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I've recently bought some cheap TowerPro SG-50 rc servos from ebay. They seem to work quite well, and despite their small size they are quite strong.

But how "strong" are they actually. The datasheet specifies:

Stall Torque: 0.8 kg / cm

What does that mean. I guess that stall torque is the amount of torque that needs to be applied in the reverse direction of the rotation, before the servo stalls. But what does the unit kg / cm mean?

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  • \$\begingroup\$ I'm pretty sure the datasheet says "kg \$\cdot\$ cm", not "kg/cm". I fixed it in the question. \$\endgroup\$ – stevenvh Sep 12 '11 at 12:37
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    \$\begingroup\$ Yes, it should be [kg*cm] however the datasheet does say [kg/cm], which apparently is a quite common mistake in (servo) data-sheets. Look at the comments in Axemanx answer below. \$\endgroup\$ – bjarkef Sep 12 '11 at 13:27
  • \$\begingroup\$ kg*cm makes sense, but on I think a Actorobotics product page on SparkFun, they listed a stall torque in kg/cm, which is how I ended up on this page. I assume it was a typo. \$\endgroup\$ – Lèse majesté Dec 8 '15 at 1:21
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The normal SI units for torque are Newton-metres. Torque is a force times the distance from the fulcrum/axis. The units that you are giving do not make sense (although I have seen them in a number of places quoted for this device - puzzling). I expect that this is a mistyping of the units on a cheap data sheet.

I would expect that the motor will stall when a weight of 0.8kg is hanging from a 1cm long arm attached to the motor spindle.

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  • \$\begingroup\$ And so there should be a linear connection, thus it will equivalent stall with a 0.4kg weight hanging from a 2cm long arm? \$\endgroup\$ – bjarkef May 25 '10 at 15:38
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    \$\begingroup\$ That's correct, assuming you're on Earth. \$\endgroup\$ – pingswept May 25 '10 at 16:05
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    \$\begingroup\$ Poor assumption. \$\endgroup\$ – Kortuk May 25 '10 at 16:55
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    \$\begingroup\$ IMHO is "Assuming a planet that has same gravity as Earth" :-) \$\endgroup\$ – Axeman May 25 '10 at 17:51
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    \$\begingroup\$ Which is why the proper SI unit has dimensions of force times distance, not mass times distance! =P \$\endgroup\$ – JustJeff May 25 '10 at 22:15
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Stall torque is the maximum torque that the unit can supply when in 'locked shaft' condition. This doesn't mean you actually have to mechanically bolt the shaft or anything, it just means the torque that can be dished out from a dead stop.

Applying a torque that opposed motion, that is less than the stall torque, will cause the rotational speed to slow down and the unit will draw more current. Applying an opposing torque that is greater than the stall torque will overcome the unit's abilities and will drive it backwards.

The value of 0.8 kg-cm means the amount torque you would get if you were to bolt a 1cm long arm to the shaft (set and right angles to the axis of rotation, and positioned dead horizontal), and hang an 0.8kg mass at the end of the arm. This is a product, so it's also the amount of torque you'd get using 0.08kg (i.e., 80g) at the end of a 10cm shaft, etc.

FWIW, the proper SI unit for torque is newton-meters, or possibly even newton-centimeters would be acceptable, as torque is defined as FORCE (Newtons) x MOMENT-ARM (distance). The english units are lbf-ft, pounds-force * feet, often written just ft-lbs, and this probably has influenced the less rigorous spec sheet writers to cast their 'metric' torques in the somewhat mistaken units of kg-cm, kg-m, and g-m.

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http://www.engineersedge.com/torque_conversion.htm

1 Kg/cm = 9.806x10^-2 N/m

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    \$\begingroup\$ that table is in Kg * cm, or N*m, not in Kg/cm. \$\endgroup\$ – Kortuk May 25 '10 at 18:35
  • \$\begingroup\$ Unfortunately use of "/" instead of "*" (or nothing) for some measure units is common in non-scientific environment (like kW/h versus kWh, or mph versus m/h) \$\endgroup\$ – Axeman May 25 '10 at 20:27
  • \$\begingroup\$ Yeah, I know that can happen, but I am not sure that it is that case. \$\endgroup\$ – Kortuk May 26 '10 at 2:12
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    \$\begingroup\$ @Kortuk, it is :-) I've used enough RC servos to tell you that that unit is really kg*cm, written almost always as kg/cm. And that, on this planet, a 3kg/cm servo will stall with a 3kg weight hung from a 1cm shaft. Talking about kg versus N as a force unit, well, "Kilogram-force" IS a (non SI) force unit: is defined as the force exerted on one kilogram of mass by a ~9.806 m/s2 gravitational field ~= 9.806N \$\endgroup\$ – Axeman May 26 '10 at 9:02
  • \$\begingroup\$ 1 Kg = 1 kelvin-gram (which is nonsensical). 1 kg = 1 kilogram. Capitals matter. Easy to remember: units named after persons are capitalised when abbreviated and lower-case when spelled out. V - volt, A - ampere, K - kelvin, N - newton, ... \$\endgroup\$ – Transistor Jun 20 '16 at 17:10
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I believe the general interpretation of "/" is that it stands for "per", and that the unit following the slash is always treated as a single whole unit.

m/s = meters (per 1) second
A/h = amperes (per 1) hour
0.8 kg / cm = 0.8 kilograms (per 1) centimeter (of lever length or gear/pulley radius)

These are also understood as inverse ratios, such that 1/2 of the second unit will be balanced by twice of the first unit, etc.

Shaft of servo represented as a lever.

The center of the servo shaft is like the fulcrum of a lever and the shaft radius is the same as the length of the lever where the mass is lifted. The maximum mass the lever can lift depends on where the load is suspended from the lever, which is the same as the radius of the shaft / gear / pulley.

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  • \$\begingroup\$ So the longer the lever length, the stronger the servo becomes? \$\endgroup\$ – bjarkef Jun 20 '16 at 13:45
  • \$\begingroup\$ No, it's an inverse ratio. Greater lever / gear / pulley radius: travel distance increases but torque decreases. \$\endgroup\$ – Dale Mahalko Jun 20 '16 at 14:18
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    \$\begingroup\$ You seem to be endorsing the incorrect "kg/cm" which is wrong. Torque is a force x distance. Can you clarify in your answer? \$\endgroup\$ – Transistor Jun 20 '16 at 17:12
  • \$\begingroup\$ I get what you are saying Dale, but the statement "0.8 kilograms (per 1) centimeter" is inherently wrong as it specifies that the servo can hold more weight as the lever length increases. I guess 1,25 centimeter / kilogram would be more correct? \$\endgroup\$ – bjarkef Jun 21 '16 at 6:20

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