Volume of cube with three holes

[Will_H]

Hello. I have been set a problem by a student. It appears to be fairly old, but I cannot find much info in the web. Here it is:

A 1" cube has three holes drilled in it, each hole connecting the centres of opposite faces, thus forming a cavity in the centre joined to the surface by six holes. The holes have diameter 1/2". What is the volume of the remaining material?

I have calculated the solution for two holes; the object shared by by each of the two holes is one twelth of a cubic inch (found by integrating for the volume of the central shape, a stack of squares with sides that varying as the cord of a circle).

I struggled to find the solution for three holes. I have found that the shape in the centre is called a Steinmetz Solid and has volume (16-sqrt(128))*r^3.

Using this I have calculated the final volume of the holey cube to be 0.567608 cubic inches, while a 3DCAD model has been constructed and the volume is 0.58773.

Does anyone know the answer, have a solution or have any help?

Thanks...

 

[Gerenuk]

I made a start, but got confused by the hassle of all the boundary conditions. Maybe you can proceed. I started with a coordinate system and the origin at the center.
The boundary conditions are
$$x^2+y^2\geq \frac{1}{4}$$
$$y^2+z^2\geq \frac{1}{4}$$
$$z^2+x^2\geq \frac{1}{4}$$
$$0\leq x\leq 1$$
$$0\leq y\leq 1$$
$$0\leq z\leq 1$$
$$x\geq y\geq z$$
I used some symmetry arguments to restrict myself to positive values only. In the end you'd have to multiply that volume by 6 to get the final answer.

Edit: First of all apparently I used a cube 8 times the volume and also I just noticed cylindrical coordinates (that I proposed before) are pointless

Two of the inequality are superfluous taking into account the last one. So we end up with
$$y^2+z^2\geq \frac{1}{4}$$
$$0\leq x\leq 1$$
$$0\leq y\leq 1$$
$$0\leq z\leq 1$$
$$x\geq y\geq z$$

This should be easy to solve, but somehow I don't manage to get a good result :(

 

[Gerenuk]

Heureka. I did a stupid mistake in my calculation. Indeed the above inequalities and symmetry considerations yield an easy solution.

$$V=1+\frac{1}{\sqrt{32}}-\frac{3\pi}{16}=0.587728072...$$

Oh, I noticed there is no pi-free solid of this kind :(

 

[blkqi]

That's a very nice calculation. I presume the $$1/\sqrt{32}$$ quantity should be equivalent to the volumes enclosed by the intersection of two of the cylinders and the intersection of three of the cylinders, but I can't seem to reckon this with calculus. What am I neglecting?

Edit: I should explain my thoughts. Suppose you have a 1" cube. You drill out a ¼" hole; a cylinder of volume π¼². Drill a second identical hole on an adjacent face and you've extracted π¼² less the intersection of the now two holes. Similarly drill the third hole to extract π¼² less the intersection of the now three holes. Done. In total we have 1in³ minus 3π¼² plus the two intersections described. But my calculations show less than $$1/\sqrt{32}$$ added back into the total volume. I reckon something closer to what the OP had.

 

[Gerenuk]

That's a very nice calculation. I presume the $$1/\sqrt{32}$$ quantity should be equivalent to the volumes enclosed by the intersection of two of the cylinders and the intersection of three of the cylinders, but I can't seem to reckon this with calculus. What am I neglecting?

Not completely sure what you mean. In any case there is a difference between the Steinmetz solid which is
$$A\cap B\cap C$$
and the solid which is
$$\overline{A}\cap\overline{B}\cap\overline{C}$$
where A,B and C are the cylinders. Both together do not add up to one (which should be $8r^3$).

 

[Will_H]

Thanks for your replies.

I've got a mental block.

Volume remaining with one hole is 1-pi/16

Volume remaining with two holes is (1-pi/16)-(pi/16-1/12)

Volume remaining with three holes is (1-pi/16)-(pi/16-1/12)-(pi/16-(16-sqrt(128))*0.25^3))

This is approx = 0.567508...

