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Station said:
Please explain to me how a differential displacement cylinder , filled with fluid, and with the ports plumbed straight to one another ( As you say is the case in the orbitrol) would be any other thing than locked up? It would be hydrolocked and you would not be able to move it even the first 1/8" in either direction . How could the cylinder possibly work " ok " in one direction? In a closed enviroment the vacuum created by a larger volume trying to draw from a smaller volume would create an instant vacuum that would be just as powerful in keeping the cylinder from moving as the pressure created in the opposite load direction.





Sean

Lets look at this cylinder http://www.surpluscenter.com/item.asp?UID=2005073009524511&item=9-5100-05&catname=hydraulic

The bore times the stroke = swept the volume of each side

So 7.06 X 5= 35.3 for the piston side

And 7.06 X 5 -1.2 X 5= 29.3 for the rod side

IF the piston is at mid stroke of the cylinder you would have 17.65ci of oil on the piston side and 14.65ci of oil on the rod side (total of both sides = 32.3ci)

For every inch you stroke the cylinder rod out you increase the volume of the piston side by 7.06ci and you decrease the volume of the rod side by 5.86ci

So from mid stroke you can connect both work ports with one hose, and lets say your perfect and can do it without loosing any fluid volume from the cylinder or hose you would then be able to pull the rod out (remember were talking about external forces acting on the steering system) swapping all the fluid from the rod side (14.65ci at mid stroke) into the area created by the piston moving in the barrel (remember at full stroke rod out you have 35.3ci volume on the piston side) so you would be pushing 14.65ci of oil into a space that has just been opened up by 17.65ci (7.06ci/inch of travel)leaving a vacuum area of 2ci (remember the cylinder was sealed that means no air in or out, if there is a force strong enough to cavitate the cylinder it will if the force is not high enough to pull vacuum the cylinder would stop moving once the 14.65ci of oil was transferred)


So you just took the single ended cylinder from mid stroke to full rod extend out now you have 32.3ci volume of oil on the piston side of the cylinder and a 2ci vacuum pocket NOW you want to push the rod in to swap fluid (the 32.3ci you have on the piston side is the total fluid you have remember the cylinder is tied together with one hose and the volume of the hose is irrelevant) from the piston side to the rod side

So we already know For every inch you stroke the cylinder rod out you increase the volume of the piston side by 7.06ci and you decrease the volume of the rod side by 5.86ci
Lets take the reverse now your pumping 7.06ci oil every inch of stroke from the piston side into a spot being created by 5.86ci/inch so math tells us the cylinder will stroke IN by 1.2 inches before the volume of fluid being pumped becomes to large for the area its being pumped into

That scenario was tossing out the area of vacuum that might be created ( I will be honest and say this theoretical vacuum pocket is the only area of this that Im sketchy on, and only on how it would affect the amount of rod movement)
 
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