to C . We
2v
get t1 = 0 y ≈ 2.4175 . Now, we calculate the approximation of t2 . Since the
g
coordinates of point C can be expressed as:
（6.148t anα +13. 157, - 6. 148t an2α - 12. 157t anα）
Since α is usually between 14o and 18o, by taking α =18o , we get the coordinates
(
84
Feroz shah Syed et al.
of point C（15.154，
-4.924）, therefore
OC = 5.314m, vc = v02 + 2 g (12.297(tan α + 2.14)) ≈ 24.329m / s.
and t = t1 + t2 = 2.685s. We determine α in a way that t =t 1 + t2 is maximal. From
O ' C = 6.148 tan 2 α + (tan 2 α + 2.14 tan α ) 2 and
241.021tan α + 684.162 + 12.976
2
2v0 y 2v0 .cos α
O'A
we get t2 ≈
and t1 =
=
and so
vo ' c
g
g
vo ' c =
t = t1 + t2 = 2.653cos α +
12.296 tan 2 α + (tan 2 α + 2.14 tan α ) 2
241.021tan α + 684.162 + 12.976
(38)
It is easy to see that t1 ≥ 10t2 . Therefore, as long as the cos α reached the
maximum, i.e., α reached the minimum, then we can obtain the maximal t [11,
12]. In the game, the player may fall several times, as a result, the velocity
sometimes becomes zero. To have a smooth gliding, the angle α must satisfy
some conditions assuring that the player can resume and continue the next jump
within the range of 30m . As shown in Fig.10, taking l = 30m, h = l sin α ,
mgh ≥
v2
1 2
, hence, we get α ≥ 16.7 .
mv0 implies si nα ≥
2
2 gl 0.2874
Fig. 10 the explanation of h,l and α
Altogether, when the angle is α = 16.7 , the longest jumping time is 2.794s .
5
Conclusion
In our studies, we have designed a model of the snowboard course (halfpipe) by
using the variational method. We claim that this well-designed model can provide
85
The Application of the Variational Method…
the gliders the maximum jumping height in the air and disturbance in the motion
of the glider on the flat plane of the course will be almost negligible. In order to
obtain the state of the playing on the flat bottom, we need the analysis of the
stress.
Considering the fact, that sliding is on the snow ground, in order to change the
direction of the motion of the player, the angle between the snowboard with
ground and the angle between the snowboard with the horizontal direction is
adjusted. To reduce the cost and expenses, the width of the plane is designed as
narrower as we can.(in ideal case).
In fact, the flat area has much importance for the players to keep their balance and
to prepare for the next jump. In our results, for a person, the total time cost in
adjusting balance and controlling the direction for the next jumping is at least .
Therefore, the minimal width of the plane is 5m. So far for the simplity, we only
considered the situation of ignoring air resistance, friction drag, and so on.
Whereas in reality ,we should have taken them into account. And if we had
enough time, we could have investgated more completed situation.
We gained the trail of the 3D figures below:
Fig. 11 3D figure of the whole course
Acknowledgment
This subject is supported partially by the National Natural Science Foundation of
China (Grant No. 10871218)
Feroz shah Syed et al.
86
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