some number crunching, showing how easy this is in theory...
If torque and angular speed are known, the power may be calculated. The relationship when using a coherent system of units (such as SI) is simply
P = \tau \omega
where P is power, \tau is torque, and \omega is angular speed. But when using other units or if the speed is in revolutions per unit time rather than radians, a conversion factor has to be included. When torque is in pound-foot units, rotational speed (f) is in rpm and power is required in horsepower:
P \text{(hp)} = {\tau \text{(ft} {\cdot} \text{lbf)} \times f \text{(rpm)} \over 5252}
The constant 5252 is the rounded value of (33,000 ft·lbf/min)/(2π rad/rev).
When torque is in inch pounds:
P \text{(hp)} = {\tau \text{(in} {\cdot} \text{lbf)} \times f \text{(rpm)} \over 63{,}025}
The constant 63,025 is the rounded value of (33,000 ft·lbf/min) × (12 in/ft)/(2π rad/rev).
seee if that posts properly, otherwise...
http://en.wikipedia.org/wiki/Horsepower
so.
lets say i have a foot long reaction arm. of course i will!
and my scales, in pounds, reads 10 when i load the thing down at 5000rpm....
so, take the equation...
10x5000/5252=
9.5hp.
rotational speed, measured torque...hp
seasy
3.6 lb at 6500rpm? 4.45 hp...etc etc...
now, my only thinking bit at the moment is mounting this spindle...
so, i need the torque reaction from the magnet disc. i guess the easy way is to just make the slot slightly oversize and let the thing spin slightly on the shaft, is how ive been picturing it so far. other option is to mount the spindle itself on bearings, the slot is tight with minimal play, and the reaction arm is clamped to the spindle... lets me use plain old spring scales but i think a proper load cell is on the horizon...