The first sentence within the post was there from the beginning – there has been no edit.

I'll take your word for it. I thought I had saved the screen I saw without refreshing it but it's gone.

I believe I have an approach that works but it’ll take me a while to get it documented here. Too many damn chores this time of year.

 

WTF? The rules weren’t there when I first answered. Do I also have to stand on my head and use only my toes to produce an answer?

The original post has been unedited since it was first posted.

 

My answer is 480m.
pi cancels out. Hence you don't need to use pi.
I cannot see how you can solve this without knowing the volumes of the sphere and cylinder.

I got 144 km in two lines of algebra, though I set the volumes equal to each other and thus my approach is not allowed. The value of pi cancels out, so that's not an issue.

Are we allowed to use the formula for the perimeter of a circle of the surface area of a sphere?

 

I got 144 km in two lines of algebra, though I set the volumes equal to each other and thus my approach is not allowed. The value of pi cancels out, so that's not an issue.

144km it is.
I made two errors in my calculation.

 

I'm going to reveal my approach without showing all the details, which are "left to the student". (Remember how much you hated to see that phrase in a textbook!) I've just got too much to do to plow through it all.

The formula for the volume of a cylinder or for a sphere can both be derived by solving the appropriate triple integrals. These integrals are much more elegant when you use spherical coordinates but you can get to the same endpoint by starting with Cartesian coordinates, x,y and z.

Normally we solve for the included volume but we could instead solve for the excluded volume - the difference between the volume of a cube and the volume of the included sphere. Would this solution violate the rules?

Converting the volume of a cube of metal to a long square wire is trivial. It's less obvious but ditto for converting a big fat cylinder to a long skinny one: The proportional excluded cross-sectional areas are the same for both configurations, so the answer is the same as for the cube. The solution for the sphere is not so obvious but there is a way to get there.

 

I'm going to reveal my approach without showing all the details, which are "left to the student". (Remember how much you hated to see that phrase in a textbook!) I've just got too much to do to plow through it all.

The formula for the volume of a cylinder or for a sphere can both be derived by solving the appropriate triple integrals. These integrals are much more elegant when you use spherical coordinates but you can get to the same endpoint by starting with Cartesian coordinates, x,y and z.

Normally we solve for the included volume but we could instead solve for the excluded volume - the difference between the volume of a cube and the volume of the included sphere. Would this solution violate the rules?

Converting the volume of a cube of metal to a long square wire is trivial. It's less obvious but ditto for converting a big fat cylinder to a long skinny one: The proportional excluded cross-sectional areas are the same for both configurations, so the answer is the same as for the cube. The solution for the sphere is not so obvious but there is a way to get there.

I would argue that in determining the formula for the excluded volume of a sphere, you have effectively found a formula for the volume of a sphere – and therefore this is not allowed.

 

I got 144 km in two lines of algebra, though I set the volumes equal to each other and thus my approach is not allowed. The value of pi cancels out, so that's not an issue.

Are we allowed to use the formula for the perimeter of a circle of the surface area of a sphere?

I would think that the use of the formula for the perimeter of a circle or the surface area of a sphere are allowed – with the proviso that they are not a surrogate for the volume of a sphere or the area of a circle (and of course you cannot use the value of pi).

 

I would argue that in determining the formula for the excluded volume of a sphere, you have effectively found a formula for the volume of a sphere – and therefore this is not allowed.

Glad I didn't bother doing the math then!

 

Assume your machine works by forcing the sphere through a series of ever smaller holes. This wouldn't really work, but this is a puzzle, not an engineering design.
The first hole is 300 mm in diameter; the resulting blob is 4 * 600 mm.
Next hole is 30 mm in diameter leaving a blob of 100 * 4 * 600 mm
Next is 3 leaving a blob of 100 * 100 *4* 600 mm.
Finally a 1 leaving a wire of 9 * 100 * 100 * 4 * 600 mm.

I don't know if I'm using an outlawed piece of knowledge by knowing the hole to volume increase is ratio of the holes squared.

Aside from yielding a wrong answer, this method would seem to claim that it doesn't matter whether the original object is a sphere or a cylinder, especially considering that you very quickly end up with a something virtually indistinguishable from a cylinder. Yet if you started off with a cylindrical disk 600 mm in diameter and of whatever length you wanted, you would not expect to always get the same result, let alone the exact same result as for a sphere.

In fact, your approach seems to apply only to a right circular cylinder whose height is equal to the diameter (sometimes referred to as a "square-bored cylinder").

