Understanding compressed air.

[mylo]

[color=#000000]Gents,

Is my understanding correct in that air at one ATM is approx 15 psi. Therefore, a 1 Litre tank with air compressed to 150 psi = 10 times (150 psi / 15 psi per ATM) the volume of air as the tank holds which, in this case, would be 10 Litres ? So.....a 1 litre tank with 150 psi air in it has enough volume to purge a 10 litre ballast tank on the surface OR ...a 5 litre trim tank when submerged at 33' ( = 2 ATM, which required twice the air pressure to purge). I use these round figures as examples only. I am trying to do the math for my 1]

[stoene]

This site is really useful

This site is really useful for that stuff, has just about everthing Unit Converter

[mylo]

Stoene,

Cool site but, it

Stoene,

Cool site but, it doesn't really answer my question about volume of air in relation to psi.

Myles.

[stoene]

Mylo,
1 psi ~= 0.068948 bar
150psi

Mylo,
1 psi ~= 0.068948 bar
150psi ~= 10.342136 bar
1 bar is ~= sea-level and one unit volume (any size) so 10 bar is 10 X 1 bar(any size). So a 1 liter bottle at sealevel that has 1 bar has 1 liter of air. A 1 liter bottle at 10 bar has 10 liters of air

bar = A unit of pressure, equal to the sea-level pressure of Earth's atmosphere; 1 bar = 0.987 atmosphere = 101,300 pascals = 14.5 lbs/square inch = 100,000 Newtons per square meter.

I guess you need to know what volume of water you need to displace and how many times you wish to do that. Then you can figure how big of a bottle you need and how much psi it will neeed to hold for a given size to get to desired volume.

Help any?

[don prince]

Hi Mylo,

I'm sure I could

Hi Mylo,

I'm sure I could dig that stuff out of my old physics books, but what's the point? Are you planning to build a normal RC sub, or are you planning some type of deep diving recovery vehicle? Most of the time, you will be running your RC sub at or near periscope depth. Besides your radio transmitter signal will be lost at 2 bars (Approx 66 ft.) and I doubt seriously a standard WTC will stay water tight at that depth. So you don't have to worry about the water pressure on your gas ballast system...

Regards,
Don_



Edited By Don Prince on 1126224614

[mylo]

Stoene,

Thanks for that. My

[color=#000000]Stoene,

Thanks for that. My calculations were correct.

Don,

Water pressure is not my concern here for the reasons that you mention. What I am wondering about is in my 1]

[don prince]

Hi Mylo,

I understand what you

Hi Mylo,

I understand what you are asking... 150 pounds sounds like a lot of weight for a 1/25 scale model U-Boat. I haven't weighed my 1/32 OTW, but without a WTC it comes in less than 10 pounds without the lead keel ballast. My understanding is that when the WTC is added it will be located just below/at the water line, and my ballast system will only have to lift what's above the water line. Therefore, most on my U-Boat's weight is offset at the water line by the entire WTC volume and trim foam ballast. I know I may be stepping way beyoug my zone of comfort/knowledge here because all this is theory to me until I complete my OTW U-Boat and it emerges from its first test dive.

Regards,
Don_

[mylo]

Don,

Hmmmm.....

I've been doing

Don,

Hmmmm.....

I've been doing some calculations and the basic premise remains true, that being, the sub has to weigh what the displacement of the WTC is in order to be close to neutral in bouyancy. I came to the 150 lb estimate only from what I figure the size of the WTC is going to be (I have plans for lots of on board goodies). Don't get me wrong, I could be way off here but I expect the weight to be 100+ pounds anyway (she's not a small boat). The weight of all the construction materials and everything else + keel weight = displacement.....from what I'm able to figure. I do expect my keel weight to be a lot, which will assist in it's stability anyway. Now, if I have to get 33% of that weight (from what I understand) in ballast, I'm trying to figure out how big of a compressed air bottle I'm going to need and at what psi it will have to be.

If your sub comes in at 10lbs, it must have a small (relatively) WTC, does it ? Those OTW 1/32 boats are awesome by the way.

[don prince]

Hi Mylo,

I didn't want to

Hi Mylo,

I didn't want to do this, but it's Saturday and I've just finished mowing my lawn....

1 Gallon of water = 8.33 lbs
1 Gallon = 231 cu in

30 lbs/8.33 = 3.6 Gallons (Your required Ballast)

3.6 Gallons x 231 ci in = 831.6 cu in (Your ballast capacity)

If you had a ballast tank with the inside diameter of 6 inches then we need to determine what the tank length must be ...

