Required Gear ratio for a DC motor

dc motorgear

I am making a dc motor vehicle, it will have 2 drive wheels each powered by a separate motor. For a desired speed and acceleration, how would I go about calculating the gear ratio needed?

Info I have obtained:
– Torque vs. Angular Speed Curve equation: y = -6E-6x + 0.0059
– Therefore Theoretical Torque Max = 0.0059 Nm and theoretical angular speed max =983.33 rad/s.
– Mass of Vehicle ~ 1.150kg
– Desired Speed 0.3 m/s
– Desired acceleration =0.1m/s^2

I am unsure which equations to use to get a proper gear ratio as to get the desired speed without overworking the motor. Note: the voltage applied is constant at about 4.5 V, it cannot be changed.
Wheel radius is 37.5mm

Best Answer

There are many contributing factors to this, most of which is related to frictional force, incline etc. If we ignore these for a moment

If we start off with \$F = ma \$ [1]

and equally \$ T = F*r\$ [2]

Where

  • m = mass of vehicle
  • r = radius of wheels
  • \$T_w\$ = torque at wheels (post gearbox)
  • \$T_m\$ = torque at motor (pre gearbox)
  • F = force to accelerate the mass
  • \$\omega_w \$ = velocity of the wheel
  • \$\omega_m \$ = velocity of the motors rotor
  • R = gearbox ratio.

\$\frac{T_w}{r} = ma \$ [3]

if \$T_m = -6\mu \omega_m + 0.0059 \$ [4]

and thus

\$T_w = -6\mu \frac{\omega_w}{R} + 0.0059 \$ [5]

substituting back into [3]

\$\frac{-6\mu \frac{\omega_w}{R} + 0.0059}{r} = ma \$ [6]

now it is stated the desired speed is 0.3m/s

The circumference of the wheel is \$2 \pi r\$ & thus

\$speed = \omega_w r \$ [7]

thus \$\omega_w = \frac{0.3}{ r}\$ [8]

substituting [8] into [6] and rearranging for R:

\$ R = \frac{-6\mu * 0.3}{0.00375*(0.1*1.150*0.00375-0.0059)} = 0.08777\$ or rounded up... 12:1 gearbox