Wednesday, January 1, 2020

Math Problem

This should be trivial.  Maybe you can see the error.  I am trying to machine this out of acetal.

Not properly scaled; I do not use LibreOffice Draw very often and it is noticeably different from the last OpenOffice Draw.  The black parts are the sections I am trying to remove.

The left side was easy.  The program I wrote starts cutting with y=0.0, then another pass at y=.136, which puts the right side of the 3/4" mill at .501".  Where should y be for the right side?  I thought .501" (the right side of the mill) + .9865" (the block I want left there) + .375" (the radius of the mill).  Yet it is most emphatically not .501" from the right side of this 2" block.  What is my math error?

Because the two sides are symmetrical, I just turned the part around and ran the same program again, and ended up with a part within a few thousandths.

5 comments:

rgr said...

If you want a .9865 block centered, then you'd want .5076 on each side (2.000-.9865)/2.
Sour your first cut should be .13175 (.5076-.375) and the second at 1.49375. This would leave .50675 on each side.
I think.

Clayton Cramer said...

Exactly what I did. Bizzare.

rgr said...

I see that you have a pass at .136, not .13175.
How far off is the second side from .5076? How did you come up with .501 on ech side? .501+.501+.9865= 1.9885 NOT 2.000. Is this within your tolerances?

Clayton Cramer said...

rgr: The difference between .136 and .13175 is not enough to explain the problem. I was getting .6" between the two sides.

rgr said...

Hmm. Then I'm stumped. At least you got it fixed by flipping the product. Only thing I can think of is the adjustment for the .375 mill radius. On the first side that'd be additive to the y of .136 - on the second it'd not be needed to be added as that's the side you're cutting out altogether.