 

In fact, your approach seems to apply only to a right circular cylinder whose height is equal to the diameter (sometimes referred to as a "square-bored cylinder").

That's a special case of the "big fat cylinder" I referenced in #25. Getting from one cylinder to another (the wire) is easy.

Here's a crude, napkin approach:

The out-scribed cube is 600^3 = 216 x 10^6 mm^3.
The inscribed cube is [600/sqrt(2)]^3 = 216/2•sqrt(2) x 10^6 mm^3 = 76.4 x 10^6 mm^3
​

The shortest 'estimate' would be the inscribed cube converted to a square wire 1mm square. This is 76.4x10^6 mm = 76.4km.
The longest estimate would be the out-scribed cube converted to a square wire 1/sqrt(2). This is 216 x 10^6 mm^3 • sqrt(2) = 305.5 km

We know the answer must lie between these extreme estimates.

 

Yes, the TS stated my solution was wrong, right after I submitted it and upon reflection I realized it was wrong. Remember, this is a puzzle; it seems clear that there is a non-algebraic (not necessarily non- mathematical) solution.

 

A sphere with a 600 mm diameter has a volume equal to the volume of a cylinder with a radius of 300 mm and a H of ~400 mm. This is because 400 is 2/3 of 600. Sphere/cylinder volume ratio.

If we take H(400) and multiply it by 4......and half R(300 mm).......we should have the same volume.

Is that true and does that work?

If so, then step R down to .5 mm. And that results in length.

Will that work?

I can not check or show work.....because I don't know math. That was a condition of problem, right?

I used play dough.

 

I don’t think we’re allowed to pull the volume ratio out of thin air, since computing it requires knowing the formula for each.

 

It did not come from thin air. It came from a proper cylinder ratio of H/R.....4/3. If you roll play dough down to R of 150 mm.........the H will be 1600 mm.

If we half R, 8 more times we get .5859 mm, a little over .5mm for R.................and 104,857,600 mm for H...which is now length of wire.

It might be a little off........my dexterity is off and my play dough expired long ago.

 

It did not come from thin air. It came from a proper cylinder ratio of H/R.....4/3. If you roll play dough down to R of 150 mm.........the H will be 1600 mm.

If we half R, 8 more times we get .5859 mm, a little over .5mm for R.................and 104,857,600 mm for H...which is now length of wire.

It might be a little off........my dexterity is off and my play dough expired long ago.

What is "a proper cylinder" and why is the H/R ratio of one 4/3?

 

The volume of a sphere(R=300)......is equal to the volume of a cylinder with radius 300 and height of 400.

Because...volume of sphere = .66X volume of cylinder with height of D(600).

Sphere volume to cylinder volume ratio.

For the 1/2R and 4xH iteration to work.......cylinder must start with 4/3 or H/R ratio.

I am using play dough. Does that match you math?

 

The volume of a sphere(R=300)......is equal to the volume of a cylinder with radius 300 and height of 400.

Because...volume of sphere = .66X volume of cylinder with height of D(600).

Sphere volume to cylinder volume ratio.

For the 1/2R and 4xH iteration to work.......cylinder must start with 4/3 or H/R ratio.

I am using play dough. Does that match you math?

Where does this 4/3 ratio come from? Is it exact? Remember, the problem asked for an exact answer. Can you really claim that your play dough experiments let you claim even three sig figs?

 

I am not using math for my premise. Using mathematics....please calculate the volume of a 300 mm R sphere.

Now calculate the volume of a cylinder with a R of 300 mm and a H of 600 mm.

The volume of the sphere should equal .66 the cylinder volume. To make the cylinder equal to sphere.....chop 200 mm of H off cylinder. Now volume of cylinder(with 400 mm H) EQUALS volume of sphere.

If I start with a "button shape" cylinder(H=400,R=300)........and roll til R reduces to 150 mm.......the H appears to increase by 4.

It's only an apparent result. A guess. Using that guess, I used elementary math for further steps.

I was just trying to solve without using pi or volume formulas. Only using known ratios. If my play dough apparent ratio(1/2R,4xH) is not right, then my answer is wrong.

It's all based on eyeballing play dough. Please tell me if my eyeballing is correct.

 

Again, where does this "known ratio" come from? How is it known? Are you claiming that you can eyeball some play dough and determine an exact ratio of something?

 

There is a known ratio of sphere volume to cylinder volume.

Edit: just like cone volume.