According to USBC regulations, a legal bowling ball must have a diameter ranging between 8.5 and 8.595 inches. Overlooking the obvious problem that a cubic bowling ball cannot roll:

If the USBC allowed cubic bowling "balls" of equal volume to spherical bowling balls:

1) What would be the maximum permissible length of a side on a cubic bowling ball?

2) Could a cubic bowling ball be squeezed between two adjacent pins in the same row, such as the 9 and 10, without touching either?

## Thursday, March 18, 2010

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Okay, Todd. This may propel you directly into the bowling geek hall of fame (do not pass Go, do not collect $200).

ReplyDeleteAnd, once again, you've stumped me-- not with the problems, but with the questions...

(maybe we could find you a special school where bowling eggheadery is not only accepted, but encouraged... I only say this because I care...)

:)

EB, Even you can solve this one. I could have made it more difficult. For example, I could have asked whether that cube while rotating very rapidly could cleanly pass between adjacent two pins. But yes, I accept your general point.

ReplyDeleteI could probably write 1,000 questions like this about bowling, so wtf, I guess I will.

You go, brother! Do that voodoo you do so well... (I'll stick to birds and bad bowling on my side of the fence.) And I wonder where the team doctor is? He must be busy at work or he would have rogered up with an answer already... He's your biggest Puzzler fan, you know.

ReplyDeleteIf the volume of a sphere is calculated by 4/3 * pi * radius then

ReplyDelete1.3333 * 3.1415 * 4.25 (1/2 of 8.5in) = 17.8013 sq in and

1.3333 * pi * 4.2975 (1/2 of 8.595in) = 18.0003 sq in

So a side of a cube is calculated by cube root of the volume or 2.6110in and 2.6207in.

So my answer to (1) is 2.6207 inches.

Since the distance between the centers of each adjacent pin is 12 inches and each pin is 4.766 inches in diameter, the distance between each adjacent pin is 12 - 4.766 = 7.234 in.

So yes, a cube with a 2.6027in side could fit in the gap between the 9 and 10 pin.

Having said all that, I can't see how a 2.6in cube can have the same volume as a bowling ball so I must be wrong.

(Psst... See what you've done to MaddysDaddy?)

ReplyDeleteMD, You are very close, but let me point out that you are saying that if a bowling ball were a cube, it be 2.6 inches long on each side? How is that possible? Check your volume calculations. In particular, look at the equation you used to calculate the volume of a sphere.

ReplyDeleteAha, I forgot the radius cubed in the formula (thank you Google).

ReplyDeleteHere's my revised answer to (1):

4/3 * pi * (4.2975)cubed = 332.4493 sq in

cube root of 332.4493 = 6.9274 in

So the max length to a side would be 6.9274in and yes it would fit between the 9 and 10 pins, as do most of my spare attempts when shooting for the 10 pin.

That looks better!

Bingo. Now, what if it were rotating on a flat side as it slid down the lane? would it still fit between the 9 and 10?

ReplyDeleteI'm with the Doc: 9.500934e-005 nautical miles (6.92747in per side).

ReplyDeleteMy "ball", however, wouldn't slip between the 9 and 10 pins, because we all know my first ball almost always takes the 10 pin out (and nothing else).

I don't think the cube would fit between the 9-10 pins because the diagonal would be:

ReplyDelete(6.9274)squared + (6.9274)squared = (diagonal)squared or 9.7968in

Right again! Well done.

ReplyDeleteP.S. Glad to hear you have a lot more material to post, keep 'em coming!

ReplyDelete