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I've that 12V DC motor with gearbox in the picture. These are the specs of that motor: Voltage = 12V No load current = 250mA Sall Current = 5 A Stall Torque = 20 Kg/cm Motor Rated RPM = 133 Shaft Diameter = 4 mm Motor Length (without shaft) = 53 mm Motor Diameter = 25 mm

I want to attach that motor directly to one of the rear wheels (4-inch diameter rubber wheel) of a toy (with 4 wheels), how to calculate the maximum weight this motor can move?

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  • \$\begingroup\$ There is insufficient information in your question to answer it. The amount of weight it can move is going to be based not only on the maximum torque at the rear wheel, but also the diameter of that wheel, and to a lesser degree, its construction. \$\endgroup\$ – Lawrence NK1G Mar 5 '17 at 1:17
  • \$\begingroup\$ OK I mentioned that it will be attached to one of the rear wheels (which will be 4 inch diameter wheel), it will be a rubber wheel moving on a flat smooth surface. \$\endgroup\$ – Ahmad Oweda Mar 5 '17 at 1:33
  • \$\begingroup\$ This still isn't enough information. Besides friction losses, vehicle weight is only limiting acceleration, given the torque and wheel diameter are set. So you have to specify the acceleration you want to achieve, and think about the friction losses if your acceleration is considerably low. \$\endgroup\$ – Janka Mar 5 '17 at 1:37
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    \$\begingroup\$ So -- thinking more about the scope of the problem -- the "maximum weight" a motor can move is going to be entirely about rolling resistance ... the acceleration you can experience is going to be related to the mechanical power. I'm gonna bow out -- there is too much outside of EE and more in the realm of mechanical engineering here \$\endgroup\$ – Lawrence NK1G Mar 5 '17 at 1:39
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    \$\begingroup\$ @Ahmad: You have to test your whole vehicle. With the real tires you want to use, on the real ground you want to drive on, with the real weight. That's the only way to get some figures about the friction losses of your vehicle. When you have those, you have the torque needed for the motor (plus a tiny tiny bit more for the nearly neglible acceleration.) \$\endgroup\$ – Janka Mar 5 '17 at 1:57
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I think you don't fully understand the specs of the motor.

Stall Torque = 20 Kg/cm

The unit is Kg*cm. And when a wheel of 4" diameter, i.e. 2"=5cm radius is directly mounted to the shaft, the "force" is 20Kg*cm / 5cm = 4kg.
This means: If you put a string around the wheel and use it as winch, it could lift a weight of almost 4kg (8.8lbs) before it stalls.

Motor Rated RPM = 133

If there is no weight attached, the wheel would spin with 133RPM. The circumference of the wheel is 2*pi*5cm=31.4cm (~1foot), so the string would move up with 133ft/min=40.5m/min=0.8m/s=2.4km/h. This would also be the absolute top speed of your cart. In reality, the speed will decrease when a load is applied.

Let's go a step further and calculate acceleration. Let's assume a constant, max torque and a mass of 100kg.

$$F=m_{force}\cdot g=m_{cart}a$$ $$\Rightarrow\quad a=\frac{4kg}{100kg}\cdot 9.81m/s^2=0.4m/s^2$$

With this acceleration, the car would reach its top speed within 2 seconds. Not bad.


Now, as said in the comments, there's lots of friction. Friction in the bearings, friction between wheels and ground (Asphalt? Grass? Sand?) And when the cart makes a turn and only one inner wheel is driven by the motor, this increases the necessary force drastically. The motor has to overcome all of this.

I'd first measure how much "force" is needed to move the cart. Let someone sit in the cart, and push it with a kitchen scale. Is 4kg always enough to move? (Acually, you should need less than that)

Finally, I'd guess it is not enough to reliably move the cart. But since you somehow have to connect the motor to the wheel, a gear with a ratio of 2:1 would half the top speed, but double the force.

And one very important point: The gearbox already attached to your motor is made to withstand the forces during normal operation plus some margin. So it is made to handle may be 30-40kg*cm. But if someone jumps into the cart, this could simply crush the gearbox!

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  • \$\begingroup\$ +1 for writing this up. But I think we will still have to deal with these "I have a motor X, will it move a weight of Y" questions again and again. \$\endgroup\$ – Janka Mar 5 '17 at 9:31
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Also note that "Stall torque" is the maximum torque at stall, and you should NEVER stall an electric motor, if fact, aim to run it at no less than 30% of max speed, stalling will very likely cause permanent damage to both the motor, gearbox, and burnout whatever motor driver board you could be using to drive it.

The other Torque number you need to get is the "running torque", which will be much lower than stall torque. Realistic Torque you can expect to get from the motor without destroying it could be more along the lines of 4KG/Cm, but that is just a guess - check your motor specifications data sheet.

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