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chameleon · Joined: 12 Dec 2009 Posts: 4

I need an equation for theta, can someone please help? This is a personal project, I assure you. I've already tried (unsuccessfully) to deduce what the equation should be in terms of the radius, r. Here's a rough rendering of what I am trying to get:

I am trying to get an angle THETA such that the second circle, when rotated around the point P(0,13.5), intersects the point P(0,0). The radii of the two circles are identical, but not r. r=13.5 and is the distance of the axis of rotation from the origin. Not that it matters, the radii of the circles are both 10m. Here are some additional notes:

AGAIN! r IS NOT equal to the radii of the circles. It's the distance of the axis of rotation to the origin.

THETA (θ) is approximately 43 degrees. I need an EQUATION, not an approximation. I can approximate to over 20 decimal places later.

The origin, P(0,0), is located in the center of the first circle

It appears from the drawing that the center of the second circle lies at some point P on the first. This may or may not be the case.

My first stab at a formula was rsinθ-x=0, which translates to θ=arcsin(x/r). my problem, of course, was that x is unknown. If this helps you, you probably don't have any more of an idea what to do than I.

This will require a pencil and paper.

I appreciate any and all responses.

CarlB · Joined: 27 Sep 2005 Posts: 121

Hi Chameleon,
Thanks for the challenge!

you're correct in that the center of rotated circle is on perimeter of first circle; it has to be for its perimeter to pass through center of first circle.

Look at the triangle formed by the 2 13.50 long sides, the base of that triangle has a length of R. Divide that triangle in half to for 2 right angle triangles. For that right angle:

sin (a/2)=(R/2)/13.5
a/2=arcsin (R/27)

solve for a/2=21.738
a=43.477

this shows R does affect the angle a, which you can envision with a small r say 1.0

Hope this is helpful and didn't steal your fun Smile

chameleon · Joined: 12 Dec 2009 Posts: 4

HA! Thanks a lot, man! Geometry 1-oh...-2, I guess, right? I didn't think to use the radius of the circle itself. I'm back in business, baby!

EDIT: Huh...Still seems a little off. Even when I knock the angle down to 43.476, the circle seems to swing just barely past the origin.

DOUBLE-EDIT: OH! Maybe it's the resolution! How do I increase it?

CarlB · Joined: 27 Sep 2005 Posts: 121

to get radians, multiply by pi/180, so a=0.7588...

EDIT...
I upped my calculator precision, a=43.47692158

DOUBLE EDIT

Used windows calculator: a=43.476921583041

chameleon · Joined: 12 Dec 2009 Posts: 4

#s fine. Just need to up the curve interpolation...but don't remember how...

CarlB · Joined: 27 Sep 2005 Posts: 121

Oh you are referring to the display resolution-
variables are VIEWRES (command) & WHIPARC (variable) I believe

chameleon · Joined: 12 Dec 2009 Posts: 4

Thanks.

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