velocity limiter / linear damper - howto?

greets all. was hoping someone could help me with a problem..
i want to lower a load down a (fixed) vertical steel cable
at a constant velocity... i'm looking to build something of a simple mechanical brake. speed isnt too critical.. i'm dont need a specific velocity.. just "slow"... maybe an inch/second? it shouldn't brake however, unless the load drops more than, say, 2 or 3 feet/second.
something similar to what mountain climbers use.. or those safety gadgets for going into high places.. if i'm not mistaken they're made with small wheels that squeeze onto the rope/cable/line and slow a fall.
the load will range anywhere from 100 to 500 lbs.
again, it doesnt need to be complicated (in fact, the simpler the better -- because i'd like to build it myself) ... i like the "squeezing wheels" idea, but i'm not sure how i know enough to crunch any numbers.
i was thinking something like this: a steel bar "X" inches long with a hole at one end, through its short side. the hole just big enough to allow the cable to slide. it would look sort of like a metal flag, perpendicular to the cable. under its own weight (ie, the moment it creates) it wouldnt move down the cable. with a little weight, it would slide down (free fall)... with more weight, however, it should kink the cable, thereby making a straight drop difficult. too heavy and it should make a bend in the cable and stop. the cable is fixed and taught.
again, it is very basic.. but i'm not sure how it would work realistically, how long the "lever arm" should be.. etc.
sorry so longwinded. thanks, -tony
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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,
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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
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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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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
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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
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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
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wrote:

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
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On Mon, 18 Aug 2003 22:21:47 GMT, snipped-for-privacy@iinet.net.au (Old Nick) wrote:

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

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On Mon, 18 Aug 2003 20:47:37 -0500, Don Foreman
......and in reply I say!:
OK. My side again. <G>
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 <G>.
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 !! <") _/ ) ( ) _//- \__/
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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 @ http://webpages.charter.net/dawill/tmoranwms
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On Tue, 19 Aug 2003 03:37:44 -0500, "Tim Williams"
......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 <G> 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.

****************************************************************************************** I could never _see_ myself as anything!
Nick White --- HEAD:Hertz Music Please remove ns from my header address to reply via email !! <") _/ ) ( ) _//- \__/
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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? <G> Because the slot car motor will destroy itself. ****************************************************************************************** I could never _see_ myself as anything!
Nick White --- HEAD:Hertz Music Please remove ns from my header address to reply via email !! <") _/ ) ( ) _//- \__/
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Old Nick wrote:

Without a slower-downer, yes because the weight will simply accelerate to the velocity that implies. If the slower-downer absorbs energy so v at the bottom is much less than what it would otherwise be, then the slower-downer must absorb P-K joules where P is potential energy at the top before anything starts moving and K is the kinetic energy at the bottom, presumably (hopefully) much less than it would be in a free fall.

Correct. Power is work per unit time - watts = joules/second.
caused by the time that the load would drop unchecked under

Nope. The actual time involved. Slow decent => low power because the energy is being dumped over a longer time.

That's right. If the decent was at velocities between 1 and 6 in/sec:
Falling Weight Problem
g32 acceleration due to gravity (feet/sec^2) (W h)500 10 weight in pounds, height in feet v1 2 3 4 5 612 final velocity in feet/second PEWh potential energy (foot pounds) KE0.5(Wg)v*2 kinetic energy (ft. lbs.) at end 3DEPE-KE energy (foot pounds) to be dissipated 4999.946 4999.783 4999.512 4999.132 4998.644 4998.047 Assuming the final velocity is small compared to the final velocity in free fall, the decent velocity would quickly approach very close to final velocity. 3vf(2gh)*0.5 final velocity, free running 25.298 So the above assumption is a reasonable approximation. 3DThv approximate decent time (seconds) 120.000 60.000 40.000 30.000 24.000 20.000 3PDEDT power (watts) to be dissipated 41.666 83.330 124.988 166.638 208.277 249.902

Too much friction in the gearing.
Ted
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(snip)

It's right for any descent velocity, even if you don't deduct K that gets absorbed by the pillow at the bottom. Deducting K just makes it more so because initial potential energy (P) converted to kinetic energy (K) is not dissipated in the brake but in the crash at the bottom.
Power is rate of energy transfer in joules per second or whatever you like. Energy (P or P-K if you prefer) is constant. Power is the definite time integral of energy over the region from 0 to T where T is the duration of the descent. If you integrate over a longer time, then it takes correspondingly less power to dissipate the same amount of energy.
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Don Foreman wrote:

You put your snip in the wrong place. The statement, "If the decent was at velocities between 1 and 6 in/sec:" gives the particular terminal velocities used in the example calculation that follows. These were chosen as covering the range of interest without attaching a graph (no binaries).

agreed.
????
Wrong. Energy is the time integral of power. Power is the time rate of change of energy. Always.

The wording bothers me a bit. I would say, "The same enegy may be dissipated with less power if spread over a longer time.
Ted
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something ......and in reply I say!:
Sorry, Ted. I was getting crossed up. I agree with what you say.
I think there were two separate streams of argument here.
The "power" I am talking about is the motor power, that will largely dictate how much current etc...force!....it will generate upon being turned at a given speed with magnetism applied to the coils.
I was getting tied up in my own pen (and don't nobody finish that word! <G>)

****************************************************************************************** Those who can, do. Those who can't, teach. The rest sit around and make snide comments.
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On Tue, 19 Aug 2003 14:06:44 -0500, Don Foreman
......and in reply I say!:
Maybe I misunderstood your original idea completely.
I understood you to base your power requirements proportional to the speed with which the object dropped.
I reckon the reverse. At any moment, the power needed is proportional to the difference between what the object "wants" to fall at, and what you want it to fall at.

****************************************************************************************** I could never _see_ myself as anything!
Nick White --- HEAD:Hertz Music Please remove ns from my header address to reply via email !! <") _/ ) ( ) _//- \__/
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On Tue, 19 Aug 2003 22:31:04 GMT, snipped-for-privacy@iinet.net.au (Old Nick) wrote:

It may depend upon how you define or understand "power" , perhaps in a way somewhat differently than is customary for us engineering and physics pukes.
Let's say I don't want the object to fall at all. That would be a maximum power situation by your reckoning, right? How much power would you reckon is being dissipated by a chain suspending a stationary load?
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On Wed, 20 Aug 2003 00:09:00 -0500, Don Foreman
......and in reply I say!:
OK. Cut to the chase...again. Ignore my and any other tangles.
Your assertion that the _motor power rating_ needed to stop the load falling depends upon what speed the weight falls at is incorrect, in my opinion. It is misleading and could be dangerous.
Your use of the energy of falling to decide the motor horsepower was flawed in the first place, and now you are coming back at me from your own seeded error.
The "horsepower" of the motor is simply some figure that is needed in order for the motor to be able to withstand the weight, without overheating its wires, or destroying itself centrifugally.
It _can_ burn out, because there, in case you had not realised in your engineering and physics pukiness, other circuits in a motor besides the windings. Also each individual windings ability to withstand current is _not_ decided upon by core temperature in all conditions.
Under your formula, using the 500lb and 2 in/sec, a weight falling ten times as fast would require a braking motor rated at 2.5 Hp to maintain its speed. A weight falling 100 times = 25 Hp etc etc
Support your original assertion, as I will stand corrected.
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