Please forgive my ignorance, your talking to an engineer who visualizes drilling when solving this problem. With each cut after the first, you must consider the volume of 'air' being removed from the cylinder being removed. In the case of the the third hole, is the object not a Steinmetz Solid as described?

 

[Gerenuk]

Volume remaining with two holes is (1-pi/16)-(pi/16-1/12)

I am not sure how you get this equation (in particular how to calculate the "air" removed). I find it hard to visualize, so I did it purely mathematically You might be right though up to here.

In the case of the the third hole, is the object not a Steinmetz Solid as described?

I think there is no Steinmetz Solid in this problem. I'm not absolutely sure and I find it hard to visualize. I cannot imagine it is as simple as you describe.

That's why I used Maths and luckily you could support my result with your simulation
Let me know if you want to have more explanation how to integrate these inequalities.

As a general formula, the total material cut out is
$$\Delta V=3\cdot\text{(volume one cylinder)}-\sqrt{2}\cdot\text{(diameter of holes)}^3$$
i.e. the air that you cut is exactly $\sqrt{2}d^3$.

 

[slider142]

The Steinmetz solid (I learned the shape as a box vault) is present as the intersection of the 3 orthogonal air cylinders.
You can use set theory to note that the volume of the remaining cube should be 1 - (3C - 2S), where C is the volume of a cylinder and S is the volume of the Steimetz solid, which is given on this page as $8(2 - \sqrt{2})\frac{1}{4}^3$. You remove 2S from the volume of the air because 3C has counted that volume 3 times, whereas we have only removed it once.
The final number should be $1 - (\frac{3\pi}{16} - 16(2 - \sqrt{2})\frac{1}{4}^3)$ which comes out to $\frac{1}{16}(24 - 4\sqrt{2} - 3\pi)$ which is about 0.557.

 

[Gerenuk]

@Slider: You should read the thread before saying something incorrect.

Will H has used 3DCAD to show that the result should be 0.5877. I can confirm this result by calculation.

Also I already mentioned set theory. We are looking for the solid
$$\overline{A}\cap\overline{B}\cap\overline{C}$$.
You however want to calculate $1-3(A+B+C)+2(A\cap B\cap C)$ which is something different.

In fact instead of adding twice the Steinmetz solid, one rather needs to add $\sqrt{2}d^3$.

 

[slider142]

@Slider: You should read the thread before saying something incorrect.

Will H has used 3DCAD to show that the result should be 0.5877. I can confirm this result by calculation.

Also I already mentioned set theory. We are looking for the solid
$$\overline{A}\cap\overline{B}\cap\overline{C}$$.
You however want to calculate $1-3(A+B+C)+2(A\cap B\cap C)$ which is something different.

In fact instead of adding twice the Steinmetz solid, one rather needs to add $\sqrt{2}d^3$.

I see no difference in your result other than notation. The volume I calculated is the volume of the cube minus the volume of the union of the cylinders, which is the same as the complement of the intersection of the cylinders with respect to the cube. Proof as follows:
We take U to be the cube:
In set theoretic terms,
$$|\overline{A}\cap\overline{B}\cap\overline{C}|$$
$$= |\overline{A\cup B \cup C}|$$
$$= |U - A\cup B \cup C|$$
$$= |U| - |A\cup B \cup C|$$
$$= |U| - (|A| + |B| + |C| - |A\cap B| - |A \cap C| - |B\cap C| + |A\cap B \cap C|)$$
Now |A| = |B| = |C|, and by Inclusion-Exclusion, we have
$$|A\cap B| + |A \cap C| + |B\cap C|$$
$$= |(A \cap B) \cap (A\cap C)| + |(A\cap B) \cap (B \cap C)| + |(A\cap C)\cap (B\cap C)|$$
$$= 3|A\cap B\cap C|$$.
Thus, the previous sum is just $|U| - 3|A| + 2|A\cap B \cap C|$.
From the information given, |U| = 1, $|A| = \frac{\pi}{16}$ and $|A\cap B\cap C| = (2 - \sqrt{2})\frac{1}{4}$, each in cubic inches.
Unless there is a flaw in the logic, an assumption being made in the derivation of the other values is incorrect.

