You would need to tell us how fast the load is to be lifted, or what the pulley or wheel diameter. One of those is needed to know how much torque is needed.
Motors are almost always much faster than 20 rpm. Anything that slow usually includes a gearbox to reduce the speed and increase the torque.
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That give a maximum speed of ~30 rpm on the output shaft, which can be lowered using a 12V speed controller or lower voltage power source.
The geared motor is rated at 20Kg-cm and with the 5:1 belt reduction the final torque would be 100Kg-cm absolute maximum, so well above the minimum needed.
As long as you use a reasonably small diameter winding drum on the output shaft, it should lift 5Kg easily and have a decent operating life. Choose the drum diameter and motor voltage to get an appropriate winding speed.
I have used ex automotive motors from a wreckers yard, I recently picked up some nice NEW surplus seat motors, complete with GB.
Regardless, I would go with DC every time, the control method choices are many.
I have used ex automotive motors from a wreckers yard, I recently picked up some nice NEW surplus seat motors, complete with GB.
Regardless, I would go with DC every time, the control method choices are many.
There was a very famous Robot Wars contestant - if I recall correctly, he was the first person to incorporate self righting.
I can't remember his name, and he's sadly dead now, but he built props and technical gear for films, museums etc.
One of his most commonly used parts were old car windscreen wiper motors. Even though he often over-drove them at 24V, he'd never had one fail, and he had a LOT out there running all kinds of moving displays.
Also, on windscreen wiper motors, even Ford repurposed them. I had a 1978 Transit with a Diesel engine. The fuel shut-off lever on the injector pump was operated by a repurposed wiper motor. It had an extra switch added to give it two park positions, so turning on the ignition would make it move to one end and stop there, and turning off the ignition would move it back to the start. A Bowden cable connected the output rod to the injector pump.
This is a current motor available at a Canadian Surplus Co, they have an outlet close to me and currently have a clearance on of this NEW geared 12vdc motor for CAN$4.00
If you want to lift 5kg up 1 meter in 10 seconds then (in theory) your doing (mass*gravity*height) 5*9.81*1 = ~50 joules worth of work. As it's in 10 seconds then it's 5 Joules per second which is 5W. Due to inefficiencies you will probably have to double that. That still seems a very low figure which makes me think I've done something wrong, can someone check this?
This is a current motor available at a Canadian Surplus Co, they have an outlet close to me and currently have a clearance on of this NEW geared 12vdc motor for CAN$4.00
The only surplus co in Canada that comes to mind is Princess Auto. The original store sold surplus is in my home town of Winnipeg. It was THE male testosterone store for tools. Now there are 50 stores and 0.1% of it is surplus.
My curiosity led me to look up Rex Garrod, who also supported the TV series with models or demos in "The Secret Life of Machines" with Tim Hunkin. So I watched the last episide of the 1st season. "The Secret Life of the Radio Set" where they discuss the primitive fundamentals of electronics with a humourous cartoon of Marconi growing up in Italy. But the episode closes with a masterful pyramid of old TVs, all operating with pyrotechnics. (lol)
The Secret Life of Machines
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If you want to lift 5kg up 1 meter in 10 seconds then (in theory) your doing (mass*gravity*height) 5*9.81*1 = ~50 joules worth of work. As it's in 10 seconds then it's 5 Joules per second which is 5W. Due to inefficiencies you will probably have to double that. That still seems a very low figure which makes me think I've done something wrong, can someone check this?
Energy = Force * Distance E = F * x + Friction power * time = 1/2 mv² ( kinetic energy at any time)
Force = Mass * Acceleration + Friction force = F = m a + Kf
This might accelerate the mass to some speed but you also have to slow it down to stop it or rely on friction.
Force = Torque / Radius , gear ratio translates motor to object such that Torque*speed is constant with losses.
Torque is also proportional to current which is initially Is= V+/ Rdc (or DCR) then reduces with Vemf
T= k I
The motor will have a curve or Torque vs speed. with max power at 50% speed perhaps..
x=1/2*(vi+vf)*t distance is average change in velocity * time
a= (vf²-vi²) / 2x acceleration is the change in kinetic energy or velocity squared over distance traveled
I used potential energy (M*G*H of the final position) to simplify the equation and to get work done. That gives 50 Joules of work done in 10 seconds. That's 5 Joules/second or 5 Watts. Even allowing for only 50% efficiency it's still only 10 Watts to lift 5kg at 0.1 m/s. Seems low to me but I can't see any error. Using acceleration and kinetic energy etc. seemed a little complicated.
I only work in S.I. units. Can this be calculated in US (imperial?) units and does it give the same answer?
I also only work in SI units as well but it's more complicated with efficiency, friction, time, speed and heat loss
for h= 1 m
m = 5 kg , g=9.8 m/s²
PE= mgh = 49 J
KE= mv²/2
But to move at a constant speed v against gravity must apply a force to oppose gravity and change it's potential energy, PE for some time duration.
I'll let you work it out. But the energy must be increased if you have to start and stop the kinetic energy.
Work done is work done. Trying to work out efficiency, friction, time(10S), speed(0.1m/s) and heat loss(friction again?) etc. is impossible and has to be a guess. Hence my estimation of ~(approximately)50J*2.
I suggest you look at 12V Automotive motors, switch and PTC thermal cutout. Such as Windshield wiper motors. ESC speed controls work best or learn how to PWM and change directions on a full FET bridge. But that requires you to have access to a car junk yard or Princess Auto as maxheadroom suggested.