Tuesday Puzzle

You can't get a true cylinder core

The ends would be rounded by the outer surface of the sphere, unless you cut them off and taped them back on.

Steve

True, but

the hole in the sphere is cylindrical, and it is the cylindrical hole that has a height of 6". Sorry if I wasn't clear on that. You are right, if the core removed were the item of interest, it would be a cylinder with two domed ends.

Re: Tuesday Puzzle

Since the diameters of both sphere and core bit are unspecified, I'll choose a carbide tipped, brad point bit of diameter 0.0 inches. To get a 6" core using this bit I'll need a 6" sphere.

Since the volume of my core is 0 cubic inches, the problem is reduced to simply finding the volume of the 6" sphere.

I'll submit that you cannot tell from info given

If you mean to imply that the length of the cylinder is 6", then it is impossible to tell. Since the domes on the end of the cylinder have to be excluded, it is impossible to determine the remaining volume without knowing the width of the bit used to bore the hole.

The answer is conceivably anthing between 0 (for a sphere that was cored by a bit of the same diameter) to the volume of the sphere (cored by a bit of 0 diameter).

If you mean to imply that the cored "cylinder" is inclusive of the two domes AND 6" in length from dome peak to dome peak, it's probably much more approachable.

Let me know before I fire up some brain cells this AM :-)

S

example

If you had a sphere of diameter 5" and you drilled it with a 3" bit, you would be left with a "donut", the rounded outer surface of which is a portion of the original sphere, and the inner surface of which is a cylinder with diameter 3" and height 4". That's the shape we are talking about, but only the height of the cylinder is given, not the diameter of the cylinder or the diameter of the sphere.

Re: Tuesday Puzzle

Excellent solution. If you accept that the original problem does indeed have a solution without any additional information, picking an extreme case gives the answer. And you are right, at least for that case, 36* pi.

But Steve's contention that there is not enough information seems pretty reasonable. Would the same answer hold if it were a larger sphere, with a larger hole?

I understand what you are looking for

but I don't think it can be defined without more information.

If there was a way to relate the diameter of the sphere and the radius of the cylinder, I could probably figure it out, but I don't think it's a trivial solid geometry problem based upon what I see.

Pi*r ^^3- pi*r(2)**2*l- (volume of the end caps)

Even with the radius defined, the volume of the end caps is not trivial.

Steve

Re: I understand what you are looking for

The relationship between the diameter of the sphere and the diameter of the bore is not so hard to calculate.

The diameter of the sphere is the length of the hypotenuse of the right triangle formed by the diameter and height of the bore. (This line starts and ends on the surface, and passes through the center of the sphere)

The height is the only known parameter (6") so:

D_sphere = sqrt(D_bore^2 + 6^2)

That said, I don't see an easy path for a general solution yet.

Then again

The formula for a Segment of a sphere is a lot easier to work with. Especially with h=6 and r1 == r2.

V = (Pi/6)(3r1^2+3r2^2+h^2)h

Re: Tuesday Puzzle

The problem cannot be solved with only the given dimension of the cylinder. I will present here a case that gives easily calculated numerical results, i.e. all integers.

I take a sphere of radius 5.

The part of the sphere that is drilled out consist of the cylinder and two identical spherical caps that are at the top and at the bottom of the cylinder.

Using the Pythagorean theorem, we can easily find that the radius of the cylinder is 4. This is also the radius of the base of the spherical cap. The height of the spherical cap is 2.

The volume of the spherical cap is pi*h*(3*a^2 + h^2)/6, where a is the radius at the base and h is the height. With our numbers this gives pi*104/6,

and since there are two of them, their total volume is pi*104/3.

The cylinder has volume pi*(4^2)*6 = pi*96.

The total removed volume volume is pi*104/3 + pi*96 = pi*392/3.

The volume of the sphere is (4/3)*pi*(5^3) = pi* 500/3.

The remaining part of the sphere is pi*500/3 - pi*392/3 = pi*108/3 =pi*36.

