Permalink Submitted by S. Burgess on April 16, 2021

Just use an O-scope in any scale to solve.
One "rotation" = 2 segments, the positive and negative peaks, of a sine wave.
The wave form is standard, of course. Any length of unit 1 is displayed as 2 unit lengths of 1 merely as the form of the wave. (+ : -) = 0 = 1 cycle. This works for any ratio where the unit length = the "set" which is to generate the "number of sets" corresponding to the larger value of the respective radius length to be used.
A = 4 units B = 1 unit
Say: 4/1 = 8 segments or 4 cycles / 2 segments or 1 cycle = A/B = 8 + : 8 - = peak to peak / 2+ : 2 - peak to peak = 4.
The number of rotations is given by unit length alone.
Graphic: unit ---- set ---- ---- ---- ---- : 4 : cycles or rotation of B about A. e.g 8 cycles / 2 cycles.
Works for me, anyway. At least it's intuitive and based on "workshop" manipulation rather than other numerical speculation.
SWB 15 APR 2021

## The 1/4 rotation problem (posted some years ago)

Just use an O-scope in any scale to solve.

One "rotation" = 2 segments, the positive and negative peaks, of a sine wave.

The wave form is standard, of course. Any length of unit 1 is displayed as 2 unit lengths of 1 merely as the form of the wave. (+ : -) = 0 = 1 cycle. This works for any ratio where the unit length = the "set" which is to generate the "number of sets" corresponding to the larger value of the respective radius length to be used.

A = 4 units B = 1 unit

Say: 4/1 = 8 segments or 4 cycles / 2 segments or 1 cycle = A/B = 8 + : 8 - = peak to peak / 2+ : 2 - peak to peak = 4.

The number of rotations is given by unit length alone.

Graphic: unit ---- set ---- ---- ---- ---- : 4 : cycles or rotation of B about A. e.g 8 cycles / 2 cycles.

Works for me, anyway. At least it's intuitive and based on "workshop" manipulation rather than other numerical speculation.

SWB 15 APR 2021