 

[Gerenuk]

Unless there is a flaw in the logic, an assumption being made in the derivation of the other values is incorrect.

3DCAD is the most direct method confirmed by two people. You should be cautious suggesting that everyone is wrong unless they can prove your proofs wrong. They always will be able to, but it is just tendious to go through every erroneous argument.

And if you really want to see the flaw here it is.

$$|A\cap B| + |A \cap C| + |B\cap C|$$
$$= |(A \cap B) \cap (A\cap C)| + |(A\cap B) \cap (B \cap C)| + |(A\cap C)\cap (B\cap C)|$$
$$= 3|A\cap B\cap C|$$

A simple Venn diagram shows that this equation is incorrect. The equation is wrong by three terms like
$$(A\cap B)\cap\overline{C}$$

 

[Will_H]

Gerenuk: That is exactly what I'm having trouble with, visuallising the cuts and objects.

The volumes calculated in previous post:

(1-pi/16) is the volume of the remaining object after the first hole is cut.

(pi/16-1/12) is the volume of the second hole removed minus the volume of the region the drill encounters as it pass through the first hole. This was found using V = (16r^3)/3.

I have modeled the two hole solid and get a small discrepancy; it gets worse when finding volume for three hole object.

There is clear problem with my mental image of the cutting process and I cannot fix it.

Please could enlighten me as to how you integrate the inequalities and how you formed them in the first case. Your help would be much appreciated.

 

[slider142]

3DCAD is the most direct method confirmed by two people. You should be cautious suggesting that everyone is wrong unless they can prove your proofs wrong. They always will be able to, but it is just tendious to go through every erroneous argument.

And if you really want to see the flaw here it is.

A simple Venn diagram shows that this equation is incorrect. The equation is wrong by three terms like
$$(A\cap B)\cap\overline{C}$$

From where do you get these terms involving complements (ie., diagram)? Inclusion-Exclusion generates the equation:
$$|(A\cap B)\cup (A\cap C) \cup (B\cap C)|$$
$$= |A\cap B| + |A\cap C| + |B\cap C| - |(A\cap B)\cap (A\cap C)| - |(A\cap C) \cap (B\cap C)| - |(A\cap B) \cap (B\cap C)| + |(A\cap B)\cap (A\cap C) \cap (B\cap C)|$$
The above simplifies to
$$|A\cap B \cap C| = |A\cap B| + |A\cap C| + |B\cap C| - 4|A\cap B\cap C|$$
$$3|A\cap B\cap C| = |A\cap B| + |A\cap C| + |B\cap C|$$
which is directly applied in the proof. This is just set algebra. No recourse to diagrams is necessary.

 

[Gerenuk]

Gerenuk: That is exactly what I'm having trouble with, visuallising the cuts and objects.
...
There is clear problem with my mental image of the cutting process and I cannot fix it.

I think you are doing much better at visualizing than me! It is just very complicating and maybe 3DCAD can help you. The discrepancy should be parts which are in cylinder A and cylinder B but not in in cylinder C for example. Maybe this is the only way to see what is missing.
When dealing with higher dimensional space it is also easier to do the Maths only.

(pi/16-1/12) is the volume of the second hole removed minus the volume of the region the drill encounters as it pass through the first hole. This was found using V = (16r^3)/3.

I am not sure how you get this 1/12 so easily? I apologize that I do not try to understand your reasoning indepth. I find it very hard to visualize and I would get back to purely mathematical calculations again - which won't help understanding either

Edit: I looked up where the 1/12 seems to come from. I believe for two cylinders only you are absolutely right, by subtracting the intersection of these two cylinders, which is 1/12. However it doesn't work for 3 cylinders.

I'll work out the numbers and post the volume integration soon.

