BALLPARK Designing a Fair Ballpark BRIEF ABSTRACT Some ball parks favor right hand hitters over left (and possibly vice versa.) This problem is concerned with the how to build the outfield fence (height) so that the ball park isn't biased for any directions. This is an involved problem accessible to students of projectile motion. GENERAL INFORMATION FileName: BALLPARK Full title: Designing a Fair Ballpark Last Update: 5/23/96 Developers: Lynn Kiaer, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA Aaron Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA Contact: Aaron Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Aaron.Klebanoff@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 A baseball park is 325 feet down the third base line and 425 feet down the first base line. The outfield fences are perpendicular to the baselines for a distance of 250 feet (left field) and 150 feet (right field). The remaining portion of the outfield fence is a quarter circle. You have been asked to design the outfield fence so that the field does not favor hitters from one side of the plate. 1. If, at its lowest point, the fence must be exactly 10 feet high, how high should the fence be at all points along the outfield boundary? Assume a launch angle (the angle the ball makes with the horizontal as it comes off the bat) of 25 degrees. You should also assume -- at least for starters -- that the wind resistance is negligible so that the only force acting on the ball after it has been hit is due to gravity. Finally, assume the ball strikes the bat 3 feet above the ground. 2. How sensitive is the height of the fence to slight changes in the launch angle? What if the launch angle is 30 degrees? KEYWORDS Projectile motion, differential calculus TEACHER NOTES ISSUES RELATED TO THE PROBLEM Prerequisites The two problems are independent of one another, but both require only differential calculus and an introduction to projectile motion. Time allotment - time management At least one week out of class with some time in class for questions. Expectations Future payoffs Extensions We have made some limiting assumptions: The easiest to consider first is the assumption of launch height. We assumed launching at 3 feet. The solution is, however, dependent on launch height. Students should determine how sensitive the solution is to launch height over the reasonable ranges of launch height, say 0 < launch height < 6. Another thing to more carefully consider is how to choose the optimum fence heights when considering all possible launch angles. There is no single correct answer to this. However, this gives students the opportunity to try their hand at modeling and arguing on a mathematical (and possibly aesthetic basis) for their particular design. Finally, wind resistance may play a role. But how much? To find out, assume that wind resistance is proportional to velocity and see how sensitive the solutions are to the drag coefficient. References and Sources POSSIBLE SOLUTION(S) First, a coordinate system: Place home plate at the origin and the field in the first quadrant. (That is, the first base line is the positive x-axis and the third base line is the positive y-axis.) Now, let's define the lines/curves that give us the outfield fence. In[2]:= LeftFieldLine = ParametricPlot[{t,325},{t,0,250}, DisplayFunction -> Identity]; In[3]:= RightFieldLine = ParametricPlot[{425,t},{t,0,150}, DisplayFunction -> Identity]; The quarter circle that defines the center field line has its center at (250,150) and has radius 175. In[4]:= CenterFieldLine = ImplicitPlot[(x-250)^2 + (y-150)^2 == 175^2, {x,250,425}, {y,150,325}, DisplayFunction -> Identity]; In[5]:= Show[LeftFieldLine, RightFieldLine, CenterFieldLine, AspectRatio -> Automatic, DisplayFunction -> $DisplayFunction] Out[5]= -Graphics- We are told that the lowest part of the fence (that is, the fence at the point in the park at which it is hardest to get a home run) is to be 10 feet high. This will be at the point furthest from home plate. That point will be somewhere on the quarter circle. In[6]:= s1 = Solve[(x-250)^2 + (y-150)^2 == 175^2, y] Out[6]= 2 300 - Sqrt[90000 - 4 (54375 - 500 x + x )] {{y -> ------------------------------------------}, 2 2 300 + Sqrt[90000 - 4 (54375 - 500 x + x )] {y -> ------------------------------------------}} 2 The second solution gives us the top of the circle, and we know the furthest point from the origin lies on the top of the circle. In[7]:= ycirc[x_] = s1[[2,1,2]]; In[8]:= d[x_] = Sqrt[x^2 + ycirc[x]^2]; In[9]:= Plot[d[x],{x,250,425}]; In[10]:= s2 = Solve[d'[x]==0,x] //N Out[10]= {{x -> 400.061}} In[11]:= xfar = s2[[1,1,2]]; In[12]:= ycirc[xfar] Out[12]= 240.037 In[13]:= d[xfar] Out[13]= 466.548 So that's the most distant point on the outfield fence, and at that point the height of the fence should be 10 feet. How much initial velocity is necessary to hit the ball over a 10 foot fence d[xfar] feet from home plate? Assuming the simplest model for projectile motion (no wind resistance), parametric equations for the position of the ball are given by In[14]:= horiz[t_] = Vx0 t; vert[t_] = -16 t^2 + Vy0 t + h0; We know that In[16]:= Vx0 = V0 Cos[theta]; Vy0 = V0 Sin[theta]; We ( reasonably ) assumed In[18]:= h0 = 3; And we are told to work with theta = 25 degrees, or In[19]:= theta = 25 Pi/180; We want to know for what value of V0 will there be some time when the distance traveled will be d[xfar] and the height be 10 feet. In[20]:= Clear[V0] In[21]:= eq1 = horiz[t]==d[xfar]; In[22]:= eq2 = vert[t]==10; In[23]:= s3 = Solve[{eq1,eq2},{V0,t}] Out[23]= {{V0 -> -141.905, t -> -3.62763}, {V0 -> 141.905, t -> 3.62763}} Clearly the second solution is the one! In[24]:= V0 = s3[[2,1,2]]; Now, to find the proper height of the fence at each point, all we need to do is determine the distance of that point to the origin, and find the height of the ball at the time when the ball has traveled that distance with this initial velocity. For instance, at the (baseline) end of right field, In[25]:= s4 = Solve[horiz[t]==425,t] Out[25]= {{t -> 3.30457}} In[26]:= vert[s4[[1,1,2]]] //N Out[26]= 26.4574 So the fence should be about 26.5 feet at the right field end. In[27]:= s5 = Solve[horiz[t]==325,t] Out[27]= {{t -> 2.52703}} In[28]:= vert[s5[[1,1,2]]] //N Out[28]= 52.3761 Yowza! The fence should be over 52 feet high at the left field end! All of a sudden, Fenway doesn't look so lopsided. We can find the appropriate height of the fence at a distance d from home plate by defining In[29]:= ht[dist_] := vert[Solve[horiz[t]==dist,t][[1,1,2]]] In[30]:= ht[325] Out[30]= 5 Pi -99.1739 + 358.598 Sin[----] 36 In[31]:= a=Plot[ht[dist],{dist,325,467}, PlotStyle -> AbsoluteThickness[2]]; Changing the launch angle requires only a few small changes. In[32]:= Clear[V0] Setting theta at 30 degrees In[33]:= theta = 30 Pi/180; And setting up the same equations as before In[34]:= eq1 = horiz[t]==d[xfar]; In[35]:= eq2 = vert[t]==10; In[36]:= s3 = Solve[{eq1,eq2},{V0,t}] Out[36]= {{V0 -> -133.038, t -> -4.04939}, {V0 -> 133.038, t -> 4.04939}} Clearly the second solution is the reasonable one In[37]:= V0 = s3[[2,1,2]]; When does the ball get down the first base line? In[38]:= s4 = Solve[horiz[t]==425,t] Out[38]= {{t -> 3.68878}} And how high must the wall be at that point? In[39]:= vert[s4[[1,1,2]]] //N Out[39]= 30.6601 Let's get the height of the wall as a function of the distance to the plate In[40]:= ht2[dist_] := vert[Solve[horiz[t]==dist,t][[1,1,2]]] In[41]:= b=Plot[ht2[dist],{dist,325,467}, PlotStyle -> AbsoluteThickness[4]]; In[42]:= Show[a,b] Out[42]= -Graphics- So we see that the steeper the launch angle, the higher the wall must be. (The thicker curve corresponds to the 30 degree launch angle.) Let's plot this wall (using the 30 degree launch angle) First, let's parameterize the wall using the angle of elevation above the x-axis. The outfield wall is a straight line from the first base line until the point (425, 150). At that point, the angle is In[43]:= a1=ArcTan[150/425]//N Out[43]= 0.339293 The other point where the function will need to be `broken' is (250, 325), and at that point our angle is In[44]:= a2=ArcTan[325/250]//N Out[44]= 0.915101 So let's define the right field wall In[45]:= x[angle_] := 425 /; angle>=0 && angle=0 && angle=a1 && angle =a1 && angle =a2 && angle=a2 && angle {-10, -10, 0}]; ISSUES IN SOLUTION