Mitre question???

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On 7/24/2005 10:45 AM Unquestionably Confused mumbled something about the following:

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.
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Odinn
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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.
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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.
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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".
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On 7/24/2005 11:35 AM Leon mumbled something about the following:

up with the cut, no matter how I positioned the board. Obviously, I was doing something wrong if it can be done.
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Odinn
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You absolutely have to do them backwards to normal thinking including the angles. Did you check my PDF file on A.B.P.W. ? That illustrates how to cut the angles for a triangle on the TS.
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On 7/24/2005 12:25 PM Leon mumbled something about the following:

for me. Not to worry, I don't have to worry about triangles very often, so it isn't that critical :)
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Odinn
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Leon wrote:

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' <g>
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news:uoOEe.381>

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.
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On Sun, 24 Jul 2005 14:58:01 GMT, "Leon"

That's just not right. The sum of the interior angles of a triangle- any triangle, is 180 degrees. What Odinn is saying is that your formula doesn't work because you're looking for the angle you need to cut, not the total interior angle of the polygon. If you're making a triangle out of planks of wood, you need to cut the miters at 30 degrees, so that when the joint goes together, you end up with a total of 60 degrees. Cutting the wood at 60 degrees will give you a hexagon.
BTW, for the other poster (and it may have been you, I forget who it was) who said that you can just flip a piece that is cut at 90-(desired angle) to get the proper angle, that doesn't work either- it has to be 180-(desired angle)
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Leon wrote:

remember Mr. Darby telling us that triangles had ony 189 degrees.
Glen
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Glen wrote:

TYPO ALERT
Make that 180 degrees!
Glen
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How do you define REGULAR polygons?
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"Leon" wrote in message

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
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Leon wrote:

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?)
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news:MjOEe.378>>

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.
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Leon wrote:

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<g>

Okay. With this one, assume it's a parallelogram. Four sides are parallel (by definition) with the opposing side. Angles differ on two corners.
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Ok Yeah with an odd number of sides shape you are going to have a problem.

Ok, what I am trying to say here is that assuming a regular shaped 8 sided table the angles are all the same for the 8 pieces to connect. Length does not have to be the same and angles are all the same. I post another picture on a.b.p.w. to illustrate what I am talking about. Perhaps I do not understand exavtly what you are talking about.

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Leon wrote:

There, you said it, sort of. Odd number of sides

I think the problem with all of this is we're (all)attempting to make one rule fit all situations. Obviously it doesn't work on all triangles and your rule that allows different length sides on a polygram but keeps the angles identical throughout will only work on a polygon with an EVEN number of sides. Once you have five, seven or nine sides to your polygon then you're right back to length of sides and angles must be identical.
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Unquestionably Confused (in 6PPEe.24$ snipped-for-privacy@newssvr31.news.prodigy.com) said:
| Leon wrote:
|| || ||| Leon wrote: ||| ||| |||| 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. ||| ||| 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<g> || || || Ok Yeah with an odd number of sides shape you are going to have a || problem. | | There, you said it, sort of. Odd number of sides | || Ok, what I am trying to say here is that assuming a regular shaped || 8 sided table the angles are all the same for the 8 pieces to || connect. Length does not have to be the same and angles are all || the same. I post another picture on a.b.p.w. to illustrate what || I am talking about. Perhaps I do not understand exavtly what you || are talking about. | | I think the problem with all of this is we're (all)attempting to | make one rule fit all situations. Obviously it doesn't work on all | triangles and your rule that allows different length sides on a | polygram but keeps the angles identical throughout will only work | on a polygon with an EVEN number of sides. Once you have five, | seven or nine sides to your polygon then you're right back to | length of sides and angles must be identical.
I don't think so. I've posted an example to abpw.
-- Morris Dovey DeSoto Solar DeSoto, Iowa USA http://www.iedu.com/DeSoto/solar.html
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