How much weight can a torque carry?

I have done a lot of research in it but I can't quite understand what they mean or talk about. This is the specification of my servo:

TORQUE 4.8 l 6.0V: 16.5 / 20.0 kg-cm SPEED 4.8 l 6.0V 0.18 / 0.16 sec/60° BEARING: 2 MAX TRAVEL: 165° GEAR TYPE: Copper & Aluminum SIZE (L x W x H): 40.7 x 20.5 x 39.5 WEIGHT: 60 g

How much weight does my 20kg torque power can carry?

Torque is measured in units which are a distance multiplied by a force. However since gravity is relatively constant on this planet, you can also use weight as a substitute for force.

The specifications seem a little mangled by posting here but it looks like when you give that servo 6V power input, it's capable of 20kg-cm torque.

If you put a 1cm arm on it, it could lift 20kg. If you put a 5cm arm on it, it could lift 4kg. If you put it into a gearbox with a 3:1 ratio (to get more turns on the output) it could deliver about 6kg-cm torque on the gearbox output.

Torque is turning force, which is measured in Nm (newton-metres) or you can think of it as
energy-per-angle (J/rad), in analogy with linear force (1 newton is 1 joule/metre).

In your specs they give torque in non-scientific units (basically engineers aren’t taught science
properly!!), the kg-cm (kilograms times centimetres), which is actually wrong as force is not
mass. They mean kgf-cm (kilogram-force times centimetres), which is better but is assuming
a particular strength of gravitational field for converting mass into force.

Basically with torque your first step is to convert to N-m as all calculations are easier in standard
SI units. 1 kgf = 9.8N (depending a little on where you are on earth and a lot on which planet
you are standing on!).

So 20 kgf-cm = 196 N-cm = 1.96 Nm (2 Nm is good enough for practical purposes).

As the previous poster has mentioned you can convert torque to force once you know the distance
off-axis of the load.

You can also convert torque to linear force for a lead-screw:

force (N) = 2 x pi x torque (Nm) / pitch (m)

So a leadscrew with 5mm pitch (advance per revolution) driven from 2Nm will give upto 2500N
force (assuming minimum friction) (2 x pi x 2 / 0.005)

Leadscrews give much more force for a given torque than a lever-arm / toothed belt.

All these torque calculations can also be done in the energy domain, using
energy = force x distance
= torque x angle

So long as you can relate angle of the motor shaft (in radians) to distance the load moves, you
can convert torque <-> force (ignoring friction though)

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MarkT: engineers aren't taught science properly

Maybe not where you are, but don't include my alma mater nor my daughter's current uni in that rude statement, thank you very much.

Its probably a lot better now, but the evidence is all over industrial practice! Stupid units are everywhere alas... Torque is one of the worst - you get all of these units routinely in datasheets: ft-lb in-oz in-lb kg-cm instead of Nm - its such a waste of everyone's time and effort not to standardize (OK there's imperial and metric, but that would mean only having to memorize one conversion factor, not the 20 you need for 5 different units)

Perhaps this is just a mechanical engineering issue (and mainly confined to imperial territories)? I am happy to be rude about it.

Imagine the nonsense if there were 5 different units for potential difference, or resistance???

MarkT:
Its probably a lot better now, but the evidence is all over industrial practice! Stupid units are
everywhere alas…

Well I was in 3rd year u/g 40 years ago this year, so it was ok here “then”.

MarkT:
I am happy to be rude about it.

I’m out…

(At least kg.cm is a bit better than the kg/cm you and I encountered here over 2 years ago, and which I continue to encounter in posts as recently as this week.)

Oh yes, I forgot the N/m thing even!

And modern engineering has to cover science much more thinking about it, so this is probably legacy - I had hoped that people have caught up with the best practices from 1960 by the 21st century - it is several generations ago!

MarkT: Actually there are 3 different units for amps volts and ohms: abamps, statamps and plain old amps. Same storey for abvolts, statvolts, abohms, statohms. This was invented by the scientists in the early days but thankfully the engineers got into the act and put some common sense into the science.

As for weight and force, the confusion seeped in in the early days before the scientists realized the difference between weight and force. The engineers made the best of a bad situation by distinguishing between pound force and pound weight and by the introduction of the slug for mass. Somewhere along the line Napoleon forced the metric system into use and eventually we got the modern SI system which works pretty well.

I still think the nautical system for distance and speed makes some sense as it is tied to the geometry of the earth. But then where is the common sense in having 360 degrees in a circle? Why not 10, 100 or 1000 ?

TORQUE 4.8 l 6.0V: 16.5 / 20.0 kg-cm SPEED 4.8 l 6.0V

In layman's terms this would indicate that a horizontal arm on the servo (being supplied with 6.0v) that is 1cm long and parallel to the earth's surface, would have neutral movement if a 20.0 kg weight were suspended from the end of the arm.

The British Association for the Advancement of Science were in on the act when the practical units
volt, amp, ohm were created - only a few years after the cgs system of electrical units was even
chosen (Maxwell, Thomson involved). That’s over 130 years ago, a nice simple standard set of units
based on the metre, second and kilogram, been used ever since. Would that this were the case in other
disciplines really! One of the early proponents of the metric system was Gauss in fact, a mathematician.

I think you’ll find the distinction between weight and force is Newton’s realisation, (after all he did
realize force = mass x acceleration, which rather implies force is not the same as mass), centuries
earlier, well before the industrial revolution.

As for the nautical units - the geometry of the earth - awful, its not a sphere, its not even a regular
ellipsoid, what basis is that for a system of measurement? The state of the art these days in interferometry
distance measurement is ~11 significant figures, that’s an error of about 100 microns in an earth-sized
measurement – orders of magnitude less that the tidal variation in the dimensions of the lithosphere even.

I’m just trying to get a chuckle here and there; don’t take it too seriously.
As for the nautical system a lot of very good work has been done with it and I’m afraid we are constrained to live on a pear shaped earth which we can blame for making it awful. Probably better we know where we are on this earth rather than on an imaginary perfect sphere.
I was always pleasantly surprised when sailing my boat by GPS in fog on Lake Ontario when the destination suddenly came out of the fog. Would the difference between a pear shaped and a perfectly spherical earth have made any difference in this???

Oh yes, quite a distinct difference. That's one of the reasons why older navigation systems were so inaccurate - they worked on a perfect sphere and you had to allow a big plus-or-minus when sailing on the ellipsoid.

GPS has a standardised ellipsoid, which describes the shape of the earth in just a few more details than just radius. It is close enough to the (also imaginary) sea level that the error is only a few feet over almost the entire globe.

Thank you all for the comments and reply. I will study what you guys have given me.

How much weight can a torque carry?

May depend on the measurements.

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