Mathematica(l) Roller Coasters

Transcription

1 Mathematica(l) Roller Coasters Selwyn Hollis Department of Mathematics Armstrong Atlantic State University Savannah, GA à Introdction Mathematica animations that simlate roller coasters provide an interesting and fn way of ptting basic ideas from mltivariable calcls and elementary physics to work. In this article we will describe how to create monorail roller coaster animations in which the track is a srface in Ñ 3 in the form of a narrow strip. The space crve comprising the center rail of the track may be specified either as a closed coordinate crve on a given parametric srface or as a closed space crve. In the first case, the orientation of the vehicle makes se of partial derivatives. In the second case, the orientation of the vehicle makes se of the crve s classical TNB frame, which consists of the tangent, normal, and binormal vectors. à The Physics The motion of or simlated roller coaster will be made to appear realistic by or se of the principle of conservation of energy. We will assme a frictionless track and that pward is the positive vertical direction. So at all times we have DH KE + PE L =, where KE = mv is kinetic energy, and PE = mg z is potential energy. Here m is mass, g is the gravitational acceleration constant, z is vertical displacement from any reference point, and v is speed. Since there is no force and conseqently no change in potential energy in any horizontal plane, we actally have KE = mv z + c where v z is the vertical component of the velocity and c is a constant. So from the conservation principle above, it follows easily that D Hv z L µ Dz. Sppose that at the highest point on the roller coaster s track we have v z = v top and z = z top. Then at any point on the track, v z - v top = kh z top - zl,

2 where k >. The reslt of all this is that if the position of or roller coaster as a fnction of time t is given by RHtL = HxHtL, y HtL, zhtll, then x HtL + y HtL is constant, and z HtL = v z = khz top - zhtll + v top. à Determining Position The center rail of the track will be a closed space crve with a given parametrization rhl = Hr 1 HL, r HL, r 3 HLL. This parameterization will only describe the path of the vehicle, not its position as a fnction of time. Since an animation will consist of a seqence of frames that show the roller coaster at eqally-spaced times, Dt, Dt,3Dt,, we need to determine the position of the vehicle at each of these times. This will be done by compting a corresponding seqence of vales of the parameter, which will entail recompting the parameter step D for each sccessive frame. The method for compting each sccessive D is developed as follows. Sppose that the actal position of the vehicle at time t is given by RHtL = HxHtL, y HtL, zhtll. Then, thinking of as a fnction of t, we have from which follows RHtL = rhhtll, R HtL = âr â â ât by the Chain Rle. We assme that â ât >, so it follows that â ât = R HtL ± âr â µ. Now, since x HtL + y HtL is constant and z HtL = v z = khz top - zhtll + v top, this becomes â ât = "################################ khz top - zhtll + c ± âr â µ, where c is a constant. Ths, the variable parameter step corresponding to the fixed time step D t can be estimated as D» "################ khz #################### top -r 3 H LL + c r H L which can be compted withot knowledge of RHtL. Since D t is fixed, we will introdce new constants k Ž and and rewrite this simply as D» k Ž "################ z ############### top - r 3 H L + r H L D t,.

3 The apparent speed of the vehicle in or animations will be controlled by adjsting the constants k Ž and. à The Vehicle and Its Orientation The vehicle will be represented by a flexible tbe connected to the center rail of the track. Its crosssections orthogonal to the track will be circles that intersect the track s center rail. At each point on the center rail there is a plane that is orthogonal to the path at that point and contains a circlar crosssection of the tbe. In that plane, imagine two coordinate axes, a horizontal axis that is parallel to the track srface and a vertical axis that is orthogonal to the track. These axes are indeed horizontal and vertical with respect to the orientation of the vehicle. So in order to keep the vehicle oriented properly at each location along the track, we need to compte a pair of orthogonal vectors that point in the directions of the axes jst described. These vectors are: (i) NHL, a nit vector that is orthogonal to the center rail and parallel to the track srface; (ii) BHL, a nit vector that is orthogonal to both the center rail and NHL. With these vectors defined, we can now describe the srface we will se to represent the vehicle at any point rhl on the track. Its parametrization is FHs, ql = rhsl +rbhsl H1 + cos ql +rsin q NHsL, - { s + {, q p. where r is the radis of the vehicle s circlar cross-sections, and { is the vehicle s length. à A Track Given by a Closed Coordinate Crve on a Parametric Srface A nice example is provided by a Möbis strip, an example of which is parametrized and plotted as follows. q@_, v_d := 8H - v Sin@ DL Cos@D, v Cos@ D, H - v Sin@ DL Sin@D<;

