The volume of the holey cube is equal to the volume of the solid silver 4 cm cube minus the quantity of three times the volume of a cylinder 3 cm in diameter and 4 cm long, minus three times the volume of a 3 cm sphere, plus the volume of 1, 3 cm sphere.

If you take the volume of one cylinder and subtract the volume of the middle sphere.....that gives you the two end cap volumes.

Repeat for remaining cylinders....add one sphere volume for the center, subtract all that from the starting volume..... and presto.......you have the volume of the remaining holey cube.

Right?

If it's not asking much... could you draw a quick sketch explaining what you've just stated?

 

No, that's not what I'm claiming.

I want you...in your mind to invert the problem.

Instead of thinking of the holey cube.....think of the orthogonal cylinder structure.

The common thing these 3 cylinders have is the center sphere.

If you take the volume of one cylinder....subtract the center sphere volume and divide by 2.......that will give you the volume of one end cap.

Now...you know the volume of all six end caps....plus the center sphere.

If you subtract those end cap volumes plus the center volume.....what you have left is the holey cube volume.

This should not be hard for anyone to understand.

 

No, that's not what I'm claiming.

I want you...in your mind to invert the problem.

Instead of thinking of the holey cube.....think of the orthogonal cylinder structure.

The common thing these 3 cylinders have is the center sphere.

If you take the volume of one cylinder....subtract the center sphere volume and divide by 2.......that will give you the volume of one end cap.

Now...you know the volume of all six end caps....plus the center sphere.

If you subtract those end cap volumes plus the center volume.....what you have left is the holey cube volume.

This should not be hard for anyone to understand.

NOW I get it.... well explained... thanks!

 

When the cube is solid....it is silver.

Then we put the first cylinder thru it....the cylinder is red.

Then we put the second cylinder in......it is yellow and it causes a green center sphere.

Then we put the last cylinder in....it is blue....and when we put it in the sphere turns white.


So now we have a silver cube with 2 red end caps and 2 blue end caps and 2 yellow end caps and a white sphere.


Got it?

 

When the cube is solid....it is silver.

Then we put the first cylinder thru it....the cylinder is red.

Then we put the second cylinder in......it is yellow and it causes a green center sphere.

Then we put the last cylinder in....it is blue....and when we put it in the sphere turns white.


So now we have a silver cube with 2 red end caps and 2 blue end caps and 2 yellow end caps and a white sphere.


Got it?

So now you are claiming that if I take a cube and turn it on a lathe in just two orthogonal axes I get a sphere. Do you really believe that?

 

With a lathe....one can turn a cube into a sphere with one axis.

 

With a lathe....one can turn a cube into a sphere with one axis.

By cutting a cylinder out of it (or off of it)? That's what we are talking about.

 

When just 2 cylinders intersect....looking from the outside......it does not look like a sphere.

But if you take the atoms that are common to both cylinders....they form a sphere.

 

WBahn.......haven't you ever cut a ball of wood on a lathe?

 

With a lathe....one can turn a cube into a sphere with one axis.

By cutting a cylinder out of it ? That's what we are talking about.

WBahn.......haven't you ever cut a ball of wood on a lathe?

Not by cutting a cylinder. Again -- we are talking about the construction techniques applicable to this problem, not techniques that you might actually use but that are not applicable to this construction -- namely three (or you claim two) intersecting cylinders.

Now, I will freely admit that I haven't worked on a wood lathe in more than 35 years and that I didn't turn very many curved surfaces even then.

 

Turning a round wooden ball in 7th grade shop was easy.

Squaring a block of wood with a hand plane was damn near impossible.

I was not thinking of this problem in anyway in relation to a lathe.

I was thinking of the easiest way to figure the volume of the holey cube.

To me, this was the easiest.

 

If I were to construct such an object......I would use a drill press.

 

Turning a round wooden ball in 7th grade shop was easy.

Squaring a block of wood with a hand plane was damn near impossible.

I was not thinking of this problem in anyway in relation to a lathe.

I was thinking of the easiest way to figure the volume of the holey cube.

To me, this was the easiest.

But that is assuming that you are correct that the intersection of the three rods is a perfect sphere. While it might be, I'm not convinced it is. What I have been saying is that IF it is, THEN using a lathe to do NOTHING but cut a cylindrical shape (i.e., moving the tool parallel to the axis of rotation) must result in a perfect sphere by starting with a cube and cutting it into a cylinder. Then turning it 90 degrees and cutting another cylinder, and then turning it 90 degrees and cutting another cylinder. I'm not convinced that this set of steps will produce a perfect sphere and, if it doesn't, then the common shape at the intersection of the three mutually perpendicular cylinders is not a perfect sphere.

 

The intersection is not a sphere....only the shared volume of the intersection is a sphere.

A 4x4x4 cm cube is 64 cubic centimeters.

A cylinder 3 cm in diameter and 4 cm long is 28.27 cubic centimeters.

A sphere 3 cm in diameter is 14.14 cubic centimeters.

The volume of an end cap is 28.27-14.14/2.

The volume of an end cap is 7.07 cubic centimeters.

Six end caps is 42.40 cubic centimeters.

The total volume of the orthogonal structure is 6 end caps and 3 cm

sphere, which is 56.53 cubic centimeters.

That leaves the holey cube with a volume of 7.47 cubic centimeters.

Verify weight before and after.

Also verify water displacement before and after.

This should satisfy you.

 

An end cap has a flat outside face.....round edge and a convex inner face that fits against the sphere.

If that helps.

 

I have a complaint.

There's no one left at AAC,

everyones's gone to Duke.

 

The intersection is not a sphere....only the shared volume of the intersection is a sphere.

A 4x4x4 cm cube is 64 cubic centimeters.

A cylinder 3 cm in diameter and 4 cm long is 28.27 cubic centimeters.

A sphere 3 cm in diameter is 14.14 cubic centimeters.

The volume of an end cap is 28.27-14.14/2.

The volume of an end cap is 7.07 cubic centimeters.

Six end caps is 42.40 cubic centimeters.

The total volume of the orthogonal structure is 6 end caps and 3 cm

sphere, which is 56.53 cubic centimeters.

That leaves the holey cube with a volume of 7.47 cubic centimeters.

Verify weight before and after.

Also verify water displacement before and after.

This should satisfy you.

It would satisfy me IF you showed math to prove it (as opposed to just stating that something is the case because you assert that it is the case) OR if you (or me, or someone else) did the weight/displacement measurements to verify it empirically.

How is the intersection -- which means that which all have in common -- different from "shared volume"?

 

Here is some mathematics concerning the interpenetrating cylinders and a reference for those who want to take it further.

If it is not clear, the chinese two cylinder formula is 2/3 d^3

 

Thanks!!

That proves that the approach of assuming that the shared volume is a sphere is wrong.

Other than cranking the calculus handle, I don't know if another way. But since the two-cylinder case has an exact solution obtainable without calculus, it seems plausible that the three-cylinder case does, too. Maybe, maybe not.

 

My old fashioned tech drawing books show how to work this sort of thing out graphically.
The right circular cylinder is, of course, developable so can be opened out and the intesection curve drawn on a plane for integration.
For right angled intersections of equal cylinders the intersection angle is 45 degrees.

They also explain how to lay it out for sheet metal work using circumscribing and inscribing spheres.
This may be what BR549 was thinking of.

The excerpt is from the Penguin Dictionary of Curious and Interesting Geometry by David Wells.

This is a fascinating tome full of all sorts of useless junk like this.

Napoleon, apparently, once discovered a maths theorem.