Drysuit vs BCD bouyancy control

[Diver0001]

....snip mind numbing stuff....

edit: steel Y ~= 200 GPa, rubber Y ~= 0.01-0.1 GPa, nylon Y ~= 2-4 GPa, from
http://www.nationmaster.com/encyclopedia/Young's-modulus

I thought that's what you were getting at. I think in practical diving terms, however, although the effect may be measurable I doubt that it would be noticeable enough to the diver to make a difference in the sloppy world of buoyancy control.

R..

 

[Lightning Fish]

Wasn't what you said. With a pressurized vessel (ie tank) will always displace the same amount of water regardless of how much gas is in it. Its buoyancy will change becuase the weight of the tank becomes less as you consume the gas durning the dive.

But if I just repeated myself I would be giving you a bigger target...

Then you would be wrong. The gas in both the drysuit and the BCD are under the same ambient pressure. The only way there could be a difference is if you create a situation where your BC or Drysuit becomes a pressurized vessel (ie filling it beyond ambient pressure.)

I've done some reading on the subject. For pneumatic envelopes such as balloons (not the rubber kind) and parachutes it is referred to as a 'wrinkled state'. So untill the BC or drysuit is completely unwrinkled, the equations I gave above wouldn't come into play.

However, try this out...
P1 x V1 = P2 x V2 = K, where P1 and P2 are the external pressures acting on the drysuit or BC and K is some constant.

Initially, the starting pressure in either the drysuit or BC is equal to the ambient (or external) pressure, so P1 = Pe, for the depth the diver is at.

Then P1 x V1 becomes K = Pe x V1. (eq'n 1)

Now add a burst of air (to either the drysuit or BC), represented by delta_P. This is equivalent to decreasing the depth until P2 = Pe - delta_P. Substituting this into P2 x V2 yields K = (Pe - delta_P) x V2. (eq'n 2)

Equate eq'n 1 and eq'n 2:

Pe x V1 = (Pe - delta_P) x V2, and solve for V2: V2 = (Pe x V1)/(Pe - delta_P)

Then the change in volume is delta_V = V2 - V1 = (Pe x V1)/(Pe - delta_P) - V1 which can be simplified to
delta_V = (delta_P/(Pe - delta_P)) x V1.

So the change in volume depends on the initial volume of the drysuit or BC (and the amount of air added and ambient pressure, obviously). So unless the drysuit and BC have the same initial volume, the change in bouyancy will be different. If there is a flaw in my logic or my math, then point it out.

Cheers,
Bill.

 

[JeffG]

I've done some reading on the subject.

I go diving

 

[lairdb]

So unless the drysuit and BC have the same initial volume, the change in bouyancy will be different. If there is a flaw in my logic or my math, then point it out.

Different, yes, but only by the energy involved in reducing the wrinkled state to the unwrinked state. IOW, 6.0000000000000000000001 of one, and a half-dozen of the other.

("First, assume a spherical cow.")

 

[wcl]

Hi,

It's a bit challenging to read the long thread on my phone browser, so if I missed something and am talking through my hat, just let me know ... but:

The P1V1=P2V2=constant only holds true for constant mass (and, of course, "ideal gas" approximation).

Adding air drives you to the more complete ideal gas law,

PV/nT=R (constant)

Adding or removing air to the bubble changes "n" (molecular count ~ mass).

That's why you add air to the bubble to compensate for increasing pressure ... the bubble pressure increases as a result of adding more mass into the volume (~constant buoyancy for the bubble <== constant volume) [edit: there is, of course, a buoyancy effect with the changing mass; it is small in this region, which is volume-change dominated, but clearly more significant at high mass changes as a tank can experience ... to be fair, I have to be picky on myself too ;-)].

Cheers,

Walter

 

[JeffG]

I think you boys should start calculating the minor adjustments to the gravitational forces that is applied to your head and feet will be when you change your attitude in the water.
That could be another concern when you are diving.

 

[Lightning Fish]

I think you boys should start calculating the minor adjustments to the gravitational forces that is applied to your head and feet will be when you change your attitude in the water.
That could be another concern when you are diving.

g = (G x m1 x m2)/r^2 and neglecting the effects of the rotation of the earth....

So how was the dive?

Bill

 

[JeffG]

g = (G x m1 x m2)/r^2 and neglecting the effects of the rotation of the earth....

So how was the dive?

Bill

It was gas fill night...dive night is tonight. Should be good

 

[wcl]

Actually, I think the Van der Waals nonlinear component of viscosity would be a more dominant factor.

;-)

I think you boys should start calculating the minor adjustments to the gravitational forces that is applied to your head and feet will be when you change your attitude in the water.
That could be another concern when you are diving.

My attitude is always better underwater. Already anticipating being happy in Lake Michigan Sunday ....

Cheers,
Walter

 

[JeffG]

I'm pretty sure the Nerd-O-Meter is fully pegged for this thread now