Power table design.

[itsmoked]

I'm tasked with building a semiautomatic lift table. I've got the controls covered,(HA!). At first blush I was going to use the classic hydro-electric scissor table. Unfortunately those completely render the space,(needed), below them unusable. So I'm using four screw jacks that I want to gang together.

The four legs will be oriented so that the shafts are parallel across the rear two and across the front two. I want to yank the existing cranks,(12.0mm), which have a bevel gear on them in the top of each jack. I'll tie each of the two sets together with a shaft. Place two gears on the front shaft and one gear on the rear shaft. Tie the front and rear shafts together with a chain drive. The second front gear chained to a gear motor mounted to one of the legs.

The jacks are rated at 3,500lbs each. The table will have about 2,500lbs on it.

One issue I see is timing each leg to be the same height and provide the same lifting force. How can I make fractional adjustments to the shaft timing on one set tied together with a shaft? I can't imagine getting the torque required thru set screws, and a flatted shaft is a 360 degree proposition.

It would be nice to be able to service the tie shafts in the future without messing with the welded together structure. I hope you can suggest a connector style/system that would allow shaft removal/adjustment.

Any other issues you can think of?

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[tygerdawg]

I've seen an press whose purpose was positioning A- and B- sides of a vacuum forming mold for automotive products. It moved the heavy platens this way:

(1) four vertical shafts. The shafts were ~300mm diameter rods with a heavy toothed gear rack machined on one side. This provided column stability and motive force for up/down. Flats on the shaft were machined on the ends of the rods for radial locating. There were additional guide rods and bushings to maintain parallelism. Travel was 2000mm or more I'd say.
(2) platen contained a housing at all four corners with matching pinion gear.
(3) Pinions were connected with a driveshaft and bevel gears around the corners.
(4) Driveshaft was driven with a gearmotor & VFD combo.

Depending upon your loading and speed requirements, you may also consider a timing belt / ball screw arrangement.

TygerDawg
Blue Technik LLC
Advanced Robotics & Automation Engineering

http://www.bluetechnik.com

 

[dvd]

First off, jacks are not linear guides. Are you planning on guiding your table?

Next, if there is any misalignment between the table and the foot the jacks are going to want to bind. I recommend that you devise a fixed-free mounting system so that the jacks don't bind when in the lowered position.

A shaft locking coupling from someone like Ringfeder would allow phase adjustment between the jacks. There are also phase adjustable sprockets, I don't know who to reference for a manufacturer.

 

[itsmoked]

tygerdawg; Wow sounds massive. Sounds pretty stout too. I am stuck with these jacks as they are sitting here on a pallet. Timing belt could work. Problem might be the length since unlike a chain I can't just add subtract a link.

dvd; Not guided. Just push the up button and it will go till it hits a limit switch and hit the down button till limit switch. Height change will be about 16 inches. The plan is a stiff square/rectangle tubing structure supporting the load.

Can you describe what you mean by "fixed-free"?

Just my luck all the slick Ringfeder couplings start at 3/4" bores.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[tygerdawg]

Yup, fairly big: vac mold was for instrument panel skins.

You could try to use the jacks you have strictly as motive force elements.

Timing belts can be had in almost any length within reason. You can put in an adjustable idler to take up the slack and provide proper tension. Or if you're really masochistic, put in a dancer arm. Check out Gates, Goodyear, Eaton PowerTransmission?, others. I prefer timing belts due to no lube reqd, no backlash, low inertia, easy maintenance.

Add some sort of guide rod and bushing. Look at Danly die sets for an idea of what I'm talking about. Done right, that scheme will take care of your parallelism issues.

TygerDawg
Blue Technik LLC
Advanced Robotics & Automation Engineering

http://www.bluetechnik.com

 

[hydtools]

Since this is a one off in-house application:
Instead of trying to time the drive gears, shim under the base plates to establish equal loading or clearance when the jacks are fully retracted. Somewhat like setting the landing leg jacks on a fifth-wheel travel trailer before running the common-drive jacks to pick up the load from the tow vehicle. Or shim between the heads of the jacks and the table for equal clearance or load.

Ted

 

[itsmoked]

Thanks hydtools.

I will adjust to within one rotation then shim any further needed distance.


I've yanked the cranks and machined the replacement shafts. At the last moment it dawned on me I can't have all the cranks face each other as then half the jacks would be descending as the other half would be ascending.

I need the torque requirements to select the motor, gears, and couplings.

Using a newly installed truck scale I jacked up a pickup by it's rear bumper with the jack sitting on the scale and the pickup's rear bumper over hanging the scale. I used another 50lb spring scale to measure the required force on the jack handle. I was careful to apply the spring scale perpendicularly to the handle and to only take a reading while the handle was in slow steady motion, avoiding stiction and static friction.

I found the results interesting. They are not monotonic!
[tt]
0.18 tons Tare (me + jack + spring scale)

TRUCK SCALE JACK LOAD TORQUE
0.33T - 0.18T = 0.15T = 300lbs 8lbs x 7" crank = 56 in-lbs
0.42T - 0.18T = 0.24T = 480lbs 14lbs x 7" crank = 98 in-lbs
0.48T - 0.18T = 0.30T = 600lbs 16lbs x 7" crank = 112 in-lbs
0.52T - 0.18T = 0.34T = 680lbs 16lbs x 7" crank = 112 in-lbs
0.55T - 0.18T = 0.37T = 740lbs 16lbs x 7" crank = 112 in-lbs
0.67T - 0.18T = 0.49T = 980lbs 19lbs x 7" crank = 133 in-lbs
0.75T - 0.18T = 0.57T = 1140lbs 22lbs x 7" crank = 154 in-lbs
0.76T - 0.18T = 0.58T = 1160lbs 24lbs x 7" crank = 168 in-lbs
[/tt]

What do you make of that 16lb plateau?

Anyway my ultimate torque calculation is as follows.

2500lbs / 4 = 625lbs

625lbs ~ 112 in-lbs per leg.

112 in-lbs x 4 = 448 in-lbs total torque required.


Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com