Hydraulic Cart

Here are the first modifications to my hydraulic cart. Bottle jack + wiper motor = easy lifting.
http://www.metalworking.com/DropBox/HydCart.txt
http://www.metalworking.com/DropBox/HydCartOverview.jpg
http://www.metalworking.com/DropBox/HydCartLeft.jpg
http://www.metalworking.com/DropBox/HydCartRelease.JPG
http://www.metalworking.com/DropBox/HydCartMech.jpg
--Winston
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Would a crank qualify as a 'rotaional to translational conversion linkage'?
The pistons move up & down but the flywheel spins 'round and 'round.

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Rich McCarty wrote:

Yup!
--Winston
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On 10 May 2004 00:28:29 -0700, too_many snipped-for-privacy@yahoo.com (Too_Many_Tools) wrote:

There's a simple eccentric arm on the wiper motor shaft, that arm converts the rotary motion to the linear motion needed to pump the jack. How do you think they get the wipers on your car to go back and forth? Same exact linkage.
Duh! ;-) Didn't even have to go look, and I knew the answer. Cart looks slick, though the battery pack is a little tacky looking.
--<< Bruce >>--
--
Bruce L. Bergman, Woodland Hills (Los Angeles) CA - Desktop
Electrician for Westend Electric - CA726700
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Bruce L. Bergman wrote:
(Snip)

Thanks!
The battery pack *is* tacky looking. Functionality was the major goal. Esthetics took a back seat (in a different car), as you can see.
I want to hear your best advice (collectively) on how this unit might be improved.
For instance, the cart only appears to be a couple percent efficient. Instead of electro-hydraulics driving a scissor arrangement, how would you design it for fewer and less wasteful power conversions?
--Winston
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Winston wrote:

http://www.metalworking.com/DropBox/HydCart.txt
http://www.metalworking.com/DropBox/HydCartOverview.jpg
http://www.metalworking.com/DropBox/HydCartLeft.jpg
http://www.metalworking.com/DropBox/HydCartRelease.JPG
http://www.metalworking.com/DropBox/HydCartMech.jpg
--Winston
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wrote something: ....and I say in return....
What do you mean by a couple of percent efficient?
Have you measured power/energy input to the motor and assumed a certain energy needed to lift a weight?
Or are you concerned that the unit is too slow? I did think that 1.5 minutes was surprisingly slow. But this could simply be underloading the motor. Is it faster or slower than the foot method?
Does the crank drive the full throw, or nearly, of the plunger of the jack? I would probably not want to speed up the plunger more, as I feel it would not be designed for such high speeds. So a larger plunger would be the next step for speed, if the motor is doing the job easily.

*******************************************************
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OldNick wrote:

Yup! Arithmetic wasn't ever my strong suit, so I went by it this way:
I figured that if the table was lifting my carefully calibrated 260 lb weight 1.69 feet in 1.5 minutes, that equated to 293 ft lb./ minute. So if it takes 33,000 ft lb./minute to make 1 horsepower, my table was exerting less than 0.0089 HP. From preliminary measurements, I guessed that the wiper motor was drawing about 10A at 24V or 240W or 0.321 HP. So, not counting losses in the battery charging circuit, I showed 0.0089 / 0.321 = 2.7% efficiency.

I am really happy with the rate of travel. It just irks me that the table is so far from optimal regarding energy use. It would be cool to be able to get twice the number of lifts out of a single battery charge at less than 2 minutes per lift. That's what I mean.

About 2 to 4 x faster, because the wiper motor doesn't get bored. :0)

It drives about half the throw, a cycle of 0.5". I originally made a crank that used the entire 1" travel of the master cylinder, but it eventually tore the wiper motor to shreds! (Fast, though.)

I expect that I am now loading the motor in excess of 200% of design power. It does get rather warm. The only thing saving it is the extremely narrow duty cycle. (Now at two operations per day = 0.2%)
Perhaps I need a wiper motor off an aircraft carrier? :0)
--Winston
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vaguely proposed a theory ......and in reply I say!: uncap my header address to reply via email

Sounds about right mathematically, but weird. Efficiency is energy, not power, but afaics, the time factor is the same in both cases, so the result is the same.
That's assuming you are pulling 10A @ 24 V.
Are you opening the valve properly?

hehe!
But seriously, something seems very wrong. How fast does the wiper shaft turn?

