Cylinder Speeds

The argument is as old as time. The roman mechanic of some far distant time trying to explain to the gallant warrior that the horse and chariot cannot go any faster and that they are doing everything they were designed to do. Of course today we do not have to worry as much about being beheaded for passing on this type information. None the less the question of whether or not a particular piece of machinery or component is operating at full capacity or not is often argued over to no end. Hopefully this article will put some of this to rest.

In dealing with Hydraulic cylinders the formula is very straight forward. Unfortunately the many other factors such as wear and friction are not taken into consideration. Of course the reason being the amount of either of these factors change from application to application. Even with these factors having their effect on the outcome we can often obtain to within 10 percent of the designed speed of a cylinder. There are many good tables out there that you can refer to or you can simply write down the formulas you use most often on the back of your clipboard and then use a pocket calculator to quickly decide where the problem if any is.

Let’s consider now the formula for basic cylinder speed.

S=V/A Where S is the speed of the cylinder, V is the volume of oil entering the cylinder to act upon the surface that is at a right angle to the direction of movement, and A is the area of that right angle. Therefore the speed of the cylinder should be directly proportional to the volume of oil entering that cylinder given the amount of right angle area. We must of course convert the oil gpm to cu in per minute. Simply take the volume in GPM and multiply it by 231 to get the cubic inches per minute since there are 231 CI in a gallon.

If you take a standard eject cylinder for example, say out of a 250 ton hydraulic press. Common bore would be 3.25 inches in diameter and a 2 inch rod. The pump that is driving this is on print an 8gpm fixed volume pump with a pressure relief valve setting of 2000 psi and a pressure reducing valve setting of 1500 psi at the eject stack or manifold. Just because the print says it is an 8 gpm make sure someone has not replaced it with the wrong pump size. This test is easiest with the eject stroking full stroke. So let’s say this stroke is 6 inches. We will examine the forward stroke first. So we have 8 gpm acting on a 3.25 diameter piston and moving it 6 inches. The question is how long this take should.

1. 1 gpm = 231 cubic inches so 8 gpm = 1848 cu in per minute or 30.8 cu in per second
2. Area of a circle is R x R x PI , so a 3.25” piston has 8.3 sq in.
Where R is equal to half the diameter and PI is equal to 3.142
3. The area at full stroke we must fill is 6” * 8.3= 49.8 cu inches
4. If you divide the available 30.8 gps into the area required to fill of 49.8 cuin you find that you need 1.6 seconds to go full stroke.

The pressure setting only comes into question if the amount of resistance to movement of the eject plate is greater than the cracking pressure of the pressure control valve with the lowest setting, in this case it would be the pressure reducing valve set at 1500 psi and probably cracking at around 1450 or so to start to allow the oil to return to tank. This is usually not of concern but if you have a large mold and cannot obtain max eject speed the pressure needed to move the eject at the desired speed must be measured and if it exceeds the lowest pressure setting then all the GPM you could divert to the circuit within reason is not going to move the cylinder any faster. You can exceed the volume of oil through a relief valve but if the circuit is designed properly it is not likely.

The Speed of the retract is the same formula but since we have less area to act upon the speed should be proportionally greater. The area that was 8.3 sq in is only 5.2 and the required cubic inches of oil to move it back approx. 30.9 cu in. So on the retract we should be down around 1 second.

There is another factor that must be mentioned and that is regenerative circuits. While I intend on making this a whole other article on the advantages and uses of regeneration, it must be taken into consideration. If you have the capability of switching the circuit in question into a regen state, then the oil from the rod side of the cylinder is blended in with the oil traveling to the blank side of the cylinder and the volume of oil is directly applied to an area equivalent of the rod itself on the blank side of the piston. Since the area the oil is applied to is smaller the cylinder will travel at a faster rate if the pressure needed to move the device does not exceed the pressure available. This is accomplished because the two sides of the piston have the same oil acting upon their right angle surfaces. Therefore the resulting force and resulting velocity is only proportional to the area of the rod. Take the square area of the rod and use it in your calculations to obtain the speed that is possible. But remember available force is also greatly reduced since the area is reduced.

The formula for FORCE is AREA*PSI .

On older machines with hydraulic clamps where a “jack ram” is employed for fast forward traverse, Cincinnati Milacron used the oil from the jack ram during the opening stroke to blend in with the oil going to the opening lands to produce faster open speeds.

REMEMBER take mechanical wear and friction into consideration. An old cylinder with leaky piston rings will not travel as fast as a newly rebuilt one. Also if the guide rods are misaligned that the plate travels on or are in need of grease you may get slow travel speeds.

These formulas work equally well for any part of your machine as long as you remember to consider the friction, mechanical wear, and pressure needed to move the device.

Follow all manufacturer recommendations on safety when doing these tests. For example do not reach under the guards to measure travel while the press is in operation.