Here's an approach that is reasonably easy to do (without too many equations) for an incline:
Make an incline that's as steep as you want to be able to handle, and take your wheel and mark one point
on the edge.
Start with the mark at the bottom of the incline, roll the wheel up one turn and mark the incline where the
wheel-mark meets it.
Measure the height (vertically) of that point on the incline. This is the amount of vertical lift the motor
does turning the wheel one turn.
Take the weight of your entire robot including payload (in kg), multiply by 10, multiply by the height you
measured (in metres). This figure is the energy to lift the robot that amount.
The theoretical torque you need is that energy divided by 2 pi. The actual torque has to overcome friction,
so double that figure for good measure.
wheel lifts robot 0.15m after one turn on incline. Robot weighs 4.5kg.
Energy = 4.5 x 10 x 0.15 = 6.75 joules.
divide by 2 pi = 6.75 / 2 / 3.14 = 1.075 (torque in newton-metres)
double the value to allow for friction and losses = 2 N-m (approx).
Note that none of this discussion concerns speed - the motor might have a reduction gear to
get the torque needed and thus be quite slow. Power = torque x angular velocity, so more
torque and more speed requires more powerful motor.
And the torque we are talking about (!) is that at the output of any gear system, direct to the