Why should it? What was happening to tear the wiper motor to shreds? Unless you were actually topping/bottoming the throw. Maybe you need to look at a supplementary bearing. The wiper simply is not built to take it?

Just from the gut, I get the feeling that a longer connecting arm would help. It feels to me as if the motor is doing a lot of work in the wrong directions???? I may be talking through my hat, but if you expect a circular motion to make a linear motion, without and sliding action or cam action, then connecting arm length will prevent some really bad angles coming into play as the circle goes around.
I am not talking about a greater leverage. I am talking about the actual con rod length.
*******************************************************
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Old Nick wrote: (Snip)

I think so. If I open the valve too wide, the platform sinks to the floor very rapidly. So normally I just crack it a bit and the platform is much more gentle in its descent. Thanks for bringing that up. Clearly I need to increase the lever arm on the valve so that it doesn't drop a heavy load too quickly.
(Snip)

By ear, I guess 70 RPM.

The gears got very stripped. I don't have a spec sheet for the motor, but if the gears strip, *that* is a serious overload.

More than 6"?

I think I understand and you are right. The original linkage on the master cylinder looks as if it would prevent a lot of side loading. The pivot points are very loose and probably take up much of the side-to-side motion. I assumed that the ~6" conn rod would be long enough to limit side motion (on the pressure stroke) to less than 2.5 degrees. I wonder if this is still too much?
--Winston
*******************************************************
I didn't eat any eleven-dimension sausage. (It is the wurst of all possible worlds.)
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vaguely proposed a theory ......and in reply I say!: uncap my header address to reply via email
I have to admit I have only fooled with this, and then not under load. See my last paras first, maybe.

Sorry, it was a dumb question.
I was thinking hydraulics; valve closed no flow, valve open job gets done.
What I meant was, is there a restriction in the system that is loading it up, preventing flow, and overloading the motor when lifting? You have no control over this.

By ear, 70 RPM? Not sure what you mean. That's quite slow. 1 rev per second. That would operate not much faster then you could push by foot.

By smooth load, or by simply bottoming out?

Shouldn't be, with that 1/2" stroke you are correct in your angles anyway. That's small. I think your setup is OK for efficiency.
OK. From here I would actually measure the force required to operate the jack lever under the 260 lb load, at the point where the con rod now connects. Or use a lever so you _can_ lift the load. Got a spring balance? Then use the stroke at the point where you push (lever or not) and the number of strokes to lift to full height, and see what efficiency you get.
They are usually surprisingly hard to push, even without load. They have seals etc, and this may be the problem.
I am surprised at the 2% efficiency, but it could be so. Nobody has argued. My only buggabboo now is whether it's 2% or 20%, because of lbs and lbs force and the whole shmoo.
Remember that to use this as a foot pump, you would have a lever and your leg strength to work on it. The whole system, if that inefficient, relies on the fact that you are not doing it very much, and have a nice long lever to work on.
This may well be why hydro pumps are usually gear type, which are very efficient.
I will do some tests this afternoon. This is relevant to me, as I have a manual-lift forklift for my shed that I was going to have my evil way with in the same manner as you have. But it is surprisingly hard to operate the lever, in my case to the point where I was looking at checking out the piston and pump.
As to alternatives? Winch, powered. I bought one the other day for Aud$99! It will (just!) lift 200Kg on a single line at 6'/ minute and 500 on double. I would use this as a direct direct lift, not tied up to the architecture of the trolley, which uses a lot of force over a short distance to get a longer lift. Screw jack (ball screws are more efficient, but co$t)
BTW. Setting the thing up using a steering pump IMO would require that you took various other parts from the steering circuit as well. The only "safety factor" as far as I know in a jack, is the fact that it runs out of oil before it hits the end. With a hydro system, that is not a good idea. So you need a limiter. I have seen the effect of overdriving a jack-powered pipe bender. I popped the bottom out of the jack, or at least cracked it right across, by having too long a handle on it, for "that bit extra". *******************************************************
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Old Nick wrote: (Snip)

Doubtless the checkvalve and associated plumbing cast into the bottle jack qualify as a hydraulic restriction. I bought the jack brand new. It appeared to be quite normal in terms of operating force before I took it apart for modifications. After modifications and reassembly, it still operated properly.
The 2% figure seems reasonable; it is the figure resulting from electrical losses plus gearbox losses plus crankshaft/con rod friction and inertial losses plus master cylinder friction plus pumping losses plus slave cylinder friction plus friction losses in the scissor mechanism plus weight of the table and associated mechanical stuff plus several other things...
I figure that the motor and gearbox are just too tiny. Back-of-the-envelope figuring showed me that the force on the end of the crankshaft is the full load on the table, plus the weight of the table per se, plus static and dynamic friction. No wonder the poor thing complains!
:0)

Yup, a smidgen over one revolution per second is my guess.

