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PostPosted: Mon Apr 12, 2010 12:12 pm 
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...3/4in thick and exactly 8 x 4ft. How many Cubes that have an internal dimension of 6 x 6 x 6 inches can be made from a full sheet after removing a circular piece that leaves a 24 inch diameter hole? All joints are mitred.

Ralph :wink:

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PostPosted: Mon Apr 12, 2010 1:35 pm 
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Hey Ralph,

Just so that we're all on the same page here, what's the kerf allowance?

Tom

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PostPosted: Mon Apr 12, 2010 2:01 pm 
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tms wrote:
Hey Ralph,

Just so that we're all on the same page here, what's the kerf allowance?

Tom


Sorry, I should have included that - Lets say 1/8 inch...

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PostPosted: Mon Apr 12, 2010 2:21 pm 
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Hey Ralph,

So here we go...
Image

That's 6" squares arrayed 7 x 15 with an 1/8" kerf. This means that complementary miters are on all sides, so you have do some flipping. It may not actually be necessary, but I wanted to make sure leave the largest possible waste pieces, just to make sure. The 3/4" boarder is to account for the miter hypotenuse (you did say inside dimension).

The hole placement takes advantage of the corner waste and leaves a perimeter (presuming that you can't have a hole intersecting the edge, and still call it a hole).

So, given six sides to a cube, that's
((15 x 7) - 15) 6 = 15 boxes

Tom

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PostPosted: Mon Apr 12, 2010 2:38 pm 
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Toms calculations look good to me!
Thanks Tom, a picture is worth a thousand words.

Rog

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 Post subject: cubes
PostPosted: Mon Apr 12, 2010 9:56 pm 
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I get 11.
Minus 1 for mistakes makes it an even 10.


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PostPosted: Tue Apr 13, 2010 12:03 am 
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I get 63 full size squares with 45 degree edge all four sides. Six sides to a cube that gives 10 full boxes with three sides to throw at the moderator when he says I'm wrong.

Just kidding, Ralph. I'll send you a 3D rendered CAD image to prove it, though.

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PostPosted: Tue Apr 13, 2010 10:16 am 
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Ten boxes.
Image
Each box is 7.5" on a side to get the 6" internal dimension

The sides are sparated by a 1/8" kerf

72 sides minus the nine obliterated by the hole yields 63 usable sides

Using six sides for each box yields ten boxes with three sides left over.

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Last edited by Winston on Tue Apr 13, 2010 8:34 pm, edited 2 times in total.

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PostPosted: Tue Apr 13, 2010 11:04 am 
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Ahw Winston,

Ya beat me to it.
I was just ready to post a correction of last night's effort, only to find you'd beaten me to the punch.
Here's my bit, essentially the same as yours.
Image
((6 x 12) - 9 = 10.5 So, ten boxes.

Tom

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PostPosted: Wed Apr 14, 2010 2:12 am 
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Dennis was the first with the correct answer so the soggy point, with a 45ยบ chamfer, goes to him. :D

Many thanks to Tom and Winston for the diagrams...

Ralph :wink:

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PostPosted: Wed Apr 14, 2010 10:49 am 
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For some reason I couldn't get this image to appear in an edited version of my first post.


Image
See ""

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