Thursday, April 15, 2010

Bowling Puzzler XIII: The Sleeper Cell

As the season is winding down, I am going to start winding down the blog as well. During the off season, I don't post with much urgency. During May, June, and July of last year, for example, I only had three posts. The summer is so short up here that it cannot be wasted away on a laptop. So, I am going to make Puzzler No. 13 the last one of the season. I will bring it back in full force in the fall. For the grand finale, I thought I would put forth a real ball buster. In fact, I have yet to solve this problem myself. If anybody out there wants to give it a go, I welcome the attempt. If the answer is too onerous to post as a comment, send it to me as an attachment to an email using the address listed above and to the right. I'll post all serious attempts to solve the problem. If nobody can solve it, I'll figure out the answer in due time. Ok, here's the problem.

The 3-9 spare leave is one of those non-split leaves that behaves like a split in that it is very difficult to pick up. The 3 pin stands directly in front of the 9, with the latter usually referred to as a "sleeper" pin. Two other leaves have identical morphology, the 1-5 and and 2-8. There are two ways sleeper leaves can be picked up. One method is bring the ball into the front pin at an angle, striking it obliquely. If the ball has enough momentum and/or spin, the ball will also strike the sleeper pin. In this type of shot, the ball takes out both pins. The other way to complete this spare is to strike the center of the front pin, so that the sleeper pin is taken out by the deflected front pin. This puzzler concerns the second approach.

If you have ever missed one of these shot attempts, you know that there is very little margin for error. It seems as if the front pin must be struck dead center in order for it to take out the back pin. So, let's start with a couple of facts and a few assumptions:

Fact 1) At its widest point, a bowling pin has a diameter of 4.766".
Fact 2) The center of the 3 pin stands 20.7846" from the center of the 9 pin.

Assumption 1) When a bowling ball strikes a pin, it strikes it at its widest point
Assumption 2) When a bowling ball strikes a pin, the pin is deflected in the direction perpendicular to the tangent of the pin edge at the point at which it was struck.
Assumption 3) When a bowling pin is deflected, it does not rotate but remains standing vertically.

With these assumptions, what is the margin of error for collecting the 3-9 spare leave measured in terms of inches perpendicular to the long axis of the lane?

Here is the essence of the question. We can agree that if you hit the front pin directly on its center, it will take out the back pin. We will probably also agree that if you miss the center of the pin by a millimeter to the left or right, you will still get the spare. What if you miss by an inch? I'm not sure. So, I am asking what is the total width of the portion of the front pin that can be hit by the ball so that it still will take out the sleeper.

I told you it was a ball buster. Have a great summer!

Here's a link to the solution.


  1. I ignored this as long as I could hoping someone else would solve. It looks to me like you could hit the 3 pin approximately .54 inches off center (.54 being the arc length). That would give you a deflection that is about 12.9 degrees off a straight line from the 3 to 9 and should cause the 3 to just brush the 9. That would be .54 either direction, or about 1.08 inches total.

  2. Hey Doc, great timing. Did you see my post from ca. five minutes ago? In short, you are correct.

  3. It might be a little weird knowing there are others out there who get a kick out of this kind of thing. Thanks for the challenge and good luck tomorrow.

  4. My wife thought it was hilarious though.

  5. I find it strangely satisfying to know that there are others like us.


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