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First, let me show what I am trying to do:

picture

As you can see I have two circles, CircleA with a RadiusB of .020, and CircleC with a RadiusD of .005. When CircleC is exactly .015 above Circle A, they meet perfectly at PointF.

What I want to do is move CircleC down .0005 and to the left until both circles touch again. What I need to find out is once I move CircleC down .0005, how much to the left do I have to move it until it touches CircleA and at what point do the circles' touch. This movement is illustrated by CircleC1 with RadiusD1.

I have been playing around with this for a while and I am assuming I have to play with radians but I'm not sure how to approach it. Any help, even a finger in the right direction would be greatly appreciated.

My end game is to find and equation that I can use to quickly find the distance CircleC will move on the X-axis given a change on the Y-Axis.

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$A = (0,0)\\ C = (0,0.015)\\ C_1 = (x, 0.0145)$

If you want your circles to have the same radius $C_1$ must be the same distance from $A$ as $C$ is from $A.$

Use the Pythagorean Theorem

$x^2 + 0.0145^2 = 0.015^2$

$x = \pm\sqrt {0.015^2 - 0.0145^2}$

If circle $C_1$ has radius $D_1$ (as described) you have added a degree of freedom that makes this a little bit more complicated.

$x^2 + 0.0145^2 = (0.020 - D_1)^2$

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  • $\begingroup$ CircleC and CircleC1 are exactly the same and have the exact same radius. so RadiusD = RadiusD1 $\endgroup$ – Jonathan Varela Mar 7 '17 at 20:34
  • $\begingroup$ That is what I suspected, however that isn't what you actually said in the text of your question. Anyway, now you have the solution for either case. $\endgroup$ – Doug M Mar 7 '17 at 20:51
  • $\begingroup$ I do! I decided to follow your method, as it was the one I was most familiar with. Thanks! It was funny, the second I read your comment that the length from the center of A from both C and C1 should be exactly the same, everything clicked in my head. Thanks again! $\endgroup$ – Jonathan Varela Mar 7 '17 at 21:03
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Hint: Using complex numbers makes things easier. Suppose the large circle has center $z$ and radius $R$, and suppose the smaller circle has radius $r<R$. If the smaller circle is tangent to the inside of the large circle, and its center lies on the radius of $A$ which is at angle $t$ (measured counterclockwise from the horizontal ray to the right of the center $z_0$), then the center of the small circle is a function of $t$: $$w(t) = z_0 + (R-r)e^{it}$$ which can be realized as $$(x(t),y(t))=(x_0+(R-r)\cos t, y_0 + (R-r)\sin t)$$ Note you can take $t=\pi/2$ to get your original configuration.

Do you see that knowing $y$ will help you determine $t$ and thus $x$?

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  • $\begingroup$ Hi! Thanks for your response. It looks pretty complicated so it'll take me a little bit to break it down and see if I can wrap my head around it. I'll post back if I can get it working. This is going into an excel spreadsheet, just FYI. $\endgroup$ – Jonathan Varela Mar 7 '17 at 19:10
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enter image description hereLet the diameter of the bigger circle be $D$, and the diameter of the small circle be $d$.

If the small circle is moved by distance $\Delta y$ down towards the center of the bigger circle, you have to move $$\Delta x = \sqrt{\Delta y (D - d - \Delta y)}$$ to the left to make contact with the bigger circle.

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  • $\begingroup$ Please upvote if you find my answer useful. $\endgroup$ – BoltzBooz Mar 7 '17 at 19:58

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