bumping holding current - how it affects torque?

[Jonathan]

Cheers Eddy, reading through the manuals, I think bumping the holding current won't change the torque in normal operation.

That's correct.

Is 240rpm the required motor speed, or shaft speed? From other threads I think it's the latter, so with the ratio you've chosen that makes the motor go at about 2.66*240=638rpm. According to the torque speed curve you've posted, the motor will only output about 2.8Nm (187 Watts) at that speed.

What I would do it make a copy of that graph in a speadsheet, by just reading off a few values and plotting them and use this to plot power vs speed. Pick the point with the highest power (which incidentally is around the corner speed) and choose your drive ratio to match that speed. To convert the torque-speed graph to power you'll need the following formula:

.

If the power you get using that approach is still not sufficient, that's the time to consider raising the motor voltage. Personally, for a coffe gringer I'd go for a generic motor with a planetry gearbox as it should be much more compact, efficient and cheaper.

Just to illustrate why this approach is better, one way that might seem logical is to run the motor at the speed with the highest torque output - i.e about 160rpm. So for instace 240/160=1.5 .. say 30T pulley on motor and 20T on output. That should get 7Nm at 160rpm, so 117W, which is clearly less power than the current setup so not a good way to approach this.