Tonight we're going to go over, very briefly, how much force we can get out of a hydraulic cylinder.

For now, we will focus on the type of cylinder I'm using which is a closed center double-ended cylinder - in other words, as one end of the rod is going into the cylinder's body, the other is pushing out at the same rate. Later on, we will breifly discuss a single-ended hydraulic cylinder and how it differs from a double-ended.

So on with the show!

Tonight's discussion will of course need a diagram so here you go

D(b) = bore diameter

D(r) = rod diameter

Going back to ye ol physics book, we are reminded of the equation

Pressure = Force / Area

or simply put: P = F/A

Rewriting this equation to solve for Force (since thats what we're interested in), we get F = P*A .. or Force equals Pressure multiplied by the Area.

Pressure is a defined number given by the hydraulic pump -- which unfortunately we don't know the exact figure but we can guestimate

. The only term we need then is Area. So of course Area = Pi * Radius^2 .. or.. in terms of diameters, Area = Pi * (Diameter/2)^2 . Plugging this Area equation back into the Pressure equation, we get

F = P * Pi * (Diameter/2)^2

Well from here we know everything so its time to plug in the numbers and finally see how much force we get out of this cylinder.

But wait, we forgot to take into consideration that the fluid does not push in the same direction on the cylinder's rod. The Force vector of the fluid pushing on the rod is 90 degrees from the Force vector of the fluid pushing on the bore. What this means is that the Force of the fluid pushing on the rod does NOTHING for us. So what we must do is subtract the cross-sectional area of the rod from the area of the area of the bore .. or A(b) - A(r) equals the new area, where A(b) equals the Area of the bore and A(r) equals the cross-sectional area of the rod.

So going back to the fundamental equation for Force:

F = P * A

We make a subsitution:

F = P * ( A(b) - A(r) )

And fill in the A's with the area equation in terms of Pi and the Diameters D(b) and D(r)

F = P * ( Pi * (D(b)/2)^2 - Pi * (D(r)/2)^2 )

Pulling out the common Pi to simplify the equation:

F = P * Pi * ( (D(b)/2)^2 - (D(r)/2)^2 )

Well then there we have it.

Now comes the part -- guestimating what pressure the hydraulic pump is rated at. I did a few minutes of searching online and I think most stock automobile pumps are around ~1,500 PSI.

With the cylinder I'm using, the bore is 2 inches and the rod is 1.25 inches (rather thick). So lets see

F = P * ( (D(b)/2)^2 - (D(r)/2)^2 )

F = 1500 PSI * 3.14 * ( (2/2)^2 inches - (1.25/2)^2 inches )

F = 1500 PSI * 3.14 * ( 1^2 inches - 0.625^2 inches)

F = 1500 PSI * 3.14 * ( 1 squared inches - 0.39 squared inches)

F = 1500 PSI * 3.14 * 0.61 squared inches

Subsituting pounds per square inch for "PSI":

F = 1500 pounds/squared inches * 3.14 * 0.61 squared inches

The "squared inches" of course cancels out leaving us with

F = 1500 pounds * 3.14 * 0.61

F = 2,873 pounds

So, assuming the hydraulic pump can flow enough gallons per minute at 1,500 PSI, then the hydraulic cylinder should be pushing just a little over 1 ton (not the "short ton", but 2,204 pounds).

So how does a single-ended cylinder differ? Well logically, since such a cylinder only has one rod, then the other side has a full bore area and therefore we wouldnt need to subtract the rod's cross-sectional area from the bore area -- which would give us more Force output.

Lets see what type of Force we would get if I was running a single-ended cylinder.

F = P * Pi * (Diameter/2)^2

F = 1500 PSI * 3.14 * (2/2)^2

F = 1500 PSI * 3.14 * 1^2

F = 1500 PSI * 3.14 * 1

F = 4,710 pounds

A difference of 1,837 pounds -- pretty significant!

Of course this extra 1,837 pounds of force is available only on extension when the hydraulic cylinder is extending. On compression, when the hydraulic cylinder is getting shorter, the amount of force is still 2,873 pounds because of the rod.