I'd had a semi-long post that got lost when I Iost my connection. When I get less PO'ed I'll retype it.
Let's assume (for the moment) that you turn on the motor to lift the load and leave it on, full blast, until just before the load hits the spring stop. You want to go ~19" (let's round that up to 20") in 2-3 secs. Let's call it 2 secs. So what kind of linear acceleration is required and what force is needed to do that ?
Time for another (semi-valid) assumption ... that is that the acceleration will be constant. It probably won't be but right now "we're" just trying to get close to see what kind of motor might be required. Given the load is starting w/zero velocity, the equation for distance vs time is :
D = 0.5 x A x t^2 ; D = distance = 20", A is the acceleration and t is the time it takes, 2 secs.
Re-arranging and solving for A :
A = 2D/(t^2) = 40/4 = 10 in/sec^2 or 25.4 cm/sec^2 or 0.254 m/sec^2
And the load will be going 20 in/sec when it hits the stops. Which sounds a bit fast.
So what kind of force is needed to make that happen ? Recall F(net) = m x A so :
F(net) = 10 kg * 0.254 m/sec^2 = 2.54 N net force
But you're fighting gravity to lift the load and the net force is the force exerted on the string - force of gravity. Just to hold the load against gravity you need 98.1 N, and so the string must exert 98.1 + 2.54 = 100.64 N. Let's just call it 100 N.
What kind of torque is then required at the shaft to do all this ? To exert 100 N of force at a distance of 15mm means :
T = F x radius x sin(angle) ; T being the torque, F being force and the angle being the angle between the applied force and the radius, 90 deg in this case.
That means the shaft torque needed is 100 N x 0.015 m = 1.5 N m.
It's also useful to figure out what RPM the shaft is turning after 2 secs. At 20 in/sec, that's ~ 50 cm/sec. The circumference of the 15 mm radius shaft is :
C = 2 x R x Pi ; so that's 2 x 1.5 cm x 3.14159 = 9.42 cm. To take up 50 cm it'll take 50/9.42 = 5.3 turns. And 5.3 turns/sec is 318.3 RPM. Let's call it 320 RPM.
So while the torque requirement sounds "high" to me (for a small motor), the RPM needed at the end of the lift sounds "low" and so some gear reduction from the motor to the shaft seems to be in order. The needed motor then torque goes down by the gear ratio and the RPM goes up. So a "high" RPM, "low" torque motor might work.
It's also useful to figure out the power needed to do the above. I'll leave it to you to look up the formula but given the above the shaft power, in watts, is 50W. And the motor power will be somewhat higher given the gear losses. And the input electrical power, which is what you need to supply, is even more. Given a 20% loss in the gearbox (SWAG !) and and 65% motor efficiency ... that means a 100W (peak) for the motor ! That's 10 A at 10 V and that's not a tiny hobby motor.
Of course a real DC motor (or stepper) has a torque vs RPM curve and so you won't get the above exactly. And you'll probably want to slow or turn off the motor so it's not hitting the spring stop so fast. Both of the above may tend to increase the lift time but that's why I choose to use 2 secs and 20" ... to be conservative.
Another question ... how were you planning to hold the load up, once lifted ? A brake of some sort ? A counter-weight will do but it doubles the mass to be accelerated,
though gravity no longer fights you on the way up (nor helps on the way down but who cares about that !). Also a worm gear can give a high ratio and generally act as a locking mechanism. I'll leave the counter-weight calcs for you to do. Remember that we've neglected any frictional forces and had a constantly accelerating load. A load already accelerated up to speed needs less force/torque to hold that speed vs getting to it ... just like getting your car to 70 MPH. And taking longer would also mean less power needed.
So what motor and gear ratio might you choose ?
ps - I guess this post was more than semi long (enough).
