velocity limiter / linear damper - howto?

You did not specify your level of capability to make one yourself, but as a guide, there is a device used on sailboats called a boom brake. It works well and is reasonably priced, but keep in mind that a device like this is an energy absorber. If the period or amount of absorbsion is excessive, heat will be an issue. Steve

realistically,

re: making one myself.. i have a small workshop, mill drill lathe welders etc and am resonably competent with them. again, it depends on how complicated the device is.. tolerance, availablity of materials, etc.

but, if its simple, i'd like to try my hand at it.

how does the boom brake work? (mechanically) anywhere i could find drawings?

the vertical drop isn't too long.. around 15 feet at most, so i dont think damage from heating should be too big an issue. though, if it ends up doing alot of braking (often) cable inspections will be more frequent.

inversely, since the drop isn't too far, the brake should respond fairly quickly to a change in speed.

-tony

You specified a mechanical brake, but perhaps you should consider a simple induction motor with DC applied to the winding(s). One with a gearbox of 5:1 to 15:1 makes the braking action even easier. Try it with a variable transformer and bridge rectifier and you will be surprised by the smooth viscous drag you will get.

Advantage is that it is controllable from a distance and has zero wear on the cable.

Earle Rich Mont Vernon, NH

what size DC motor would one use to limit the freefall velocity of a 500# load? what size diode?

and is a gearbox required, as the other post suggests?

not a bad idea, i rather like it. but i was getting keen on making my own mechanical limiter.

-tony

A gearbox helps a lot to minimize the size of the motor. It also depends on the size of the payout pulley.

When your only tool is xxx, all your problems look like yyy.

Earle Rich Mont Vernon, NH

I'm interested in this too.

The motor ideas would work, but there would need to be enough gearing so the motor spin at maybe 1/2 rated speed while the load was falling at desired speed. Then, the motor must be large enough that it could produce enough torque (at half speed) to lift the load. If you use an induction motor, then I'd go by rated current and HP even though you'll be exciting it with DC.

500 lb falling at 2 in/sec is 0.152 HP or 112 watts, so a 1/4 HP motor would be plenty -- but you'll definitely need some gearing.

Another approach would be a "Weston brake" as used in boatlifts and manual chainhoists. You could make that. One version of that mounts a winch spool (for chain or cable) on a threaded shaft connected to the actuating crank, wheel or whatever. The spool has a braking surface that mates with a ratcheted disc free to spin on the shaft but restrained from moving axially. When you turn the crank one direction (lift), the threads act to screw the brake closed -- but the ratchet allows the brakedisc to turn. Let go of the crank, load stops.

If you turn the crank in the "down" direction, if the load doesn't move then the crank screws the spool away from the brake until it does turn, thus re-tighening the brake until the load stops.

The only force required for lowering is enough to loosen the brake, which depends on the coeff of friction in the brake, relative radius of brake to spool, and friction in the screw that must be overcome to get it moving -- like a tightened nut. In a well-designed and well-made device this force can be quite small -- as in lowering a

1200-lb boat with one finger on the spoke of the wheel. A small gearmotor could then run the load down at a fixed rate, or just spin the wheel (or crank) by hand.

I think the "Skyhook" 500-lb toolpost hoist works this way.

If you find out how the boom brake works, I'd like to know also by email or rcm post, please.

Don Foreman

For a purely mechanical device, what you need is a centrifugal brake. If you take an old dial phone apart, you will see a small example of it. It kept the dial from overspeeding when returning to rest. Just 2 spring loaded counterweights inside of a brake drum. The faster it spins, the more braking force on the drum. Just scale it up to the load you're dealing with. Probably find the parts to make it at the local auto recycler.

Fred

How about a squirrel cage fan (in most gas or oil heaters) with appropriate pulleys? All Mechanical and one can control the air intake for fine speed control. An other advantage is cooling is automatic.

Dan Cl>

On Sun, 17 Aug 2003 09:03:29 GMT, "tony" wrote something ......and in reply I say!:

The heating damage will also, and probably more so, be in the unit, which cops the heat all the way down.

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Nick White --- HEAD:Hertz Music Please remove ns from my header address to reply via email !!

No power is dissipated if it falls unchecked -- just a whole bunch when it hits. If restraining force is zero, so is dissipated power.

I just thought of another way. Use a governor like old wind-up toys had. It was an un-balanced flywheel geared to the clock spring. No frictional losses in a brake, easy to fabricate and, by changing the size of the flywheel easy to tune. In the toys the flywheel was a semicircle but it could be any shape as long as it is unbalanced.

Richard Coke

On Mon, 18 Aug 2003 20:47:37 -0500, Don Foreman wrote something ......and in reply I say!:

OK. My side again.

