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post #1 of 14 (permalink) Old 02-03-2010, 04:05 PM Thread Starter
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Default Help With Math (no not the new math)

Ok, I have a hex shaped box (in this case an aquarium) to build. If the dia is 18" (measured from one side to the other) what is my length on each of the six sides and what angle would I route to?

I just went as high as High School Algebra, the rest of the time I spent in Shop Class all through high school)

thanks Charles
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post #2 of 14 (permalink) Old 02-03-2010, 04:20 PM
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search "circumscribed hexagon", you should find the formulas

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post #3 of 14 (permalink) Old 02-03-2010, 05:06 PM
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D=18
PI=3.14159
DXPI= 56.54
Divide 56.54 by the number of facets required

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post #4 of 14 (permalink) Old 02-03-2010, 05:18 PM
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The short answer is .577 times the distance across the flats. Cut your pieces at 30 degrees. When the parts are assembled you will have 60 degree joints. Six sides equal 360 degrees for a circle.

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Last edited by jlord; 02-03-2010 at 05:37 PM.
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post #5 of 14 (permalink) Old 02-03-2010, 06:20 PM Thread Starter
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Thanks for the help everyone
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post #6 of 14 (permalink) Old 02-03-2010, 07:03 PM
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Quote:
Originally Posted by lemonyx View Post
Ok, I have a hex shaped box (in this case an aquarium) to build. If the dia is 18" (measured from one side to the other) what is my length on each of the six sides and what angle would I route to?

I just went as high as High School Algebra, the rest of the time I spent in Shop Class all through high school)

thanks Charles
Hi Charles - I haven't got a clue what the math is so I drew it out in DeltaCad and it comes out 10 3/8". You wanted 18"dia across the inside of the flats, right?

John Schaben

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post #7 of 14 (permalink) Old 02-03-2010, 09:36 PM
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Quote:
Originally Posted by jlord View Post
The short answer is .577 times the distance across the flats. Cut your pieces at 30 degrees. When the parts are assembled you will have 60 degree joints. Six sides equal 360 degrees for a circle.
+1, This is the easiest method to find the length of the flats.

18 x .577 = 10.386, which is within .006" of the exact flat length.

18 x .5773 = 10.3914, which is within .0009" of the exact flat length.
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post #8 of 14 (permalink) Old 02-03-2010, 10:48 PM
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Quote:
Originally Posted by Noob View Post
+1, This is the easiest method to find the length of the flats.

18 x .577 = 10.386, which is within .006" of the exact flat length.

18 x .5773 = 10.3914, which is within .0009" of the exact flat length.
OK, I gotta ask, where did the .577 come from???

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post #9 of 14 (permalink) Old 02-04-2010, 12:59 AM
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Quote:
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OK, I gotta ask, where did the .577 come from???
It is the tangent of 30 degrees, actually 0.577350269 (according to Google calculator).
See Tangent - Wikipedia, the free encyclopedia
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post #10 of 14 (permalink) Old 02-04-2010, 03:47 AM
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Originally Posted by MarcoBernardini View Post
It is the tangent of 30 degrees, actually 0.577350269 (according to Google calculator).
See Tangent - Wikipedia, the free encyclopedia
Guess I should have spent more time in geometry class..
Now that I understand that part, I think we only answered half the question.
It doesn't look like we accounted for stock thickness. Gonna need to add 2x stock thickness to the diameter.

John Schaben

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