Ok. You don't have to calculate the volume of the point or the missing parts around it.
View attachment 153707
2D diagram:
All sides 'a' are 0.5cm.
The missing triangles '3' and '4' are the same size as triangles '1' and '2' - so must be the same as the equivalent '1' and '2' at the other end of the hole.
Therefore if you deduct just one 'a' from the vortex to vortex length (17.32 - 0.5 = 16.82), you actually have the area of the missing part
Therefore, by extension, this must also apply 3D cubes.
In cuboids, though , where the entry and exit angle will not be 45°, you would have to calculate 'a' (well, we have in your example; it was just very easy).
Do we have an answer ?
Hmm not sure, for me Pip's is certainly on the right track, and could well be the right answer.
And his answer was?
95.048% of original volume remains.
No it 95 %
The key is where the two holes cross each over these will shave 0.25 cm x 4 = 1cm + 4 x 1cm each hole drilled = 5cm
That's what pip saidThat’s way too low.
Yes, you are right.Sorry, it's a cylinder going through it, more complex I am afraid.
fret ye not. Understanding the problem itself is quite challenging, let alone the answer!Yes, you are right.
I just had an Archimedes moment while in the bath.
I did ,though, get the same answer as Magicmushroom and his CAD.
More thinking.
calculate a cylinder