Hydraulic Motor/Pump Question

[ebme96]

I have an old 52" grasshopper mower that the engine is shot, and the hydros are weak. I checked into new hydros and OUCH, $1,500/peice. So I thought that if I could find a pump and wheel motors I could build my own system. (I have built several go-karts from old mowers so this doesn't seem out of line) I found a pump on surplus center, and some wheel motors (see links below) I am looking to power it with a 24-27 hp engine. By my figures, this should give me a top speed of 13.4 mph. Anyway, to my question. Is there any reason you hydraulic gurus see that this setup wouldn't work? Am I missing anything? (obviously I will have to use many more parts, resevoir and such) Will this pump and wheel motor set-up handle the load I am imposing on it?

So if anyone can help me out I would be glad to hear from you.......

wheel motors (6.1 CU.IN. | Wheel Motors / Wheel Hubs - Quadra Trading Corp)
pump (0.43 cu in HYDROSTATIC KANZAKI DUP-7A TANDEM PUMP)

(Yea, I know I could just buy a new mower, but that doesn't have the cool factor of "I built it myself" and did you check prices recently on a commercial grade mower?)

 

[J_J]

One pump requires about 6 HP. X2 - 12 HP

Your motors will turn at 227 rpm. torque is 2,9i4 in lbs or 76 ft lbs

Will that 227 rpm give you the speed you want?

You have enough HP for a larger pump.

The motor you selected, will run on 20 GPM.

GRASSHOPPER LAWN MOWER HYDRO PUMPS

 

[ebme96]

If my calculations are right, 225 rpm with a 20" diameter tire should give me 13.4 mph (approx.). And as far as the torque, anybody have any idea if that is enough? I know the engine I was looking at is rated at 35 ft-lbs but obviously that is a much higher rpm. I am totally green on this torque/horsepower/rpm stuff...... (other than my go-kart/garden tractor with a 10hp motor and bigger pulley flat moves)

On the capacity for a bigger pump, I am also going to be running a deck (has 52" thinking about a 61" or 72" deck in the future) so I definitely want to keep enough power for that.

Basically being my first project involving hydraulics I want to make sure there are no obvious mistakes in my product choices.

 

[oldnslo]

Vehicle drive fomulas:

travel speed: MPH = RPM x wheel dia. inches / 336

Axle torque in inch lbs: T = F x r
T = troque inch lbs
F = force or draw bar pull in lbs
r = whee radius in inches

Grade resistance: F = GR x W
GR= grade resistance in percent (20% is written .20)
w = vehicle gross weight

Most lawn mowers use very little power for travel since they are typically rather light and the power is being used for running the mower deck.

 

[Cpjlube]

I have a Woods 5215 (grasshopper 721) diesel. The hydros are getting weak on mine also. New hydros are 2500 each for mine. The motor is dead reliable and I think it can be adapted to new pumps and motors.
I looked at new pump / wheel motor style ztrs. The larger machines use bigger pumps. Most use two pumps. Each turning a wheel motor. I have been looking at Surplus center to try and get a pump/motor combo that will move my machine properly.
From my amateur and quick calculations you will need to run two pumps that are at least 1 cu in ($120 each) And wheel motors that are around 14 cu in. ($180 each). Both are available in the catalog.
Getting the pumps and wheel motors mounted will require additional pulleys and hoses and mounting brackets.
This is a big heavy mower with a 60 or 72 inch deck. I think this combo is more expensive but much more durable.
Of course I could be wrong and your solution might work perfectly. I hope so. It will be much cheaper if I can use your idea. Be sure to post pics. I'm already planning a winter rebuild.

 

[ebme96]

Thanks alot to all who have replied! oldnslo those formulas have been stowed in my "need to remember folder."

Cpjlube: after looking and figuring last night, I am leaning heavily to the dual pump set up with 1.1 cu.in. pumps from surplus center, and some larger wheel motors in the 12-14 cu.in. size. That will give alot more torque for a little more money. I have a neighbor that has a full machine and weld shop so the fab work most likely will be done partly by him. I think this setup will allow me to put a snow blade on it (for light use).

Pictures, I def. will try to post pics! I am just tearing down the mower right now, gonna spray a little paint on the frame where it is chipped, and rewire the whole thing. (did I mention my dad is an electrician)

And if anyone has any more comments/advice, chime in! I will be glad to hear it even if you tell me I am all wrong and my setup is a failure waiting to happen.......

Merle

 

[oldnslo]

ebme96,
I don't have a clue what your mower weighs in at but figuring 1800 lbs of tractive effort, 20" diameter wheels and 2500 PSI system pressure I came up with requiring 23 cu.in motors and 1.1 cu.in pumps. These motors seem rather large so please double check my numbers. I have made other mistakes today....

 

[Cpjlube]

ebme96, I don't have a clue what your mower weighs in at but figuring 1800 lbs of tractive effort, 20" diameter wheels and 2500 PSI system pressure I came up with requiring 23 cu.in motors and 1.1 cu.in pumps. These motors seem rather large so please double check my numbers. I have made other mistakes today....

The diesels w 60 decks are around 1500. What formulas were you using to derive your numbers? I'm still learning about hydraulic motors. Any insight is appreciated. I was using torque of the current hydros (eaton 771) and trying to match requirements to available motors in the Surplus Center Catalog. I think a 14-15 cu in will work.

I'm just trying to wrap my mind around how to match up pumps and wheel motors and then getting the drive pulleys on the engine and the pumps to match up properly.

 

[oldnslo]

Cpjlube
I used 1800 lbs x 10 inch wheel radius to arrive at 18,000 in lbs torque required to propel this mower.

divide the 18000 in lbs by 2 to get 9,000 in lbs required per drive wheel.

Hydraulic motor size formula:
Torque in ft lbs = displacement (CIR) x pressure drop across the motor x mechanical & hydraulic efficiency ( I typically use between .8 & .9) divided by 75.4

9,000 in lbs dived by 12 = 750 ft lbs

750 x 75.4 divided by (2500 x .9) = 25.13 CIR (Cubic Inch / Revolution)

still seems like a large motor for that load but these are the formulas that I have used for figuring propulsion systems on equipment

 

[ebme96]

I am going to make a quick guess that the mower with operator will weigh in at approx. 1200 (that is shooting high). Both the pump and the motors say they can handle 3500 psi int. if I do this build I will probably put a pressure gauge in the system just 'cause I wanna know what pressure I am pushing. (drive a turned-up diesel pickup, and love gauges)

Another question, if I do this should I be adding any relief valves? (I just now see the pump has this built into it, is this enough?)

Here are the links to the pumps/motors I am considering.
Motors: 12.2 CU.IN. | Wheel Motors / Wheel Hubs - Quadra Trading Corp
Pumps: 1.10 cu in WHITE HYD PISTON PUMP 101018000012

oldnslo, So the force/draw bar pull in your formula is the weight of the machine? It would seem to me that is overkill but...... Because when you move something on wheels you are at an advantage.......(I am green to all of this so.... :ashamed