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 ) ) 76 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 References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] Snowboard Resort Information Sheet, http://wenku.baidu.com/view/ e78ef328647d27284b735100.html. J.W. Harding, C.G. Mackintosh, D.T. Martin, A.G. Hahn and D.A. James, Automated scoring for elite half-pipe snowboard competition: important sporting development ortechno distraction? High Efficiency Silicon Solar Cells (Automated Scoring for Snowboard Competitions), DOI: 10.1002/jst.69. J.W. Harding, J.W. Small, D.A. James, Feature extraction of performance variables in elite half-pipe snowboarding using body mounted inertial sensors, BioMEMS and Nanotechnology III, Proceedings of SPIE, Canberra, ACT Australia, Bellingham, WA: SPIE, (2007). X.J.G. Ning-ning, Research of the Technical Characteristics of Half-pipe Snowboarding, China Academic Journal Electronic Publishing House, (2009). Yan-qiu Zhang, A comparative study and development countermeasures on Half-pipe snowboarding skill in China, Journal of Beijing Sport University, 31(6) (2008). H. Yan, P. Liu and F. Guo, Factors influencing velocity away from decks in snowboard Half-pipe, Journal of Shenyang Sport University, 28(3) (2009). C. Chuan-miao, The Introduction of Scientific Computation, Science Press, Beijing, 2007. Hui-qiu Zhang, Shi-ping Guo and Bao-heng Wang, China Winter Sports, 3(6) (2008). Chang-you Li and Yan-long Li, Journal of arb Institute of Physical Education, 26(5) (2008). Bo Mi and Hong-wei Sun, China Winter Sports, 30(2) (2008). Li Fu, Journal of Hebei University, 19(4) (1999). Bao-heng Wang, Journal of Shenyang Physical Education Institute, 24(2) (2005). Shou-qiu Wang, Fu-bin He, Chang-fu Liu, China Winter Sports, 31(3) (2009). Yun-song Tang, Lin-bin An and Peng-cheng Wang, China Winter Sports, 3(2005).