Inclusion-Exclusion generates the equation:
$$|(A\cap B)\cup (A\cap C) \cup (B\cap C)|$$
$$= |A\cap B| + |A\cap C| + |B\cap C| - |(A\cap B)\cap (A\cap C)| - |(A\cap C) \cap (B\cap C)| - |(A\cap B) \cap (B\cap C)| + |(A\cap B)\cap (A\cap C) \cap (B\cap C)|$$
The above simplifies to
$$|A\cap B \cap C| = |A\cap B| + |A\cap C| + |B\cap C| - 4|A\cap B\cap C|$$
$$3|A\cap B\cap C| = |A\cap B| + |A\cap C| + |B\cap C|$$

I personally find it a bit cocky ignoring all direct and simple proofs and clinging to a (wrong?) complicating one. I can again point out the mistake. For no reason you equated
$$|(A\cap B)\cup (A\cap C) \cup (B\cap C)|\neq |(A\cap B)\cap (A\cap C) \cap (B\cap C)|$$
Venn diagrams are a basic visualization tool in set theory. Please look it up so you can find mistakes in equations yourself.

 

[Gerenuk]

Gerenuk: That is exactly what I'm having trouble with, visuallising the cuts and objects.
...
There is clear problem with my mental image of the cutting process and I cannot fix it.

I think you are doing much better at visualizing than me! It is just very complicating and maybe 3DCAD can help you. The discrepancy should be parts which are in cylinder A and cylinder B but not in in cylinder C for example. Maybe this is the only way to see what is missing.
When dealing with higher dimensional space it is also easier to do the Maths only, so I wouldn't be worry if it is hard to visualize such intersection and one uses Maths instead.

(pi/16-1/12) is the volume of the second hole removed minus the volume of the region the drill encounters as it pass through the first hole. This was found using V = (16r^3)/3.

I am not sure how you get this 1/12 so easily? I apologize that I do not try to understand your reasoning indepth. I find it very hard to visualize and I would get back to purely mathematical calculations again - which won't help understanding either

I'll work out the numbers and post the volume integration soon.

Inclusion-Exclusion generates the equation:
$$|(A\cap B)\cup (A\cap C) \cup (B\cap C)|$$
$$= |A\cap B| + |A\cap C| + |B\cap C| - |(A\cap B)\cap (A\cap C)| - |(A\cap C) \cap (B\cap C)| - |(A\cap B) \cap (B\cap C)| + |(A\cap B)\cap (A\cap C) \cap (B\cap C)|$$
The above simplifies to
$$|A\cap B \cap C| = |A\cap B| + |A\cap C| + |B\cap C| - 4|A\cap B\cap C|$$
$$3|A\cap B\cap C| = |A\cap B| + |A\cap C| + |B\cap C|$$

I personally find it a bit cocky to ignore all direct and simple proofs and clinging to a (wrong?) complicating one. I can again point out the mistake. For no reason you equated
$$|(A\cap B)\cup (A\cap C) \cup (B\cap C)|\neq |(A\cap B)\cap (A\cap C) \cap (B\cap C)|$$
or maybe
$$|(A\cap B)\cup (A\cap C) \cup (B\cap C)|\neq |A\cap B \cap C|$$
Venn diagrams are a basic visualization tool in set theory. Please look it up so you can find mistakes in equations yourself.

 

[Gerenuk]

I set a coordinate system (x,y,z) with origin at the center of the cube.

You want to integrate a volume element
$$V=\iiint_A \mathrm{d}x\mathrm{d}y\mathrm{d}z$$
in a region A which is described by
$$x^2+y^2\geq \frac{1}{16}$$
$$y^2+z^2\geq \frac{1}{16}$$
$$z^2+x^2\geq \frac{1}{16}$$
$$-\frac12\leq x\leq \frac12$$
$$-\frac12\leq y\leq \frac12$$
$$-\frac12\leq z\leq \frac12$$

Thanks to you I learned that transforming such kind of equation into suitable integral boundary conditions is called cylindrical algebraic decomposition (use Mathematica if you ever have a complicating problem). Luckily in this particular case symmetry considerations simplify a lot.

We restrict ourselves to
$$0\leq x\leq y\leq z$$
This will give us a 1/48 of the total volume since due to symmetry all other missing coordinates would be found from permuting x, y and z (factor 6) or changing the sign of the coordinates (factor 8).

