Cannon ball problem

Bob Itnyre · May 13, 2007

A friend of mine gave me this problem. If you’ve seen it please tell me where. I’ve searched without luck. Anyway, this was supposed to have been a problem at the Naval academy quite a few years ago.

You have four perfectly spherical cannon balls. Let’s say they are 12 inches in diameter. They could be any size as long as they are all the same. You place three of them on a flat surface, blocked, so they won’t roll and the three are all touching each other. Then you place the fourth one in the space at the top. All four are now touching each other. What would be the size of a smaller ball that could be placed in the center of the four, so that it touched all four balls and did not keep them from touching each other?

A couple of math teachers I know are wrestling with it, but I figure machinists do this stuff every day.

johnoder · May 13, 2007

What would be the size of a smaller ball

2.697 for 12" balls

John

Bob Itnyre · May 13, 2007

John, thank you but could you tell me how you got the answer. I've been moving angles around in my head until I'm not sure where I'm going with it. Bob

johnoder · May 13, 2007

Less than ten minutes in Autocad

Drew a plan view of the three, then a section view thru one of the apexes on the plan view - so I could then draw in the top ball. The rest was to see what ball would be on center and touch the two balls in the section view (which of course represent all four of them)

The 12" "connector" between any of the three and the top ball stands at an angle of 54.7356 degrees above the "floor" the three sit on. Bisceting this connector with a line that intersects assembly centerline establishes the center point of the 2.697 dia. "small" ball.

Plainly, this could be reduced to some handy formula, which I will leave to someone else to do

.

John

Bob Itnyre · May 14, 2007

Again, Thank you, Bob

Fuzzbean · May 14, 2007

Just think of all the space I can save now, if I plan ahead!

Damien W · May 14, 2007

Sorry, I got the wrong idea. When the topic was "cannon ball problem" I thought the solution might be Preparation H. Well ointment or cream anyway.

gwilson · May 14, 2007

No,Damien,for cannon ball problem,the treatment is a visit with the ladies.

smokepolesc · May 14, 2007

Given a 12" diameter ball, the ball diameter that will fit is 3.464101. Calculator for this is located at:

http://www.1728.com/polygon.htm

Formula:
r = .5 * s * Cotangent(180/n)
r = .5 * s * Cotan(180/3)
= s * .28868
n = is the number of sides
R = radius of big ball ball
r = radius of small ball (inscribed circle)
s = is the length of each side (for n=3, R = s!)

SmokepoleSC

Twinhit · May 14, 2007

I think this same process could be calculated using a compass, triangle and precision CE scale.

I mean... you know...

jabezkin · May 14, 2007

so which answer is right?

johnoder · May 14, 2007

so which answer is right?

2.697 or very close to that number. Autocad is incapable of telling fibs.

John

jabezkin · May 14, 2007

Thanks, you both seemed like you were sure.

Fuzzbean · May 15, 2007

Assuming 2.697" is correct, wouldn't the formula simply be D multiplied by .22475 where D is the diameter of the big balls?

2.697/12 being .22475. Seems like the proportion would always remain constant. Though it's hard to type when I'm constantly blinded by these flashes of insight.

agrip · May 15, 2007

John is correct.
In paraphrase.
The centers of the four balls define a tetrahedron, 12 inches on any edge.
Establishing a plane with the centers of any two balls and the centroid of said tetrahedron, allows one to draw the two 12" dia circles, and determine that a third circle centered on the centroid, and tangent to the 12s is 2.696938" dia.

Now is the time to remember the tall tale of the brass monkey that stored cannon balls. ( as in cold enough to freeze the balls off of)

If anybody cares, any dihedral of said tetrahedron is 70.528779 deg.

Ag

[ 05-15-2007, 11:06 AM: Message edited by: agrip ]

Old Bill · May 15, 2007

bsamc2000 · May 15, 2007

I think a picture is worth a 1000 words. From the math that I did, I believe that johnoder is correct. This image was created using 3 12" diameter spheres and a 2.697" diameter sphere. I hope this helps a little.
AutoDesk Inventor is great!

Brian

Twinhit · May 15, 2007

umm Brian, the size of the ball in the center doesn't fit.

Dirtdobber · May 15, 2007

It has to be above centerline to touch all 4, and it looks like it is.

bsamc2000 · May 15, 2007

The center ball is larger than just calculating the 3 tangents between the larger spheres. As the smaller sphere grows to contact all 4 larger spheres it's center moves away from plane created by any 3 spheres.

Brian

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