Ball bearing offshoot

[Charlie Velasquez]

Ok, after reading the ball bearing problem it brought back memories of a problem my dad gave me in 6th grade. I had just mastered the pythagorean theorm and used it to square a concrete form for a parking curb to our property line. I was all 'know-it-all'

So, he says, "You have a 12" cube with a 12" sphere inscribed in it. What is the largest sphere you can inscribe in the gap in the corner?"

2D was simple, but adding that third dimension really confused me.

Forgot about it for the last few decades. Any help?

[Dale Coons]

You were probably closer than you thought with the 2D. You just need to apply the pythagorean theorem twice to get started
You need to find the distance from the center of the cube to a corner. There are two triangles that can get you there--the first lays on the 'floor' of the cube and is the green hypotenuse which becomes the base of the second right triangle. So square root of 36+36=72 is approx 8.485 inches. You can use that to get the red hypotenuse, which is what you need to start, so square root of 36 + 72=108 is 10.392 approximately, the value you need. Now, a 12 inch sphere centered in this cube has a radius of 6 inches, and would extend 6 inches along that 10.392 dimension. So 10.392-6=4.392, the diagonal of the largest cube that will fit there in the corner between the sphere and the outside cube. You can use the new 'inside cube' to find the center of the sphere that will fit in the corner. You know the angle between the green and red line, because the sine of that angle is 8.485/10.392 or .816, so the angle is about 54.74. So the angle between the red line and the vertical of the cube is 90-54.74=35.26, the same as the angle between the red line and the black vertical which is the length of the radius of the largest sphere that will fit in the corner, since they are parallel. Half of 4.392 is 2.196, the length along the red line between the center of the new inside cube and it's corner. We can rearrange cosine= adjacent/hypotenuse and use the formula cosine(35.26)*2.196=adjacent, or 1.793 to find the length of the little black line. Twice that is approximately 3.59, the diameter of the largest sphere that will fit in the corner.

Somebody sharper than me may find a math mistake, but I think this is right. The picture will hopefully help you follow the words.

cubeprob.GIF

[Eric DeSilva]

We can calculate the length of a corner to corner diagonal through a cube (a diagonal that would go through the center of the cube). For a unit cube, a face diagonal is 2^(1/2) the side. The diagonal we want is the hypotenuse of a right triangle with sides that are the face diagonals we just calculated, so the diagonal we want is 2^(1/2) x [2^(1/2)], or 2. That is a linear relationship, so the length of the long diagonal is always twice the length of a side (interesting... never knew that).

Thinking about that long diagonal and a cube with an inset sphere, we can calculate the distance (along that diagonal) from the corner of the cube to the surface of the sphere. That is 1/2 the diagonal minus the radius of the inset sphere, or 1/2. Call that the "corner void distance" for a cube.

Think about the big cube. Our corner void distance is 6". A smaller sphere in that corner will touch that point on the diagonal where the diagonal intersects the large sphere. Which means a small cube containing the small sphere will actually cut into the big sphere. But if we imagine that smaller cube, the corner void distance for the big cube (6") is actually two times the radius of the small sphere (r) plus the corner void distance for the little cube. But we can figure out the corner void distance for the small cube--the side of the small cube is 2r, so the diagonal is 4r, so 1/2 the diagonal (2r) minus the radius (r) is... r. So, 6" is 3r. So r is 2", so the diameter of the little sphere is 4".

Anyone want to check my math?

[Eric DeSilva]

I think Dale made a mistake, which is that the largest sphere you can put in the corner is bigger than the largest cube, but I think I made a mistake because the length of the long diagonal is actually the hypotenuse of a right triangle made from an edge and an edge diagonal (not two edge diagonals), so for a unit cube, [(2^(1/2))^2 + 1^2]^(1/2), or 3^(1/2) (20.78" for a 12" cube). That means my corner void distance is actually (3^(1/2)/2) - 1/2, or (3^(1/2)-1)/2 (4.39" for a 12" cube). And therefore 4.39 = 2r + [(3^(1/2)-1)/2]r, or r = {2 + [(3^(1/2)-1)/2]}/4.39 = 0.53", so diameter is 1.06".

[Dale Coons]

Eric--I think I'm ok. I made the assumption the corner sphere must also fit completely inside the big cube, which means it cannot extend outside the small inset cube--hence the last bit of fiddling to find a sphere that will fit inside the smaller cube. If one wants to let the smaller sphere touch the big one AND let it extend beyond the original square then the diameter of the smaller ball is just the length of the red diagonal (10.392) - radius of large sphere (6) or 4.392, which would be the diameter of the small sphere that touches both the original sphere and the corner of the original cube. But it wouldn't fit inside the original cube. If you stopped there we would agree--no?

[Charlie Velasquez]

My head hurts already.... Now I remember why it lay dormant in my brain all these years....

