On Tue, 19 Aug 2003 14:06:44 -0500, Don Foreman wrote something ......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.
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On Tue, 19 Aug 2003 21:13:13 GMT, "tony" wrote something ......and in reply I say!:
that's where gearing comes in to play. the experiment was the simplest way to check. the drag of the motor is prop to the square of its speed, I think it was said, all things being equal.
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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
g{is}32 {rem} acceleration due to gravity (feet/sec^2) (W h){is}500 10 {rem} weight in pounds, height in feet v{is}1 2 3 4 5 6{div}12 {rem} final velocity in feet/second PE{is}W{times}h {rem} potential energy (foot pounds) KE{is}0.5{times}(W{div}g){times}v*2 {rem} kinetic energy (ft. lbs.) at end 3{fmt}DE{is}PE-KE {rem} 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. 3{fmt}vf{is}(2{times}g{times}h)*0.5 {rem} final velocity, free running 25.298 So the above assumption is a reasonable approximation. 3{fmt}DT{is}h{div}v {rem} approximate decent time (seconds) 120.000 60.000 40.000 30.000 24.000 20.000 3{fmt}P{is}DE{div}DT {rem} power (watts) to be dissipated 41.666 83.330 124.988 166.638 208.277 249.902
On Wed, 20 Aug 2003 00:38:27 GMT, Ted Edwards wrote: (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.
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?
I agree. The induction motor experiments were fascinating, but a friction brake may well suit your need better if space/size/weight is an issue.
A centrifugal brake would work if there's enough gear ratio to get it spinning fast enough to provide the desired/reqired braking force. A go-cart centrifugal clutch might fit this job very nicely if you can provide the gearing to make it work at your speeds of interest.
You said at the outset that you'd like a mechanical solution you could build in your shop. You can buy a centrifugal go-cart clutch from Northern Tool for $22.99. Then off with you to the shop to determine the (unspecified by Northern Tool) "contact speed" and then devise and make, scrounge or buy gearing to match your desired descent rate to the clutch's "contact speed." Sounds like a fun project.
Keep us posted, please! I'm looking forward to seeing photos and a success story on the dropbox.
Don Foreman -- who is still amazed at how well an induction motor works as an eddycurrent brake.
Use a disc brake from a motorcycle. Squeeze the handlebar lever as the speed increases. These are used on zip lines and cable cameras for movie filming purposes
I'd be a little nervous about the motor drag idea, if your diode fails open or your brush sping breaks or something like that you have nothing. My suggestion isn't fail proof either, but I think it may be a bit more predicatlble and also more heat tolerant. Oh, and pretty simple to build.
Get a hydraulic pump, most any gear pump will do. and a little resivouir and plumb it so the oil can come out of the tank, into the pump, through a restriction, and back into the tank. There you have it.
A few details, if you want this adjustable you can just use a needle valve after the pump. For a fixed resistance you can put a restrictor (little plate with a small orifice) in the out port of the pump.
If this isn't going to cycle much you the resivouir will be able to dissipate enough heat. otherwise you could run the oil through a heat exchanger between the restriction and the resivour. (this can be as fancy as an automatic transmission oil cooler with a fan on it or as simple as a coil of copper tubing, on the low pressure side ot the restriction you don't have to worry about blowouts.
You can probably scrounge all the components, or, if you feel like spending a little cash you can probably get everythign you need with one trip to any hydraulic supply place.
Oh, if you only want resistance going one way you could plumb in a check valve going the opposite direction parallel to the needle valve If you are using fixed orifice restrictors you can get them that work as a "leaky" check valve and give nearly no restriction in the other direction.
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.
Pretty close. It isn't quite linearly proportional because, although rotor inductance is small the current is huge so inductance does have an effect. Rotor current frequency increases with speed so rotor impedance |R + jwL| also increases with speed.
