# Selection of correct DC gearmotor with given load

I am designing an Actual working extendable welding station for my college project. The idea is to lift the unit up/down with the help of 2 lead screws and 2 DC gear-motors, one on each side. The total weight needed to lift will be around 500 lbs. I want to select suitable dc gear-motor for that. Lead-screw will be 3/4" diameter of 0.2" pitch.

I want the torque calculations done to select a dc motor.

I have done this much so far:

$F = m \cdot a = 227 \mathrm{kg} \cdot 9.8 \mathrm{m}/\mathrm{s}^2 = 2224.6 \mathrm{N}$

But, we are using 2 motors to drive this, so i just divide the force/2 = 1112.3 N. I need the table to lift 15". So, the lead screws can be around 20" long. Considering 20" long lead screw:

$T = F \cdot r = 1112.3 \mathrm{N} \cdot 0.508 \mathrm{m} = 565 \mathrm{Nm}$

But, that is the absolute minimum torque just to keep the system in steady state right?

So, I need assistance in finding the real torque required and how to assume the minimum acceleration needed to find the real torque.

Also, how to calculate Power for the motor. Based on this calculations, I will be buying the motors to actually install it in my design so I need the calculations to be accurate! Thanks to all in advance.

• What is the speed at which you want to lift your weight, and how does that translate to RPM at the motor output shaft for your chosen lead-screw? This will directly affect the Power of your motor. What about acceleration? Remember F=m*a – Jonathan R Swift Dec 6 '17 at 10:45
• Using a separate motor for each leadscrew is almost certainly a non-starter. Unless you have a sophisticated control circuit that keeps them in lock-step, they will run at slightly different speeds, causing the the mechanism to skew and jam. It would be far simpler to drive both screws from a common motor through a toothed belt or chain drive. – Dave Tweed Dec 6 '17 at 13:48
• Also, what you have computed is not a torque, but rather the total work required to do the lift. It is unfortunately confusing that torque and work use the same units -- force x distance. They are not interchangeable. – Dave Tweed Dec 6 '17 at 13:55

No, it does not work that way ;) 565 Nm is much more than you need for this. The error is in the second part of your calculation.

The relationship between force which a lead screw can work against and the torque which has to be applied to the screw is depending on some variables. Assuming that you have steel on steel ($\mu$ = 0.25), using your diameter ($d$ = 19 mm), pitch (which is the same as the lead for normal screws, $l$ = 5.08 mm) and this formula (Wiki):

$T = \frac{F \cdot d}{2} \cdot \frac{l + \pi \cdot \mu \cdot d}{\pi \cdot d-\mu \cdot l}$

gives you a required torque of 3.6 Nm.

However, consider that this does not include acceleration. After correcting your $F$ by:

$F = \frac{m \cdot (g + a)}{2}$

where $g = 9.81 \frac{\mathrm{m}}{\mathrm{s}^2}$ and $a$ the required acceleration ($1 \frac{\mathrm{m}}{\mathrm{s}^2}$ should be more than sufficient). I would furthermore add a security factor of 2 before selecting the motor. Keep in mind that you have to include all moving parts into your $m$, meaning that

$m = m_{table} + m_{everything~positioned~on~the~table} + m_{lead~screw}$