Distance between two offset unequal circles

Jan 2012
14
0
For several days I have been brewing on this problem at work. I can't seem to figure it out. I have solved it using a CAD design program, but I want an equation that I can use.

The situation is described by two circles, with an unequal radius. The smaller circle is placed within the bigger circle and is at one point tangential to the bigger circle. The radius of the bigger circle crosses the smaller circle twice.

I am interested in the distance X between the the circles, measured at the the radius of the bigger circle, as a function of the angle theta. I have shown an illustration of the problem below. Anyone interested can download the word file that has the illustration.

One of the ways of solving, I thought, is to express THETA2 as function of THETA1. But I didn't even manage this and I'd love to get some help on this!



Of course, I have attempted many additional things to try and solve my problem, but in order to keep the start post simple and quick, I will not go into detail of these attempts. But all I used so far is based on simple trigonometry, but as my knowledge of maths is limited, I might have kept it too simple.
 

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Feb 2010
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162
GsJosh said:
I am interested in the distance X between the the circles, measured at the the radius of the bigger circle, as a function of the angle theta.
A solution just using simple right triangle trig is

\(\displaystyle X = R_1 - \dfrac{R_2 \cdot \sin(\frac{1}{2}\theta_2)}{\sin(\frac{1}{2}\theta_1)}\)

but I don't know which theta is the one you want.
 
Jan 2012
14
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Thank you mrtwhs for this part of the solution. I had come this far too. The problem though, I only have theta 1, not theta 2. But since theta 2 is definitely related to theta 1, I figured that part of the solution lies into finding the relationship between these two.

Do you think you would be able to help me find the relationship between theta 1 and theta 2?
 
Jan 2012
14
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Please refer to the illustration below.


Beta = theta1/2 so in order to know X as a function of theta1, one either needs to know d or e.

d can be described as follows: d = (chord1-chord2)/2
e can be described as follows: given the that the height of a triangular portion is r, e = r1-r2

 

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
from the diagram...it doesn't seem possible e = r1-r2
 
Jan 2012
14
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agentredlum said:
from the diagram...it doesn't seem possible e = r1-r2
Ah! Of course not!

But e = r1-r2+R1-R2
 
Oct 2011
522
0
Belgium
Letting \(\displaystyle a=R_1; b=R_2; c=\cot \frac{\theta}{2}\)

large circle:\(\displaystyle x^{2}+y^{2}=a^2\)
small circle:\(\displaystyle x^2+(y-a+b)^{2}=b^2\)
line:\(\displaystyle y=cx\)

intersection large circle - line:
\(\displaystyle x^{2}+(cx)^{2}=a^2\)
\(\displaystyle x=\frac{a}{\sqrt{1+c^2}}\)
\(\displaystyle y=\frac{ac}{\sqrt{1+c^2}}\)

intersection small circle - line:
\(\displaystyle x^2+(cx-a+b)^{2}=b^2\)
\(\displaystyle (c^2+1)x^2+c(-2a+2b)x+a^2-2ab=0\)
\(\displaystyle x=\frac{2ac-2bc+\sqrt{c^2(-2a+2b)^2+(8ab-4a^2)(1+c^2)}}{2(1+c^2)}\)
\(\displaystyle x=\frac{2ac-2bc+\sqrt{4b^2c^2+8ab-4a^2}}{2(1+c^2)}\)
\(\displaystyle x=\frac{ac-bc+\sqrt{b^2c^2+2ab-a^2}}{1+c^2}\)
\(\displaystyle y=c\frac{ac-bc+\sqrt{b^2c^2+2ab-a^2}}{1+c^2}\)

distance x:
\(\displaystyle \sqrt{(\frac{a}{\sqrt{1+c^2}} - \frac{ac-bc+\sqrt{b^2c^2+2ab-a^2}}{1+c^2})^2 (1+c^2)}=a - \frac{ac-bc+\sqrt{b^2c^2+2ab-a^2}}{\sqrt{1+c^2} }\)
 
Jan 2012
14
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wnvl said:
distance x:
\(\displaystyle \sqrt{(\frac{a}{\sqrt{1+c^2}} - \frac{a-b+\sqrt{b^2+2ab-a^2c^2}}{1+c^2})^2 + (\frac{ac}{\sqrt{1+c^2}} - c\frac{a-b+\sqrt{b^2+2ab-a^2c^2}}{1+c^2})^2}\)
Wow, that is brilliant wnvl!

But I have tried to verify it and it doesn't add up... It should work from \(\displaystyle theta = 0\) but it doesn't.
 
Oct 2011
522
0
Belgium
I updated and simplified my formula in my previous post. Please check again.

My formula is working for \(\displaystyle \theta=0\).
When \(\displaystyle \theta=0\), c becomes infinity and when you take the limit of my formula for c going to infinity, it returns 0, which is correct.
 

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
GsJosh said:
agentredlum said:
from the diagram...it doesn't seem possible e = r1-r2
Ah! Of course not!

But e = r1-r2+R1-R2
where is little r1 and little r2 in the diagram?