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* = 0.1547*R*. Or *d* = 0.1547*D *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* = 0.414*R*. For *n* = 5, *r* = 0.701*R. *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*)).

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