Looking for High Torque Servos.

Hi! Im working in a 6DOF Robot Arm and I need a High Torque servo for de DOF in the base. The higher one I have found is a 40kg/cm or 555 oz/in one, do you know any higher servo.

Thanks.

Where are you located, and where have you looked so far?

joseto: kg/cm .... oz/in

What are those units?

:stuck_out_tongue_closed_eyes:

Powerful servos are available for some $$$.

http://www.robotshop.com/en/hs-1000sgt-heavy-duty-giant-scale-digital-servo.html

http://www.robotshop.com/en/invenscience-torxis-i00600-12v-high-torque-servo-motor.html

http://www.robotshop.com/en/dynamixel-mx-106r-serial-servo-rs-485.html

JimboZA:

joseto: kg/cm .... oz/in

What are those units?

:stuck_out_tongue_closed_eyes:

Quite! For the uninitiated torque is in units of length times force. The tangential force times the distance from the axis to be specific. SI units N-m (newton-metres, ie newtons x metres). kg are not a unit of force, although you can say "kgf" for kilogramme-force, or in imperial units ozf for ounce-force.

Using newton-metres for torque has the great advantage that you can directly use

power = torque x angular-velocity, note).

Torque can also be (to my mind much more logically) quoted in units of joules/radian, just as force can be described in joules/metre (which is the definition of the newton).

Perhaps the world would be simpler if there was a named unit of torque :)

equation without conversion constant (although angular velocity is in radians/second

MarkT: Perhaps the world would be simpler if there was a named unit of torque :)

Well, a toque [sic] is a kind of hat, popular with chefs, so I'd go with chefhats as the unit. Of course, we would then have to decide if that was a Metric (specifically, SI) or Imperial (read as: American) unit.

I can live with kg.cm since kgf is an accepted Metric (but not SI) unit of force, but it really grinds my gears to see a torque described as force / distance not force x distance.

Imperial is not American - that's more of a UK thing http://en.wikipedia.org/wiki/Imperial_units

We're still going our own way in many things. http://en.wikipedia.org/wiki/United_States_customary_units

The base of a robot arm should have a lower torque requirement that the elbow of the robot arm.

The elbow ( and the shoulder ) have to hold up the entire weight of the arm and provide torque to prevent the end of the robot arm falling down.

On the other hand, at the base of the arm, the weight of the arm should be carried by some kind of bearing, and the servo only has to turn it, not hold it up.

michinyon: On the other hand, at the base of the arm, the weight of the arm should be carried by some kind of bearing, and the servo only has to turn it, not hold it up.

He's probably talking about the tilt of the arm, not the slew...

MarkT: Quite! For the uninitiated torque is in units of length times force.

While you may be technically right, it is easier for the layman to think about torque in "how much weight or mass will this shaft lift, given a lever extending from the center of the shaft" - which is why we see torque expressed as kg-cm and lb-ft (though for the latter, it is almost always said, and sometimes written, as "foot pounds" or "ounce inches" or similar). Sometimes a dash is used, sometimes a slash, rarely an "x" for some reason, though.

But I am sure you already knew this.

The thing is - most sites (not all) selling motors or other actuators don't list the information in the manner you prefer and indicate. Many of those sites aren't selling to scientists or even engineers, but rather mechanics, hobbyists, and other "laypeople" - that is, those generally without the formal training in such matters.

What I would personally like to see, is a web page or something similar that could relate and explain your method in a clear and concise manner, along with graphical examples - while comparing and contrasting that methodology to the more common "layperson" method of understanding torque. You yourself have explained several times how you view torque and how it works (and I commend you for that!). I haven't seen, though (maybe I missed it?) any explanation from you how that method compares (and translates to) the colloquial layperson method.

Furthermore, a site or page (or something) that could take all of that information, and show how to calculate the needed torque given a wheel on a robot platform - both for flat ground and for an angle (climbing a slope), and perhaps even for a joint on a robot arm (and maybe for something being slewed in a circle on a flat plane) - well, that would just be icing on the cake. Ideally it would show you how and why it all works, then have a section which you could plug numbers in and get a result back (via javascript or similar).

michinyon: The base of a robot arm should have a lower torque requirement that the elbow of the robot arm.

The elbow ( and the shoulder ) have to hold up the entire weight of the arm and provide torque to prevent the end of the robot arm falling down.

On the other hand, at the base of the arm, the weight of the arm should be carried by some kind of bearing, and the servo only has to turn it, not hold it up.

All the joints have bearings, of course, so that argument seems odd.

The base will have the largest torques as it handles more mass and more reach. However a rotation-only base is protected from gravity forces by a thrust bearing, perhaps that is your point? Not all arms are like that though.

Remember dynamic torques have to be handled too so a fast moving arm will have the largest torques at the base, whatever DoF it has, as dynamic torques can be far larger than static in such a machine.

Static torque loads are straightforward to calculate, just do the sums once you've got a good estimate of weights and sizes. Dynamic torques depend on speed/acceleration profiles.

You need something like a “crane” and not an “arm”. One servo to raise/lower the boom and a second servo to pan the boom between tanks (where the tanks are arranged in a semicircle). Personally I’d just hang the board off a wire hook instead of trying to use a gripper, and if you want agitation you could put a continuous rotation servo on the end of the arm so that it raises/lowers the board as it spins.

You don’t need much torque for the pan, and because you can counterbalance the boom you don’t need much torque there either. Any standard size servo would handle it, but always a good idea to shop around and try to find ball bearing / metal gear servos. Hobbyking.com has plenty of generic servos that would meet those specs for not much dough (<$10).

Late reply, but I wanted to simplify the Definition of "Torque."

Torque is how much you "weight" you can pick up by wrapping around a shaft. (the amount of Horsepower determines how fast you can move it).

Torque is Measured in "distance/weight" -- "inch/lbs" or "cm/kg" or "foot/lbs" or even "mile/tons" (but that is kind unrealistic, but theoretically possible) The metric world has a N/m or Newton/meters, but I've never felt it was practical.

Torque is a pretty straightforward equation: (think of a cable wrapping around a drum on a shaft)

Torque is [how far you are from the center of the shaft] divided by [how much you are trying to lift by rotation]

So if torque = 10 cm/kg, then you can pick up 10kg that 1cm away from the center of the shaft OR>>OR>OR>> 1kg that is 10cm from the center of the shaft OR>> any "math" that makes the equation true: 2kg that is 5cm away 3kg that is 3.33cm away .01kg that is 1000cm away

since the farther you get away from the center of the shaft, the "linear speed" increases, so (if HP stays constant) then the weight you can move must decrease