thinking about trying this
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Farmerford
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On my PTO pump driven mower, I wanted the mower and the tractor engine to turn at the same RPM as when the mower was on the TPH. That required that the displacement per revolution of the PTO pump and the mower hydraulic motor be the same; actually, the pump should be 5-10% larger than the motor to make up for pump and motor leakage (the leakage is not that much when pump and motor are new, but it increases as they age).
So if the displacements are to be the same (or nearly the same), the next step is to determine the displacement needed to transmit the power available. I am working with rough figures and doing the math in my head, so the following figures are only approximate; precise formulae are available in handbooks or on line. A 6CI/REV gear pump at say 2,000PSI takes about 19HP to drive at 540RPM and produces about 13GPM , which in turn results in about 16HP being applied to the input shaft of the mower gearbox.
So, if your tractor produces 19PTO HP at 540PTO RPM, a 6CI pump will handle all the power that can be produced. A larger pump and motor will not provide any more power to the mower; and a smaller pump and motor will not transmit the full PTO power. And a larger pump and motor will increase the GPM that flows through the system, which in turn increases the requirements for hose and fitting sizes, quick disconnect sizes, and reservoir size. So nothing is to be gained by increasing the pump and motor size.
If your tractor produces more power than required to drive the mower, you can achieve a relative RPM reduction in the tractor engine by increasing the size of the pump relative to the hydraulic motor. In the above example, if your tractor produces say 38PTO HP at 540PTO RPM, but you only need the same 16HP to the mower input shaft, you could increase the pump size to 12CI/REV, run the engine at 1/2 standard speed, get 19HP at the PTO, and still produce the same 13GPM needed to drive the mower (I know that HP is not necessarily linearly related to engine RPM, I am just assuming that for simplicity).
So if the displacements are to be the same (or nearly the same), the next step is to determine the displacement needed to transmit the power available. I am working with rough figures and doing the math in my head, so the following figures are only approximate; precise formulae are available in handbooks or on line. A 6CI/REV gear pump at say 2,000PSI takes about 19HP to drive at 540RPM and produces about 13GPM , which in turn results in about 16HP being applied to the input shaft of the mower gearbox.
So, if your tractor produces 19PTO HP at 540PTO RPM, a 6CI pump will handle all the power that can be produced. A larger pump and motor will not provide any more power to the mower; and a smaller pump and motor will not transmit the full PTO power. And a larger pump and motor will increase the GPM that flows through the system, which in turn increases the requirements for hose and fitting sizes, quick disconnect sizes, and reservoir size. So nothing is to be gained by increasing the pump and motor size.
If your tractor produces more power than required to drive the mower, you can achieve a relative RPM reduction in the tractor engine by increasing the size of the pump relative to the hydraulic motor. In the above example, if your tractor produces say 38PTO HP at 540PTO RPM, but you only need the same 16HP to the mower input shaft, you could increase the pump size to 12CI/REV, run the engine at 1/2 standard speed, get 19HP at the PTO, and still produce the same 13GPM needed to drive the mower (I know that HP is not necessarily linearly related to engine RPM, I am just assuming that for simplicity).
Xfaxman
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Have you checked here for a motor? https://www.surpluscenter.com/home.asp
mj007
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Yeah - they do :shocked: There seems to be some good info here though - you're just gonna have to 'winnow the wheat from the chafe'
mj007
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RE: post & pic by XxLT250RxX
So, you set the angle and then run the loader in 'Float' or ........ ?
So, you set the angle and then run the loader in 'Float' or ........ ?
Last edited:
classacthollow
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I am also interested in doing the same thing, I am purchasing an LS 4047H and adding an additional set of remotes for a front mount grapple. my tractor specs. show a 8.2 gpm implement flow. I went on to surplus outlet and they have several pumps but I don't understand the formula to figure out the correct Hydraulic motor. could you advice me? thanks, Bruce
Farmerford
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Bruce:
I am sensitive to DT86's comment that "You guys sure know how to crush a mans dreams", but think understanding the pitfalls may save you time and money in the long run.
I don’t know the pressure of your hydraulic system, but assume it is 2,250psig. The rule of thumb is that 1gpm @1714psig=1HP. So 8.2gpm x 2250psig/1714 = 10.76HP; that is HP available in the fluid as it leaves the pump. Several things will reduce the horsepower ultimately produced at the blade shaft: hose, fitting, and valve losses between the pump and the tank; leakage in the motor; friction losses in the motor and gear box. Those losses can be estimated fairly precisely, but it would not be unreasonable to assume a total of 20%. So you will have perhaps 8HP to turn the blade holder. If you use a small, lawn mower type blade, perhaps 36” long and limit your cutting to grass, weeds, and very light brush, in each case not very tall or thick, it might work.
The machine I made has a 42” cut and 14HP at the blade shaft; it works fine for my purposes, which are cutting pond banks, ditches, between trees, etc. where you go forward a few feet and then back up and move over for another bite, but it would not be a very good pasture mower. We have about 80 ac just outside the city with a 3 acre pond, several small intermittent streams, and 15 ac of woods that are just thick enough with desirable mature trees that it is difficult to maneuver a rear mounted mower without damaging the trees. So our FEL mower gets used most weekends, and saves a tremendous amount of labor. It is clearly worth the time, money, and effort it has taken to get it functional (not to say even close to perfect).
Another consideration is the effect on your other hydraulic systems. I assume, but do not know, that the 8.6gpm fluid flow also powers the front end loader and three point hitch. Most motors tolerate only limited backpressure in the exhaust port unless you have a case drain line to the tank. So the motor must either be the last device in the hydraulic circuit (so that nothing downstream can block flow and build up pressure in the exhaust port) or it must have a case drain connected to the tank.
