<html><head><meta http-equiv="Content-Type" content="text/html; charset=UTF-8"></head><body><div>Dan,</div><div>You are brilliant!</div><div>Thanks,</div><div>Scott Waters</div><div><br></div><div><br></div><div><br></div><div><br></div><div><div style="font-size:75%;color:#575757">Sent from my U.S. Cellular© Smartphone</div></div> <br>"Dan H." <jumachine@comcast.net> wrote:<br>How to solve the problem of depth figuring in on the situation is to use <br>double ended cylinders. With a rod of the same size sticking out of each <br>end of each cylinder the effect of water pressure balances out. The force <br>is the same in either direction.<br><br>Dan H.<br><br>----- Original Message ----- <br>From: "hank pronk" <hanker_20032000@yahoo.ca><br>To: "Personal Submersibles General Discussion" <br><personal_submersibles@psubs.org><br>Sent: Saturday, April 12, 2014 8:58 AM<br>Subject: Re: [PSUBS-MAILIST] new submarine inside to outside hydraulic<br><br><br>Hi Carsten,<br>I am not sure about your math, I get a much different figure.<br>A 1/2 in rod has an area of .19in<br>multiply by 1,000 foot depth salt water 445psi = 87.33 lbs<br>With a 10 to 1 lever that would be 8.7 lbs to push with your arm.<br>Also you want a smaller cylinder inside with a longer stroke and you get an <br>even lighter load on your arm.<br>Hank<br><br> Subject: Re: [PSUBS-MAILIST] new submarine inside to outside hydraulic<br> To: "Personal Submersibles General Discussion" <br><personal_submersibles@psubs.org><br> Received: Saturday, April 12, 2014, 2:56 AM<br><br><br><br><br><br> E-Mail Software 6.0<br><br> Hi Scott,<br><br> 1 bar = 0,1<br> N/mm2<br><br> 1000 meter depth = 100 bar or 10 N/mm2<br> (or 1<br> kg/mm2)<br><br> A Hydraulic stamp of a 1/2 inch has a surface of 126<br><br> mm2<br> means the outside waterpressure on the cylinder stamp is<br> 1267<br> N or<br> 127 Kg or 0,13 ts.<br><br> If you make a drawing of your schematic you<br> will see<br> that it is not selfcompensating.<br> Means you need a inside force of<br> that<br> amout just to compensate.<br> If you asume you can take a pressure<br> of 0,013<br> ts<br> with some comfore by hand you inside zylinder piston has to<br> be<br> the 10<br> times more diameter thn the outside one.<br><br> By the way the same<br> force works<br> on your troughulls cables of the same diameter.<br><br> vbr Carsten<br><br><br><br><br> <swaters@waters-ks.com> schrieb:<br><br> I have a question maybe someone can answer.<br> If you have two hydraulic cylinders that are completely<br> filled with<br> oil (no air pockets anywhere in the system) one in a<br> submarine and on<br> outside of a submarine. Each cylinder has the rod side<br> connected to the<br> head side of the other cylinder so when on rod extends, the<br> flow of one<br> makes the other cylinder do the same exact thing. Would the<br> one cylinder<br> that does the same as the other cylinder on the surface<br> function the<br> same way at depth? Or would the deeper you go the more force<br> you would<br> have trying to push the rod into the cylinder?<br><br> Thanks,<br> Scott Waters<br><br><br><br><br> -----Inline Attachment Follows-----<br><br> _______________________________________________<br> Personal_Submersibles mailing list<br> Personal_Submersibles@psubs.org<br> http://www.psubs.org/mailman/listinfo.cgi/personal_submersibles<br><br><br>_______________________________________________<br>Personal_Submersibles mailing list<br>Personal_Submersibles@psubs.org<br>http://www.psubs.org/mailman/listinfo.cgi/personal_submersibles <br><br>_______________________________________________<br>Personal_Submersibles mailing list<br>Personal_Submersibles@psubs.org<br>http://www.psubs.org/mailman/listinfo.cgi/personal_submersibles<br><br></body>