Four Dimensional Bowling Alley & Billiard Hall

[Hornbein]

Usually I would post this to the 4D bulletin board but it ain't working now.

I finally got around to doing the geometry and got a surprise.

As preliminary, I figure that in 4D bowling the pins would be arranged in a tetrahedron with three of the vertexes in the wx plane and the fourth closest to the bowler. Have a pin at each vertex and in the midpoint of each edge. With four vertexes and six edges that conveniently makes for the standard ten pins. The bowling alley is a triangular prism surrounded by a gutter. The question remained, which pins would occlude other pins? That is, which portions of which pins would be hidden from the bowler?

The standard layout is one foot between the center of each pin and its nearest neighbor, and 4.75" diameter for the pins. It turns out this this is the maximum pin size without any occlusion. That is, the pins would appear to be osculating. Golly.

So how to do this (perhaps wrongly.) It's convenient that it reduces easily to a 3D problem. Imagine you have spheres with 4.75" diameter in that tetrahedral arrangement. Six spheres are on a flat surface, the other four suspended above them, and you are looking straight downward at them. I drew a diagram and found that if you are far away enough that perspective is negligible -- such as would be a would-be bowler -- then no sphere is occluded and what's more they would appear to be osculating. Diagrams aren't precise so the answer isn't exact but close is good enough for me.

With the standard measurements the game would be too difficult. I figure the distance between pin centers has to be reduced in order to get the "domino effect" necessary in bowling. Eight inches would do it.

Billiard balls, similar setup except the balls really are osculating. They occlude one another. But I couldn't get an arrangement with the standard 15 balls. Adding one more layer to the tetrahedron gives 20 balls. The best I could do was two tetrahedra with a common base, like our nineball. That gives 14. Both of these two arrangements have all outer balls, no inner. I tried to figure out the minimal number of balls that would have an inner ball. That is, it can't be struck by the cue ball without first touching an outer ball. The number is at most 13 but that's as far as I got. This is similar to the notoriously difficult sphere packing problems. If I really wanted to approximate this I'd buy some pingpong balls and stick them to each other.

 

[Hornbein]

So I bought a dozen pingpong balls and gave it a shot. At first I tried sticking them together with rat glue. Rat glue never dries, but it couldn't hold a pingpong ball still so it just made a very sticky mess. Next was superglue but it was too old and wouldn't set. How about black duct tape. That was ugly as could be but worked pretty well. I concluded that the 3D arrangement of balls such that there is an inner ball can be done with seven balls. It's a regular arrangement -- octahedral with a central ball -- so would be easy to prove if so. Six might be possible but I'd had enough and threw the ugly mess in the trash.

 

[Hornbein]

The answer is seven. The cue ball will still be at least 0.18 away from the inner sphere when it contacts an outer sphere. With six spheres the cue ball can reach the ostensibly inner ball while still about 0.10 away from any outer ball.

 

[Hornbein]

Building models superglue works very well for white pingpong balls but for the orange ones it fails to gel. For orange pingpong balls one must instead use yellow contact cement. Should such things be considered obvious?

My favorite now is a truncated tetrahedron with 15 balls. It looks the most traditional. Start with a triangle with four rows instead of the canonical five. Add another layer of six balls on top of that for a total of fifteen. It means there is no head ball and no inner balls but I don't mind that.

The other candidate is take the 13 dense pack balls -- twelve osculating balls surrounding an inner ball -- and add two more at mutual antipodes to get 15 balls. It looks exotic, highly untraditional, but better than the the spherical arrangements that are evocative of atomic nuclei. And it does have the advantages of having all of a headball, an inner ball, and maximal connectivity, all of which are traditional.