Object Ball Contact Areas

[Patrick Johnson]

Thought somebody might be interested in this...

I drew these to get a visual idea of where and how big the object ball contact area is for various shots. The idea is to place a ball anywhere on the table with lines stretching from its center to each of the pockets, showing the contact area "projected" onto the surface of the ball for each shot (the red lines), then blow the ball up to a size that makes the contact areas easier to see.

These drawings show oversized object balls on the center spot and the foot spot with contact areas for shots into each pocket. It's amazing we manage to hit these small target areas on the object ball as often as we do.

Center Spot

Foot Spot

pj
chgo

 

[mosconiac]

Few people ever consider the extreme skill required to actually make a ball. We know it takes a precise hit, but just how precise?

Here's some math to go along with your diagrams.

Let's take the example of the OB sitting dead center of the table and being cut into the corner.

The ball is sitting approx. 76" from the hole (4.5' x sqrt of 2 x 12"/foot).

We have a span of approx. 2.75" of pocket to shoot the ball into (assuming a 4.5" pocket and about a 1/4" on each side where you can make the ball even though you contacted the rail slightly). From your diagram, center-to-center of the balls in the pocket = 2.25" (I added 1/4" + 1/4" to get the total span).

This gives us the two legs of a triangle we need to determine the angle available to us, which is about 2 degrees (arctan of 2.75"/76").

Translating that to the surface of the CB, you get a contact area of 0.04", which only slightly larger than 1/32"...put that on your caliper and see how small that is...pretty small, huh?

In the case of cutting the ball down a rail, the contact area is only 0.024" or significantly smaller than 1/32".

I looked around the house for items that small (that everyone would have around their house) and the only thing I found was 7 sheets of newspaper.

 

[Patrick Johnson]

By the way, these contact areas are also the "margins for error" that are mentioned in the S.A.M. and other similar aiming systems. You can see that they don't really add significantly to the shotmaking potential of any system.

pj
chgo

 

[Patrick Johnson]

The ball is sitting approx. 76" from the hole (4.5' x sqrt of 2 x 12"/foot).

That seems long to me... I think Pythagorus would say:

sqrt(25^2 + 50^2) = ~56 inches

Maybe a typo?

This gives us the two legs of a triangle we need to determine the angle available to us, which is about 2 degrees (arctan of 2.75"/76")

My calculus is long gone, so I have to take a simpler approach:

360 x (2.75 / (pi x (56+56))) = ~2.8 degrees

This translates to a contact area of a little less than 1/16 inch (1.75 32nds) on the object ball, not quite as small as you calculated, but close enough for government work.

A spot shot on the same table has a margin of error of about 3.6 degrees, which translates to a little more than 1/16 inch (2.3 32nds) on the object ball.

pj
chgo

 

[Jal]

Thought somebody might be interested in this...

I drew these to get a visual idea of where and how big the object ball contact area is for various shots. The idea is to place a ball anywhere on the table with lines stretching from its center to each of the pockets, showing the contact area "projected" onto the surface of the ball for each shot (the red lines), then blow the ball up to a size that makes the contact areas easier to see.

These drawings show oversized object balls on the center spot and the foot spot with contact areas for shots into each pocket. It's amazing we manage to hit these small target areas on the object ball as often as we do.

Center Spot

Foot Spot

pj
chgo

Nice diagrams...a very elegant way to show the margins!

It is surprising that the balls can be made to enter the pockets with any consistency. And I think it's downright amazing that this can be accomplished even after running the gauntlet of squirt, swerve and throw (sometimes).

(I agree with Mosconiac that the balls will generally drop with the 1/4" overlap he indicated, if not hit too hard.)

Jim

 

[mosconiac]

Patrick's right about the 76"...I posted first thing in the morning and I made a mistake. I assumed an isosceles triangle and it certainly is not.