We are trying to _stop_ it falling unchecked, which gravity will make it do otherwise.

To maintain a constant speed we are stopping acceleration due to gravity.

The energy dissipated in the brake during the entire journey is surely the difference between what happens at collision at 2in/sec vs what happens at the speed according to gravitational acceleration(?) The power to make the difference happen is what we want, not the power to produce 2 inches / sec.

Your statement that some gearing would be needed makes some difference. But the problem as I see it would be that too small a motor would exceed its rated max speed, and probably current when shorted, under the needed gearing, in order to generate enough power to slow the load sufficiently.

The motor/gear setup needs to be sufficient that it can lift the weight against gravity without overloading, not just to lift it at the speed you want it to fall at. This is also useful to get the weight back up again I suppose .

If we reckon we need power based on the _required_ speed, then the higher the required speed, the higher the power needed to fight gravity acceleration. That does not make sense.

3rd party opinion? Preferably about the subject and not about me personally.

Out of interest, is it possible, with a sufficiently large motor and gearing, to completely stop the thing, because the gearing friction and motor torque at short are such that the device will simply not even start? As soon as the weight moves the friction and torque oversome the movement, so it cannot even creep. Seems reasonable. But it may be easier to simply place a mechanical brake on the motor shaft, and use a much smaller setup.

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Nick White --- HEAD:Hertz Music Please remove ns from my header address to reply via email !!

On Sun, 17 Aug 2003 15:11:28 -0400, "Jon" wrote something ......and in reply I say!:

I do have to say that testing with a 500lb load 15' up will certainly be a fast teacher if you get it wrong though

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Nick White --- HEAD:Hertz Music Please remove ns from my header address to reply via email !!

Nope, strictly speaking -- eddy current loss (shorted turns, whatever) is a dynamic effect. As RPM increases, breaking torque increases proportionally; thus total losses follow a square law. However, if you use really, *really* crappy bearings, where static friction is greater than the torque produced by the weight, then yes it will not fall. :)

Tim

-- In the immortal words of Ned Flanders: "No foot longs!" Website @

I don't like it. The starter motor doesn't have permanent magnets for the field, so the field would have to be seperately excited.

If you want to use the shorted generator idea, the motor choosen should have permanent magnets. The best braking action occurs when the rotor is turning fast, therefore, a gearbox is desirable.

If you want to use an induction motor with DC on the windings, a gearbox is still a good idea. The faster the rotor turns, the more braking action you will get. A large motor, 2 hp or so will give you more braking action. I think you need to try some variations on all the ideas you've been given. Experimentation will give you a much better feel of what's possible.

Earle Rich Mont Vernon, NH

Not quite that simple. A starter motor is a DC motor and often has two field windings. You want to short the armature and apply some DC power to one of the fields, preferably the one with the smallest diameter wire. Do a bit of experimenting. It should work.

Ted

On Tue, 19 Aug 2003 03:37:44 -0500, "Tim Williams" wrote something ......and in reply I say!:

Thanks Tim. I realise that braking force is dynamic. But I figured if you had a motor shorted out that was able to generate quite high force for a very small movement of the load (big motor, lots of gears), then simply the gear friction and the force would be sufficient. In other words you had high gearing, so there was quite a bit of gear friction and the motor turned fast for a small load movement, then any gravitational attempt to move the load, while the motor was shorted, would be thwarted.

I do realise, as I said to myself, that it's probably more practical to place a brake......or crappy bearings on the motor shaft, and let that and the gears do the work. the gearing and motor would be ridiculous to obtain anything like what I am talking about.

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Nick White --- HEAD:Hertz Music Please remove ns from my header address to reply via email !!

20' / min is perfect. ~45sec drop across 15' thats the normal rate of descent for the lift!

problem: the motor was excited. not something i can do as i have no way of telling the motor a "problem" is occuring.

also, 5HP induction is huge! (for where this motor would sit)

sounds like its time for me to fire up the lab and start running the mad scientist experimentations

'evening,

-tony

On Tue, 19 Aug 2003 18:04:49 GMT, Ted Edwards wrote something ......and in reply I say!:

Sorry. Won't P-K be 0? The energy to be absorbed = P, or K minus whatver energy is released when the object impacts at its contolled speed.

However, I take your point and I agree with it. It puts figures on what I am trying to say.

The _power_ (read brake motor size) required to do this is surely proportional to the energy to be absorbed, and the time taken to absorb it, caused by the time that the load would drop unchecked under G, not to the speed required. There seems something wrong with saying that the slower I make the thing drop the less power I need to slow it. This is what I understood Don Foreman to be saying.

Why don't we have massive gearing and a slot car motor to do the job? Because the slot car motor will destroy itself.

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Nick White --- HEAD:Hertz Music Please remove ns from my header address to reply via email !!