Mace Canute, I hear you, but using your reasoning how does a 2 ton bottle jack move 1000 lbs. 4 in. in very much less than the low time you talk about?
Since it doesn't, and I never claimed it did, what is the basis for that question?
I can easily pump a hand pump 4 times in 1 second. That's just 1/4 inch per pump. I think you are considering the mechanical advantage correctly, but don't understand the situation.
Since when does pumping a hand pump 4 times a second equate with 1/4 inch per pump? That's 1/4
second per pump, not 1/4
inch per pump.
First of all I have a handle on the pump with a pivot. Say 2 inches to 16 inches. That's an 8 x advantage right there. So with a 20 pound force on the handle I have a 160 pound input to the pump. If I move the end of the handle 16 inches I have a movement of 2 inches. So the lever itself will move 160 pounds 2 inches with 20 pounds of input.
I know that is correct.
I agree also.
I need only a 4x (4x160) advantage from the pump to get to 720 pounds force out. That drops the movement of the ram from 2 inches to 1/2 inch per pump.
I also know this is correct.
If you still have a 20 lb force acting on a 16" handle with an 8:1 mechanical advantage, that isn't correct. Just shortening the stroke of the master cylinder itself will not give you any increase in MA unless it is done by altering the fulcrum point. If you have a 20 lb force acting through a mechanical advantage of 36:1, then it is correct.
Also I don't need to have 720 lbs./sq. in of force on the piston. I need 720 pounds of weight. The size of the piston is not material. Can be 1/2 inch or 2 inches. And before you get your drawers in a knot, what the ram is pushing on is what is important. A two inch piston pushing with 720 pounds on a 1/2 inch object applies the same force on the 1/2 object as a 1/2 inch piston pushing with 720 lbs. What this amounts to is that in my calculations I can ignore volume, except as concerns the total mechanical advantage.
So all I have to be concerned with is the total mechanical advantage to any size piston out.
So the hand pump only has to pump 1/4 the volume of fluid I displace with my 1/2 inch stroke. I'm unsure here, but looks like 8 pumps with the handle will get me the desired results. More like two seconds, but remember I said, no less than 1 second and more was ok.
My math before may have been incorrect, likely was, but not now.
You did state in Post#5 "I need to move a ram about 3 inches with 500-700 lbs of force within about 1 second no faster." and "The diameter of the cylinder less than 1/2 inch perhaps."
You
absolutely cannot ignore volume. Mechanical advantage is just half the equation.
You can divide up the volume over a number of pumps, but you still have to make up a finite amount of volume if you want the slave cylinder to travel 3".
I doubt a pump made from a hydraulic jack would have that amount of volume at that pressure even with a lot of modification.
A bottle jack is a hand pump with a SA piston. If a two ton bottle jack will lift 2000 lbs. 4 inches, why wouldn't it have enough volume to lift 700 lbs./3inches.???????
The jack will, but you don't want the whole jack, you want to use just the high pressure piston from the BJ. It's volume is very small, hence the very large number of strokes it takes to move the BJ ram.
My math was likely wrong.
Your math as long as it goes is correct, but the advantage of a hydraulic system lies not the in length of the handle but the the difference is volume of fluid pumped each time. A 2 ton bottle jack wouldn't work at all using your math. In addition to the mechanical advantage of the handle there is also the mechanical (fluid) advantage of pumping a large amount of fluid to move a small amount of fluid and moving the handle a large number of times to accomplish this.
ljg
The handle length make's a
large difference in the mechanical advantage.
The amount of fluid pumped each time has nothing to do with MA. The amount is determined by the swept volume of piston stroke, and two pistons with the same MA can have different strokes and hence different swept volumes. It's the relative areas of the two pistons (the large ram that you see extending from the jack body and the small piston that the jack handle is connected to) that the hydraulic fluid acts upon that determines the MA there. In a bottle jack, you have two separate MA that when multiplied by each other give the total overall MA.
I think that validates my position. Two seconds would be 1/8 hp and about 4 pumps per second. Pretty fast, but I stated my max requirements. Less than 2 inches movement at a 200 lb. requirement would be more normal.
You did state in Post#5 "I need to move a ram about 3 inches with 500-700 lbs of force within about 1 second no faster." and "The diameter of the cylinder less than 1/2 inch perhaps."
I may try it because the bottle jack is a closed system, basically just take the ram out, weld up the hole left and drill and tap a hole for the fitting in the inner and outer reservoir. And a clearance hole for the inner reservoir fitting.
And while the $50.00 for the Northern Tool pump is doable, $20.00 for a bottle jack and a couple of fittings is better.
By the way a bottle jack is a pump with a SA piston. There are two holes in the cast iron bottom. One opens into the inner cylinder and one opens into the outer cylinder. So the pump which is the base is basically a manifold pump.
ljg
There is only one reservoir. That's the "outer" one. The "Inner" one is the high pressure area of the ram. They are separated by a valve that allows the oil to bypass the piston pump when it's open and allows the jack to be retracted.
All I can say now is buy your bottle jack and modify it however you want. I'm sure it will be an interesting learning experience for you. Good luck.
P.S. Don't think I am busting your chops here, it's just that I see you are under some "misconceptions" and possible lack a bit of understanding of basic hydraulics. Hopefully you will try to rectify both of those conditions and have success with your endeavor.
