# angles: trig help

I want to double check some wedges I've made for angled cuts. In a 24" long wedge, how much gain on the square corner is there for each degree of angle?
I know the method is rattling around back there amongst the unused and as yet unkilled brain cells but I can't seem to come up with it.
Thanks to the more mathmatically inclined.
jc
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Joe wrote: | I want to double check some wedges I've made for angled cuts. In a | 24" long wedge, how much gain on the square corner is there for | each degree of angle? | | I know the method is rattling around back there amongst the unused | and as yet unkilled brain cells but I can't seem to come up with it. | | Thanks to the more mathmatically inclined.
It's not the same for each degree of angle (the gain going from 0 to 1 degree is a lot larger than the gain when you go from 88 to 89).
Since the tangent of an angle (a) is equal to the rise (y) divided by the run (x\$), you can calculate:
tan(a) = y / 24
y = 24 * tan(a)
If a1 is your starting angle then
y1 = 24 * tan(a); and y2 = 24 * tan(a+1)
and the gain for that 1 degree difference is
y2 - y1 = 24 * (tan(a+1) - tan(a))
-- Morris Dovey DeSoto Solar DeSoto, Iowa USA http://www.iedu.com/DeSoto
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Morris Dovey wrote:
| It's not the same for each degree of angle (the gain going from 0 | to 1 degree is a lot larger than the gain when you go from 88 to | 89).
I think I said that backward. :-(
larger --> smaller
-- Morris Dovey DeSoto Solar DeSoto, Iowa USA http://www.iedu.com/DeSoto
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I am not going to attempt ASCII art.
If by the "square corner" you mean the two sides of a right angle triangle which are orthogonal (90 degree angle between these sides, then the formula is
Height = Length * Tangent(opposite angle).
For Length = 24 inch and opposite angle = 1 deg, then Height = 0.418922
If you have MS Excel, then you can enter a formula Cell for Height = Îll for Length*TAN(RADIANS(Cell for angle))
For some strange reason MS designed Excel to use RADIANS in angle functions instead of degrees where 180 deg = PI Radians.
Dave Paine

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As Morris Dovey warned in a separate message, the height is not linear for each degree. Hence you do need to use the formula.
Dave Paine

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Assuming the 24" side is included between the right angle and the angle to be incremented, and the 24" side and the 'gain' side are at right angles to each other:
Gain in inches = 24*tangent(incremented angle).
For some angles: 1 deg-> .4189516" 2 deg-> .8380985 15 deg -> 6.4307806" 22.5 deg-> 9.5941125" 30 deg -> 13.8564061" 45 deg -> 24.0000000" 89 deg-> 1,374.9590791" etc.....
Using Windows calculator makes the computation easy once you know the equation.
Tin Woodsmn
Happy New Year to all of the Wreckers....
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What do you mean by "gain" ? If you mean, for a right triangle with angel A between the hypotenuse and a 24" side, that the "gain" is the length of the side opposite angle X, then the change in the length of that side is not a linear function wrt the angle X. However, it will be equal to 24 X sine(A)
--
Often wrong, never in doubt.

Larry Wasserman - Baltimore, Maryland - snipped-for-privacy@charm.net
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OOPS, maybe a little doubt would be good on this one. G/HYP does = SIN X, but you have two unknowns, since the base is fixed at 24. G = HYP*SIN X doesn't help, since we don't know the hypotenuse. As Tin said, TAN X = G/24, so G = 24 TAN X. I hope I'm not too sleepy to get this right. WL

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You are right, I was mixing up my sides, the tangent is correct.
--
A man who throws dirt loses ground.

Larry Wasserman - Baltimore Maryland - snipped-for-privacy@charm.net
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Funny story on that subject:
On of my aspiring architectural drafters was trying to make a model of his project with a 6/12 roof slope. The roof just wouldn't work out.
After lots of investigation, I found out that he had reasoned that if a 12/12 slope gives a 45 degree angle, a 6/12 should give a 22.5 degree angle. He just couldn't see my arguments to the contrary.
The only way I convinced him was to ask him to draw a roof slope at 24/12, and see if it produced a 90 degree angle....
Old Guy

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Thanks everyone for the help. In the original post, I neglected to mention that I was calculating this for the first couple of degrees. I've got it now.
Have a very happy and healthy New Year.
jc

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