EQ6 gear ratios

[pmrid]

I'm perplexed by the mysteries of the EQ6 motors and gears.
I'm attempting to replace the motors in my EQ6 in order to deal with some erratic behaviour in one of the original motors.
When I started to transfer the motor gear and transfer gear from the old motors to the new, I found that the transfer gear on one motor had 36 teeth while that on the other had 32. The motor gear on each had 12 teeth.
I remember seeing in another thread a comment by Bojan that the number of teeth in the intermediate gear didn't matter. What mattered was the number of teeth in the motor gear and on the shaft of the worm.
Is that right? I am mathematically challenged I'm afraid, so getting my head around this is proving to be very difficult. I am hoping someone can help me understand how you get to a gear ratio of 705:1 for the standard EQ6 assuming a 200 step motor and 64 microsteps - or is it 32?

Peter

[tlgerdes]

Thinking it through and the function of it, then it shouldnt matter. As both input and output gears connect to the transfer gear, then one "tick" on input equals one "tick" on output regardless of how many teeth are on the transfer gear. The gearing then happens as the ratio of input and output teeth.

[bojan]

180*47/12=705

(180 teeth on worm gear, 47 on worm and 12 on motor shaft).

[pmrid]

Still not getting it. Here's a rough schematic of the gears. The motor has 2. One on the shaft has 12 teeth and the other (which I have called a transfer gear) has 32 on one motor and 36 on the other. These connect to the worm and thence the main gear.
Everything else being the same, I'm struggling to understand why the number of teeth in the transfer gear can be ignored. What am I missing?
Peter

[hikerbob]

Peter don't forget that the ratio between the intermediate gear and the worm changes proportionally as well. As the ratio on one side of the intermediate gear goes down it goes up on the other side.

Bob

[bojan]

Exactly.

I should have presented the math as follows:

180 * (47/36) * (36/12) = 180 * 47 * 36 /(36*12) = 705
or
180 * 47 * (36/36) /12 =705


This is the same as

180 * 47/12 = 705 because 36/36 =1... so you see the middle gear could in principle have any number of teeth (even only single one.. mathematically this is OK but in reality/practice of course this doesn't make sense).


BTW, your drawing is not correct.
The main worm gear has 180 teeth (as you indicated,) but it is driven by worm (like screw, resulting in 1:180 ratio at this tage) and not by 47-teeth gear as on your drawing.
Perhaps this is also one of the reasons you have been confused.

[pmrid]

I'm beginning to see the light. Does it also make no difference what the diameter of the transfer gear is? As a matter of interest, I just threw a micrometer across them and the 32-tooth was 27.3mm and the 36 tooth was 30.4mm.

Peter

[bojan]

Diameter of transfer gear is also irrelevant of course, but it is linked to number or teeth and teeth pitch - which must be the same as other gears to mesh with them properly.

[pmrid]

My thanks to you all. Older and wiser, both.
Peter