Lift Math Equation

OK, I am trying to find how much we can lift with a certain gear ratio and motors.

Here are the equations I have right now that I think have to do with it, but could someone map out exactly what I have to do to get my answer?

Weight Needed Lifted - 11lbs = 48.93newtons
Length Of Arm - 14in = 0.3556m
Stall Torque (2 393 Motors) - 3.34 newtonian meters
Arm Gearing - (15:1)

Weight(newtons) x Arm Length(meters) = (Needed)Torque

48.93newtons x 0.3556m = 17.4 newtonian meters

(Needed)Torque x Arm Ratio = ?

Where does the Stall Torque fit in? Do I have the equations right?

Normally we use inches instead of meters, but this works I think. What you need is:

Needed Force/Actual Force(With no gear ratio) = Gear ratio required

That would solve for the required gear ratio to lift your intake and sacks.

That would probably be more efficient than solving for random gear ratios to figure out what you want to gear the arm at.

There are also efficiency issues that occur with the motor/lift. We don’t actually calculate that… we just add a little bit more than we need, which isn’t hard as the gear ratio that normally comes out is some obscene ratio that can’t actually be made/made easily in the space available.

So what do I not have that I need? Do I have everything?

Can you write out a specific equation?

This is one way I would approach it:

We start with the equation: F=T/R (A little simplified but works in this case).
Where F is the force (in this case the weight being lifted), T is the torque, and R is the radius (in this case arm length)

Assuming you already have a motor and gearing setup T=Tm*Gr
Where Tm is the motor torque (often 1/2 stall torque to prevent overheating) and Gr is gear ratio

Putting it together gives F=(Tm*Gr)/R
This equation will give you a very basic idea of the force that can be lifted by a motor with a gear ratio. In addition if you know the force you can calculate a gear ratio that meets the requirements (remember to build with a margin of safety since this equation is very simplified).

For your example:
Tm=(1.67 N/m)2(1/2) (For safely lifting the load)
R=(0.3556 m)
F=(1.67*15)/(.3556) (N/m)*m
F=70 N
So your arm can lift about 70 N, or about 15 pounds

The torque of a 393 motor running at 100 rpm is 5.17 in-lbs

No, the torque with full power applied at 65 rpm is approximately 5.17 in-lbs. The torque at max rpm is theoretically 0.