Designing a Fair Home Run Stadium BRIEF ABSTRACT For a given playing field outline how high must the outfield fence be at each point in order to make a homerun equally likely in all fair directions? GENERAL INFORMATION FileName: STADIUM Full title: That Ball Is Going, Going, Gone! Designing a Fair HomeRun Stadium Developer: Brian J. Winkel, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Contact: Brian J. Winkel, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8412. Email: winkel@rose-hulman.edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. STATEMENT OF PROBLEM Consider a baseball field with prescribed irregular outfield and fence configuration. Suppose a player can hit a homerun which just clears the fence at one given point if the ball leaving the bat (2 feet off the ground) has initial velocity v0 and launch angle at 45 degrees. Assume no wind resistance. Redesign the fences so that this player (and hence any player) has the same chance of hitting a homerun to anywhere in the outfield (i.e., in the outfield and fair!) We consider Hypothetical Stadium (HS) with the following outfield wall configuration. Home plate is at (0, 0); left field foul pole is straight up the positive y-axis; right field foul pole is straight down the positive x-axis. Left field outfield fence runs parallel to first base line for 135 ft at a distance of 360 ft from homeplate. Right field outfield fence is a straight line from right field foul line at (409, 0) to the point (300,252). Center field outfield fence is a circular wall going through deepest center field point out at (200 Sqrt[2], 200 Sqrt[2]), point LC = (135,360) in left center, and point RC = (300,252) in right center. (a) Write equations which completely describe the outfield fences and plot them with homeplate at (0, 0). (b) Determine the farthest point of the outfield fence from home plate, measured along the ground to the base of the fence. (c) Place a 10 ft high fence at this farthest point found in (b) and determine the initial velocity a hitter must impart to a ball hit 2 ft high over homeplate to get it to just clear the 10 ft high fence here. We shall assume there is no air resistance. Hint: It is known (and you should be able to prove it is true!) that a hitter should always try to launch the ball from the bat at a 45 degree angle with the horizontal for maximum range. (d) Now determine for every other point at the base of the outfield fence just how high the fence must be so that this player can hit a homerun by launching a ball from his bat 2 feet above homeplate with exactly the same initial velocity at a 45 degree angle with the horizontal. (e) Discuss the reasonableness of your answers. How would resistance alter (if at all) the shape of your wall? KEYWORDS Projectile motion (without resistance), parametric equations. TEACHER NOTES ISSUES RELATED TO THE PROBLEM Prerequisites Students need to know how to describe two projectile motion in parametric or rectangular form and use this in context. Time allotment - time management This is a big project and could be set up in one class hour period with several visits to the problem over a few day (or weeks) project period. Expectations Future payoffs Extensions Do the same problem with resistance. This would require a great deal of numerical work in the solving of differential equation model for two dimensional projectile motion as a system of such differential equations.. References and Sources POSSIBLE SOLUTION(S) (a) We determine the equations which describe the location of the outfield fences. We first go after the circular centerfield section which must go through three points: (135, 360), (200 Sqrt[2], 200 Sqrt[2]), and (300, 252). First task is to determine the equation of the center field wall as a portion of a circle of the form (x-h)^2 + (y-k)^2 = r^2 through these three points. We enter the general equation of the circle with center (h, k) and radius r. eq = (x-h)^2 + (y-k)^2 == r^2; We stipulate that the circle go through the point (135, 360). eq1 = eq/.{x->135,y->360}; We stipulate that the circle go through the point (300,252). eq2 = eq/.{x->300,y->252}; We stipulate that the circle go through the point (200 Sqrt[2], 200 Sqrt[2]). eq3 = eq/.{x->200 Sqrt[2],y->200 Sqrt[2]}; We solve for h, k, and r. sol = Solve[{eq1,eq2,eq3},{h,k,r}]//N {{r -> -179.337, h -> 135.462, k -> 180.664}, {r -> 179.337, h -> 135.462, k -> 180.664}} We determine y as a function of x, i.e. cf(x) for the centerfield fence position. Since we are in the first quadrant we are interested in only one branch of the circle, namely x> 0 and y > 0. soly = Solve[eq,y]; Finally, we have the equation of the centerfield fence. cf[x_] = y/.soly[[2]]/.sol[[2]]//N 0.5 (361.327 + Sqrt[130557. - 2 4. (18827.4 - 270.923 x + x )]) Now to the left field fence: lf[x_] = 360; And the right field fence: rf[x_] = y/.Solve[(y-252)/(x-300) == (0-252)/(409-300), y][[1]] -252 (-409 + x) --------------- 109 Now we plot the outfield fence(s): leftField = Plot[lf[x],{x,0,135}] rightField = Plot[rf[x],{x,300,409}] centerField = Plot[cf[x],{x,135,300},AspectRatio->Automatic, PlotRange->{{0,400},{0,400}}] Now we examine the ball park's outfield fences. Show[leftField,centerField,rightField,AspectRatio->Automatic, PlotRange->{{0,450},{0,450}}] (b) Now to determine the farthest point we need to see if it is in centerfield or right field, as it obviously is not in left field. We consider the distance function to the center field fence, plot it, and determine