What Will The Cylinder Do ?

To start from the same point, let's say the cylinder is at the midle of its stroke.

Nothing. What's your point?

The point is the loss of area on the backside of the piston due to the rod.

the piston will extend!

Pressure=Force/Area i.e. Force= Pressure times area.The extend port has the area of the full back of the piston = 0.25 x pi x 2"^2, while the retract port has the area of the back of the piston, less the area of the shaft =0.25 x pi x (2"^2-0.75"^2). However there is still the atmospheric pressure to take into account on the shaft of the culynder acting to retract the piston, but this is neglected by assuming that the incoming pressure is with respect to atmosphere! Therefore the net force is acting to extend the piston since, the pressure on either side is the same.

Some may ask then, why at atmospheric pressure, when the lines are not presurised doesn't this occur? It is because only at atmospheric pressure, the external dimensions of the piston come into play, and at this point the effective areas are equal!

However, if the lines were not fed by the same source, it is possible that after some motion, the pressure in the retract side will increase due to the compression by the extend line and the system would reach equilibrium.

The only case that this theory falls down is when the extend side also has a shaft of exactly the same dimensions as the retract side, which was not presented in your description.

Regards,

Guest

Guest #1.5

http://books.google.ca/books?id=VhVUa997dbwC&pg=PA85&lpg=PA85&dq=hydraulic+cylinder+reduced+area+rod&source=bl&ots=yFwXTcVB1J&sig=rKscmadW3c6LfTDGqUwo_rsHHPU&hl=en&ei=ZIIxSo_wE5iuNbuV9NkH&sa=X&oi=book_result&ct=result&resnum=7

"the piston will extend!" Yes, but the question was, "What the cylinder will do ?"

The really simple answer is "Uneaqual Areas at the same pressure have unequal force. Unequal Force always tries to move in the direction the Highest Force wants to go.

In actual practice a cylinder with the dimensions for your stated bore and rod diameter will probably not move since the Force Difference is not high enough to overcome the frictional resistance of the mechanism's construction. I've tried this demonstration many times on an 1.5" Cylinder with a 5/8: Rod and never see movement on a Vickers Trainer setup at the 750 PSI limit of the Motor/Pump combination. However, using the available 1.5" cylinder with a 1" Oversize Rod, the cylinder extends and retract at almost the same speed and force.

That is part of the reason that most training books recommend using a 2:1 Area ratio Cylinder Area when using a Regeneraation circuit to reduce cycle on circuits that need an extended travel stroke with a short High Force stroke such as when Punching holes in an shoped a part that needs a long stroke to allow for loading and unloading the part.

Brilliant demonstration that reality is far more relevant than daft contrived catch questions. GA

Del
_{BTW what happens if a cat an a squirrel chase a slinky down the up escallator?}

Force = pressure x area. Since pressure is equal on both ends, the cylinder will tend to move slowly towards the end where the actuator rod is since the piston surface area is smaller on that end. Smaller area x same applied pressure = less force.

Why don't we increase the rod size to 1.5" and get some useful work from it.

It will extend. The area on the full piston face is greater than the other face (difference is shaft area). so there is unbalanced force. Also the cylinder volume is greater when extended than when retracted.

Cheers.........Codey

Perhaps I am unclear on the concept.

Your question refers to a cylinder and a shaft, there is no mention of a piston.

At what location is the shaft? Does this "shaft" enter the cylinder?

Is the shaft entry sealed?

Where, in relation to this shaft are the "extend" and "retract" ports?