Motor selection for vertical lift door with balanced counterweights

[davegmech]

I'm having trouble selecting a suitable motor to lift a single leaf 1000kg door. To cut down on power costs i plan to use perfectly (or as close as i can get) balanced counterweights. The door is to be lifted 4m vertically using wire rope hung over two sheaves connected to the respective counterweights (each weighing 500kg). The door is to be accelerated from rest at 0.05m/s^2 to a speed of 0.1m/s, kept constant until an equal decceleration occurs. As the weigth of the door is balanced by that of the counterweights it seems that the power required to operate the door is absolutely minimal. The only power input neccessary seems to be that needed to overcome the inertia of the sheave (which will weigh 50kg and have a diameter of about 0.6m) and the friction associated with the roller bearings which i plan to fit the sheaves on, which according to my calculations seems negligible. The only additional source of friction i can think of is the anti-friction cam followers which will guide the door.
All of this seems to reuire a Very small motor,only a fraction of a HP at most, which seems a bit odd to me! Any insights would be greatly appreciated!!

 

[desertfox]

Hi davegmech

Your right, if you have a system of counterbalances your energy you need to supply would essentially be any shortfall between the counterbalances, maybe due to the position of the door centre of gravity at some point during movement.
Hope this makes sense to you.

desertfox

 

[davegmech]

Hey desertox

Thanks for the reply! The thing that still gets me is what will occur in the (unlikely) situation that the counterweights and the door are exactly balanced? Am i right to assume that power loss via friction in the system are the only thing the motor must work against? If this is the case the roller bearing friction is so minimal the motor size i will need seems absurdly small on the order of 1/50 of a hp. Would you suggest playing it safe and looking into maybe a 1/5 or1/4 hp motor?

Thanks again,
Dave

 

[desertfox]

Hi davegmech

Well if they completely balance your motor needs to be able to overcome the friction etc in the system and possibly some inertia if it slows down to bring it back to speed.
Can you post some sketches of the door and weights etc.
Yes to looking for a higher powered motor you need some leeway.

desertfox

 

[davegmech]

Unfortunately i wont have acces to CAD drawings for a number of days. The following link provides a VERY simplified outline of the setup

http://www.airportdoors.com.au/Drawings/Vertical_Lift.pdf

This is the closest match I could find qucikly! Note that a number of guide rollers will be located on each side of the door (about 4 each side). I think i gave most of the dimensions in my first post. The door is 4m high by 3m wide.

Dave

 

[desertfox]

hi davegmech

Thanks for the diagrams, it looks like the door is always in balance, so your motor needs to overcome the friction in the system and overcome the inertia initially to accelerate the door upto working speed.
If the door is perfectly balanced what keeps the door from moving when in either the open or closed positions?
A brake on the motor or the pulley system?

desertfox

 

[davegmech]

Hey,

Would that inertia include the mass of the door and counterweights about the sheave, or just the sheave itself? Yes, there will be an electro-mechanical brake that can be applied against the sheave as well as a mechanical locking mechanism which will keep the door in position when fully opened.

Dave

 

[desertfox]

hi davegmech

Well it will include the inertia of the door and counterweights I suppose, because technically the door and weights would be in perfect balance so they would need a kick to get them moving.Think of it another way if the weights could get the door moving, unless they can accelerate the door upto working speed, its going to need an external force to accelerate the system
ie the motor.

regards

desertfox

 

[davegmech]

Thanks for the advice. I appreciate it!

Dave

 

[desertfox]

Hi davegmech

Your welcome

desertfox

 

[desertfox]

Hi davegmech

My apologies,I have just done a freebody diagram of the door and weights and it appears that if the door and counterweights were balanced then you only need overcome the torsional inertia and friction in the system with you motor.
I have uploaded the file for your reference.
However if your system is not perfectly balanced then your motor will need to supply the difference in energy.

desertfox

 

[desertfox]

Hi davegmech

If the door and counterweights are balanced then you only need to overcome the torsional inertia of the pulleys and friction in the system.
Sorry for the error above.

desertfox

 

[davegmech]

hey desertfox

I was thinking of that myself, it seems counter-intuitive that such a large door could operate with a small motor but the math makes sense!Thanks again for the time you've taken and for putting my mind at ease!!

Dave

 

[desertfox]

Hi davegmech

Your welcome, yes it does seem counter-intuitive but once I had done the freebody it came clear to me.

desertfox

 

[MikeyP]

Whoa there! If that were true then under the condition of a massless pulley and zero friction, any applied force would result in infinite acceleration.

A system in static equilibrium still requires a force to accelerate it from rest. Admittedly that force will be relatively small because your rate of acceleration is small, but it is not insignificant. There is no getting away from the fact that 2 very large masses have to be accelerated and that doesn't happen by magic.

I suggest you go back to your free body diagram

     ____
    /    \
   /      \
  |        |
  |\      /|
 T| \____/ |T
  |        |
-----    -----
| m |    | m | 
|   |    |   |
-----    -----
  |       | |
  V       V V
  mg     mg F

Apply a downward force F on the right hand mass and it will accelerate down at acceleration a and the left hand mass will accelerate up at acceleration a

define "up" as positive
T is the tension

Sum forces on LH mass
ma = T-mg
so T = mg+ma ---(1)

Sum forces on RH mass
-ma = T-mg-F
so T = mg+F-ma ---(2)

equate (1) and (2)
mg+ma = mg+F-ma
so F = 2ma

As you would expect

The counterweight negates the effect of gravity, but you can't negate the effects of the inertial mass!

M

--
Dr Michael F Platten

 

[desertfox]

Hi MikeyP
Thanks
Yes I am wrong I should have followed through with the maths I just looked at the forces each side oh yea they balance but they don't.
My earler post was right, so you need to include the door and weights as well as friction and torsional inertia of pulley's

desertfox

 

[davegmech]

Some pretty basic stuff there - can't believe I didn't see that! Goes to show the importance of taking things slow and not jumping to conclusions! It's a small but significant force.

Cheers mate!
Dave

 

[desertfox]

hi davegmech

Well I made the same mistake, just looked at the forces and direction and thought yeah they cancel.
Its sorted now thats the main thing.

desertfox