Maths puzzle four

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
 
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They're are probably people at our place who can solve this.

I wouldn't ask them to do anything useful, productive, or customer-facing though.
 
In mm

100x100x100 = 10(6) (1000000)

Volume of the hole, V=Pi 10(2), Height=Pi 100(2)
31415.92654

Material removed is 3.14%, 96.86% remains.
 
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It must be 99.something remaining. The round hole at 10mm wouldn’t remove a full 10mm cube so removes less than 1% of the total volume.
 
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55BF745E-723F-4540-A008-06F9E784963B.png


100-78.54=21.46

=99.2146% of the cube remaining.
 
I meant “wouldn’t”, the 10mm drill bit is round not square so wouldn’t take out the full 10mm
 
I meant “wouldn’t”, the 10mm drill bit is round not square so wouldn’t take out the full 10mm

Yes but it's still less than 99% accounting for the radius.

Just for arguments argument's sake (not accurate) but we divide the intersection by 4 and remove 1 part to account for the radius, that still leaves 97.5% there abouts.
 
Take a roll of cardboard such as the centre piece of a toilet roll - just as an example.
Set the tube over a corner of a box or cube imagine the 'roll' as a flat ended cutting tool machining along the line between vertices.
At first a pyramid shape will be removed - until the full diametre is cutting on the three cube edges, at this point the tool begins to bury itself into the 'box'.
After a while the tool fully enters the cube visible by the curved line along three faces of the cube, at the point where the tool is completely buried - the cut is merely a cylinder until the tool appears at the opposite faces.
The final shapes at the corners in question will be crown like.
This can be modelled with Google sketchup, set a solid cube to dimensions given, add a solid cylindrical shape on common centre being line through opposite vertices of cube.
Extract the cylindrical shape - find the volume of cube remaining.
Can also be calculated - need volume of pyramid shape - maybe use http://mathworld.wolfram.com/CylindricalWedge.html for the parts twixt pyramid and full cylinder.

My idea on shapes involved :-
full

You are almost there.
As we were back here :- https://www.diynot.com/diy/threads/heres-a-maths-one-for-ya.344532/
-0-
 
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Yes but it's still less than 99% accounting for the radius.

Just for arguments argument's sake (not accurate) but we divide the intersection by 4 and remove 1 part to account for the radius, that still leaves 97.5% there abouts.

I think you mean 75%.

That’s 75% of the 1% your driling. 99% of the whole cube never gets touched. So that would leave a total of 99.25% remaining.
 
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