We can place a lower bound on the volume by the following reasoning.

Each face starts a volume that is drilled right through by one cylinder only.
This volume is always 0.5cm thick.

So allowing for the overlap strips between faces this means that the minimum volume untouched is

\(2x0.5x\left( {4x4 - \frac{{9\pi }}{4}} \right) + 2x0.5x\left( {4x3 - \frac{{9\pi }}{4}} \right) + 2x0.5x\left( {3x3 - \frac{{9\pi }}{4}} \right)\)

\(8.93 + 4.93 + 1.63 = 15.79c{m^3}\)

It can be seen that on two faces alone there remains more material than allowed by BR549 in his estimate in post#34.

 

We can also place upper bounds on the volume.

As you've noted, we have a static volume consisting of that portion of the volume far enough from the center such that each face only has one cylinder cut into it. So we can start with a cube that has an inner cube already removed from it whose face is equal to the diameter of the cylinders. This will place a lower bound on the volume of the real holey cube since not all of this inner cube will be removed by the cutting process -- but we know that nothing will be removed that is not within this inner cube. So the lower bound by this very simple approach is

\(
V_{lower} \; = \; \(4cm\)^3 \; - \; \(3cm\)^3 \; - \; 6\(\pi\(1.5cm\)^2\(0.5cm\)\) \; = \; 15.79 cm^3
\)

We also know that the a sphere of diameter 3cm definitely WILL be removed from this inner volume, so we can remove this instead of the 3cm cube to get an upper bound on the volume

\(
V_{lower} \; = \; \(4cm\)^3 \; - \; \frac{4}{3} \pi \(1.5cm\)^3 \; - \; 6\(\pi\(1.5cm\)^2\(0.5cm\)\) \; = \; 28.66 cm^3
\)

While these bounds aren't as tight as we might like, they are very simple to arrive at and they ARE bounds. Assuming I cranked the math right, any answer that is outside those bounds is wrong.

 

It'salways good to discuss these things as it develops and promotes engineering thinking.

There is plenty of room for other in this discussion as well.

I look forward to a joey solution.

 

It'salways good to discuss these things as it develops and promotes engineering thinking.

Right... if only all heated arguments between people were like this one... with an objective solution in the end.

 

A tighter and simpler upper bound would be to subtract one complete 4cm long cylinder plus 4 off short stub cylinders in the 0.5cm margins, from the cube.

This comes to

\(64 - \left\{ {\left( {\frac{{9\pi }}{4}} \right)\left( {4 + \left( {4 \times 0.5} \right)} \right)} \right\} = 21.6c{m^3}\)

 

A tighter and simpler upper bound would be to subtract one complete 4cm long cylinder plus 4 off short stub cylinders in the 0.5cm margins, from the cube.

This comes to

\(64 - \left\{ {\left( {\frac{{9\pi }}{4}} \right)\left( {4 + \left( {4 \times 0.5} \right)} \right)} \right\} = 21.6c{m^3}\)

Yep, that's definitely better.

 

I wish one of you braniacs would solve this thing so I can see how close my answer was.

@joeyd999 you've been called out already but here it is again...

 

@strantor
Didn't Studiot show the answer in post #58?

 

@strantor
Didn't Studiot show the answer in post #58?

Yeah but then he went on to join wbahn in establishing upper and lower boundaries and inviting joey to solve it, which led me to believe he wasn't 100% convinced of his answer either.

 

Yeah but then he went on to join wbahn in establishing upper and lower boundaries and inviting joey to solve it, which led me to believe he wasn't 100% convinced of his answer either.

I think those are separate strands of the discussion. I still hope to do a numerical computation or an integral-based solution, but who knows when it will be.

 

So far as I am concerned the maths is established and the proof of the formulae I used published in the Wolfram link.

A good engineer should always have a 'feel' for his answer and establishing bounds as close as possible is a really good way of doing this so he can 'know' if something is wrong as well as generating confidence in his answer.

I didn't do that in the probability thread a while back with disastrous consequences so the bounds discussion here is worthwhile.

We should also not forget the few dozen other good questions in cmartinez link.

 

I would expect the volume of the holey cube to equal the following (V_2 and V_3 are defined in equations #8 and #20 on the Wolfram Alpha page about Steinmetz solids linked to in post #42):

\(V_{final} = V_{cube} - (3 \times V_{cylinder} - V_2 - V_3)
= L^3 - (3\pi r^2L - \frac{16}{3}r^3 - 8(2 - \sqrt{2})r^3)
= 4^3 - (3\pi (1.5^2)4 - \frac{16}{3}1.5^3 - 8(2 - \sqrt{2})1.5^3)\)

With these calculations I get a value of 12.99 cm^3, which is less than the lower bound conjectured above. What's up here?

