Finding coincident angle between 2 equally spaced sets

[Telosian 7]

Hi Math people. Budding math wiz here (Aerospace Engineer / Coordinate measuring machine programmer) , need help with a formula.

I have 2 external gears on one shaft, the top gear has 12 equally spaced teeth and bottom has 14 equally spaced teeth. At some location there will be 2 gears visually/nominally in-line or at the same rotation angle around 360°. Assuming a part is loaded and a top tooth or angle value is randomly measured, (this value will be positive from 0-360°). Assume we then measure an arbitrary adjacent tooth on the gear set below the top one. I have these values, and can store these as variables. My program I am using for this purpose is fairly basic, has some algebra and trig, but barely an if/then statement. I need to map or calculate all other tooth positions (angle position) for each gear set, and ideally 'wrap' them to 360 (e.g. 395 = 35). From there I need to find the smallest angle difference between any one tooth on top and adjacent tooth on bottom. Keeping in mind a value of 7 on one gear and a value of 355 on the other is a circular distance or delta of only 12, and not 348.

The trouble is in my experiment (assuming 12T has a start value of 60 (2 pitches), and 14T has a start value of 51.428 (also approx 2 pitches), how could I calculate and find out that 12T gear # 11 and 14T #13 are the teeth 'in-line' or rotationally coincident around 360°? my functions available are : + - / *, MIN ,MAX ABS, < or > , ATAN, ASIN, COS, SIN, and I dont have if/then, else, for, while, loops or patterns, best I have is 'jump to step', I can only jump ahead but I have discovered a clever way of testing things with a boolean ... e.g. STEP1 : (A>B*1)+2 will go to step 2 if false but step 3 if true. Anyways I need less help with programming and more with a simplified formula. Thank you math Team - I have exhausted my options for days on ChatGPT. Obviously, I can compare gear 1 tooth 1 to all 14 teeth on gear 2, and so on for all 12 teeth on one gear for 168 comparisons, would take lots of manual coding I think there must be a way with angular pitch ratios etc. Cheers !

<Moderator note: moved from a technical forum>

 

[haruspex]

If one gear has X teeth and the other Y teeth, Y>X, and we start at an aligned pair, stepping to the next tooth of each in the same direction discovers a separation of ##2\pi/X-2\pi/Y##.
Repeating N times, the separation grows to ##N(2\pi/X-2\pi/Y)##, though at some point this will exceed ##|2N\pi/X-2(N+1)\pi/Y|##.
If this brings us to the randomly chosen pair of near-adjacent teeth, with a separation of ##\theta##, then ##N=\frac{\theta}{(2\pi/X-2\pi/Y)}##.
You can obtain a similar expression for the other case. The challenge then, if I understand the task, is to find the lower of the two values for N.