How did I decide on a #4 machine screw to go in the middle of the Ranger cluster for motor retention? Trial and error would have worked, but actually what I did was to draw it to scale in a vector drawing program and see how big a circle I could fit in the gap.
But a little easy geometry and trig will do it, too:
Here are three motor tubes — of radius R, let’s say — arranged in a triangle, with the largest possible motor retention screw, call its radius r, in the middle. Triangle AOB obviously is isoceles, with angle AOB = 360°/3 = 120°. So triangle ADO is a right triangle, with angle ɸ = AOD = 60°. AD is of length R, and AO’s length is R+r.
But sin ɸ = AD/AO = R/(R+r) = √3/2. Solve for r: r = (2/√3–1)R = 0.1547R. Or d = 0.1547D to work in terms of diameters D and d. For D = 18.69 mm (BT-20), d = 2.89 mm = 0.114″. A #4 screw will fit, just barely.
You can generalize this for a ring of n tubes, if you want. Then ɸ = 180°/n and r = (1/sin ɸ–1)R. For n = 4, sin 45° = √2/2 so r = 0.414R. For n = 5, r = 0.701R. For n = 6, r = R, but we all knew that, right? Six tubes will fit around a seventh of the same diameter. Nice to know the formula gives the right answer in that case.
This is starting to get ridiculous for motor retention purposes: It’s telling us if the motor tubes are BT-20 then 5 can fit around a threaded rod of 5/8″ diameter, but I don’t think you’d want to do that. You might want to know how many tube fins of one diameter you can fit around a body tube of another diameter, though, or how many small motors you can fit around a big motor, and by inverting the formula to get n as a function of r and R you can do that. R/(R+r) = sin (180°/n) so n = 180° / arcsin (R/(R+r)).