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  #1  
Old Tue 09 February 2010, 06:40
sailfl
Just call me: Nils #12
 
Winter Park, FL
United States of America
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.
Attached Files
File Type: dxf Calculate Example.dxf (26.2 KB, 56 views)
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  #2  
Old Tue 09 February 2010, 07:00
bradm
Just call me: Brad #10
 
Somerville(MA)
United States of America
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.
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  #3  
Old Tue 09 February 2010, 10:49
bradm
Just call me: Brad #10
 
Somerville(MA)
United States of America
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
Attached Files
File Type: dxf ForNils.dxf (24.1 KB, 21 views)
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  #4  
Old Tue 09 February 2010, 10:55
lumberjack_jeff
Just call me: Jeff #31
 
Montesano, WA
United States of America
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..
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  #5  
Old Tue 09 February 2010, 11:51
sailfl
Just call me: Nils #12
 
Winter Park, FL
United States of America
Brad and Jeff,

Thanks for your help.
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  #6  
Old Tue 09 February 2010, 12:03
lumberjack_jeff
Just call me: Jeff #31
 
Montesano, WA
United States of America
Nils, verified in CADD, Brad's results are exactly right.

The difference is due to rounding.
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  #7  
Old Tue 09 February 2010, 15:18
sailfl
Just call me: Nils #12
 
Winter Park, FL
United States of America
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.
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  #8  
Old Tue 09 February 2010, 17:34
bradm
Just call me: Brad #10
 
Somerville(MA)
United States of America
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
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  #9  
Old Wed 10 February 2010, 03:38
sailfl
Just call me: Nils #12
 
Winter Park, FL
United States of America
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.
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  #10  
Old Wed 10 February 2010, 08:50
Sergio-k
Just call me: Sergio #61
 
Athens
Greece
hmmm

Big round object top secret project

Is your customer from Area 51 Nils ???
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  #11  
Old Wed 10 February 2010, 10:17
lumberjack_jeff
Just call me: Jeff #31
 
Montesano, WA
United States of America
If Nils starts asking questions about ballistics, I'll get worried.
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