 

I would expect the volume of the holey cube to equal the following (V_2 and V_3 are defined in equations #8 and #20 on the Wolfram Alpha page linked to in post #42):

\(V_{final} = V_{cube} - (3 \times V_{cylinder} - V_2 - V_3)
= L^3 - (3\pi r^2L - \frac{16}{3}r^3 - 8(2 - \sqrt{2})r^3)
= 4^3 - (3\pi (1.5^2)4 - \frac{16}{3}1.5^3 - 8(2 - \sqrt{2})1.5^3)\)

With these calculations I get a value of 12.99 cm^3, which is less than the lower bound conjectured above. What's up here?

What is V2 and V3?

Consider what they physically represent and then ask of the values that you get for them specifically makes sense.

 

What is V2 and V3?

Consider what they physically represent and then ask of the values that you get for them specifically makes sense.

V_2 = the volume common to two intersecting cylinders
V_3 = the volume common to three intersecting cylinders
as explained on the Wolfram Alpha page about Steinmetz solids that I linked to.

 

V_2 = the volume common to two intersecting cylinders
V_3 = the volume common to three intersecting cylinders
as explained on the Wolfram Alpha page about Steinmetz solids that I linked to.

Isn't the volume common to all three intersecting cylinders also common to any two of them?

 

Isn't the volume common to all three intersecting cylinders also common to any two of them?

I think so, but the reverse isn't true. To find the total volume of the three intersecting cylinders, I thought about it this way:
The volume of one cylinder equals the volume of one cylinder. Duh.
Add a second intersecting cylinder (of equal volume), and the volume of the entire object is increased by the volume of the cylinder minus the volume common to the two cylinders.
Add a third intersecting cylinder (of equal volume), and the volume of the entire object is increased by the volume of the cylinder minus the volume common to the three cylinders.
Thus, I came up with 3*V_cylinder - V_2 - V_3

 

I think so, but the reverse isn't true. To find the total volume of the three intersecting cylinders, I thought about it this way:
The volume of one cylinder equals the volume of one cylinder. Duh.
Add a second intersecting cylinder (of equal volume), and the volume of the entire object is increased by the volume of the cylinder minus the volume common to the two cylinders.
Add a third intersecting cylinder (of equal volume), and the volume of the entire object is increased by the volume of the cylinder minus the volume common to the three cylinders.
Thus, I came up with 3*V_cylinder - V_2 - V_3

You might go back and ponder Posts #57 and #58 where Studiot and I both showed how to account for the overlaps. Keep in mind that you need to subtract overlapped volumes exactly once.

 

Add a second intersecting cylinder (of equal volume), and the volume of the entire object is increased by the volume of the cylinder minus the volume common to the two cylinders.

Add a third intersecting cylinder (of equal volume), and the volume of the entire object is increased by the volume of the cylinder minus the volume common to the three cylinders.

The first statement is true (sort of) but the second is not.
Remember the 'object' is a cube an we are working out the volume of a hole not an object!
So your first statement would read better if you distinguised between them. this is what I meant by sort of.

The second statement is not true because there hole comprises three cylinders.
Each cylinder has parts in common with both other cylinders (the c in my Venn diagram) and also parts in common with only one of the other cylinders, but not the other (the b).
Volumes labelled there are two such volumes labelled b per cylinder since there are two other cylinders.

I will happily expand on the Venn diagram, I think it is the best to show the breakdown of the parts of the hole with or without maths.

 

The first statement is true (sort of) but the second is not.
Remember the 'object' is a cube an we are working out the volume of a hole not an object!
So your first statement would read better if you distinguised between them. this is what I meant by sort of.

@studiot: You're right that we're working out the volume of a hole. I was just visualizing the object made up of three intersecting cylinders that fits in the hole. The object that fits in the hole has the same volume as the hole itself.

 

You might go back and ponder Posts #57 and #58 where Studiot and I both showed how to account for the overlaps. Keep in mind that you need to subtract overlapped volumes exactly once.

I see I was wrong, but now I'm certain that I know the correct answer.

I came across a thread on PhysicsForums.com where the exact same sort of problem was discussed. The starter of that thread apparently thought exactly like I did, because my formula in post #72 of this thread yields the same incorrect answer that he originally calculated. In post #17 of that thread, someone mentioned the following formula:

\(V = V_{cube} - 3 \times V_{cylinder} + 3 \times V_{bicylinder} - V_{tricylinder}\)

which must be correct because it yielded the same answer that was computed by 3DCAD software for that problem.

Applying this formula to the problem we've been discussing here, I get the answer of 17.36 cm^3, which falls between the lower and upper bounds that others have calculated above. So my formula in post #72 made use of the correct variables, but with the wrong coefficients.