View Full Version : Torque in DC motos

03-16-2012, 12:37 PM
Can any one explain to me the unit used for torque "Kg.cm" or some times "Kg/cm" in dc motors. It is used by most robotic webshops in the specs.

in Robokits (http://robokits.co.in/shop/index.php?main_page=product_info&cPath=2&products_id=50) the unit used is kgcm while in Nex-robotics (http://www.nex-robotics.com/products/motors-and-accessories/300-rpm-centre-shaft-economy-series-dc-motor.html) the unit used is kg/cm.

03-17-2012, 09:38 AM

kgcm or kg/cm is the same things. The above tutorial explains it in detail.

03-19-2012, 04:20 AM
Remember the formula for torque (T)

T = r x F
where, r is the moment arm and F is the force applied. To look at it another way, imagine you have a lever of length r attached to a pin joint,
if you apply force F at the far end of the lever, the torque at the pin joint is r x F.

Now, let's attach numbers and units, if you applied a force of 1Kg at a distance of 1cm from the pin joint on the lever, the torque experienced at the
pin joint would be 1 Kg.cm. Though we usually multiply gravitational constant (g = 9.8m/s^2) to a weight to convert it to Newton force depending on its direction.

Inversely, in case of a motor, If you have some object attached to a lever connected to the shaft of a motor, assuming the distance of load from shaft is 1cm, then a 1Kg-cm motor would be able to apply a force of 1Kg to that load. So on and so forth. That value will change based on moment arm.

So, if you want to lift a certain weight using the motor, based on its moment arm, or perpendicular distance of the line of action of force on the load from the center of the load, you can calculate the applied force using the formula
T = r X F = rFsin(theta)
or F = T/(r sin(theta))
as forces are not always perpendicular ,thus r sin(theta) signifies the distance of line of action from the center of load multiplied by the sin of the angle between the line of action and the radius to the point of tangency of the line of action on the load.

Thinking about it, Stiffness K = F/x where F = applied force and x is the change in length. For a force of 1 Kg causing a displacement of 1cm, the stiffness would be 1 kg/cm. So Kg/cm is a wrong unit for torque, just think of it as Kg-cm. (Original units are N-m in SI and lb-in in imperial)