Understanding compressed air.

[mylo]

Fantastic debate / discussion on

[color=#000000]Fantastic debate / discussion on a very important aspect of R/C subs. Don't think for a second I don't appreciate everyone's points of view.

Kevin, your explanation is basically what I have it figured to be in my head.

It would seem that everyone is right. In construction of my sub, I am sticking to the concept of]

[Guest]

ok can i pass my

ok can i pass my problem past you.
my current project uses a small diamter piston tank inside a 3.5" WTC, being used in a Thor Permit. now the WTC just fits, its very tight. The Piuston is 48mm with a 100mm stroke. At present i guess part of the WTC is above the waterline in normal running, (i need to check against the drawing as im at work right now) and the piston causes the sub to lower but not submerge. Im guessing that if i was to put the piston into a 3.25" diameter WTC that the wTC would be completely submerged at the surface so now i only have to displace the wet hull weight / volume above the water and so it should all work. does this sound correct?

[mylo]

Alb,

If your sub

Alb,

If your sub is not neutrally bouyant with a full ballast tank, then your sub is not heavy enough. Add lead keel weight. If this results in too much of your sub going below the water line, your ballast tank isn't big enough.

....that's the way I look at it.

Mylo

[JWLaRue]

Albion,

Possibly......

There isn't enough information provided

Albion,

Possibly......

There isn't enough information provided to be sure. For example, how far down does the sub go when the piston tank is full of water? How much working volume is inside the piston? How much of the WTC is above the water...and what, if any, internal components are also above the waterline?

One thing does appear to be true - if the surfaced state of the sub has the WTC above the waterline, then you are lifting more weight than is (normally) necessary.

What Mylo said definitely applies and it could very well be easier and more expedient to try adding some ballast to get the sub to a proper submerged trim....and then re-trim for surface use.

-Jeff

[safrole]

What if your conning tower

What if your conning tower were made of solid ice? (And ice was neutrally bouyant!) A tank of the same volume as the tower would not suffice

Taking water on board is actually reducing the boat's displacement which is I believe the nature of the miscommunication.

To dive, you reduce your displacement first by the "weight" of the superstructure, in grams of water of course, but then the superstructure going under will itself INCREASE your displacement again, so you have to further reduce your displacement by that amount.

Ballast Tank Grams of Water = Grams of Superstructure + Displacement of Superstructure in Grams

A significant superstructure displacement (balloon or a foam- filled conning tower) would require a much larger variable displacement, because it does increase the overall displacement of the boat upon diving. But most of the time the volume of the tower is negligible because there's a margin for error.

Doing volumetric calculations alone is only incorrect by the difference in density from your material of choice to water. Again the margin of error probably covers you, as evidenced by 25 years of success.

Then again, I could be wrong.

[yabbie1]

I'd like to appeal to

[color=#000000]I'd like to appeal to a higher authority on the displacement vs weight discussion (finally found the book!).

Norbert Bruggen no less, in his book "Model Submarine Technology" states in the chapter "Design", p. 18 of my edition]

[stoene]

First, it seems we have

First, it seems we have two groups here (the Crips and the Bloods so to speak ).

1. One side looks at this from the perspective of: how to lift the sub up to its proper waterline i.e. how much water to push out.

2. The other from the point of view of: how to sink a sub below the waterline i.e. how much water to take in.

But before we go on let's rewrite these famous words, "Archimedes' principle states that a body immersed in a fluid is buoyed up by a force equal to the weight of the displaced fluid."


Another analogy, if I have a plastic cup (neutrally buoyant) floating upright just below the waterline (full of water """""U"""") what would I have to do to raise the cup so that the bottom is at the same level of the waterline ( U ). I would have to displace some volume of water below the cup. How much to displace? It would be an equal weight of water as in the cup. And, in this case both the volume and the weight of the cup and displaced fluid would be the same.
Now again, if I have a 1g ball of lead below the waterline (""""o""""), how much water would I have to displace to raise the lead to just above waterline? ( o ) It would be ......... 1g of water. However, *note* the weights are again the same but the volumes are different!
1g of lead has a volume of 0.08818342151675486 cm^3
1g of water has a volume of 1.003009027081244 cm^3

Here is were I got that info, another good website for finding volumes and densities. AllMeasures

It now takes a larger volume of water to be displaced, obviously, because lead has a higher density than water.