Volume = Pi x R sq x h

831.6 cu in = 3.14 x 3 x 3 x h
831.6 cu in = 3.14 x 9 x h
831.6 cu in = 28.26 x h

831.6 cu in/28.26 = h

H = 29.42

Therefore, given that your ballast tank has a cylinder shape; if the inside diameter is 6 inches, then the length must be 29.42 inches to equal a ballast weight of 30 pounds.

I think???

Mylo, that's a lot of ballast for one extra large U-Boat. I guess it will have everything onboard except the crew of 45 men. My friend, your project sounds like it's gonna take awhile... good luck! Post photos during your build.

Regards,
Don_



Edited By Don Prince on 1126385006

[mylo]

Don,

I already have over

[color=#000000]Don,

I already have over a month simply researching the project....I haven't even lifted a tool yet. This is typical of me though. I never said I would be finished quickly. My brain is constantly thinking of this project and how it can be done. I'm still set on 1]

[JWLaRue]

Hmmmm, a couple of things.....

You

Hmmmm, a couple of things.....

You posted ..."If I require approx 30% of that weight in ballast, (say 50lbs = 22 litres) for surface running...." but it's not exactly clear what you mean by that.

This ties to Don's calculation or comment of "30 lbs/8.33 = 3.6 Gallons (Your required Ballast)"

It's not obvious what any of this has to do with calculating the amount of ballast required to move from the submerged state to the surfaced state.

Let's try this approach to see what size the ballast tank should be....

If we assume that your boat will be built with similar materials and to similar construction standards as the OTW boat, then at 1.28 times the size of the OTW Type VII, your boat would require a working ballast volume that is 2.1 times that required for the OTW boat. (1.28**3 to calculate weight) The OTW Type VII has a ballast tank that provides 5.65 pounds of lift...so your boat would require 11.85 pounds of lift (5.65 * 2.1).

In general, the amount of working ballast(*) that will be required needs to be (at least) equal to the weight of everything above the waterline. All other components that are never raised out of the water would be covered through the use of floatation foam.

So...even for a 1/25th scale Type VII, 30 pounds sounds high.

My OTW Type VII only required approximately 8-10 pounds of lead ballast. I've never weighed the boat without the dive module, batteries, and ballast weights...but using Don's 10 pound number sounds right as I estimate that fully loaded the boat weighs around 25-30 pounds.

-hope this makes sense,

Jeff

(*) working ballast - depending on the type of ballast system you are using, there may be part of the tank volume that does not contribute to providing lift. It may not be possible to get all of the air or water out of the tank.

[don prince]

Hey Jeff,

Thanks for bailing me

Hey Jeff,

Thanks for bailing me out... I Thought I was giving Mylo good advice/information, but I haven't completed my sub as of yet. So it's still mostly theory to me...

Regards,
Don_

[mylo]

Whoa !!, I need to

Whoa !!, I need to go back to the drawing board. It would appear as though I've got things screwed up. Looks like there is going to be a delay in the project until I get my math figured out.

.....now where in the heck did I go wrong I could have swore that the weight of the sub had to equal the displacement of the WTC.

Myles.

[JWLaRue]

Myles,

The variable flotation (e.g. the

Myles,

The variable flotation (e.g. the ballast tank) of the WTC needs to be at least equal to the weight of that part of the sub that is at or above the waterline.

You do not need (or want) the ability to raise the sub higher than it's waterline. If you made the total displacement (in weight) of the ballast system equal to the total weight of the boat....that would be overkill.

Remember, the ballast system only needs to provide lift for what the flotation foam cannot provide lift for....the parts above the waterline. The flotation foam, of course, only provide lift if it stays in the water....

-Jeff

[mylo]

Jeff,

I think you are

Jeff,

I think you are misunderstanding me. I was under the impression that you needed approx 30 % of the total sub's dry weight in variable flotation, which would, in effect, raise the boat to the waterline with an empty ballast tank. With a full ballast tank, the sub would be close to neutral. This part all makes sense to me.

What I seem confused about is, I was under the impression that whatever the displacement of the sub's WTC was, the total dry weight of the sub was going to have to be equal to this in order, basically, for it to be heavy enough to pull the WTC under water (running submerged). Displacement being the force upward that water pushes, I was only assuming I needed an equal force pushing downward in order to obtain close to neutral. I'm figuring on quite a large WTC in order to house all the things I want to put in my 1]http://www.subcommittee.com/forum/icon_smile.gif[/img]

Myles.