A torque-speed curve would be shaped like the torque-speed curve of an induction motor plotted against slip speed -- which isn't too amazing given that it *is* an induction motor!
i could add a pump to my gear rack, so it would move up and down with the lift. the pump is basically only cycling oil through at a fixed rate, depending on my needle valve setting.. so if i can tweak the needlevalve to about the same speed as the lift, i can get it to ascend and descend at the same speeds even if there is a failure in the main (electric) motor?
as far as dissipating heat: i think i'm building this thing robust enough so NOTHING WILL GO WRONG. (more on this misconception below) -- the hydraulic pump should only see a real load if this thing falls -- which should be never.
i have a 1HP motor running a 30:1 worm gear so if the motor goes, the worm gear should be able to hold the load on its own. (ie the lift will stop where it breaks and not come plumeting to the earth)
i wanted a motor with an electromagnetic break, to take load off the gearbox when the lift reaches the top.. but they're hard to find out here. would have to wait 2 months for the order.
on the worm gear i have two sets of rack/pinions (one on each side) so if one rack/pinion were to break, i still have one on the other side (which, on its own, is more than capable of doing the work)
but, being realisitic, something obviously could happen and thats why i'd like to add the mechanical brake. or "slower-downer" (thats a prize winner, there)
back to the hydraulic pump... (sorry this is so long)... is the resevoir necessary? can i just plumb the inlet to the outlet and fill it up with oil? and what happens when you run a hydraulic pump backwards? (the case when the lift is ascending)... and when its "slowing-downing" how can i be sure i wont blow the needle valve off the pump?
sizing: will any pump do? and is the relationship between pump rpm and gallons/min linear? that is, i obviously am not going to be running the pump at rated speeds.. but can i bet on the fact that if i'm at 1/4 the rated rpm i'll be getting 1/4 the oil flow? i would need to know this to ensure that i'm not causing any resistance to the motor/wormgear when it is ascending.
On Wed, 20 Aug 2003 00:38:27 GMT, Ted Edwards wrote 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! )
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On Wed, 20 Aug 2003 00:09:00 -0500, Don Foreman wrote something ......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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On Wed, 20 Aug 2003 18:17:20 GMT, Ted Edwards wrote something ......and in reply I say!:
My only reference is memory of a post on this thread.
Actually, it was a misquote from Tim Williams. "The total losses follow a square law" (meaning, I assume, gear losses * motor losses.
I should have said "the drag of the _system_"
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My original assertion that the power dissipated is proportional to descent speed is correct, has been defended once already and has been corroborated by another. Feel free to disbelieve if you like. That doesn't make it untrue. Calling it misleading and dangerous does no one a service .
The original assertions I made about the implied motor power were indeed partially misleading, though not for the reasons you've cited. Max torque capability is a function of motor rated HP, nominal speed at that HP and rated field current, as was asserted. But that just determines the max braking torque one can safely expect the motor to exert with DC on the field. Torque will continue to increase with speed up to a point. But at some speed above the rated slipspeed of the motor (75 RPM for a 1725 RPM motor) torque would peak and then diminish with further increase in speed. That's easily explained but I'll try to stay relevent to your comments here.
The part that was erronious was the amount of braking *power* (not torque) one might expect in continuous operation, though I doubt if a load brake sees continuous operation. In this situation the motor is not exporting any electrical power from the field as does an overspeeded induction motor with AC excitation. Essentially all of the energy absorbed from the load is converted to heat in the rotor as I^2R loss in the rotor. Induction motors typically dissipate about 15% of their rated shaft output power as heat, so that would probably be the safe continuous limit. However, there is significant thermal capacity (mass * specific heat) present, so a given motor in intermittent braking service could probably absorb rated power for minutes before reaching rated temperature rise which is typically 40C above ambient.
I think the case at hand was 500 lb for 15 feet, That's 7500 lbf-ft, 10170 Joules (wattseconds) or 13.636 HP-seconds per controlled descent. That amount of energy would raise the temperature of a 6 lb rotor (steel and copper) only about 15 deg F or
8.3 deg C assuming no air cooling at all. Rotors of less mass would have correspondingly greater temp rise for the same energy input. So for low dutycycle use (much more time spent cooling than braking), a surprisingly small motor would suffice. A motor with a 1.25 lb rotor could safely manage this load for one descent with temp rise of
40C. It would then need to cool a while before braking another descent. That's not a very big motor! A rotor 2" dia and 1.4" long would be about that weight.
So the rack is going up and the pinions on the worm driven shaft are mounted on the stationary part right? you will probably want the check valve in parallel with your needle valve to let it go mostly unrestricted on lifting, no reason to tax your system (actually in the reverse direction on lifting this would be the low pressure suction side and too much restriction would cause cavitation and eat your pump, so you really need that check valve.). you would want to set it to fall probably a little faster than your system will normally go so it isn't dragging and producing a lot of heat in normal operation.
