Mrblaine, Thank you for taking the time to post this info; I'll be using it to sort out the springs on my LJ . So far this info is the most straight forward and succinct I have come across, and I appreciate seeing/learning how this is figured out. The tips and insight you provide, like accounting for the small amount of lift from the reservoir, are invaluable.
Thanks again for this and all the information you share with this forum.
Someone smarter than me needs to work out the math for that one. I think I understand it but I'm not 100 % sure that I do.
I know how to calculate pressure and in the case of the reservoir it would be the charge pressure x the surface area of the piston.
If we take the theoretical of 200 psi charge pressure x the piston area of a 2.5" reservoir which is roughly 5 square inches, the math says that is 1000 lbs of force. That would extend the shock fully and not let it move. My theory is the force is acting against the diameter of the fluid equal to the cross section of the hose. In that case, the area of that is .19 square inches or 39 lbs additional lift which is about what we see.