Nitrile O-ring belt drive - suggested stretch percentage..?

[Jonathan]

I'm guessing that this wouldn't be ideal as there'd be very little traction or transfer of torque - so my question is: how much stretch percentage should I introduce to the o-ring for it to do its job as a conduit between the two motors? 10%? 20%?

Motor power rating is 12V*0.35A=4.2W at 20RPM so available torque=4.2/(20/30*pi)=2Nm. Roughly.

The torque is transferred due to friction between the belt and 'couplers'. So the way to work it out is to calculate the friction force and multiply by the coupling radius to get the torque transmission capability. The force depends on the spring constant of the o-ring, which you could measure by handing weights from it and measuring the deflection, or use the young's modulus to calculate it ... but that gets involved as it's an elastomer so doesn't care too much about Hooke's law (the modulus is only constant for small deflections).

So use the co-efficient of friction between rubber and aluminium (about 0.5 - citation needed!) to find the required force - I'm guessing the couplings are 25mm diameter, so F=2Nm/25mm=80N which means tension is 80/0.5=160N. That's about the weight of four domestic cats ... according to Wikipedia the tensile strength of nitrite rubber is (at least) 10Nmm^-2, so 0.25cats/mm^2 which means you need at least 4/0.25=16mm^2. That's a (16/pi)^0.5=2.3mm radius, so 4.6mm diameter ring. Roughly. Add a bit more to be safe etc...

Feel free to experiment, but I think you'd be better off with something more substantial.