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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.
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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.
-
Using trigonometry,
find that the center of mass of the pin needs to be raised .279 cm in order
to fall over.
-
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.
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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.
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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.
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Links
Finally, here
are some links that might be interesting.
My Powerpoint Presentation of
this Project in HTML Form
California Lutheran University
(the univeristy I attend)
Math
Modeling at CLU Homepage
My Homepage (desparately
needing an update)
A
Good Reference for the Pin Layout
Interesting List
of Bowling Terms
Animation Library,
where I got the animated gif below
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