Due to our new restriction your inequalities simplify to
$$x^2+y^2\geq \frac{1}{16}$$
$$0\leq x\leq y\leq z\leq \frac12$$
(everything else drops out)

This can be written
$$0\leq x\leq\frac12$$
$$\sqrt{\frac{1}{16}-x^2}\leq y\leq\frac12$$
$$x\leq y\leq \frac12$$
$$y\leq z\leq \frac12$$

We need to split into two cases so that we have to take into account only one of the middle two inequalities for y. (depending on the interval one inequality will dominate).
$$0\leq x\leq \frac{1}{\sqrt{32}}$$
$$\sqrt{\frac{1}{16}-x^2}\leq y\leq \frac12$$
$$y\leq z\leq \frac12$$
and
$$\frac{1}{\sqrt{32}}\leq x\leq \frac12$$
$$x\leq y\leq \frac12$$
$$y\leq z\leq \frac12$$

Therefore
$$V=48\left(\int_0^{\frac{1}{\sqrt{32}}}\mathrm{d}x\int_{\sqrt{\frac{1}{16}-x^2}}^{\frac12}\mathrm{d}y\int_y^{\frac12}\mathrm{d}z+\int_{\frac{1}{\sqrt{32}}}^{\frac12}\mathrm{d}x\int_x^{\frac12}\mathrm{d}y\int_y^{\frac12}\mathrm{d}z\right)$$

Please check it for yourself, as initially I used slightly different numbers for myself.

 

[Gerenuk]

I got good news for both of you!

In one of sliders equations there is actually some truth.
$$|\overline{A}\cap\overline{B}\cap\overline{C}|$$
$$= |U| - (|A| + |B| + |C| - |A\cap B| - |A \cap C| - |B\cap C| + |A\cap B \cap C|)$$

Therefore you need to calculate
V=(Cube)-3(Cylinder)+3(Bicylinder)-(Tricylinder)

(see http://mathworld.wolfram.com/SteinmetzSolid.html)

So it seems sliders mistake was that he thought the bicylinder and the tricylinder have equal volume.

 

[slider142]

For no reason you equated
$$|(A\cap B)\cup (A\cap C) \cup (B\cap C)|\neq |(A\cap B)\cap (A\cap C) \cap (B\cap C)|$$

This statement is nowhere in my proof. Where did you find it?

or maybe
$$|(A\cap B)\cup (A\cap C) \cup (B\cap C)|\neq |A\cap B \cap C|$$

Again, please highlight where you have found this equation in my post. This nonsense is not there either.

In aside, it's strange that you find an analytic triple integral that you used Mathematica to evaluate simpler than a "complicated" set theory equation with three terms that can be done on paper. However, this is a personal opinion, and has nothing to do with mathematics.

I see I will have to post a diagram of the situation. This will be posted later to clarify matters.

 

[Gerenuk]

This statement is nowhere in my proof. Where did you find it?

I really don't get why you are against everything the world around you knows.
This set theory statement is my best guess of what you were trying to do in your derivation. Your "simplication" is wrong and comes out of nowhere.

And I am shocked that despite my second proof of the result 0.5877 with your own set equations, you still don't accept it. Feel free to explain yourself in more detail, but please do not post wrong rearrangements again, but state which rule you used.

 

[slider142]

I really don't get why you are against everything the world around you knows.
This set theory statement is my best guess of what you were trying to do in your derivation. Your "simplication" is wrong and comes out of nowhere.

And I am shocked that despite my second proof of the result 0.5877 with your own set equations, you still don't accept it. Feel free to explain yourself in more detail, but please do not post wrong rearrangements again, but state which rule you used.

What? I still don't know what line you have a problem with, as the statement you posted appears nowhere in my proof. Also, I have yet to rearrange any of my lines, so I'm not sure what "rearrangements" you are referring to.