Smooth load. I reduced the stroke of the master cylinder to 1/2" from 1" so I expect that I have plenty of room on both ends of piston travel.
(Snip)

It will exceed ~300 lbs. with our 260 lb test weight. There is a 5:1 mechanical disadvantage in the scissor mechanism and about a 5:1 hydraulic advantage in the bottle jack. They cancel to leave the force of the entire load plus weight of the table plus friction to show up on the poor little motor crank.
(Snip)

I have seen a couple different approaches to this problem. They were pricey. Even used! 6' a minute is really nice!

Oops! I noticed that the bottle jack in my cart does not over drive at all. At maximum extension, it just sort of oscillates. Doesn't do anything dramatic at all. I guess they have a valve to bypass the check valve in that case.
--Winston
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vaguely proposed a theory ......and in reply I say!: uncap my header address to reply via email

I can see what it results from, but that is a poor system at 2%.

.8
.9
99.9
.8 (but here I wonder)

.8
.8
should be very small. .95

50/260 = .8

well,,,probably! <G>
But I still see 30% above.

No. The force is not that at all IMO. Sorry if I am lecturing. That's the whole idea of the jack. The force will vary as angles vary, but the force will be the weight of the table, multiplied by the area of the master plunger / area of the slave cylinder. Then the angle at which it acts will cause many variations.
I can't see the "architecture" of the lifting mechanism of the cart from the pictures. It's hidden in shadow and of course I can't get measurements. Is it "fighting itself" somehow? Are the angles so acute at some stage that the forces are OTT?
Are you using the full throw of the jack to lift the table to max height?

5:1 in a bottle jack? The ones I have have a master cylinder about , say, 1/2" diam, and the slave about 2-3". Say 2.5" That's a _25:1_ advantage.
Surely the scissor system's mechanical disadvantage will alter as the system changes shape?

Which would _never_ handle 300 lb.

Well, the thing is, it's probably too fast! So you could use a system of pulleys to vastly underload the winch.
I got this Aud$99 thing from the local SuperCheep Auto shop. It's a boat trailer winch. The plastic bits are crappy and brittle, but the gears appear well built. I had to drill one yesterday, and it was not mild steel!

Either a valve, or as I say they simply run out of oil. If you take some of the oil out, it behaves the same, only lower down <G>. DAMHIKT *******************************************************
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Old Nick wrote:

Agreed.
However, not too bad considering that the 'design' process never took place...

Gotta be much lower efficiency than that. 300 lbs on a 3/8" diameter journal bearing... It has grease but is still spalling. Clearly this needs to increase in diameter and work with an oillite in the con rod, at least.
I would be surprised if it exceeded 0.7

Me, too. There is a lot of 'sticktion' apparent in the movement of the master cylinder piston.

Should be. Isn't. The rollers on the bottom of the traveling leg act square, rather than round. I have never seen the roller on the left side of the unit even make contact with the surface it is supposed to roll upon.. Zero friction, right? :0)
Perhaps 0.25? Oh. Huh. Maybe we're on to something...

Call me pessimistic at ~5.1%

(Snip)
Envision a lever. Fulcrum on one extreme end, free to pivot but bolted to the ground. One unit away from the fulcrum is the business end of our bottle jack. Free to pivot. Other end of jack is on the ground. Continuing in the same direction along the lever and five units away from the fulcrum is a weightless table that magically remains parallel to the horizon frictionlessly. On that table is a 260 lb. weight. At 100% efficiency, the force on the slave cylinder is 1300 lbs. The slave cylinder of the bottle jack measures about 1.1" diameter and the master cylinder measures about 0.43". So 6.5:1 advantage at 100% efficiency. Best case is that the master cylinder sees 199 lbs. of force. No wonder the jack handle was so long!

Yes.
(Snip)
That makes sense intuitively. My performance reading encompassed the entire shape change from bottomed out to topped out. Is that what they call integration?

