Yes. On a three square-inch surface area of the sub, that would be perfectly true. But the pressure around the sub would not be changed: It would still be 267psi.
The recent delay in the posting of emails (that, long with a few cups of coffee) has me rethinking this whole thing. And Frank, if I've come across in my emails as sounding condescending, I sincerely apologize. My intent is not to challenge, merely understand.
Let me see if I can ask this a different way, and see what kind of answers I get (and yes, I am looking for answers)
If I had a square tube welded to the side of my hull, and the opening was exactly one inch wide by one inch high, and I had a square rod that fit perfectly inside this tube, then when the sub is 600 feet down, them amount of pressure I would have to exert on the rod (to keep it in place) would be 267 pounds because the end of the rod that is facing the ocean has one square inch of surface area.
Now, instead of a 1 square-inch example, I perform the same experiment with a ¼? tube & rod combination. The total amount of surface area facing the ocean is a quarter of the surface area on the one-inch rod, so the pressure I would need to keep this rod in place would be about 67 pounds.
Now, let?s set up a third experiment: Let?s use that same 1? square tube, but this time we?ll weld a plate over the end (facing the ocean) that has a ¼? square hole in it. Now? how much force is it going to take to keep that 1? rod in the tube? My money would be on 267 pounds.
Here?s my idea of thinking: If you?re using a piston-type pump to force water out of the sub into the surrounding sea, the amount of force required is equal to the external pressure times the surface area of the top of the piston. I don?t believe it has anything to do with the diameter of the pipe between the pump and the hull of the sub.
Could someone tell me if I?m right or wrong on this one?
-- NP
From: bobtravis@comcast.net
Reply-To: personal_submersibles@psubs.org
To: personal_submersibles@psubs.org
Subject: Re: [PSUBS-MAILIST] RE: payload
Date: Fri, 08 Dec 2006 19:12:06 +0000
Wouldn't that be 267 pounds per squair inch? Which would mean that if you had a surface area of three inches, then the force against it would be 801 pounds?-------------- Original message --------------
From: "Nomdae Plume" <nomdae@hotmail.com>
Frank:
I read this post first thing this morning, and my brain still isn't in gear, so let me see if I understand this correctly:
I'm in my sub, 600 feet down, and the outside pressure on the hull is 18atm.
(600 / 33) = 18.18atm or 267psi
Now let's say I have three pipes entering the hull, and all three have pressure gauges on them. The first pipe is 3" in diameter, the second is 1", and the third is 1/4" in diameter. Won't all three gages read the same pressure? Will a pump have to generate 267.0001psi in order to pump anything out of the hull?
I think I can understand what you're saying about smaller pumps. A small pump can be geared down so that the volume of water it's moving is decreased while the pressure it creates is increased. It would just take more time to pump out all the water you were trying to move. However, hooking a pump up to a 1/4" hose is not going to give you any more of a mechanical advantage than hooking it up to a 3" pipe... or directly to the hull of the sub.
Am I right? Or should I get a few more cups of coffee in me before I start posting? ;-)
-- NP
From: ShellyDalg@aol.com
Reply-To: personal_submersibles@psubs.org
To: personal_submersibles@psubs.org
Subject: Re: [PSUBS-MAILIST] RE: payload
Date: Wed, 6 Dec 2006 21:30:09 EST
Just a quick note on this pressure thing. If you are pushing against the pressure in a 1/4 inch high pressure hose/line, a small pump would work. Frank D.
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