 

[Gerenuk]

I wrote which line a couple times. Well, this is getting to silly for me. It's like you skipping most the lines people write and make your own crazy story (with wrong Math) from the rest. For me there is no point anymore to explain where you went wrong, if it seems you ignoring anything that doesn't suit your needs. Maybe someone else can help you. Good luck.

 

[slider142]

I wrote which line a couple times. Well, this is getting to silly for me. It's like you skipping most the lines people write and make your own crazy story (with wrong Math) from the rest. For me there is no point anymore to explain where you went wrong, if it seems you ignoring anything that doesn't suit your needs. Maybe someone else can help you. Good luck.

So instead of actually quoting a line from my post, you will continue to claim that you have found a problem that you do not describe but have quoted "a couple of times" and post instead erroneous equations found nowhere in my posts, ascribe them to me and claim that is what I meant. You then ask me to stop posting "rearrangements". This is a bit tiring, so I will save both of us the trouble and contribute to other threads instead.

 

[Gerenuk]

In this point you are right. I'm guilty in not quoting it again and I actually was going to edit my post. I apologize. Here is the line. And please check it carefully and do no invent another unchecked proof to justify it.
$$|A\cap B| + |A \cap C| + |B\cap C|$$
$$= |(A \cap B) \cap (A\cap C)| + |(A\cap B) \cap (B \cap C)| + |(A\cap C)\cap (B\cap C)|$$
This line is no logical consequence of any of the previous lines.

 

[Gerenuk]

And btw, to check this line just take an example
A={1,2,3,4}
B={2,3,7,8}
C={3,4,6,7}
which yields in your equation
2+2+2=1+1+1
So clearly this set theoretic equation is wrong.

 

[blkqi]

There is clear problem with my mental image of the cutting process and I cannot fix it.

I first visualized the volume in the same way. Here is the resolution to this problem:
In the last hole drilled, you imagine the volume removed to be the volume of the cylinder drilled (call it cylinder $$C$$, the last one) less the intersection of this cylinder and the two cylinders previously drilled--that is $$(A \cup B)\cap C$$, NOT $$A\cap B\cap C$$; this is the flaw in your thinking...here is a Venn diagram to see this clearly.

The total volume of the cube, less the three orthogonal cylinders is thus
$$|\overline{A}\cap\overline{B}\cap\overline{C}|$$
$$= |U|-\bigl[ |A|+(|B|-|A\cap B|)+(|C|-|(A\cup B)\cap C|) \bigr]$$
$$= |U| - \bigl[|A| + |B| + |C| - |A\cap B| - |A \cap C| - |B\cap C| + |A\cap B \cap C|)\bigr]$$
(as Gerenuk and slider pointed out above)

 

[Bill Crean]

I have just scanned the above set of posts. I have applied simple tests to prove many have wrong answers. The Steinmetz solid is not the solution to the question asked.

Try the following tips: for convenience chose the radius of the 3 intersecting cylinders as 0.5 units. This means the intersecting cylinders are contained in a 1x1x1 cube. Start your thinking by having a solid cube and drilling the first face through the centre to the opposite centre. You will have removed a volume pi*r^2*h, which is for this problem

= pi*0.5*0.5* 1.0 or pi/4 (= 0.785398163)

With two more orthogonal holes to drill, we know that the volume of the space created by 3 crossing cylinders, within the cube, will be greater than the above number (pi/4). Effectively, therefore, after first hole is drilled, 78.5398163% of the volume is removed. So any final answer given above with less than than this ratio is wrong.

A Steinmetz solid is the mathematical intersection of the 3 cylinders. That is, those sections of the drilled round holes, which do not reach either of the two other holes will be excluded. That is why the volumetric result for Steinmetz solid will yield the wrong answer.

In any case it is possible to solve this problem without using calculus or any advanced maths. The maths known by the ancients is sufficient. One just uses the methods of chopping things up until the primative shape you are left with can be calculated by elementary maths.

I have a really good solution to this problem, which I wrote down 19 years ago, but there is not enough room to write it down here.

 

[Bill Crean]

In my previous post I forgot to give the numerical answer to the 3 cylinder problem presented; it is about 0.94 cubic units.

