Question about torque of motors

[Nezhul]

Ok, I was never good at physics, and google can't answer me what I'm looking for.

I'm looking at a motor with a torque of 7 kg*cm. I found out that it meants that it can lift 7 kilogramms vertically on the arm of 1 centimeter.
What I need to know is - what weight will it be able to lift on a 2 cm arm? and so on. Or, in other words, when extending the arm, the weight it lifts will drop or will increase?

[computergeek]

I don't know the exact formulas for gear ratios but the longer the arm, the greater the speed and less the weight it can lift. If I were to guess and try to pull a rough formula out of my ass it would be something like:

Torque * Pie (3.14159) / Length in cm * Pie (3.14159)

Only other thought on this is if you are really good with calculus you might look into gear ratios:

http://en.wikipedia.org/wiki/Gear_ratio

[pjh776]

You meant 7kgf*cm (the f is for force; without it, the value does not measure torque, but something else).

Suppose you have a point O, the origin, in which you place your motor. Then you have a point P, P != O. Then suppose a force is applied on point P in a direction perpendicular to the line define by O and P. the torque of that force is the "strength" of that force (which is N (newton, and yes, the capital and small letters are correct here) or kgf, but not kg which is a measure of mass) multiplied by the distance between O and P.

Now, let's assume that gravity is perpendicular to the line defined by O and P (iow, O and P are at the same height, the form an horizontal line.

Put a body os mass M on P. The weight of the body is W = M*g. The torque of the body with respect to the point O is defined by T = W*d, where d = distance between O and P.

If your motor provides you a torque of 7kgf*cm, you have a constant T, and the masses of the bodies it can balance at different distances are given by:

7kgf*cm = M*g*d

So, let d = 2cm. You have:

7kgf = 2Mg.

Take out the gravity on both sides:

7kg = 2M

And you obtain your mass: 3,5kg.

Note, however, that things might not work out as perfectly as simply as this analysis shows: motors are usually designed to ... move things, not hold things still. Also, motors usually have a curve ofmtorque, ie, the torque provided varies with the speed the motor is turning.

In any case, there's your math.

[Nezhul]

so I basically want to lift some weight up and down repeadedly on about 3cm arm (6cm total height difference between lower and higher points).
The weight is about 3 killogramms, arm is 3 cm, as I said, so I need torque of about 9 kg*cm? 10-11 to be sure, right?

[PiJoy]

so I basically want to lift some weight up and down repeadedly on about 3cm arm (6cm total height difference between lower and higher points).
The weight is about 3 killogramms, arm is 3 cm, as I said, so I need torque of about 9 kg*cm? 10-11 to be sure, right?

Nezhul,

There's an important consideration for doing what you want that I haven't seem mentioned so far, continuous torque vs. stall/peak torque:

A motor at stall (for instance, holding a position against a torque) will suck lots of power and get hot. Motors actually have two ratings (assuming a full and honest data sheet),1. a continuous torque -- namely the torque the motor can do all day without burning out and 2. the peak/stall torque -- the maximum torque the motor can produce, when not spinning, typically only for short periods of time (varies with motor, but usually a minute or less -- typically shorter for smaller motors.) If a motor gets too hot, it'll burn out. This typically happens by melting the insulation on the rotor wires. Once this melts, it essentially glues the rotor to the stator -- and the motor is toast.

A common way around this is to use a motor that's heavily geared. This helps in two ways: first, a geared-down motor (whose output shaft spins more slowly than the motor itself) can generate ballpark of the gear ratio times the motor's torque. (Ballpark because you loose some torque to friction in the gears.) Second -- and for you I suspect more importantly -- a sufficiently high gear ratio generally makes the motor/gear combination non backdrive-able. What this means is that a torque applied to the output shaft can't make the shaft turn (the frictional forces jam/lock the shaft); only the motor on the other end of the gears can make the shaft turn. So a non-backdrive-able motor is great for holding a position against a steady torque. Once the motor/gears reach the desired position, friction holds the load, so you don't need to continue running high currents through it to hold a position. R/C servo motors are usually geared well into the non-backdrive-able range (for just this reason, I suspect.) So an R/C servo rated to at least your calculated torque (I'd add a margin of at least 30% -- 50% if you can) may be a good choice for your application.

Hope this helps!

[Nezhul]

I actually wasn't even considering stopping/holding motor position, but still thanks for clarification.

[Nezhul]

Hey, I was wondering recently, if there are electric motors (low-voltage) that are already made to do forward-backward motion suitable to build a mini sex-machine? Or all there is is standart rotational motors?