Distance between two origins

• Jul 5th 2005, 11:19 PM
Vector
Distance between two origins
Hello =)

Here's my problem:(attached)

Basically I have two circles set up. I have the location of the origin of the circle to the left.
I also have the location of P1 and P2, and also the rotation of those points from the origin of the first circle. The two circles must intersect such that they intersect at the given points. The angle of the red line is exactly half way between the angle of the two given points.

I can make the radius of the second circle (r) whatever it needs to be, but it will always be larger than the radius of the first circle (R).

I'm trying to discover d, or the origin of the circle to the right (which can be derived from d), so that I can draw an inwards arc between two points.

I've wracked my brain for hours over this, any help would be very much appreciated. =)
• Jul 5th 2005, 11:46 PM
ticbol
If I undertand what you mean, then
>>>by "origin" you mean the center of the circle.
>>>You want to find the center of the second (and bigger) circle by finding the disdtance, d, between the two centers.

Try giving us the locations of the center of the first (and smaller) circle, and of the two points P1 and P2, and the radius of the second circle, so we might help you find "d" and the location of the center of the 2nd circle based on the location of the 1st circle.

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I can show you the way to find "d" even if you don't show us the locations of the said 3 points and the dimension of the bigger radius. But it is easier to explain with numbers.
• Jul 5th 2005, 11:59 PM
Vector
Actually this is to be written algebraecly, so I can change it as I need. Each time i need to apply this it could be up to or more than 7 different different sets of points.

As much as I could try, the most I could figure out was that I needed the radius to figure out the distance of how far away it needed to be to get it to intersect with both of the points.... The rest of it is a bit beyond me until it's explained :P

Ummm... I'll throw you some values... oh yeah I should probably mention that the rotations are in degrees from 0 degrees vertical going clockwise, but if you need the points then that shouldn't be a problem.

(From now on the left circle will be C1 and the right circle will be C2)

C1_origin = (50, 50)

P1 degrees from origin = 0 * (360 / 7) -- 0/7 of the way around
P2 degrees from origin = 2 * (360 / 7) -- 2/7 of the way around

C2_radius = 100 (just to make things simple)

Thankyou very much :)

<edit>Sorry, forgot to include the numerical locations of the points...
P1: (50, 100)
P2: (61.126, 51.254) -- <edit again> Fixed that second value... :o
• Jul 6th 2005, 01:46 AM
ticbol
Umm, fast.

Too many origins.
So, "origin" of circle is center of circle?
And, "origin" of angles is the usual origin of angles---the righthand or positive half of the horizontal axis through the center of the circle, where the center is the origin, [ :-) ], of the positive half?

Hope I got what you mean.

You gave the dimension of R too. That lessens the explanation.

My understanding is
P1 is 0 degrees from the "origin"
P2 is 2(360/7) = 102.8571429 degrees from the same "origin"
So, the central angle "P1-C1-P2" is 102.8571429 degrees also? (C1 is the center of circle C1.)

Are the locations of P1 and P2, as shown on your posted figure, not reversed?
If not, then the "origin" of the angles is the vertical axis through center C1?

Anyway, let me continue as I understand your question now. (If my understanding is wrong, then the following explanation is wrong too.)

Since "d" is exactly halfway between the angles of P1 and P2, then "d" bisects the central angle "P1-C1-P2", and so "d" bisects the chord "P1-P2".

["P1-P2" is not "P1 minus P2". If A were P1, and B were P2, then "P1-P2" is line segment AB.]

Then we can solve for the chord "P1-P2"
For simplicity, let us call "P1-P2" as chord "c"

d bisects c, and d is perpendicular to c.
Let us call the midpoint of c as point A.

----------
In right triangle "C1-A-P2",

angle "A-C1-P2" = (1/2)(102.8571429 deg) = 51.42857143 deg
sin(51.42857143deg) = (c/2)/R = c/(2R) = c/(2*50) = c/100.
c = 100*sin(51.42857143deg) = 78.18314825 ----***

cos(51.42857143deg) = "A-C1" / R = "A-C1" / 50
"A-C1" = 50*cos(51.42857143deg) = 31.17449 ----***

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In right triangle "C2-A-P2"

sin(angle "A-C2-P2") = (c/2)/r = (78.18314825 / 2) / 100 = 0.39091574
So, angle "A-C2-P2" = arcsin(0.39091574) = 23.01149142 degrees

Then, cos(23.011491242deg) = "A-C2" / r = "A-C2" / 100
So, "A-C2" = 100*cos(23.01149142deg) = 92.04264687 ----***

------------
Therefore,
d = "C1-A" + "A-C2"
d = 31.17449 +92.04264687

And the location of center C2 based on/from center C1 is:

50 +d*cos(angle "A-C1-P2)
= 50 +(123.2171369)*cos(51.42857143deg)
= 50 +76.82462827
= 126.8246283 ----***

50 +d*sin(angle "A-C1-P2)
= 50 +(123.2171369)*sin(51.42857143deg)
= 50 +96.33503681
= 146.33503681 ----***

Therefore, C2 is at (126.8246283,146.33503681) ----answer.

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See if you can follow that, and if my understanding is correct.

After you understand the above solution, then
I can give algebraic solution---no numbers, if you want.
Only, this time, make sure P1 and P2 are in the correct places.
• Jul 6th 2005, 03:51 AM
Vector
Thank you very much indeed, this explains it all :)

Sorry I was so ambiguous with my wording... I really should be more carebul :D

I made a slideshow while I was figuring out what you'd given me to try to help me understand, I think I've got it all right. I left it in algebraic form so I can translate it into a function for use... :) I put that here if you wanna have a peek:
http://www.vector-sector.cjb.com/junkyard/problem2.swf
Thanks again mate :D
• Jul 6th 2005, 04:19 AM
ticbol

Everything is correct, except that r and R are reversed, although they are more correct now: R should really be the radius of the bigger circle, and r is for the smaller circle.

Also, except
c/(2r) = [sin{(thetaP1 +thetaP2)/2}] / r

No.
The right side should not be divided by r.

The correct one is
c/(2r) = sin{(thetaP1 +thetaP2)/2} ----***
Then,
c = 2r*sin{(thetaP1 +thetaP2)/2} ----***

---------
Mate ?

You are Australian.
We have the same Time Zone.
Midnight is about an hour and half there and here now.

'Night, Mate.

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Oopps.

Also, not (thetaP1 +thetaP2)/2.
It should be (thetaP2 -thetaP1)/2
It should be half of the difference of the two angles.

(thetaP1 +thetaP2)/2 is the angle of point A from the same "origin"

Investigate that, to see my point.
Assign non-zero values to thetaP1 and thetaP2, and see the difference from using (thetaP1 +thetaP2)/2 and using (thetaP2 -thetaP1)/2 in your formulas.