Good evening gents – I’m new to this forum (came across it when searching for something else) but I believe I can throw a little more light on the issues – apologies beforehand for a long post.
This seemingly simple project is actually quite problematic. Firstly you have to recognise that as long as the number of layers of rope on the winch drum doesn't change, that there are no other pulleys or sheaves between the drum and the thing to which the rope is attached, and that the drum has no friction …. then a "constant tension" in the rope equates to a “constant torque” on the drive to the drum. These aren’t realistic scenarios but just bear with me for a while. The control strategy now boils down to one of “constant torque”.
Secondly, you need to pay particular attention to the characteristics of the device which is converting hydraulic pressure to mechanical torque and vice-versa, i.e., the “hydraulic motor”. The wording is unfortunate in this instance because that component with “motor” engraved on its label will sometimes be working as a motor: “you push oil in and make the shaft turn against some mechanical resistance” and it will sometimes be working as a pump: “you turn the shaft and make it push oil out against some hydraulic resistance”. But syntax aside, let’s keep calling it the motor.
Imagine for a moment that the tension in the rope was 1500 newtons, the drum had a diameter of 400 millimetres, the rope was infinitesimally thin and we were on the first layer. That 1500 N force would be acting at a radius of 0.2 metres and the resultant torque on the drum would be 300 Nm.
Now imagine that there was no friction in the drum bearings and we had a direct drive to a perfect hydraulic motor (and no gearbox). Suppose we wanted to haul in the rope with a motor differential pressure of just 100 bar, then the [theoretical] motor size would be worked out like this…
A perfect 63 cc/rev motor (actually 20 x pi cc/rev) would deliver 1 Nm per bar. We want 300 Nm from 100 bar so our motor needs to deliver "3 Nm per bar" which means it needs to be 3 times bigger than the one delivering just "1 Nm per bar". Answer = 3 x 63 = 189 cc/rev.
In practice, the actual output torque is less because of various pressure drops and friction losses in a real motor - so we can describe the motor as having a “hydro-mechanical efficiency” of, say, 92% ... which means we would only get 276 Nm for our 100 bar.
When the same “motor” is running as a pump because the rope is pulling the drum round, we should expect that a [theoretical] torque of 300 Nm would be needed to turn the shaft if the pressure difference across the ports was just 100 bar, but the hydro-mechanical losses in the [imperfect] device mean that we have to put in even more torque. It is fair to assume that the efficiency when working as a pump will be the same as the efficiency when working as a motor so our drive torque of 300 Nm will only generate a pressure of 92 bar (92% of 100 bar). If we had to get 100 bar out of the “pump” before the shaft could turn then the input torque would have to be higher, i.e., ~ 326 Nm
This is the result: if we could design a hydraulic circuit that held the difference between the motor ports at exactly 100 bar (regardless of how this was achieved) then the maximum tension when the winch was hauling in the rope would be 1380 newtons but we would need that tension to rise to 1630 newtons before we could pull the rope back off the drum. So it’s not really “constant tension” at all. Incidentally the ratio of these two numbers is the square of the efficiency: 0.92² = 0.846, 1380/1630 = 0.846.
Now let’s consider the effect of some of the things we previously ignored: There is friction in the drum bearings which means that some of the drive effort we put in when hauling isn’t converted into rope tension (tension is lower than expected) and the friction also means that, when rendering, some of the rope tension is lost to the friction in the bearings so you have to pull even harder to get the motor shaft to turn (tension is higher than expected). In fact, anything which decreases the efficiency of the winch when hauling will work the other way when rendering – increasing the gap between the two tension figures.
What can you do about it? If you want to take the “constant pressure” solution then you must maximise the efficiency of the winch:
1) Make sure the drum bearings are low friction.
2) Use a good quality radial piston motor if you can because these have a high hydro-mechanical efficiency and a very good starting efficiency (if the rope is changing direction then you are continually ‘starting’ the motor)
3) Don’t use a gearbox, increase the size of the motor instead and run it at low speed (radial piston motors are low speed devices)
4) Restrict the number of pulleys and sheaves between the drum and the load (or, if you must have them, make sure they have good quality bearings)
5) Make sure the drum brake doesn’t drag at all.
6) Be generous with the hydraulic pipework sizes between the pressure controlling valve(s) and the motor port and similarly on the outlet/suction side of the motor and the compensator.
7) Use a pressure control valve (or valves) with very little difference in pressure between the reducing function and the relieving function.
8) Size the valve(s) generously to minimise the pressure underride (apparent loss of pressure when hauling at full speed) and the pressure override (apparent increase in pressure when rendering at full speed).
If you want to allow the electrical boys some input then don’t bother trying to get them to hold the pressure constant (using some form of pressure transducer and closed loop control of an electro-hydraulic pressure modulating valve) because, as explained above, perfect control of pressure does not result in perfect control of tension. So get them to fit some sort of load cell and measure rope tension directly – then use the closed loop control system to modulate the pressure in order to keep the tension constant. (Note that this scheme will also compensate for changes in the number of layers on the drum ... but for the most precise control avoid changing layers while the system is active.)
Other issues to bear in mind with your hydraulic circuit:
A) The hydro-mechanical efficiency is a function of motor speed. The figure is lower at very low speeds (when poor hydro-dynamic lubrication regimes apply) and also lower at high speeds (when churning losses, flow related pressure losses and mechanical friction losses predominate). The efficiency is best at speeds about the middle of the motor’s speed range – so picking a low speed motor for a low speed application is a good choice.
B) Some motors need a boosted inlet pressure to allow them to work as pumps. The value of this pressure depends on the speed and any pressure on the inlet port (when pumping) will affect the available pressure differential – there may be an advantage in connecting the drain lines of the pressure control valves to the low pressure port of the motor rather than back to “tank”. If you can’t get a figure from the manufacturer then measure the inlet pressure needed to run the [bare shaft] motor at the maximum speed you want in service and use half of this figure as the boost pressure.
C) Most hydraulic winches will have some form of counterbalance valve on the motor port(s) – you need to find a way of piloting this valve open or circumventing its operation so that a constant tension function can be brought into play.
Sorry for such a long and detailed contribution – hope you made it to the end.
DOL
