|
Gui:
Suppose we have a motor with these specs:
....no load speed at 12 V is 9600 rpm, drawing 1.2 A.
....stall torque 2 N*m Stall current is 37.2 A.
Suppose we want to use wheel of 10 cm (= 0.1 m, ~4") radius, and the box is propelled at 3.14 m/s (about 10 feet per second).
Suppose we want maximum power out of the motor to do this.
(This may or may not be a good asumption! You may decide that you want max power at some other speed.)
-+-+-+-+-+-+-+-+-
Max power is found close to the midpoint of the speed range, giving about half the stall torque, at about half the stall current. Electrically and mathematically, the product of voltage (=k1 * rpm, graph \) times current (= k2 * torque, graph /) reaches a maximum for speeds and torques in the mid-range. Let's conjure with 4800 rpm, or 80 rps.
The driven wheel is 0.1 m radius: to go at 3.14 m/s, it must turn at 3.14 m/s / 2piR = 3.14 / 2 * 3.14 * 0.1 = 1/.2 = 5 rps or 300 rpm.
Our speed ratio must be 4800:300 = 16:1. Since ratios of 8:1 were supposed (on this board) to be the highest practicable for efficiency, let's use two drops at 4:1 each. It was also mentioned that every gear interface loses something, lets assume 25% for our spur gears.
The first drop finds the driven gear going 4800/4 = 1200 rpm, but the torque available is not 4 x 1 N*m, but due to losses, only (4 - 25%of 4) * 1 N*m, = 3 N*m. The next gear interface turns the output shaft at 1200/4 = 300 rpm, and the torque available is 9 N*m.
The wheel pushes along at 3.14 m/s, but the torque available is 9 N*m * 0.1 m = 90 N.
The torque could also be figured as 16 * 1 N*m (1-0.25)(1-0.25).
If the friction your box encounters is less than the available force, 90 N here, you're home free. If there is more friction than available force, it ain't gonna work: either revise your speed expectations and change the ratios for enough torque, or choose a more powerful motor.
To measure your Force requirements, disconnect the gears and pull the robot with a force scale. To read the force required to turn the gears, reconnect them, and repeat the experiment. Don't forget that the force is multiplied at the upper end of th echain, nearer the motor.
Power efficiency:
At this speed, the motor is drawing ((37.2 - 1.2)/2) + 1.2 A, or 19.2 A at 12 V for a power input of 230.4 W or 230.4 W/746 kW/HP = 0.308 H.P.
The 25% losses in two gear interfaces leave us with about 60% of the power we put in, as an approximation.
The wheel moves at 3.14 m/s, with a force of 90 N. Power out = Work out/time = Force x Displacement/time = 3.14 m/s x 90 N = 202.6 W, about 60% of what we put in.
Fine Print:
If we decide that max power should occur at a lower speed, we'll need to gear it down farther, and the torque available will be more, but the top speed will be lower. If the max power floor speed is to be greater, then we won't gear it down as much, and maybe, whem we need to accelerate from a standstill (1..2..3..go!) there will not be enough torque: then the motor will stall, the current will go way up, the motor temperature rises, and "pop" goes the breaker.
If the friction with the floor that you obtain with your wheels is greater than the force available, locked rotors will be common. Many FIRST teams try to make the wheels break free before the stall is reached.
We could go at this from desired torque output, as well. (Ex., How fast will the output shaft turn if we need 270 N of force on the edge of a wheel 5 cm in diameter ?) (No, I don't want to answer this one :-) !)
HTH
|