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size=2>Hi Scott, <BR><BR>1 bar = 0,1
N/mm2 </font> <BR><BR>1000 meter depth = 100 bar or 10 N/mm2
(or 1
kg/mm2)<BR><BR>A Hydraulic stamp of a 1/2 inch has a surface of 126
mm2<BR>means the outside waterpressure on the cylinder stamp is <BR>1267
N or
127 Kg or 0,13 ts. <BR><BR>If you make a drawing of your schematic you
will see
that it is not selfcompensating. <BR>Means you need a inside force of
that
amout just to compensate. <BR>If you asume you can take a pressure
of 0,013
ts <BR>with some comfore by hand you inside zylinder piston has to be
the 10
times more diameter thn the outside one. <BR><BR>By the way the same
force works
on your troughulls cables of the same diameter. <BR><BR>vbr Carsten
<BR>
<BR><swaters@waters-ks.com> schrieb:
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<span style="FONT-SIZE: 10pt; FONT-FAMILY: Verdana; COLOR: #000000"
><div>I have a question maybe someone can answer.</div>
<div>If you have two hydraulic cylinders that are completely filled with
oil (no air pockets anywhere in the system) one in a submarine and on
outside of a submarine. Each cylinder has the rod side connected to the
head side of the other cylinder so when on rod extends, the flow of one
makes the other cylinder do the same exact thing. Would the one cylinder
that does the same as the other cylinder on the surface function the
same way at depth? Or would the deeper you go the more force you would
have trying to push the rod into the cylinder? </div>
<div> </div>
<div>Thanks,</div>
<div>Scott Waters </div></span>
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