Drawing Large Angles Using a Ruler, Yardstick or Tape - Router Forums
Old 02-04-2015, 06:16 AM Thread Starter
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Drawing Large Angles Using a Ruler, Yardstick or Tape

With your line ran between the two legs, mark eight inch or other chosen increments on it for the forty-five degree marks.

When marking degree points, mark critical positions, such as 90, 72, 60, 45, 30 and 22-1/2 degrees for easy location in the future.

Using this, you can mark any angle accurately. You can use a chalk line to mark the angle, if desired.

I'd like to see how others do it.

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Old 02-04-2015, 06:37 AM
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If possible I use rise and run instead. If I needed a really big angle I would probably just do the trig calculations for the legs. For small to mid size 45*s I usually a (triangle type) speed square. For large ones I measure equal legs. There are a lot of methods I imagine so it's whatever makes sense and is the most comfortable for the individual.

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Old 02-09-2015, 09:47 AM Thread Starter
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So, if you would, ramble a bit more about your rise and run method.

After some checking, my way only works for certain angles.

I suppose one could build a chart, using rise and run, for any angle from zero to ninety and using a leg length that would be easy to multiply to calculate for, say, either a ten inch long layout or a ten foot long one.

Please keep in mind, many of us never [knowingly] used trig or calculus, in school or otherwise, just as most have never used many of the things I or others use and take for granted.

Quote:
Originally Posted by Cherryville Chuck View Post
If possible I use rise and run instead. If I needed a really big angle I would probably just do the trig calculations for the legs. For small to mid size 45*s I usually a (triangle type) speed square. For large ones I measure equal legs. There are a lot of methods I imagine so it's whatever makes sense and is the most comfortable for the individual.

The reason I have what you want is, I never lent it out before.

Scraps are a myth.
Dejure is offline

Old 02-09-2015, 11:17 AM
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C√=aČ+bČ

ex.
a= run, b= rise, C= hypotenuse.
a=10', b=10'
aČ=100, bČ=100
aČ+bČ=200
C√=14.142135xxxxxx
solve C by squareroot
This method gives you the length of the hypotenuse, because a&b are equal lengths the a/C and b/C angles are 45° If you're looking to discover the angles in degree you need a diff formula
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Last edited by Ghidrah; 02-09-2015 at 11:21 AM.
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