Gonna digress here. There is a thing they call a flow fuse, we used them in the lift cylinder ports on forklift masts. you can get two kinds, lock up, or bleed down. They basically do like the name implies, they let fluid flow through unimpeded until a certain flow rate is exceeded and it either locks up (shuts down the flow completely) or bleeds down (lets things settle at a slow rate). The idea is you buy a flow fuse that is set for just over the flow you see in normal operation, but if a hose blows and oil starts gushing out of the cylinder (and your 6000lb of stuff up 20 feet in the air starts comming down in a hurry the flow fuse trips and the thing either locks in place or bleeds down slowly (you reverse flow (fix hose and lift again) to reset them). you could probably put a flow fuse in the port of your pump and you would never have any restriction unless something happened and you got more than normal flow (like when your load falls). The ones we purchased were from a german outfit called VonBurg or VonBorg or something like that. Actually, I'm not sure what kind of lift you are making but you may have been better off to go the hydraulic route all along as this is an easy nearly foolproof way to add a safety to a hydraulic lift system.
you mean a clutch to keep the motor from crashing into the ends of the pinion? sounds like a good idea. may not hurt to have a sacrificial joint that will break before major components if this fails (like a pair of flanges with two soft bolts holding them together, slam it hard enough it just shears the bolts and don't break anythign. Or maybe just a belt drive that could slip it things went south.
Carefull here, you have both the pinions on the same shaft? if say a bearing on this shaft goes out is it possible that the pinions could ride away from the racks? Is the rack fully supported so it can't buckle away from the pinions?
Another thought, consider what would happen if somehting (bolt falling off somethign, pin, washer, some kind of junk) falls in the works and goes (driven by the descending load) between a pinion and the rack could it: a: bend the shaft away from the rack enough to make both pinions not engaged enough and let it fall. b. break the shaft mount/bearing and let the pinions out of the rack.
May want to put some sort of shirld over the works too.
I'm sure you have looked at this, just wanted to mention it.
It would be easiest to attach your slower-downer to the pinion shaft, but that assumes no fault is possible in the pinion/rack system, if a shock load (something big gets dropped on the platform) and a tooth gets sheared off each pinion (or shears the keys...) you could have a situation where the pinion shaft doesn't turn and the racks go crashing down. So I'd suggest putting the pump (or electric motor/generator resistance) on it's own pinion shaft to the racks.
not a good idea, your system will not make much heat, but oil expands with heat and you need to let it vary in volume. That said, you don't need much of a resivoir it just has to be vented or have some give.
good question. :) some gear pumps are designed to have the plates on the sides of the gears driven in tight by the pressure of the pump to give you higher efficiency. some of these designs may not be good to use. Your simple old gear pump with nothign fancy may not have quite the same efficiency both ways, but I think would be ok.
ignoring some leakage, yes, linear, they are a constant volume device.
I'm thinking your best deal would be to put an appropriate flow fuse int eh pump port. That way you don't have any real resistance (and that means no heat) unless the thing actually falls
Diodes almost always fail short rather than open unless subjected to HUGE overloads. A shorted diode would simply slow things down even more and would be pretty obvious. The rare case of a diode failing open is easily solved by putting two diodes in parallel, each sized to be capable of handling the full load - diodes are cheap.
How many brush springs have you seen that broke? The failed brush springs I've seen have all been destroyed by excess heat long after the brushes had exceeded their useful load.
Note that hydraulic systems have lots of springs in them too - e.g. the check valve.
Personaly, I like the idea of an electric motor - either induction or PM. I think I prefer the PM (if a suitable motor can be found cheap) since it's all passive and won't leak. :-)
Heating sources in a motor come from I^2 loss in rotor, I^R loss in field, hysteresis and eddycurrent losses in the steel. If these are about equally distributed then I'd expect about 13 deg C rise if rated current was supplied to the field. Designers tend to make things that way for economical reasons.
Rated current on the 1/3 HP 120 volt motor was 6 amps. 12 VDC produced 6 amps of DC in the winding.
In the 220 volt 5 HP motor, 12 volts produced about half of rated currrent, which I think was about 20 amps.
On Thu, 21 Aug 2003 18:22:19 GMT, Ted Edwards wrote something ......and in reply I say!:
The comment I would make here is that the project will _probably_ use a wrecker's yard starter, which could be on its way out. I have seen brush srpings that have lost their spring simply because the brushes were worn, not just because of load. This creates poor contact and heat, and softens the springs. I have had a motor that has done this. Mind you, it did not result in catastrophic failure, but simply poor performance.
I agree that the motor will be far cheaper, cleaner and simpler. Any of these systems can fail, very fast. So realy a backup complete braking system would be need for greater safety.
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