Re: I understand what you are looking for

"but I don't think it can be defined without more information."

Actually, it can, which is surprising , or even counter-intuitive (to me). Bill gave the answer for a sphere of diameter 6, and a cylinder of diameter 0. Surprisingly, the volume is the same if the sphere is earth-sized, and the "hole" takes away most of the earth, leaving a ring 25,000 miles across (or whatever the earth's diameter is), 6" wide, and very very thin.

"I don't think it's a trivial solid geometry problem based upon what I see."

I'll grant you that. The piece that is not part of elementary geometry is the volume of a spherical dome (by which I mean the portion of a sphere like the endcaps. The volume of a spherical dome is pi*A*(3r^2 +A^2)/6, where A is the height of the dome (in this case, the sphere's radius, R, minus 3) and r is the radius of the dome (the radius of the cylindrical hole, NOT the curvature of the top).

"Pi*r ^^3- pi*r(2)**2*l- (volume of the end caps)"

Correction: First term needs a 4/3 multiplier

And we know that l=6.

"Even with the radius defined, the volume of the end caps is not trivial."

Yeah, not a formula that comes readily to mind (ok, not a formula I'd ever seen before!).

Combine your solution with Bill's observation

about the relationship of the sphere's radius and the hole's radius. So instead of a 3:4:5 right triangle, work with a 3: sqrt(R^2-9):R right triangle, and you will have the answer.

I must be missing something?

If I read it right the length of the CYLINDER is 6", not the diameter of the sphere. It could be a very narrow core all the way out to a very large core, but the cylinder solid minus the end caps is 6" in either case?

I'll need to see a drawing at some point...

Steve

Re: I must be missing something?

"If I read it right the length of the CYLINDER is 6", not the diameter of the sphere. It could be a very narrow core all the way out to a very large core, but the cylinder solid minus the end caps is 6" in either case?"

Yes, you are reading it correctly.

"I'll need to see a drawing at some point..."

I can't figure out how to do that in Sketchup. Maybe someone with a solid modeler or 3D CAD can draw some examples.

To try to visualize it, consider two examples already given Bill and Nikos, along with a third:

Sphere Diameter = 6", Hole diameter =0", hole height =6"

Sphere diameter = 10", hole diameter = 8", hole height = 6"

Sphere diameter = 30", hole diameter = 29.4", hole height = 6"

Here's 2D drawing, the cross-section of each, showing the original sphere and the part left when the hole is drilled.

I see what I was missing now...

I didn't realize that the hypotenuse of the triangle would have to pass through the center of the sphere. The diameter of the cylinder is always defined by:

D=SQRT(Radius of Sphere^^2-36)

Using this, I think everything can be related back to a single variable- the size of the original sphere, since only one diameter cutting bit will produce a cylinder with a 6" length?

Am I thinking clearly?

Steve

Re: I see what I was missing now...

Yes, but mixing radius and diameter in your formula.

Cylinder radius = sqrt ( Sphere radius^2-9)

or

Cylinder diameter = sqrt (sphere diameter^2 - 36)

Answer

The pieces have been given by other posters, but to tie them all together:

The volume of the sphere is 4/3 pi R^3, where R is the sphere's radius

The volume of a cylinder is pi r^2 h, where r is the cylinder's radius and h is the cylinder's height (6 in this case)

The volume of a spherical dome is pi*A*(A^2+3r^2)/6, where A is the height of the dome, and r is its radius.

We know that h=6

From Pythagorean theorem, we get that r^2 = R^2-9

And we see that A = R-3

Now we can express the volumes of all three shapes in terms of R:

Sphere: 4/3 pi R^3

Cylinder: pi (R^2 -9)*6

Dome (each): pi (R-3)*((R-3)^2+3*(R^2-9))/6

Now, taking the sphere - the cylinder - 2x the dome gives us the volume of the remaining shape. And working through the algebra, all the R terms fall out, leaving us with 36*pi, independent of the radius of the sphere!