We agree that I've seriously overloaded the motor.

I wonder where I can find a 12V 4 HP 150 RPM motor for US$20?
(G)
--Winston
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vaguely proposed a theory ......and in reply I say!: uncap my header address to reply via email

Oh yeah. The other factor is the ratio of the distance that the motor acts on the pump lever, over the distance from fulcrum to plunger. Then there is the _angle_ of the plunger lever!
There's a bit to take into account!
If you are using the full throw of the jack, then you can't get away with less force at the working end, unless there is really something wrong with the layout angles.
When the table is down, the angles _will_ cause a lot of force to be required to start the lift. You are pushing almost straight along the piece you are trying to turn.
If you let the motor act on a longer lever on the master cylinder plunger (as you do with a jack under a car), you get an even slower lift (although if the motor is labouring, maybe you will not lose it all), but without needing a tougher motor. *******************************************************
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Old Nick wrote:
(Snip)

Now I know why you engineers make the big bucks!

This is less a design problem than an execution problem. See my previous post...

Yup. Amazing it works at all.

Agreed.
--Winston
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vaguely proposed a theory ......and in reply I say!: uncap my header address to reply via email
Sorry. I was trying to help. *******************************************************
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Old Nick wrote:

Where is your tag line ".........no I'm not!"?
Seriously, thanks for the feedback.
--Winston
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vaguely proposed a theory ......and in reply I say!: uncap my header address to reply via email
Ok. I stopped jabbering and mulled it over.
You are lifting 118 Kg (1180 N) 0.5 metres = 590 N-M (/ joules.) To do this you are using 240 * 90 = 21600 watt-seconds (/ joules.)
As you say, the apparent efficiency is around 2-3%.
A bigger motor will of course not solve the apparent efficiency problem. There still seems something wrong to me. Any system that requires you to put in _40 times_ as much work as it would have taken to "simply" lift the weight is not right IMO.
Even though the crank system is ineffective, in that for most of the revolution of the motor the crank is not doing max speed, because of rotation angle, during that time the motor should also be loaded less.
I jacked up one side of my light truck, using a "kangaroo" jack, with the wheels on that side _just_ off the ground. The end of the handle of the jack moved through 2 strokes of 600mm, with a force required of 30kg at the end. That lifted 930 Kg 40mm
1.2 * 30 = 36 Kg-m 930 * .04 = 37 kg-m
Hey 103% efficiency! <G> I was measuring pretty roughly...But you get my point.
But. My thoughts.
When you said you had a 5:1 disadvantage, I assume that means that you have the jack acting at a point 1/5th of the way up the lifting leg?
Your figure of 300lbs on the plunger is not far off, actually, even though there may be 25:1 advantage at the jack.
When the table is down, the actual force required to lift that load could easily be around 9000 lbs, because of the angle of the lifting leg. You have (260 lbs * (5:1)) / (sin of the angle between the jack cylinder and the lifting arm). I worked on 9 degrees for a rough start.
So if there is 25:1 advantage in the jack, you have 360 lbs needed to force the master cylinder in.
I have to believe that the jack has more than 25:1 advantage. I have a 12 tonner out there and I just checked. It has a plunger about 3/8" and a cylinder maybe 3". That's 60:1 plus. We are still talking maybe 150 lbs on the end of the plunger.
Whatever, when that table is down, there are some serious forces to deal with. As the table rises, you approach your 5:1 ratio as defined by beam lengths, ad the job gets easier and easier. The jack will be walking it in near the top of the lift.
You need a way to alter the action of the motor's crank on the jack pump as the table rises. It would need to have greater mech advantage when down, gradually causing longer and longer throws as the table rose. Maybe "walking" along a lever toward the plunger as the table rises?
The table would rise faster and faster as it rose. It's not out of the question. Given what that motor appears to be handling now, you may even average out to the same or less rise time!
*******************************************************
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Old Nick wrote:
(Snip)

Correct!
Closer to 6.5:1 advantage. It's a toy jack, Jack. (G)

Ah. I didn't think of that. Should have.

So, 1400 lbs with a 6.5:1 jack? Whoa.

That's not a toy jack.

This could get complicated. Just now, I had a vision of *two* bottle jacks arranged in parallel, but at different angles to the lever arm. One would get the table "off the ground" and the other would take it to full height. Hmmm.
Thanks
--Winston
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