simple (for some) geometry questions :)

When I was in grade 12, I could do this in a snap. I guess I ain't as smart as I was 30 years ago...
I have three points equidistant on a circle, spaced 120 degrees apart. They
form an equilateral triangle. I know the length of the legs of the triangle. How do I determine the diameter (or radius) of the circle?
I have two points on the diameter of a circle. I know the diameter of the circle and the distance between the points. How do I determine the angle between the points?
Lessons in simple trigonometry welcome...:)
Thanks, brian
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Brian wrote:

Two points on one diameter don't have an angle between them. Unless you count a straight line as a 180 deg angle. ...lew...
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the
angle
I think he means the angle between the points and the centre of the circle.
For a circle of unit radius the length of the chord subtended by the angle x is 2sin (x/2).
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Lewis Hartswick wrote...

I gathered that he meant "on the circumference."
Jim
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I told you that it was a long time since high school...:)
Thanks all for the excellent help here. I've got some formulas that I can work with, and a calculator with an arcsin button, so I should be good to go.
Thanks again, Brian

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Brian wrote:

The length of a chord ( one side of the triangle ) is C= 2 r sin A/2 Where A is the included angle.
So for 60 deg ( an equalateral triangle )
C= 2 r 0.866/2 or just 0.866 r
...lew...
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Try this.
Sin (A/2) = 0.5C/r
Where A = 120 degrees this simplifies to
Sin 60 = 0.5C/r ==>
0.866r = 0.5C ==>
r = 0.577C
Dave Baker - Puma Race Engines (www.pumaracing.co.uk) I'm not at all sure why women like men. We're argumentative, childish, unsociable and extremely unappealing naked. I'm quite grateful they do though.
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Brian wrote...

Let s be the length of a leg. The radius r is then
r = (1/2) s / cos(30) ~= 0.5774s
To understand why, look at half of the isosceles triangle formed by the center and one leg.
Jim
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wrote:

OR, manually. Construct the perpendicular bisectors of the three sides. At the point of intersection set your compass point and open to one apex of the triangle. Draw your circle and measure its diameter.
OR, for an equilateral triangle only! Construct the bisectors of the interior angles. At point of intersection set compas point and open to one apex of triangle. Draw circle and measure diameter
Mike in BC
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Micheal, that manual method is what I've been doing :) But what I'm actually doing is putting three dowels in a flywheel to locate a clutch housing, and with the clutch housing weighing more than the flywheel by about 50% (don't ask, it's a "performance" clutch, not a racing clutch like it should be), I really need to get the dowels in the right place. I figure with the BP and a rotary table I should be able to do better than trying to line up over some scribed lines and a center punch mark:)
Customer bought the clutch and me telling him that a double plate Tilton is what he "should have" bought didn't go over all that well...
Brian

They
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wrote:

Brian, take a look at:
http://www.hp-h.us/p/littlemikey/hobbies/machining/jig%20borer%20co-ordinates.gif
where I've posted a page from "Machine Shop Theory and Practise" by Wagener and Arthur. Exactly what it says, jig borer coordinates for drilling from 3 to 12 holes on a circle of any diameter. I posted it _very_ large for clear reading and good definition when printing out.
HTH, Mike in BC
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wrote:

Sorry, better use:
http://www.hp-h.us/p/littlemikey/hobbies/machining/jig_borer_co-ordinates.gif
for this picture. Mike in BC
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Thank you very much. That table is excellent.
Brian

es.gif
http://www.hp-h.us/p/littlemikey/hobbies/machining/jig_borer_co-ordinates.gif
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Brian wrote...

With a diameter d, and the distance s between the points, the angle t (at the circle's center) is:
t = 2 * arctan( s / (2 * sqrt(d^2 / 4 - s^2 / 4)))
It seems there should be a simpler equation, but this is what I get from half the isosceles triangle with two sides of length d/2 and one side of length s.
Jim
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Jim Wilson wrote...

Based on the formula given by Lewis,
t = 2 * arcsin(s/d)
That's certainly simpler, and it does produce the same result as the formula I found above.
Jim
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wrote:

Personally, I'd draw it in Autocad and use the dimension function to answer the questions.<GRIN>
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Ron Thompson wrote:

TCad for me. But if my boots were awash in swarf and I didn't want to catch hell for tramping into the house, I'd take the circumference, divide it by the distance between the two points, and then divide 360 by that result. That's assuming I understand the question. :-)
Wayne
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Brian wrote:

Draw picture with dots for points known, circle center is one of knowns. Connect the dots with lines. Draw perpendicular legs from existing lines to circle center. Fill in all known information. Get a trig book and match up the pictures. Simple. Nobody has books these days, formulas can be found online, with pix.
mj
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