But there is another issue that even the case drain does not solve: suppose the motor is supplied from the power beyond port in the FEL valve. If the motor valve is activated so that fluid is flowing through the motor, and you then activate the boom spool to raise the TPH, you will stop the flow from PYB to the motor, but the momentum of the motor, gear box, and blades will keep turning the motor, causing it to cavitate; that may not happen often enough to damage the motor, but it is hard to predict. The motor will also slow down because no pressurized fluid is driving it; if at the same time you are in heavy cutting the blades may quickly slow down. It is difficult to predict the extent of these problems; they may be merely annoying or they may seriously impair the usefulness of the machine.
These problems are to a large extent curable with check valves, pressure relief valves, and the associated plumbing. But they add to the cost and complexity of the system. If you plan to use the mower regularly, you should consider a PTO hydraulic pump with a separate reservoir. That will probably add $1,500 to the cost.
It will be difficult to start with the tractor hydraulic system powered arrangement to determine if it works well enough for you and then shift to PTO pump power if the former is unsatisfactory. With 8.2 gpm supply, to obtain from the motor the typical 540 rpm used to drive a mower gear box, the motor will need to be about 3.5 cu in/rev ( Motor cu in = 8.2 x 231/540). That should produce the roughly 8HP at the blades discussed above. If you then decide that you must have more power and use a PTO pump, the 3.5 cu in/rev motor will not work. In order to produce more power you must have more hydraulic flow (assuming pressure stays the same), but that increased flow will turn the motor faster, which will over speed the blades.
Suppose you select a PTO pump that produces 20gpm. Now the motor will need to be 20 x 231/540 = 8.5 cu in/rev. Your initial 3.5 cu in/rev motor will not be usable in the system. And you will probably have to upgrade from ½” hydraulic lines and fittings to ¾” lines and fittings.
I have made some changes (improvements?) to mine in the past few years and will try to take some pictures while I am using it this weekend.
I am sensitive to DT86's comment that "You guys sure know how to crush a mans dreams", but think understanding the pitfalls may save you time and money in the long run.
I don’t know the pressure of your hydraulic system, but assume it is 2,250psig. The rule of thumb is that 1gpm @1714psig=1HP. So 8.2gpm x 2250psig/1714 = 10.76HP; that is HP available in the fluid as it leaves the pump. Several things will reduce the horsepower ultimately produced at the blade shaft: hose, fitting, and valve losses between the pump and the tank; leakage in the motor; friction losses in the motor and gear box. Those losses can be estimated fairly precisely, but it would not be unreasonable to assume a total of 20%. So you will have perhaps 8HP to turn the blade holder. If you use a small, lawn mower type blade, perhaps 36” long and limit your cutting to grass, weeds, and very light brush, in each case not very tall or thick, it might work.
The machine I made has a 42” cut and 14HP at the blade shaft; it works fine for my purposes, which are cutting pond banks, ditches, between trees, etc. where you go forward a few feet and then back up and move over for another bite, but it would not be a very good pasture mower. We have about 80 ac just outside the city with a 3 acre pond, several small intermittent streams, and 15 ac of woods that are just thick enough with desirable mature trees that it is difficult to maneuver a rear mounted mower without damaging the trees. So our FEL mower gets used most weekends, and saves a tremendous amount of labor. It is clearly worth the time, money, and effort it has taken to get it functional (not to say even close to perfect).
Another consideration is the effect on your other hydraulic systems. I assume, but do not know, that the 8.6gpm fluid flow also powers the front end loader and three point hitch. Most motors tolerate only limited backpressure in the exhaust port unless you have a case drain line to the tank. So the motor must either be the last device in the hydraulic circuit (so that nothing downstream can block flow and build up pressure in the exhaust port) or it must have a case drain connected to the tank.
But there is another issue that even the case drain does not solve: suppose the motor is supplied from the power beyond port in the FEL valve. If the motor valve is activated so that fluid is flowing through the motor, and you then activate the boom spool to raise the TPH, you will stop the flow from PYB to the motor, but the momentum of the motor, gear box, and blades will keep turning the motor, causing it to cavitate; that may not happen often enough to damage the motor, but it is hard to predict. The motor will also slow down because no pressurized fluid is driving it; if at the same time you are in heavy cutting the blades may quickly slow down. It is difficult to predict the extent of these problems; they may be merely annoying or they may seriously impair the usefulness of the machine.
These problems are to a large extent curable with check valves, pressure relief valves, and the associated plumbing. But they add to the cost and complexity of the system. If you plan to use the mower regularly, you should consider a PTO hydraulic pump with a separate reservoir. That will probably add $1,500 to the cost.
It will be difficult to start with the tractor hydraulic system powered arrangement to determine if it works well enough for you and then shift to PTO pump power if the former is unsatisfactory. With 8.2 gpm supply, to obtain from the motor the typical 540 rpm used to drive a mower gear box, the motor will need to be about 3.5 cu in/rev ( Motor cu in = 8.2 x 231/540). That should produce the roughly 8HP at the blades discussed above. If you then decide that you must have more power and use a PTO pump, the 3.5 cu in/rev motor will not work. In order to produce more power you must have more hydraulic flow (assuming pressure stays the same), but that increased flow will turn the motor faster, which will over speed the blades.
Suppose you select a PTO pump that produces 20gpm. Now the motor will need to be 20 x 231/540 = 8.5 cu in/rev. Your initial 3.5 cu in/rev motor will not be usable in the system. And you will probably have to upgrade from ½” hydraulic lines and fittings to ¾” lines and fittings.
I have made some changes (improvements?) to mine in the past few years and will try to take some pictures while I am using it this weekend.