I have Three Equal Inscribed Circles Into Big Circle.
Diameter of internal circles is 1.19mm.
How I calculate diameter of outer circle.
Thanks
Let R denote the radius of the outer circle. Then you know:
$\displaystyle (R-0.595) \cdot \sin(60^\circ) = 0.595$.
Since $\displaystyle \sin(60^\circ) = \frac12 \cdot \sqrt{3}$ you'll get:
$\displaystyle \frac12 \cdot \sqrt{3} \cdot R = 0.595(1+\frac12 \cdot \sqrt{3})$ Solve for R.
I've got $\displaystyle R \approx 1.282$
Hi Jonny,
The diameter (denoted by do) of a big outer circle that is circumscribed around 3 identical smaller mutually tangential circles (each with a diameter denoted by di) is
do = di * (3 + 2 * sqrt(3)) / 3 ,
which is approximately equal to
di * 2.1547005384 .
If di = 1.19 mm as in your example, then do = 2.5640936407 mm.
In case you are interested, I just finished solving the more general (and much more difficult) problem of how to calculate the diameter and center coordinates of a circumscribed circle around three mutually tangential circles with any given diameters if such a circumscribed circle exists, or else how to calculate the line or the exterior circle that is tangential to all three given pins in a triangular arrangement.
Several attempts all failed when I tried to attach a file called hole3pins_090814.zip to this reply, so please contact me directly by e-mail at DonCBraun@aol.com if you want it. That zip file includes my exhaustive mathematical derivation concerning this problem as a 125-page MS-Word 2003 document called circle3pinsMath.doc .
That zip file also includes two computer programs I wrote in C and compiled into executable files for a Windows PC, a newer Mac with an Intel processor, and an older (G5) Mac with a PowerPC processor. The first program called "circle3pins" solves for a circumscribed circle, a tangent line, or a tangent exterior circle from given diameters of three mutually tangential circles. The second program called "hole3pins" essentially solves the inverse of that problem. It is intended to help machinists check the intended diameter of a round hole (say, 1.525 inch) by inserting three round pins from a set of precision gauge pins (for example, a set with 1002 pins whose diameters range diameters from 0.004 inch to 1.005 inch in steps of 0.001 inch). Program "hole3pins" computes the sizes of the three largest gauge pins (given the increment between consecutive gauge pin diameters like 0.001) that barely fit inside a hole with a given diameter. In this example, the diameters of the largest gauge pins to fit in the hole would be 0.706, 0.708, and 0.709 inch, which have a circumscribed circle of diameter 1.524813858 inch, leaving a clearance of about 0.0001861 inch.
-- Don Braun
Hello Don,
Better late than never..
My task actually was to select correct shrink tube which after heating will shrink and will close ob three wires. I knew diameters of wires, and needed to know the diameter of the shrink tube.
Anyway thank you for your reply.
Some time ago, I came across the problem of finding the radius of the circumscribed circle around three mutually tangential circles with radii $\displaystyle r_1$, $\displaystyle r_2$, $\displaystyle r_3$. The solution is that the radius of the circumscribed circle is $\displaystyle \frac R{2\sqrt{PR}-Q}$, where $\displaystyle P = r_1+r_2+r_3$, $\displaystyle Q = r_2r_3+r_3r_1+r_1r_2$ and $\displaystyle R = r_1r_2r_3$. The proof is not too hard if you use the technique of geometric inversion (inverting with respect to the point of tangency of two of the given circles, thus transforming them into a pair of parallel lines). If the denominator of the fraction is negative then the circumscribed circle becomes an exterior tangential circle. If the denominator is zero then there is a common tangential line for the three circles. The same analysis gives the radius of the inscribed tangential circle as $\displaystyle \frac R{2\sqrt{PR}+Q}$.
If all three radii are equal then my formula agrees with Don Braun's formula for d_0.
The problem of finding the centre of the circumscribed circle ought to be solvable by the same technique of inversion, but I haven't tried to do that.