Looking to order a steering valve for my home-built articulated tractor. Lots of options and want to get it right so asking for those of you smarter then I to check out my math and logic.
Have been operating the system with a simple 2.5 X10 inch hydraulic cylinder with a 1x3/8 rod. Controlled by a simply hydraulic valve which is hooked by a rod to the steering wheel it has done a good, lets us get some time on the unit and see what we have. It does get a little dicey, twitchy, though on the road and it is time to go with an orbital steering valve. Also, would like it to be able to turn if we lose power like a tractor does. Tons of options on both size and features on these valves.
I have pulled the old cylinder off and replaced it with a 2.5X ten inch stroke steering cylinder with a 1.5 inch rod which comes out both ends of the cylinder. This is currently working well with the aforementioned hydraulic valve but really no different then what we had before. Having the steering cylinder though does equalize the cubic inches required to turn the unit left or right.
So now to get it fixed up with a steering valve.
I think I need an open center. This means the fluid will flow through the valve and back to the tank when not turning. Correct me if I am wrong.
My pump is fixed displacement. I have a priority valve which directs the first 4 gallons to the steering system.
I think I want non load reactive. this means if I let go of the wheel it will stay put, it will not return to center. Again, correct me if I am wrong.
I believe I want it equipped with a manual steering check valve that will allow me to turn the wheel and steer manually if the motor or pump quits.
The math. Cylinder diameter 2.5 inches divided by half is 1.25. 1.25x1.25 =1.5625 Square inches 1.5625 X 3.14= 4.90 square inches.
Rod dia =1.5 1.5 divided by 2=.75 .75x.75=.562 .562x3.14=1.76 square inches.
Subtract 1.76 from 4.90 should give us 3.14 square inches which times cylinder stroke of ten inches should give us 31.4 cubic inches of displacement.
If I figure I would like to have 3.5 turns of the wheel lock to lock I should take 31.4 and divide by 3.5 to get 8.97 cubic inches which is what I believe I need for a size on the orbital steering motor.
Hydraulic Steering Unlimted, HSU has one I think would fill the bill, part #9.76CI-160ML
Certainly would appreciate everyone's feedback, thank you in advance. Roy
Have been operating the system with a simple 2.5 X10 inch hydraulic cylinder with a 1x3/8 rod. Controlled by a simply hydraulic valve which is hooked by a rod to the steering wheel it has done a good, lets us get some time on the unit and see what we have. It does get a little dicey, twitchy, though on the road and it is time to go with an orbital steering valve. Also, would like it to be able to turn if we lose power like a tractor does. Tons of options on both size and features on these valves.
I have pulled the old cylinder off and replaced it with a 2.5X ten inch stroke steering cylinder with a 1.5 inch rod which comes out both ends of the cylinder. This is currently working well with the aforementioned hydraulic valve but really no different then what we had before. Having the steering cylinder though does equalize the cubic inches required to turn the unit left or right.
So now to get it fixed up with a steering valve.
I think I need an open center. This means the fluid will flow through the valve and back to the tank when not turning. Correct me if I am wrong.
My pump is fixed displacement. I have a priority valve which directs the first 4 gallons to the steering system.
I think I want non load reactive. this means if I let go of the wheel it will stay put, it will not return to center. Again, correct me if I am wrong.
I believe I want it equipped with a manual steering check valve that will allow me to turn the wheel and steer manually if the motor or pump quits.
The math. Cylinder diameter 2.5 inches divided by half is 1.25. 1.25x1.25 =1.5625 Square inches 1.5625 X 3.14= 4.90 square inches.
Rod dia =1.5 1.5 divided by 2=.75 .75x.75=.562 .562x3.14=1.76 square inches.
Subtract 1.76 from 4.90 should give us 3.14 square inches which times cylinder stroke of ten inches should give us 31.4 cubic inches of displacement.
If I figure I would like to have 3.5 turns of the wheel lock to lock I should take 31.4 and divide by 3.5 to get 8.97 cubic inches which is what I believe I need for a size on the orbital steering motor.
Hydraulic Steering Unlimted, HSU has one I think would fill the bill, part #9.76CI-160ML
Certainly would appreciate everyone's feedback, thank you in advance. Roy