# Mitre question???

I want to make an 8 sided table top. What would he mitre angle be for the skirting?
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JAKE wrote:

Tell me you are kidding.
Dave
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22.5
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Here is the way to figure any of these questions
For every side add 180 degrees, after 4 sides. example 6 sides....360(for the four sides then 180 (for fifth side) + 180 (for sixth side) add up...360 + 180 +180 r0 Now divide by number of sides....720/6 0...divide 120 in half because you are mitering 60 degrees.
Now let's try it on the 8 sided table.
360 for the square + 180 + 180 + 180 + 180 + 1080
1080/8= 135 135/2g.5 degrees If you flipp the board over it equals 22.5 degrees.

-
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Geez you made that complicated. The simple equation to any sided table is simply take the number of sides multiply by 2 and divide that number in to 360. Period.
More simply put, 360 divided by double the sides.
360/(4 sides x 2) = 45 360/(8 sides x 2) = 22.5 360/(60 sides x 2) = 3
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Wrong
A 6 sides table would require 30 degree cuts. 360/ (6 sides x 2) = 30
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Leon wrote:

I'm not sure which is more intriguing: the fact that the question was brought up by a "woodworker", or the confusion it has engendered in this thread...
Dave
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sixth
180 / (number of sides) has always worked for me. What's with all the complicated equations??
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That will work. The math is how to get there.
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| > > Here is the way to figure any of these questions | > > | > > For every side add 180 degrees, after 4 sides. example 6 | > > sides....360(for the four sides then 180 (for fifth side) + 180 (for | sixth | > > side) add up...360 + 180 +180 =720 Now divide by number of | > > sides....720/6 =120...divide 120 in half because you are mitering 60 | > > degrees. | > | > Wrong | > | > A 6 sides table would require 30 degree cuts. 360/ (6 sides x 2) = 30 | > | > | | 180 / (number of sides) has always worked for me. What's with all the | complicated equations?? | |
Rube Goldberg ring a bell???
-- PDQ
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That will work also but for me it is easier to remember 360 as that forms a complete circle vs. a straight line. Actually 360 divided by the total number of cuts works also.
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Which of course is the same thing. 30 and 60 degree angles are complimentary.
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Secret Squirrel wrote: ...

And on top of liking each other, they're "complementary" in forming a right angle... :)
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Actually you probably are not going to find a 60 degree setting on any saw. So uh you would set to 30 degrees.
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of course not. The point was that someone, I'm not sure who, said that the math was wrong. The math was absolutely correct. Understanding that 30 and 60 degree angles are complementary is necesarry for practical application, but doesn't make his original statement incorrect.
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Well in one of his examples his math was wrong.
He went on to give another examples of Now let's try it on the 8 sided table.
360 for the square + 180 + 180 + 180 + 180 + 1080
1080/8= 135 135/2g.5 degrees If you flipp the board over it equals 22.5 degrees.
That does not add up. He adds 360+180+180+180+180+1080 which would normally = 2160 not 1080.
A reasonable person would gibe an answer to the saw setting to come up with the end result. He simply made this way too complicated for some one that could not determine the answer in the first place. Typically and or at least I was always taught to make the equasion as simple as possible. Why not use a formula that gives the angle setting found on the saw?
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Why not just divide 360 by the number of sides. This gives you the angle of each joint. Mitre each side joint at half this angle and assemble.
8 sides = 360/8E 45/2".5
There, just mitre at 22.5 degrees.
Oldun
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If you read higher up in the thread I made that same BASIC suggestion a few days back. Actually I divide the sides by the number of end cuts needed.
More simply put, 360 divided by double the sides.
360/(4 sides x 2 end cuts) = 45 360/(8 sides x 2 end cuts) = 22.5 360/(60 sides x 2 end cuts) = 3
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On 7/24/2005 9:58 AM Leon mumbled something about the following:

This only works with shapes of 4 or more sides, it doesn't work for a triangle.
--
Odinn
RCOS #7
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Odinn wrote:

Actually it ONLY works for equilateral triangles and REGULAR polygons.
Try your method or his (same thing really) on a parallelogram, trapezoid or rhombus and get back to us.
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