Understanding cylinder speeds

Editor's note: Woody Woodrell is president of Woodrell Project Management, an injection molding consulting firm based in Anadarko, OK. You can reach him on the Internet at www.woodrell.com. This column is his fourth in a series of troubleshooting articles.

The argument is as old as time: The Roman mechanic of some distant age tried to explain to the gallant warrior that the horse-drawn chariot cannot go any faster and that it's doing everything it's designed to do. Today, the question of whether or not a particular piece of machinery or component is operating at full capacity is often argued to no end. Hopefully this article will help put such arguments to rest.

Cylinder Speed Formulas

How do you know if you're getting all you can out of your hydraulic cylinder? For hydraulic cylinders, the formula for calculating speed is straightforward. Unfortunately, factors such as wear and friction are not taken into consideration, mainly because these factors vary from application to application. Even with such factors in effect, most molders can often obtain the designed speed of a cylinder to within 10 percent. There are many good tables to which you can refer, or you can simply write down the formulas you use most often on the back of your clipboard and then use a calculator to quickly decide where the problem 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. Remember, you must convert the oil gal/ minute to cubic inches/ minute. Simply take the volume in gal/minute and multiply it by 231 to get the cubic inches/minute (there are 231 cu inches in a gallon).

Take a standard eject cylinder, for example, out of a 250-ton hydraulic press. Common bore diameter is 3.25 inches with a 2-inch rod. Check the machine prints for the capacity of the pump driving this rod. The standard is an 8-gal/minute fixed volume unit with a pressure relief valve setting of 2000 psi and a pressure reducing valve setting of 1500 psi at the eject stack or manifold.

Though 8 gal/minute may be the standard, be careful to make sure someone has not replaced the pump with another that is the wrong size. This test is easiest with the full eject stroke. For our purposes, let's say this stroke is 6 inches; we will examine the forward stroke first. So we have an 8-gal/minute pump acting on a 3.25-inch- diameter piston and moving it 6 inches. The question is how long should this take? These calculations show us.

The pressure setting only comes into question if the resistance to movement of the eject plate is greater than the cracking pressure of the pressure control valve with the lowest setting. In our example, it would be the pressure reducing valve set at 1500 psi, probably cracking at around 1450 psi to start to allow the oil to return to the tank. This is usually not a concern, but if you have a large mold and cannot obtain maximum eject speed, the pressure needed to move the ejector at the desired speed must be measured.

If it exceeds the lowest pressure setting, then diverting all the gal/minute you can 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 this is unlikely.

The speed of retraction is calculated with the same formula, but since there is less area to act upon, the speed should be proportionally greater. The area that was 8.3 sq inches is only 5.2 sq inches on the rod side of the piston. The required volume of oil to move it back is about 30.9 cu inches. So the retract time should be down about a second.

Regenerative Circuits

There is another factor to consider: regenerative circuits. If you have the ability to switch the circuit in question to a regenerative state, the oil from the rod side of the cylinder is blended with the oil traveling to the blank side of the cylinder. The volume of oil is directly applied to an area equivalent to the volume of the rod itself on the blank side of the piston. Since the area the oil is applied to is smaller, the cylinder travels faster if the pressure needed to move the device does not exceed the pressure available. This is accomplished because both sides of the piston have the same oil acting upon their right angle surfaces.

The resulting force and velocity is only proportional to the area of the rod. Take the 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 times psi.

Some older machines with hydraulic clamps where a jack ram is employed for fast forward traverse use 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.

But don't forget to take mechanical wear and friction into consideration when checking capacity. An old cylinder with leaky piston rings will not travel as fast as a newly rebuilt one. Also, if the guide rods on which the plate travels are misaligned, or 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.

Remember to follow manufacturer recommendations on safety when testing. For example, never reach under the guards to measure travel while the press is in operation.