[David Epperson]

OK. I'll give this one a shot. I think Dale was on the right track...in the beginning. So looking at his diagrams, we have a 12" sphere in a 12" cube. His 10.392" line from the center of the sphere to the corner of the cube needs to be extended to the far side of the sphere - making the extended line 16.392". This is the line that runs through the diameter of an inscribed sphere to the corner of the containing cube. 16.392 minus 12 = 4.392 which is the length of the smaller circles diameter (from a point touching the larger sphere) extended to the corner of the containing cube. Since these two shapes are 3 dimensional proportional, the same ratios should apply. So since 16.392/12=0.73205, then 4.392 * 0.73205 should equal the diameter of the largest sphere that will fit in a corner. Which is 3.215".

[ray hampton]

My head hurts already.... Now I remember why it lay dormant in my brain all these years....

charles,a square one foot wide will support a circle one foot wide with a 2 " space at all four corners,
a 12 " square are just over 16 " from corner to corner

[Eric DeSilva]

I was gone all weekend. Unless I'm missing something, the problem with your calculation is that if you size the inset cube in the corner such that the inset cube is the largest cube that can be put in the corner without touching the big sphere, you will end up with space between the big sphere and the small sphere inside the inset cube. The point at which the large and small spheres should touch is on the line which extends from corner to corner and hits the center of both the big and little spheres. If you think about that line, radiating from the center of the large sphere outward, it will be inside the large sphere, then it will hit the point where it contacts the corner of your inset cube--but it should go from being inside the large sphere to being inside the small sphere. So you have a gap.

[David Epperson]

I was gone all weekend. Unless I'm missing something, the problem with your calculation is that if you size the inset cube in the corner such that the inset cube is the largest cube that can be put in the corner without touching the big sphere, you will end up with space between the big sphere and the small sphere inside the inset cube. The point at which the large and small spheres should touch is on the line which extends from corner to corner and hits the center of both the big and little spheres. If you think about that line, radiating from the center of the large sphere outward, it will be inside the large sphere, then it will hit the point where it contacts the corner of your inset cube--but it should go from being inside the large sphere to being inside the small sphere. So you have a gap.

Eric, that sounds a lot like the premise I worked my proposed solution upon. Since the large "sphere in a cube"(corner), is exactly proportional to the smaller "sphere in a cube corner" then the portion of the line from "corner to corner" which is also a line from "diameter end to corner", would also be proportional to the line from "diameter end to corner" of the small sphere - and both would have identical proportions of "diameter to remainder".

The next largest sphere to fit between these two and also in the cube, looks like it would be much harder to find.

[Dale Coons]

Two thoughts. I think the proportional approach is correct--and is certainly simpler than what I started with. And Charlie's use of the unit cube is also simpler, I just started with the op's pythagorean theorem. I'd make just one change to the proportional approach--which is that you need the ratio between the full diagonal (rather than the 'far side' diagonal) and the edge of the cube. That ratio represents the largest ball bearing that can fit *within* the box. Since we have the diagonal of the largest box that will fit in the "void" corner, that would be the correct ratio to use in order for the new bearing to fit completely withing both the original cube and the new smaller one.

It is true there will be a gap between the two ball bearings. But the little bearing that fits within the small cube will touch all the sides of the original cube. I don't believe it is possible to make both touch the big bearing AND touch all the sides of the original cube AND fill the corner (or have no 'diagonal' space) AND not extend beyond the original cube. The original large bearing does not touch the corner of it's bounding cube, the way I read the problem the little one doesn't have to either--but it does have to fit inside of it.

[Eric DeSilva]

David nailed it. Much better explanation. I think his solution is what I was trying to describe, but his use of the proportions makes the solution a lot easier to describe. Still haven't figured out where I messed up my math.

[Jason Roehl]

Obviously, the smaller sphere can't have a diameter of half the cube diagonal minus the radius of the large sphere--that would put the bounds of the small sphere outside the cube.

However, it is entirely possible to have a small sphere touch the large sphere as well as touch 3 sides of the large cube. You just have to find the radius of the small sphere such that the radius is also orthogonal to the cube on the three sides that form the corner in which the small sphere resides. I'd take a stab at it, but I just ate--too little blood flow to the brain right now.

[Eric DeSilva]

And therefore 4.39 = 2r + [(3^(1/2)-1)/2]r, or r = {2 + [(3^(1/2)-1)/2]}/4.39 = 0.53", so diameter is 1.06".

Gah! That's what I messed up. Multiple times. root3/2 - 1/2 is the ratio of the *edge* length to the corner void distance, not the radius (which is half the edge length). So for the little sphere, the diameter (edge of cube) is related to 4.39 as 4.39 = D + (root3/2 - 1/2)*D. So 4.39 = D * (root3 + 1)/2, which gives me the same 3.2 David gets.

[Jason Roehl]

Okay, after drawing a 2-D image on my white board, I have to concur with David's explanation using the ratio of the half-diagonal of the cube plus the radius of the large sphere to the diameter of the sphere is proportional to the ratio of the half-diagonal of the cube minus the radius of the large sphere to the radius of the small sphere.

I bet someone more skilled than I could generate a 3-D image of this in SketchUp.