Mitre question???

[Leon]

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.

[CW]

sides....360(for

via Encryption =----

[Secret Squirrel]

"Leon" wrote in news:qKeDe.1063$ [e-mail]:

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.

[Leon]

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= 135 135/2=67.5 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?

[Oldun]

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 = 360/8=45 45/2=22.5

There, just mitre at 22.5 degrees.

Oldun

[Leon]

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

[Odinn]

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.

[Unquestionably Confused]

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.

[Odinn]

Okay

360 / (3 x 2) = 60 An equilateral triangle has 60 degree angles. I must not be getting it, because I can't seem to match 2 60 degree angles and end up with a 60 degree angle.

[Leon]

It certainly does work with a triangle. However you will be hard pressed to find a saw that will cut at 60 degrees. The solution if using a TS would be to set the saw at 30 degrees and put the board on end and run through the saw.

[Leon]

How do you define REGULAR polygons?

[Leon]

An equilateral triangle has 60 degree angles at the end of each peice. That is because you end up with a 120 degree angle when matching up 2 60 degree angles. Add 3 120 degree angles up and you get 360 degrees. An equilateral triangle.

[Odinn]

On 7/24/2005 10:58 AM Leon mumbled something about the following:

and how do I come up with 120 degrees? 360 ( 3 sides X 2) is the equation we were using doesn't come up with 120.

[Swingman]

There is only one definition: all angles in a REGULAR polygon are equal.

The sum of the _exterior_ angles of ANY polygon is 360 degrees.

The sum of the _interior_ angles of ANY polygon = 180° (n-2) where n=number of sides

There shouldn't be much more any one needs. LOL

[Unquestionably Confused]

sides all equal length and angles all the same. Rectangle would also fit here even though the sides differ so _I_ would define a rectangle with its 90 degree corners as a regular polygon but that might be stretching it.

Also, an "Oops, my bad" is owed to Odinn. I got hung up on the angle feature and what I was thinking was equilateral triangle, three equal angles, divide them by 2 and that's your cut. Won't work on any other triangle AFAIK.

None of the methods discussed here will work on parallelograms, trapezoids, or rhombi (sp?)

[Unquestionably Confused]

No, Odinn's right. ANY triangle is a three-sided polygon. The sum of the angles of a triangle is 180 degrees.

The END of each side of an equilateral triangle is a 30 degree angle which meets another 30 degree angle to come up with 60 degrees.

I'm not Unquestionably Confused for nuttin'

[Leon]

Ok that is not correct in reference to the equasion only working on Regular polygons. All sides do have to be equal angles but all pieces DO NOT have to be the same length. With the formula that I use all pieces do not have to be the same length. All parallel pieces have to be equal length.

I agree.

[Leon]

Well. I will admit that it does become a bit tricky at this point. Because most saws will not tilt the blades much past 45 degrees, a 60 degree bevel has to be accomplished by rotating the board 90 degrees and has it to be cut at the 30 degree setting. This leaves you with a complimentary result angle of 60 degrees. I'll post a PDF file to alt.binaries. picture.woodworking with a CAD drawing of the set up and how the angles are "conjured up".

[Leon]

Yes I agree to a point. My formula of dividing the total number of cut, "2 on each end of 3 pieces" into 360 still works on a triangle but the set up on the saw has to be bit different since a TS will not tilt its blade 60 degrees in relation to the fence. You have to rotate the piece of wood to be parallel to the fence and the 30 degree setting on the bevel will give you a complementary 60 degree angle result. If the blade would tilt to 60 degrees in relation to the fence you could cut it in the traditional mannor.

Correct but again my formula works for the saw bevel setting or perceived saw bevel setting. Since the saw will not tilt past 45 degrees you have to rotate the board parallel with the fence to end up with what the same angle that the saw would cut if it could tilt to 60 degrees during a normal setting with the work laying flat on the TS top. I posted a CAD drawing in PDF format to a.b.p.w. You will have to open the file to see the drawing.

[Unquestionably Confused]

But if you have a polygon (more than four sides), how can you have equal angles AND differing sides. I submit that you can't.

Imagine a pentagon with four sides being 2' long and 1 side 3' long. Show me the equal angles

Okay. With this one, assume it's a parallelogram. Four sides are parallel (by definition) with the opposing side. Angles differ on two corners.