I was looking at these dc motors and I converted the torque to force at a meter distance and I got that two of them should be able to lift over 130 pounds, is this right? http://www.active-robots.com/high-torque-dc-servo-motor-10rpm-with-step-dir-drive There has to be some catch i am not seeing
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Where did you get that number?
The first page of the datasheet posted on the page you linked states, "Max 15VDC and 7A" (near the top).
So the maximum input power for the motor is $P=IV$, or $P=(7)(15)=105\mbox{W}$.
Mechanical rotating power can be calculated by $P=\tau \omega $, where $\tau$ is torque in Nm and $\omega $ is rotational speed in rad/s.
The motor's top speed is 10 rpm. Multiply by 2$\pi$ to get to rad/m, then divide by 60 to convert to seconds and you get $\omega=10*(6.28/60) = 1.047 \mbox{rad/s} $.
Now, divide input power by rotational speed to get the theoretical maximum torque: $\tau = 105/1.047 = 100\mbox{Nm} $.
Note that this is Newton meters; a Newton is about a quarter of a pound, so the maximum force is about 25 pounds.
However, an electric motor is generally only about 80% efficient, so I wouldn't expect more than 20 pounds at 1m, max.
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$\begingroup$ I would also add that the electrical numbers looked kind of more like an absolute rating that an operating/continuous rating, so I would expect actual torque to be even lower. But again, I didn't see an actual torque given anywhere in the datasheet. $\endgroup$– Chuck ♦Commented Jan 22, 2016 at 0:42
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1$\begingroup$ Here is a quote from the product details: "Motor is a Industrial grade 10RPM high torque motor with a massive torque of 120kgcm" that's where I got my number from, yours makes a lot more sense though $\endgroup$ Commented Jan 22, 2016 at 5:39
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$\begingroup$ Using the result @Chuck posted, you could add a gear train of about 10:1 (with margin built in for inefficiencies) if you really need to lift 130 lbs using that stepper motor. Nema 23 gear reducers are readily available from multiple sources. $\endgroup$– SteveOCommented Jan 22, 2016 at 14:40
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$\begingroup$ @user65909 - multiply your rating, 120kg-cm, by gravitational constant 9.81 to get to Ncm, then divide by 100 to get from cm to meters. This converts the number from 120kg-cm to 11.7Nm. This is far, far lower than my estimate. The theoretical maximum from the input power is 100Nm, so you're burning about 88% of the input power. Maybe the motor torque you state is for the motor before reducing gears? $\endgroup$– Chuck ♦Commented Jan 22, 2016 at 20:00