 

[Gerenuk]

I have just scanned the above set of posts. I have applied simple tests to prove many have wrong answers. The Steinmetz solid is not the solution to the question asked.

That's good for you, but this whole post is about that it's not the Steinmetz solid, but if you do it right, you can use the Steinmetz solid for the solution.

 

[Bill Crean]

There is no doubt numerous ways to solve this problem. When I was given it in 1970, there was a stipulation to only use elementary maths, that is, no calculus.

I have just gone into my attic to find my old manuscript of the solution. With all the hand drawings it covers many A4 pages. I will try to shorten it by leaving out lines of working and then photocopy it for posting.

I may be some time.

 

[Gerenuk]

Ths solution is written in this thread and it coincides with the CAD simulation. It's not saying that the solution is the Steinmetz solid! It's rather saying how to use it correctly.

 

[Zamaster]

I'm sorry but I looked this problem up on google, because I remember this problem being thrown at me by a tyrannical 8th grade algebra teacher trying to show what kind of problems 'he thought were difficult.' Now, I've taken a vector calculus class I noticed the problem is really easy -> if you take it from a geometric calculus standpoint. I say this because as soon as I learned the general domain version of a double integral, an analytic solution is obvious, as i simply wrote it out on paper while my professor was talking.

It comes out to be, given d the diameter of the hole and l the side length of a cube, l^3 - (3/4)*pi*l*d^2 + 2^(1/2)*d^3. Its quite elegant really. And you get this with four integrals, that is to say using other method appear to give way overblown answers. I'll only post my work if anyone near this (should be) dead topic is interested enough.

 

[Zamaster]

Haha reading this thread, yes, that's how I got the answer

Cube - ThreeCylinders + WeirdCircledCylinderThing + EvenWeirderCircledDoubleCylinderThing

 

[Bill Crean]

Zamaster's formula seems promising, but Zamaster does not state whether the formula is for the volume of the material removed from the cube or the volume remaining. For the case L=3 and D=1, we can easily calculate the volume of 6 cylinders with diametre = 1 and height = 1. It is of course 6*pi*D^2*L/4, which is, for D=1 and L(height)=1, 4.71238898 cubic (A) units. This is already greater than the value given by zamaster's formula and I have yet to include the intersecting part in the centre of the 3*3*3 cube. See below

L^3 - (3/4)*pi*L*D^2 + 2^0.5*D^3 = 9 - 7.0685833471 + 1.414213562 = 3.345630215.

The intersecting 3 cylinders, which occurs within the centre cube of the 3*3*3 occupies a volume (according to my calculations) of about 0.94 cubic units (B). Hence, the total volume removed from a 3*3*3 cube when a hole of diametre 1 unit is drilled through the centre of 3 orthogonal faces is A + B = 5.65 cubic units (2 decimal places).

The method I used was a non-calculus method and it was how the ancient Greeks would have solved it, if they were given the problem (namely section until a primative shape is established). I have a formula for my solution and it is given by:

3*pi*r*r(L-2r) + 8*r^3(2^0.5 - 1/2) + 12*r^3(pi/4 - 2^-0.5)(1 - 2^-0.5) - E (.05*r^3)

where r is the radius of the hole, L is the length of an edge of the cube, and E is an error term, which can be calculated depending how many decimal places of accuracy is required.

I stated Zamaster's solution looked promising: it might be a coincidence but 9 minus my answer (5.65) = 3.35, which is close to Zamaster's solution for the L=3, D=1 case.

 

[Bill Crean]

Did anyone spot my error - L^3 = 27, and so Zamaster's formula gives 21.34563009 (= 21.35 to 2 dec. places) for L=3 and D=1. This means that the Zamaster's formula yields the volume remaining after the holes are drilled, because my formula yields the volume removed namely 5.65 (2 dec. places) i.e. 27 - 5.65 = 21.35.

This means the problem has been solved successfully by two different methods.

 

[Zamaster]

Volume remaining! The fact that it comes out to square root of two boggles my mind! Yeah, my solution is entirely analytical without any approximations, but yes I did use calculus.