A Modeling Problem: Counting Ping Pong Balls Suppose you have a cylinder of height h with base diameter b (perhaps an empty pretzel jar), and you wish to know how many ping-pong balls of diameter d have been placed inside the cylinder. How could you determine this? This problem, along with the strategy for computing the lower bound on the number of ping-pong balls, is adapted from (Starfield 1994). A lower bound for this problem is found as follows: N L -Lower bound on the number of balls that fit into the cylinder. V cyl -The volume of the cylinder. V cyl h π b 2 2 V cube -The volume of a cube that encloses a single ball. V cube d 3 The lower bound is found by dividing the volume of the cylinder by the volume of the cube enclosing a single ball. N L V cyl V cube The interactive approach You are given the following values: d 1.54 in b 8 in h 14 in Type commands at the command line prompt to compute N L . The following shows the commands typed at the >> prompt and the output produced:


>> d = 1.54

d =

    1.5400

>> b = 8

b =

     8

>> h = 14

h =

    14

>> vcyl = h*pi*(b/2)^2

vcyl =

  703.7168

>> vcube = d^3

vcube =

    3.6523

>> nl = vcyl/vcube

nl =

  192.6796

Using an M-File Create an m-file to solve . We created the following file named PingPong.m:


% PingPong.m - computes a lower bound on the number of 
%   ping pong balls that fit into a cylinder
% Note that most lines end with a ";", so they don't print
%   intermediate results

d = 1.54;
h = 14;
b = 8;
vcyl = h*pi*(b/2)^2;
vcube = d^3;
nl = vcyl/vcube

When run from the command line, this program produces the following output:


>> PingPong

nl =

  192.6796

To complicate your problem, suppose that you have not been given values for d , b , and h . Instead you are required to estimate the number of ping pong balls for many different possible combinations of these variables (perhaps 50 or more combinations). How can you automate this computation? One way to automate the computation of N L for many different combinations of parameter values is to use a for loop. (Read Programming with M-Files: For Loops if you are not familiar with the use of for loops.) The following problems ask you to develop several different ways that for loops can be used to automate these computations. Use a for loop Add a for loop to your m-file from to compute N L for b 8 in , h 14 in , and values of d ranging from 1.0 in to 2.0 in. This solution is by BrieAnne Davis.


for d=1.0:.05:2.0
    b=8;
    h=14;
    vcyl=h*pi*(b/2)^2
    vcube=d^3
    nl=vcyl/vcube
end

Can you still use a for loop? Modify your m-file from to plot N L as a function of d for b 8 in and h 14 in . This solution is by Wade Stevens. Note that it uses the command hold on to plot each point individually in the for loop.


clear all
hold on
for d=1.0:0.1:2.0;
    b=8;
    h=14;
    C=h*pi*(b/2)^2; %volume of cylinder
    c=d^3; %volume of cube
    N=C/c; %Lower bound
    floor(N)
    plot (d,N,'g*')
end

This solution creates the plot in .

Plot of N L as a function of d ; each point plotted individually. This different solution is by Christopher Embrey. It uses the index variable j to step through the dv array to compute elements of the nlv array; the complete nlv array is computed, and then plotted outside the for loop.


clear
dv=1.0:.05:2.0;
 
[junk,dvsize] = size(dv)
for j=1:dvsize
    d=dv(j)
    b=8; %in
    h=14; %in
    vcyl=h*pi*(b/2)^2;
    vcube=d^3;
    nl=vcyl/vcube; 
    nlv(j)=nl;
end
plot (dv,nlv)

This solution creates the plot in .

Plot of N L as a function of d ; points plotted as vectors. And finally, this solution by Travis Venson uses vector computations to perform the computation without a for loop.


%creates a vector for diameter
dv=1:.02:2;
b=5.5;
h=12;
 
%computes volume of cylinder
vcyl=h*pi*(b/2)^2;
 
%computes volume of cube
vcube=dv.^3;
 
%computes lower bound
lowerboundv=vcyl./vcube;
 
%plots results
plot(dv,lowerboundv)

More loops? Modify your m-file from to compute N L for d 1.54 in and various values of b and h . This solution is by AJ Smith. The height, h, ranges from 12 to 15 and the base, b, ranges from 8 to 12.


for h=12:15; %ranges of height
    for b=8:12; %ranges of the base
        d=1.54; %diameter of ping pong ball.
        Vcyl=h*pi*(b/2)^2; %Volume of cylinder
        Vcube=d^3; %volume of a cube that encloses a single ball
        Nl=Vcyl/Vcube %lower bound on the number of balls that fit in the cylinder
    end
end