Opening Bifold Doors

The counterweight idea is excellent. However, you would be accelerating a larger mass than the typical garage door, so it might take a more skookum door opener than usual. Still the right idea, though.

While it does not provide help with your question...

Similar to this top drive setup:

Or, the bottom drive setup:

Or, something else entirely?

Images pilfered from http://www.bifold.com/

It appears these guys will sell components and parts.

Why will the door get heavier as it gets higher?

anxiously awaiting a reply to that one.

Thank you all who replied and sorry I have not gotten back to you sooner but, about 12 hours after writing the question, I had my appendix removed and have been off the computer for a week until today. All is well now and I am slowly getting back to work.

In response to TCMTECH: Wonderful idea for a smaller door but for a door weighing at least 2K pounds and then having a 2K counter weight is not practical. Also to take into account would be the wind-age when the door is open. 460 sqr.ft. of sheet metal would an extremely powerful sail just hanging with a counter weight. Some type of a locking mechanism would have to be employed to a door that, when open, was 16' off the ground at its lowest point.

DOORMAN: Exactly the design that I have already built but my question still remains and that is how to convert the torque rating given in a gear box to lifting a weight (ie: door).

Tom Consulting:I do believe the door will become somewhat heavier as it is raised because of the following factors. In the closed position, all of the weight hangs on the top hinges. As the door is raised by lifting the bottom, the panels both start pushing out. As rolling wheels at the bottom of the door rise, the pressure on the wheel increases inward as the top hinge is now being loaded in an outward horizontal direction. I believe the lifting force needed will reach its peak when both panels reach an angle of 90 degrees when referenced to each other. After that, the lifting force will decrease as the door gets higher. The horizontal outward pull of the hinge and the inward push on to the rolling wheels will continue to climb as the door rises.The point of component failure will be when the bottom of the door is less then 36 inches from the top hinge. The failure point would probably be the wheels in this design which is why my maximum opening will only be to the 48 inch mark.

This seems like it should be a simple mathematical formula with a logical answer. I am trying not to be just a "parts changer" until I get it right. Below are the specs of one unit I am concidering. Any other input would be greatly appreciated.

Speed Reducer, 56c, 60:1

• Power Transmission
• > Speed Reducers
• > Speed Reducers

C-Face Right Angle Speed Reducer, Nominal Output Speed 29 RPM, Input RPM 1725-1750, NEMA Frame 56C, Overhung Load 1500 Pounds, Nominal Ratio 60:1, Output Torque @ 1/2 HP 619 Inch Pounds, Output Torque @ 3/4 HP 995 Inch Pounds, Output Torque @ 1 HP 1371 Inch Pounds, Worm Center Distance 3.25 Inches, Double Output Shaft Assembly, Mounting Any Position Except Input Motor Shaft Up

As the door moves from closed to open :

The force required to lift the top leaf will increase (applied by the bottom leaf)

The force required to lift the bottom leaf will decrease

The increased force transfered from the top to bottom leaf will mostly be seen in the horizontal direction against the door frame.

My point is that the frame must be designed to withstand the considerable inward pressure against it as the door rises.

6" inches a second ! that's pretty aggresive for this type of door!

Are you buying this door or building it?

If you're buying it maybe you could go to a local airport and check out similar doors on T hangers. Look at the motor/gear reducers they use.

Careful whit the gear reducer selection, even the worm type will creep backwards if the ratio is high and enough force is applied to the output. May want to get a brake motor.

Thank you for your response. The speed of opening the door is faster then I would have preferred but it is based on a 1750 rpm motor attached to the above listed gearbox with a 60:1 ratio. This seems to be the lowest gear ratio I have been able to readily find. Since the door will be opened to the 16' mark, an opening time of 60 - 70 seconds would be ideal. That would also solve any braking problems of a creeping door.

I have already built two of the four doors for my shop and currently I just use my forklift to open them as needed. Going to a local airport is also a possibility although I would like to come up with a plan on paper first rather then trial and error.

" I would like to come up with a plan on paper first rather then trial and error."

I couldn't agree more.

Can you come up with a weight of each panel

I think you could neglect the friction of the hinges and roller/vertical interface

You can get larger ratio gear reducers, up to ~ 100:1. Then you have to go to a "double reduction" type. of course the price goes up!

You could get a 900 rpm motor

The max torque needed will be when the door is starting to open. If you use cable you could finrd a stepped sheave and install it such that the diameter about which the cable runs, increases as the door rises.

I know I said this allready but have you accounted for the horizontal forces on the frame? The door will be pulling out at the header and pushing in at the rollers

The weight of each door panel is about 860 pounds. The building was stressed for a door weight of 3000 pounds and I also have two doors built and hung. I have raised both doors to there full opening hight with my forklift so I am comfortable with my design.

I have not looked into a lower RPM motor and I do not believe I would want a double reduction gear because of the extra cost involved. Another option to reducing RPM would be to use a chain and sprocket to reduce gearing from the output shaft of the gear box to the shaft that rotates the cable drum.

It has been quite a few years but torque is the force required to rotate an object about a point. So if the output shaft is capable of 619 @ 1/2 HP in lbs the force exerted on the cable would be a function of the radial distance as measured from center of shaft to outer surface of pulley/pipe drum where cable connects/seated. You could also look at this problem from a moment arm (distance) times a force thus the units of torque in inch pounds. We would first need to convert the stated torque output into a force value which is t=rf or 619/.375 for a 3/4 inch dia. shaft which equates to f=1650lb. So if the radial distance, as measured above, is two inches, than the problem becomes t=rf or t=1650 x 2 or t=3300 in-lb. The key thing to inote here is that while the torque increased the force remains the same. So then the force exerted from system to cable is 1650 lbs. In this application, a change in radius of cable drum will not change the force exerted on the cable rather a change in open and close time will occur. This is a single stage pulley system so the force remains one to one. There is possible force required reductions in a two stage pulley system however this may not be easily implemented within your design. Also, the number of cable/strap locations will not affect the force realized on the output shaft rather on or at each cable. Only benefit is redundancy in design for safety should a cable break if cable is sized accordingly. I suggest one or two cables required to hold weight of the door in an application where four or more cables are in operation. So, the net result is for a single stage pulley system as you have described, the gear box output torque should be 860lb x 2 panels x 1in = 1720 in-lbs. The actual torque is t=rf or 1 x 1650 or t=1650 in-lb. Not ok. Move up to the 3/4 hp. t=rf so 995/0.375=f which results in f=2653lbs. 2653 x 1 = 2653 in-lbs. OK. This results in a factor of safety of 1.54 which makes me feel more comfortable and will ultimately result in the motor using less horsepower which equates to less power consumption and longer motor life. Hope this helps. By the way, given 2 in dia drum, circumference is pie x d or 2 x 3.1415 = 6.283 inches. 20 feet opening minus 3 feet leaves about 17 feet of cable travel. 17/(6.283/12)=32.5 revolutions to either open or close. The listed nominal output rpm is 29. 32.5/29=1 minute and 7 seconds open/close time.