4 track = ParametricPlot3D@Evalate@q@, vdd, 8,, p<, 8v, -.,.<, PlotPoints -> 86, 3<, Boxed -> False, Axes -> FalseD; Ÿ The Vector Comptations The track s center rail is the coordinate crve parametrized by r@_d = q@, D 8 Cos@D,, Sin@D< (By coordinate crve we mean a crve with parametrization given by setting one of the coordinates of the srface parametrization eqal to a constant.) The corresponding nit tangent vector is nittan@_d = r Simplify 8-Sin@D,, Cos@D< A vector that is parallel to the track srface and not parallel to the crve is given by q v v =. vec@_d = v q@, vd.v Simplify 9-Cos@D SinA E, CosA E, -SinA E Sin@D= evalated at One step of the Gram-Schmidt process prodces a vector that is parallel to the track srface and orthogonal to the crve: nrml@_d = vec@d - vec@d.nittan@d nittan@d 9-Cos@D SinA E, CosA E, -SinA E Sin@D=

5 The corresponding nit vector is = nrml@d.nrml@d TrigExpand Simplify SinA E, CosA E, -SinA E Sin@D= This is the vector denoted above by NHL. (Note that in this example, the last two steps prodced no change. These steps are needed in general, however.) Finally, the cross prodct THL NHL prodces the vector denoted above by BHL: binrml@_d = nittan@d nitnrml@d Simplify 9-CosA E Cos@D, -SinA E, -CosA E Sin@D= Ÿ The Vehicle The following prodces the vehicle at any point along the crve. Below that is a sample plot sing = 1.5. tbe@_d := ParametricPlot3D@Evalate@ r@sd +.15 HH1 + Cos@vDL binrml@sd + Sin@vD nitnrml@sdld, 8s, -., +.<, 8v,, p<, PlotPoints -> 85, 9<, DisplayFnction -> IdentityD; Show@tbe@1.5D, DisplayFnction -> $DisplayFnction D ; Ÿ The Animation Now we are ready to compte the frames for the animation. The first line below defines the fnction that comptes the parameter step D. The nmbers. and.67 there were chosen experimentally. Forty-eight frames will be prodced, the first of which is shown below.

6 ztop = ;!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ztop - r@d@@3dd +. Step@_D =.67!!!!!!!!!!!!!!!!!!!!!!!!!! = ; While@ 4 p, = + Step@D; Show@tbe@D, track, Axes -> False, PlotRange -> 88-.5,.5<, 8-1.1, 1.1<, 8-.5,.5<<, Boxed -> False, DisplayFnction -> $DisplayFnction DD; Clear@D à A Track Given by a Closed Space Crve An interesting track is obtained from the crve parametrized by r@_d := 8-H + Cos@DL Sin@ D, H + Sin@DL Cos@ D, Cos@D< ParametricPlot3D@Evalate@r@DD, 8,, p<, ViewPoint 83, -, 1<D;

7 Ÿ The Vector Comptations In this sitation the vectors of interest comprise the classical TNB frame at each point along the crve. Becase even fairly simple crves give rise to very large and complicated symbolic expressions for the necessary vectors, we will not calclate them explicitly. The tangent vector of interest here is = TrigExpand Simplify Cos@D- 3 Cos@3D, - Cos@D + 3 Cos@3D-4 Sin@D, - Sin@D= The corresponding nit tangent vector THL is nittan@_d := ModleA8v = tanv@d<, v!!!!!!!! E v.v We will approximate the derivative of this nit tangent vector by a central difference approximation: dt@_d := HnitTan@ +.5D - nittan@ -.5DL.1 Now, the principal nit normal vector NHL is nitnrml@_d := ModleA8n = dt@d<, n!!!!!!!! E n.n Finally the cross prodct THL NHL prodces the binormal vector BH L: binrml@_d := nittan@d nitnrml@d Ÿ The Track and the Vehicle To prodce the track from the crve, we se the vector NHL to create a srface parametrized by GH, vl = rhl + vnhl, p, - w v w, where w is the width of the track.

8 track = ParametricPlot3D@r@D + v nitnrml@d, 8,, p<, 8v, -.,.<, ViewPoint 83, -, 1<, PlotPoints -> 81, 3<D; The following prodces the vehicle at any point along the crve. Below that is a sample plot sing =. (Note the absence of the Evalate fnction inside of ParametricPlot3D here. This is becase of the way the nit normal and binormal vectors were defined above with delayed evalation.) tbe@_d := ParametricPlot3D@ r@sd +. binrml@sd H1 + Cos@qDL +. Sin@qD nitnrml@sd, 8s, -.15, +.15<, 8q,, p<, PlotPoints -> 85, 9<, DisplayFnction -> IdentityD; Show@tbe@D, DisplayFnction $DisplayFnctionD; Ÿ The Animation This animation is created in essentially the same way as the previos one. Forty frames will be prodced with the particlar parameter vales sed. The GraphicsArray that follows contains frames 1, 14,, and.

9 ztop = ;!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ztop - r@d@@3dd +. Step@_D =!!!!!!!!!!!!!!!!!!!!!!!!! = ; While@ p, = + Step@D; Show@tbe@D, track, ViewPoint 83, -, 1<, Axes -> False, PlotRange -> , 3<, 8-3.3, 3.3<, 8-.1,.4<<, DisplayFnction -> $DisplayFnction DD; Clear@D