[PSUBS-MAILIST] new submarine inside to outside hydraulic

[hank pronk]

Hi Carsten,
I am not sure about your math, I get a much different figure.  
A 1/2 in rod has an area of .19in
multiply by 1,000 foot depth salt water 445psi = 87.33 lbs
With a 10 to 1 lever that would be 8.7 lbs to push with your arm.
Also you want a smaller cylinder inside with a longer stroke and you get an even lighter load on your arm.
Hank 

 Subject: Re: [PSUBS-MAILIST] new submarine inside to outside hydraulic
 To: "Personal Submersibles General Discussion" <personal_submersibles at psubs.org>
 Received: Saturday, April 12, 2014, 2:56 AM
 
 
 
 
  
 E-Mail Software 6.0
 
 Hi Scott, 
 
 1 bar = 0,1 
 N/mm2  
 
 1000 meter depth = 100 bar  or 10 N/mm2
 (or 1 
 kg/mm2)
 
 A Hydraulic stamp of a 1/2 inch has a surface of 126
 
 mm2
 means the outside waterpressure on the cylinder stamp is 
 1267
 N or 
 127 Kg or 0,13 ts. 
 
 If you make a drawing of your schematic you
 will see 
 that it is not selfcompensating. 
 Means you need a inside force of
 that 
 amout just to compensate. 
 If you asume you can take a pressure
 of 0,013 
 ts 
 with some comfore by hand you inside zylinder piston has to
 be
 the 10 
 times more diameter thn the outside one. 
 
 By the way the same
 force works 
 on your troughulls cables of the same diameter. 
 
 vbr Carsten
 
 
 
 
 <swaters at waters-ks.com> schrieb:
 
 I have a question maybe someone can answer.
 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? 
  
 Thanks,
 Scott Waters 
 
 
  
 
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