Hydraulic motor

[jbelokur]

Here is a link to a hydraulic motor that I have been looking at. (White Drive Products)

It says that it produces a max torque of 23,550lb/in at 3,000 psi. How does this translate to real world work power? If I attached this motor to one end of a say 6 ft lever which weighs say 250# how much weight could this thing move around if I hung additional weight at the end of the opposite end of the lever. (See attached file for a simple illustration)
Also what would I need to run this motor? Say this system runs 2 of these motors listed above, and four other smaller motors. These four other smaller motors would be paired off, ie two motors working in tandem. I am talking everything from what kind of engine would be required, to hydraulic hoses, to possible computerized controls (something that I could write a program for and it would execute.

Also, with a motor like this, if you use two of them in tandem, is their net force potential greater than their individual potential forces combined?

Does anyone know of any other hydraulic motors that are more powerful?

 

[dfkrug]

It says that it produces a max torque of 23,550lb/in at 3,000 psi. How does this translate to real world work power? If I attached this motor to one end of a say 6 ft lever which weighs say 250# how much weight could this thing move around if I hung additional weight at the end of the opposite end of the lever.

First, do not confuse force with power. Power is consumed or dissipated
when work is done, while forces can be exerted with no motion at all. No
motion means no work done, so no power used. Your 250# weight exerts
250 lb of vertical force, or 250 x 6 = 1500 ft-lbs of torque (force) about
your pivot, but if it does not turn, no power.

In designing your hydraulic system, you need to not only determine the
max force you need to deliver to the object of your work, you also need
to know how fast you need to go. A hydraulic system requires a pump,
a motor, and some controls. The pump is driven by some power source,
and it delivers a certain amount of power, measured by its flow rate and
pressure. The motor can be linear (a cylinder) or rotary, like a White Roller-Stator unit. Your motor should be matched to the pressure capabilities
of your pump, but you can deliver any amount of force to your work,
depending on the leverage of your design. You can only deliver the amount
of Power to your work that your pump can put out (minus innefficiencies).

I hope this helps. I wish there were more online resources for hyd
design out there.

 

[dynasim]

23,550 inlbs X1ft/12 inches = 2000 ft lbs

For simplicity, imagine you a picking up the bar in the middle, therefore, the bar will take 3 ft*250 lbs = 750 ftlbs of torque to move, leaving you 1250 ftlbs for the end of bar load.

Since you have 1250 ftlbs of torque left, and you want to lift 6 ft away, you can lift 1250 ftlbs/6 ft = 208 lbs at he end of the bar.

That being said, this isn't a good idea, as such a motor turns fast, and you will bust someones noggin. If you want to lift something like that, use a cylinder.

Note that as the arm lifts, you would be able to lift a lot more, since the load gets a lot closer(horizontally) to the pivot point as it turns. Unless you have a specific need for doing things this way, this isn't a good idea.

Chris

 

[jbelokur]

23,550 inlbs X1ft/12 inches = 2000 ft lbs

For simplicity, imagine you a picking up the bar in the middle, therefore, the bar will take 3 ft*250 lbs = 750 ftlbs of torque to move, leaving you 1250 ftlbs for the end of bar load.

Since you have 1250 ftlbs of torque left, and you want to lift 6 ft away, you can lift 1250 ftlbs/6 ft = 208 lbs at he end of the bar.

That being said, this isn't a good idea, as such a motor turns fast, and you will bust someones noggin. If you want to lift something like that, use a cylinder.

Note that as the arm lifts, you would be able to lift a lot more, since the load gets a lot closer(horizontally) to the pivot point as it turns. Unless you have a specific need for doing things this way, this isn't a good idea.

Chris

Gearing this motor down would increase the torque potential and the turn radius right?

 

[J_J]

If you are considering using this motor, you will need a pump that can provide 28 GPM at least 2000 psi, and to drive it will take about 45 HP. Now the hyd motor working with an efficiency of 85%, will give about 38 HP at the motor. The hyd motor should be relieved at 1750 psi. The motor can only turn at 50 rpm. Now depending how you set up the levers, you can multiply torque, as has been mentioned. You could automate this setup with electric valves, and energize them in a sequence of events, using a computer.

As far as additional motors, you have to figure the GPM required on all of them in case they may all operate at the same time. Otherwise, each hyd motor will develop so much HP. Add all the HP's together and that will relate back to the size of the pump required. A hydraulic motor can only develop 85% or the Hp it takes to drive the pump, because of losses. .