Maths puzzle four

[Cocat2517]

Edit: I've calculated from one flat surface to another, OP specified from one vertices to an opposite vertices, in which case OP can feck off.

I make it....

10x10x10 = 1000cm³

Drilled out:
Area = π r² = 3.142 x (0.5 x 0.5) = 0.7855

Amount drilled out = 0.7855 X 10 = 7.855cm³

Left with 1000 - 7.855 = 992.145 as a percentage = 99.2145%

Maybe.

 

[rsgaz]

That's from one flat face to another, the OP wanted diagonally, from one vertex to the opposite vertex.

What we need to calculate is the volume of a 1cm diameter rod, length 17.3205cm, but where both ends of the rod have a three sided 90° corner, like...



That will give us the volume of material removed, but I don't know how to do it!!

 

[EddieM]

Trust me, I claim absolutely no superior knowledge on solving this as it's an absolute doozy. Trying to understand the problem statement is pretty hard TBH! (I have seen a solution to it, but I didn't understand it!)

 

[EddieM]

The breakouts at the vertices are the problem, the rest is pretty straightforward.

 

[Ian H]

Left with 1000 - 7.855 = 992.145 as a percentage = 99.2145%

Maybe.

I got the same but didn’t know a vertices was a corner and that the drill is going diagonal, now I know that I reckon it’s more like 98.58% left.

I get that because I reckon the drill is travelling 18.2cm and removing 78% of that material so:

18.2 x 0.78 = 14.2 (1.42% of the material is removed)

Which leaves 98.58% of the material there.

 

[EddieM]

You have to imagine what the hole at the vertices looks like, imagine a 3 pointed crown.

 

[Cocat2517]

Why was my answer using weighing method deleted? It answers the question by giving the % of material remaining.

MOD: A thousand apologies, it was a bit of finger trouble, there was nothing wrong with your post. Again sincerest apologies.

 

[magicmushroom666]

98.7%, OK i cheated and used CAD!

 

[Cocat2517]

98.7%, OK i cheated and used CAD!

Your only crime is being caught.

 

[EddieM]

98.7%, OK i cheated and used CAD!

Sounds believable, but 2 things :-

1) I want to know the maths behind it.

2) I would like to know how to do CAD drawings.

 

[empip]

95.05 % of the original volume remains.
Started with 1000 000 mm³, removed 49515.008 mm³, leaving 950484.992 mm³.
950484.992 ÷ 1000 000 = 0.95048'ish hence 95.048% of original volume remains.

At one corner of the cube.
Need to visualise the triangular based 'pyramid'.
Then the three odd shaped 'cylindrical wedges' a little like a tilted cylindrical tank having an open end with fluid just dribbling out, the fluid is in the shape of a cylindrical wedge.
In this case the wedges represent the fresh air being cut as the tool works its way into the cube, the curves slowly disappearing as the tool fully enters the cube.
So calc the pyramid vols, then the distance between the pyramid bases use this to calc a solid cylinder vol from which we subtract the six cyl' wedges, this being the material removed between the pyramids.

Using a 10mm radius or 2 cm dia' cutting tool- as in the original Q in 2012 !

-0-

 

[EFLImpudence]

Sounds believable, but 2 things :-

1) I want to know the maths behind it.

Didn't I tell you on page 1?

 

[EddieM]

Didn't I tell you on page 1?

Well apart from quite rightly correting me from using d instead of r, no.

 

[EddieM]

worried about the CAD now !

 

[EddieM]

95.05 % of the original volume remains.
Started with 1000 000 mm³, removed 49515.008 mm³, leaving 950484.992 mm³.
950484.992 ÷ 1000 000 = 0.95048'ish hence 95.048% of original volume remains.

At one corner of the cube.
Need to visualise the triangular based 'pyramid'.
Then the three odd shaped 'cylindrical wedges' a little like a tilted cylindrical tank having an open end with fluid just dribbling out, the fluid is in the shape of a cylindrical wedge.
In this case the wedges represent the fresh air being cut as the tool works its way into the cube, the curves slowly disappearing as the tool fully enters the cube.
So calc the pyramid vols, then the distance between the pyramid bases use this to calc a solid cylinder vol from which we subtract the six cyl' wedges, this being the material removed between the pyramids.
-0-

Sounds suspiciously like it may well be right, problem domain is well defined, and solution probably correct. Right then Mr. Pip looks like I owe you another pint at the Plough in Nempnett.