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#1
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Strong Math Person
I need help with circle, tangent line and how to determine the angle. I forget how to calculate this.
Lets say I have a circle that has a radius of 15'. I have a tangent line that is 12" long. I want to know the length of the line that runs from the end of the 12" line that is perpendicular but touches the circle and I want to know the the angle were they meet. Example Attached. It is not to scale. Thanks for your help. |
#2
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As everybody probably knows, I do words better than drawings ...
In your example, the original radius is side A. The 12" Tangent is side B. Draw out the rest of the rectangle. Side C runs parallel to side A for 15'. Side D runs parallel to side B for 12", and hits the center of the circle. Now, you have a right triangle formed by Side D, part of Side C, and another radius of the circle as the hypotenuse (the one you're looking for). You know what Side D is (12"), and you know the radius (15'). Trig here, then: the Sin = opposite / hyp (nope), Cos = adjacent / Hyp ( yep ). Cos(ang) = 12" / 15' (Use acos to get the angle). You want 90 degrees minus that angle for your answer, since the sum of those angles is one of the 90 deg corners of the rectangle. Similarly, you'll want 15' - X, where x is what we calculate that remaining triangle side to be. You can get this either out of a sq + b sq = c sq, or from the sin of the angle we found above. Let me know if that's not clear enough, or maybe one of the graphically inclined can sketch it. |
#3
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Hmm, there was a post here from Jeff that understood the question better; I needed to get to a machine where I could view that .DXF.
Here's another take, including a drawing: R = 15' = 180 " Tn = 12" Sin = Opposite / Adjacent Sin(Ang1) = Tn / R Ang1 = arcsin(12/180) = 3.8225537 Cos(Ang1) = AJ / R Aj = 180 * Cos(3.8225537) = 179.5995545 - or - AJ = Sqr(R*R - Tn*Tn) = 179.5995545 X = R - AJ = 0.400445 Tan(ang2) = X / Tn ang2 = arctan(0.400445 / 12) = 1.9112768 |
#4
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Sorry for the initial confusion. I deleted my original post because it was wrong.
Here's how I would do it. Not necessarily the right way. I am a lumberjack after all. A circle centered at x0 y0, with a radius of 180 can be described by the formula x^2 + y^2 = 180^2 For an x value of 12, you want to know the y value. (12^2)+(y^2)=(180^2) 144+(y^2)=32400 y^2=32256 y=179.600" (or -179.600") "this length" in your drawing is (180-179.600) or .400" In other words, a vertical line 12" off center, crosses the circle 179.6" above and below the vertical center. "This angle" is tan Θ = .400/12 tan Θ = .03333 Θ = 1.909° Last edited by lumberjack_jeff; Tue 09 February 2010 at 11:14.. |
#5
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Brad and Jeff,
Thanks for your help. |
#6
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Nils, verified in CADD, Brad's results are exactly right.
The difference is due to rounding. |
#7
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Brad and Jeff,
Thanks again for your assistance and the verifying. I put the formula in Excel where the arcsin and arctan are calculated in radians but I converted them to degrees. I knew that I would get the help I needed from the forum. Lots of bright people on this site. Now when I need to do these calculations, I change two numbers and I have what I need. I pulled out one of my old college text books. There are a couple I never tossed. Thanks again. |
#8
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Nils, you're quite welcome. Actually, I did a special conversion out of radians into degrees above
I enjoyed Jeff's alternate and more shopworthy approach. Now, I want to know what the heck you're doing with a 15 FOOT circle. Especially with these rumors flying around about your secret indexer cabal |
#9
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Brad,
Thanks for your help. I am sorry it is for a client and some thing that will be sold so I can not tell you about it. I am looking to add indexer capability for myself so that is not a secret. Sorry no rumors.... no top secret. |
#10
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hmmm
Big round object top secret project Is your customer from Area 51 Nils ??? |
#11
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If Nils starts asking questions about ballistics, I'll get worried.
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