Find the radius of a circle given a known smaller circle and other information.

Question

There is a large circle with two smaller circles on the inside edge (each $r=6$), the distance between each circle being $50$ (that is directly not along the curvature of the outer circle) and the base of the circles to the 'top' of the outer circle being $15.2.$

I need to find the radius of the outer circle. How would I go about doing that? Cannot use the Intersecting Chord Theorem BTW.

This is based off a bearing. The outer circle being the outer raceway, the inner circles being the balls and the the block in the middle being a gauge block.

Please ignore the crudity of the diagram. Units unimportant but all the same (being a bearing, there're probably in mm).

Thank you in advance.

LLAP & DFTBA

Answer 1

Let's call $O$ the center of the circle you seek, $I$ the center of the right small circle, $R$ the radius of the large circle, $H$ the height between base and top of circle, $P$ midpoint between the two small circles and $d=PI$.

You have the triangle $PIO$ which has a right angle at $P$.

$PI=d$, $OI=R-r$ and $PO=R-H+r$

Then you have $(R-r)^2=d^2+(R-H+r)^2$ or $(R-r)^2-(R-H+r)^2=d^2$

That is $(R-r+R-H+r)(R-r-R+H-r)=d^2=(2R-H)(H-2r)$

$R=\dfrac{d^2}{2(H-2r)}+\dfrac H2$

Answer 2

Shrink the two small circles to a their centers while you keep the large circle tangent.

The new radius is $S:=R-r$ and the height between the centerline and the apex of the circle is $G:=H-2r$.

By Pythagoras, this height is also $S-\sqrt{S^2-D^2}$, where $D$ is the half distance between the two centers.

You solve with $$(S^2-D^2)-(S-G)^2=0=2SG-D^2-G^2,$$ $$S=\frac{D^2+G^2}{2G},$$ or $$R=r+\frac{D^2+(H-2r)^2}{2(H-2r)}.$$