Thanks for your response, it is very thorough. I went over my design and determined that 4.7. in (.12 m) would be the preferred diameter. Following your calculations, this would mean the wheel circumference is about 14.8 in. Therefore, a rpm of 300 would give about 4.2 mph. According to the website it would be reasonable to assume it operating at 60% so a 500 rpm would be enough. Does 60% sound right?
My current concern is regarding how much motor power I'll need. My guess is that this would be caused partly by friction, slowing the rpm, and partly from inefficiencies. According to https://saferroadsconference.com/wp-content/upl…Grass-Types.pdf, the coefficient of friction of wet grass (the worst case) would be roughly .2. I do not have an exact weight for my model, but will set an upper bound of 10lbs (4.5 kg). If I want to go from 0 to 1 mph (.5 m/s) in a second then that will require an acceleration of .5 m/s^2. Since Tnet = Tmotor - Tfriction = Tmotor - r * Ffriction = Tmotor - r * m * g * mu = m*a => Tmotor = m * a + r * m * g * mu = 4.5 * (.5) + (.12) * (4.5) * 9.8 * .2 = 3.3 (roughly). This mean I would require at least 3.3 Nm of torque. Now, I'll probably need to calculate for things like hills and inclines too, which I can adjust my calculation for, however, my question is, assuming 2.5 Nm of torque is needed, how much inefficiency should I assume. In other words, if I theoretically need 2.5 Nm, would it be safe to assume 3 Nm of torque would be enough in the real world to get 2.5 Nm? Is there a general rule of thumb for it?
The link you sent on amazon says that the rated torque for the 500 rpm one is .4 Kg*cm. I would think the units would be Kg * cm * (m/s^2) but assuming these are the units, the torque would be 40 Nm. Does this sound correct to you?
My last question is how important is motor power? I didn't see any information about the motor power on the website.