Hydraulic jack capability ?

[Externet]

Hi.  Have a biggish hydraulic jack with no markings. Looks like the image below.

Its pumping rod diameter is 12mm,  its piston is 55.5mm   I do not know what is the typical operation pressure of these things, or the typical force-length-leverage applied to the lever that plunges its pumping rod.

How do I calculate its rating ?

[TheVat]

The cheap ones - they're called bottle jacks - are usually 2 - 4 tons.  Mine is 2, iirc.  Haven't used in years.  

[mistermack]

I think you could get a fair idea of the lifting force by comparing the movement of the handle to the movement of the piston. Calculate the ratio, then use it in reverse, to calculate what force a forty pound push on the handle would produce from the lifting piston. It would involve some precise measuring of the movement of the piston and handle. Or to improve accuracy, you could measure the rise of the piston for ten measured pumps of the handle. 

[Externet]

2 hours ago, mistermack said:

you could measure the rise of the piston for ten measured pumps of the handle

Thanks.   Would the ratio of pistons give me that ?   

Assuming 40 poundsfoot force as you mention,  -if that is the typical operation force for a 1 foot long lever- (or torque?)... would the ratio of pistons also give me the force ?

The pistons area ratios is 113.1 to 2419.2   =  1 : 21.4

Similar sized jacks seen on the web are rated 20 ton.  If that means tonnes I do not know 

[mistermack]

9 minutes ago, Externet said:

Thanks.   Would the ratio of pistons give me that ? 

No, I wouldn't go by that. It won't include the length of the handle, and I wouldn't be confident that what I can see and measure externally is the true size of the internal pistons. I would just do as I suggested. Fit the handle, decide the point in the handle that your force can be considered act through. Measure the distance a full stroke passes through. Then pump it five, or ten times, and get the ration of movement of the top of the jack, to movement of the handle. 

I just plucked 40 lbs force out of the air as a typical force you could apply by hand. But you can do the same calculation for 20 or 30 lbs. 

For most cars, a 2 ton jack is ok. You're only lifting one corner of a normal car, which might weigh 2 tons in total. So one corner probably only weighs about half a ton. 

Those bottle jacks are not great. You need a very solid flat surface, they are not very stable at all, but they will lift the weight, but slowly. A trolley jack is much better, faster, safer and more stable, but a bit more bulky. Most trolley jacks are rated at two tons for car models, which is plenty for wheel changing. 

20 tons is massive overkill for a car.

[John Cuthber]

17 hours ago, Externet said:

 

How do I calculate its rating ?

 

To a good approximation, weigh the jack.
Find jacks of similar design on the net and then look for one that weighs the same.

[mistermack]

A friend of mine, who has done his own maintenance for years, bought an inflatable car jack recently. It's got an electric pump that runs off the battery. He swears it's a game changer, for changing a wheel. I haven't seen it, but he was well impressed.

I've got no idea what weight they will lift, I expect there is a range. 

[sethoflagos]

On 11/14/2023 at 8:21 PM, Externet said:

Its pumping rod diameter is 12mm,  its piston is 55.5mm 

Not entirely clear on what you mean by 'pumping rod'. Bottle jacks have two pistons: a pumping piston to generate hydraulic pressure and a ram piston to actuate the hydraulic ram.

On 11/15/2023 at 3:18 AM, Externet said:

The pistons area ratios is 113.1 to 2419.2   =  1 : 21.4

This is reasonable for a x-section ratio of pumping piston : ram piston and indicates that 1kgf applied to the pumping piston will generate 21.4 kgf lifting force at the ram in theory (there's always going to be a little leakage of hydraulic fluid).

But the load applied to the pumping piston is also subject to the leverage (mechanical advantage) of the handle which is the ratio of overall length to the distance from the pivot to the shaft of the pumping piston. Using the handle extension supplied, you may get a mechanical advantage of perhaps 20 : 1 which would give you an overall mechanical advantage of 20 x 21.4 = 428 : 1.

So a force of 1 kgf at the handle should generate a theoretical lifting force of 428 kgf. 

On 11/14/2023 at 8:21 PM, Externet said:

How do I calculate its rating ?

The jack maximum load rating is set by an internal pressure relief valve that limits the pressure difference between the pumping pressure and low pressure reservoir.

In principle you can measure this by extending the ram and then gradually increasing the load on it until the safety valve lifts and the ram lowers. 

 

Edited November 17, 2023 by sethoflagos
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