Now again but in reverse, let’s displace 1g of water and place 1g of cork on top (not in the water).

1g of water has a volume of 1.003009027081244 cm^3
1g of cork has a volume of 5.000000000000001cm^3

So, when we are talking above the waterline (were it no longer contributes to buoyancy) the volume is not our concern. The 1g of cork has higher volume but works the same as 1g of lead with less volume to balance out the system.

So when talking about buoyancy (below the water line), volume does count. We had to increase the volume of displace water to get the same weight of the lead.

So lastly, let’s say the superstructure is the cup above the waterline that is buoyed up by a force equal to the weight of the displaced fluid. Archimedes. This means we have to displace an equal weight of water below the waterline to that of the weight of the superstructure above the waterline.

In conclusion, this discuss only pertains to lifting a superstructure above the waterline. This, in turn, speaks to the size of the displacement tank needed to lift the superstructure out of the water. Keep in mind, the superstructure has its own buoyancy characteristics but do not come into play when out of the water. That means one should both make the superstructure as light as possible (means having a smaller displacement tank) and using materials that are as neutrally buoyant as possible (if not floatation or weight can be add but the weights of this ballast or floatation then has to be addressed by the displacement tank).

**A note about volume. Volume is a scalar quantity (one dimension). That means by itself it can do no work. Work=force*distance Weight is a vector quantity (two dimension {magnitude and direction}) which means it can do work. Force=Mass*Acceleration for us Weight=Mass(water)*Acceleration(gravity) W=M*g
So, when thinking of volumes say to yourself "volumes of what?"

**A note about displacement (buoyancy) it is the same force as W=M*g but pointing up.

[gotland]

ouhwouwou.....

Stoene, one question]http://www.subcommittee.com/forum/icon_question.gif

Regards

Lothar

ouhwouwou.....

Stoene, one question]http://www.subcommittee.com/forum/icon_question.gif[/img]

Regards

Lothar

[safrole]

I'm unfortunately going to have

I'm unfortunately going to have to go with the people who say that volume is all that matters, much as it pains me. As your tower weighs more, you pull more lead out of your keel to compensate.

This could be wrong, so feel free to fire away, but I think it's right.

[gotland]

Dear safrole,

great Idea to make

[color=#000000]Dear safrole,

great Idea to make it with drawings! It makes this discussion easyer to understand.

Look at drawing 1]

[gotland]

Sorry I forgot...The drawings are

Sorry I forgot...The drawings are nearly right! The displacement of the Ballast tank is not allowed to add or to count to the whole displacement, because You made Your boat 500 ml smaller by flooding the tank. The displacement of the dived boat is again 3 Liters.

Gotland

[safrole]

Gotland, I think I agree

Gotland, I think I agree with you completely]http://modelbrass.com/images/BallastTankStudy2.jpg[/img]

[JWLaRue]

....let's try this mental exercise

....let's try this mental exercise (I'll leav eit to someone else to create a grahpic!)......if the only factor is volume, then it would make no difference how much the item actually weighs. That doesn't make any sense...intuitive or otherwise!

Try thinking of a slightly modified version of the graphics. One with the 'extra' 500cc/1.5Kg solid inside the 3 liter boat and the other with no additional solid. The volumes are the same. They require different amounts of flotation force (ballast for example) to achieve the same level of trim.

We're still talking about a volume that is equal to the weight differential.

Yes/no?

-Jeff

[stoene]

LOL, you guys will like

LOL, you guys will like this

fluids interactive display

[yabbie1]

Safrole, your pictures are worth

[color=#000000]Safrole, your pictures are worth a thousand of my words! Well done (and I'm still figuring out how to underline something).

Things are not always immediately intuitive. It was not intuitive to me for a 400 tonne Airbus (sorry, I mean Boeing) to pick up its skirts and soar off the ground, seemingly lighter than a feather. (Note]