Pin Action!

Welcome To My Mathematical Modeling Project's Home on the Web.
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The Project

So, what was my project?  Well, I created a mathematical model to determine which bowling split is more difficult, the 4-6 split or the 6-7-10 split.

Difficult: When I say that one split is more difficult than another, I mean that there is a smaller range of angles of incidence with which the ball can hit the 6 pin and still pick up the spare.

Assumptions:  In order to make this project doable, I needed several assumptions.  OK, it's a lot of assumptions.  But they're necessary.  Really, they are.

• The collision between the ball and pin is elastic.
• The bowler can throw the ball as needed (within reason).
• The mass of the ball is 5.4 kg (12 lbs., like my bowling ball).
• The mass of the pin is 1.5 kg (3 lbs. 6 oz.).
• Nothing can bounce out of the gutter.
• The lane is frictionless.
• The bowler is right handed.
• The ball is thrown parallel to the gutter.
• The center of mass of the pin is 12.25 cm above the middle of the base.
• The ball hits the pin at the same height as both the ball's and the pin's centers of mass.
• The 6 pin stays standing after it has fallen over.
• If two pins touch, they fall.
The last two assumptions go together.  The first basically says that instead of traveling on its side, the 6 pin slides over still standing.  That way I don't have to account for rotation effects.  The second allows me to ignore velocity and kinetic energy to determine if a pin hit by the 6 will fall.  Hopefully, it counteracts the previous assumption.  The rest of the assumptions should be pretty straight forward.  Just go with it.

Facts:  Here are some facts that are necessary.

• The distance between 2 adjacent pins, center to center, is 30.48 cm (12 in.).
• The radius of the base of the pin is 2.63 cm.
• The radius of the widest part of the pin (at the height of the center of mass) is 12.12 cm.
Equations:  Finally, here the equations used with two dimensional elastic collisions.
The last equation is merely a simplification of the first two.

Falling Pin
Before I could work with the splits, I had to determine what was necessary for a a bowling pin to fall over.  This took two steps.
1. Using trigonometry, find that the center of mass of the pin needs to be raised .279 cm in order to fall over.
2. Using (1/2)mv6f^2 = mgh, find that the velocity of the 6 pin, v6f, must be at least .23 m/s for pin to fall over (m is the mass of the pin, g is gravity, and h is the .279 cm found in the previous step).
.23 m/s = .51 mils/hr.  Clearly, the pin doesn't have to move very fast at all.

The Splits
Now I will analyze the two splits.  First, here's what the pins look like before the first throw.

The 4-6 Split:  In the 4-6 split, all the pins above are knocked over except for the 4 and the 6.  The strategy is to hit the 6 with the ball and have the 6 then hit the 4.  Here's my diagram for the 4-6.

Using that diagram and some geometry, I find that theta, which is the angles the 6 pin can travel and still hit the 4 pin, ranges from 0 to 11.24 degrees. Alpha, the angle of incidence, equals ninety degrees minus theta.  Also, the ball's initial velocity must be greater than or equal to .23/cos(alpha).  Since .23/cos(alpha) goes to infinity as alpha goes to 90, I made the highest alpha value 89.26 degrees, which corresponds to throwing the ball 40 miles per hour (very fast).  So the angle of incidence range was from 78.76 degrees to 89.26 degrees.

The 6-7-10 Split:  In the 6-7-10 split, all the pins have been knocked over except for the 6, 7, and 10.  The strategy is to hit the 6 and 10 with the ball and have the 6 hit the 7.  Here's the diagram.

Using the diagram and geometry, I found that theta, the angles at which the 6 pin can move and hit the 7 pin, ranges from 10.58 to 26.28 degrees.  Again, alpha, the angle of incidence, equals 90 degrees minus theta.  I had to introduce two new assumptions because I had difficulty dealing with the ball bouncing off the 6 pin:  a) the change in the ball's path after hitting the 6 pin is insignificant and b) if the ball hits the 10 pin, it will knock it over.  To back up these assumptions, I had to say that the initial velocity of the ball is greater than or equal to 1.2/cos(alpha).  The angle of incidence range in this case is 63.72 to 79.42 degrees.

Conclusion
Because the range of angles of incidence for the 4-6 split is only 10.5 degrees wide, while the range for the 6-7-10 is 15.7 degrees wide, the 4-6 is more difficult!  That's the conclusion.  It agrees with what I had expected, which is always good.

And, it's important to mention that I couldn't have ever gotten to this conclusion without the help of Dr. Cynthia Wyels and Dr. Paul Stanley.