where its derivative is zero and hence where the distance is greatest as well as its greatest distance value. dcf[x_] = Sqrt[x^2 + cf[x]^2]; Plot[dcf[x],{x,135,300}] maxcf = x/.FindRoot[dcf'[x]==0,{x,250}][[1]] 243.045 dcf[maxcf] 405.145 Thus it would appear that the farthest point is NOT centerfield fence but at the left field foul pole - 409 ft. (c) We place a 10 ft high fence at this farthest point found in (b) i.e. the point along the left field foul line: (409, 0) and determine the initial velocity a hitter must impart to a ball hit 2 ft high over homeplate to get it to just clear the 10 ft high fence here. We shall assume there is no air resistance. We enter the x and y coordinats as functions of time by the usual projectile motion equations for an initial velocity of v0 and a launch angle of Pi/.4 radians or 45 degrees. x[t_] = v0 Sin[Pi/4] t; y[t_] = -1/2 32 t^2 +v0 Sin[Pi/4] t + 2; z[x_] = y[t]/.Solve[x[t] == x,t][[1]] 2 32 x 2 + x - ----- 2 v0 And thus we have the parabolic path of the ball in terms of distance x from the plate. We seek the velocity v0 which will make z[409] = 10, i.e. just clear a 10 ft fence out over the point (409,0). solv = Solve[z[409] == 10, v0]//N {{v0 -> -115.538}, {v0 -> 115.538}} Hence the velocity we are talking about is 115.538 ft/sec. Remember we do not have air resistance in this model. In the case of resistance the velocity would have to be greater. vel = v0/.solv[[2]] 115.538 (d) Now determine for every other point at the base of the outfield fence just how high the fence must be so that this player can hit a homerun by launching a ball from his bat 2 feet above homeplate with exactly the same initial velocity at a 45 degree angle with the horizontal. We have the three different outfield walls to consider, lf[x], (0 <= x <= 135) cf[x], (135 <= x <= 300) and rf[x] (300 <= x <= 409), and the respective set of points at the base of these walls: (x, lf[x]), (x, cf[x]), and (x, rf[x]). We thus define the outfield wall with a conditional function. of[x_] = If[x<135,lf[x], If[x<300, cf[x],rf[x]]]; Plot[of[x],{x,0, 409},PlotRange->{{0,450},{0,450}}, AspectRatio->Automatic] Now what we wish to do is pick a specific x value (in the range of the outfield fence - [0, 409]) and its respective distance from home plate d[x_] := Sqrt[x^2 + of[x]^2] Plot[d[x],{x,0,409},PlotRange->{0,450}] We plot the distance from the plate as a function of the angle made with the right-field foul line. ParametricPlot[{ArcCos[x/d[x]],d[x]},{x,0, 409}, AxesLabel->{"angle - theta","distance"}, PlotRange->{{0, Pi/2},{0,450}}] and determine how high a ball is when it goes over the wall at the point at the foot of the outfield fence (x, of(x)) Now we recall from part (c) that we know the height of a ball at distance x from the plate when it is hit with initial velocity v0 = vel. w[x_] = z[x]/.{v0->vel} 2 2 + x - 0.00239716 x We plot this to confirm this fact and we also verify that at x = 0 the ball is 2 ft over the plate and at x = 405.1 it is exactly 10 ft high, just clearing the fence. Plot[w[x],{x,0, 409}] w[0] 2 w[409] 10. Thus we now wish to find out how high the fence must be at each point of the nearer distances on the fence (d(x) - where x is the distance down the right field foul line) so that a ball hit with initial velocity of vel =115.538 ft/sec will just clear the fence at each point. height[x_] = w[d[x]]; Plot[height[x],{x,0, 409}] We now construct a view of the outfield fence. First the base of the outfield fence. base[x_] = {x,of[x],0}; Next the outline of the top of the outfield fence. fence[x_] = {x,of[x],height[x]}; And we plot the base pbase = ParametricPlot3D[base[x],{x,0, 409}, ViewPoint->{-3.910, -3.340, 0.880}, AxesLabel->{"right","left","height"}, PlotRange->{{0,450},{0,450},{0,150}}] And we plot the top of the fence pfence = ParametricPlot3D[fence[x],{x,0, 409}, ViewPoint->{-3.910, -3.340, 0.880}, AxesLabel->{"right","left","height"}, PlotRange->{{0,450},{0,450},{0,150}}] We fill in some boards on the outfield fence - for effect. boards = Graphics3D[Table[ Line[{{x,of[x],0},{x,of[x],height[x]}}], {x, 0, 409,2}]]; We offer a view of the outfield fence for this park from behind home plate. field = Show[pbase,pfence,boards,Axes->False] We offer an animation of a ball being hit out of here as a function of the angle, theta, off the right field foul line. We call it homer[theta]. We need the distance along the line of flight - that is x; the height at any distance x from the launch site - that is zh[x]; and the angle - theta. z[x_] = y[t]/.Solve[x[t] == x,t][[1]] 2 32 x 2 + x - ----- 2 v0 In this animation, once we fix theta by calling the function homer[theta] we simply move along the line in the direction {x Cos[theta],x Sin[theta]} while also giving the ball height zh[x]. zh[x_] = z[x]/.{v0->vel}; homer[theta_] := Do[Show[field, Graphics3D[ {PointSize[.02], Point[{x Cos[theta],x Sin[theta],zh[x]}]}]], {x,0,409,10}] We animate the homerun over the left field fence. LeftCenterFieldHomeRun = homer[Pi/3] We animate the homerun over the other field's fences. CenterFieldHomeRun = homer[Pi/4] RightCenterFieldHomeRun = homer[Pi/6] RightFieldFoulLineHomeRun = homer[0] ISSUES IN SOLUTION One does not have to push students to animate anything, but the concept of getting a function defined which will tell you how high the fence is to be at a given position in the outfield for the homerun to be equally likely everywhere in the outfield is a complex one to envision and to compute - hence this problem.