What are the size ratio's of this simple mechanism?

Because you don’t define the size of the smallest gear, you can’t determine the size of the transfer gear.

That is, you need the transfer gear to make the two hands spin the same direction.

So the transfer gear could be any size.

ahh good point. as it is my model let us say that it is approx. 5 in. is it possible to determine both the size of the lower and upper part of the transfer gear?
thanks

Well, the ratios are just the ratios of the diameters—so if the pitch diameter of the smaller is 5”, the PD is 20” of the bigger, then you’d have 1:4 ratio. You need 1:60.

I assume your two gears are mounted to concentric shafts? So you need to get back to the Pitch Diameter of the smaller one, while also maintaining the gear ratio and direction that you need?

Ignoring the 5” assumed gear, because that might not be the right size, now that I try to work it out.

I’m thinking a 2” gear off the 20” gear, which gets you 1:10. Shaft that to the plane of the smaller gear, and you need to get a total of 1:6 to that shaft. You have 11” between the centers of the shafts. 1:3 on the first stage and 1:2 on the second stage could be 3” to 1” gear (2.5”, leaving 8.5”) 5⅔” diameter gear shafted to the 1” gear, and meshed to an 2 5/6” gear on the shaft I think would get you the 1:2 speedup.

I think.

thanks i think i understand now, the mathematics just completely confused me

ok so i entirely screwed up, i actually need a 1\12 ratio, ive figured that if i base it on the # of teeth i should end up with the correct ration no matter the sizes of the trasitional gear.

@shadowofdeath Yes—number of the teeth work just fine.

The diameter is a function of # of teeth, yadda yadda =)