Yet another ingenious UW breathing device

[almostDIR]

I am unsure if it's a GOOD thing if that thing is a scam less Darwin awards to be expected in that case

I don't see any reason why they would not be delivering by now so looks like a perfect scam to me and someone did good money out of it

 

[BlueTrin]

I am unsure if it's a GOOD thing if that thing is a scam less Darwin awards to be expected in that case

I don't see any reason why they would not be delivering by now so looks like a perfect scam to me and someone did good money out of it

I would have stopped at the idea stage: I never thought you could sell for a million $ of this product ?

I guess it is all in the hype and marketing .

 

[60plus]

I have to respectfully disagree.
PV=NRT

Does the greatest pressure change occur in the top 10 ft? To some extent it depends on what exactly you mean. the pressure at the surface is 1 bar (1 Atm give or take). The pressure at 10 ft is 1.25 Bar so in descending the first 10 ft the pressure has increased by 25%. If you are at 75 ft the pressure is 3 bar, descend 10 ft and it is 3.25 bar so it has increased by 8% much less than the % increase in the first 10 ft.
I am aware my figures are near estimates, not absolutely exact.

 

[Nathan Doty]

I could see using something like this for shallow lobster hunting like in the keys. 5-10 feet but you need a couple minutes underwater to get a bug...

 

[tridacna]

Does the greatest pressure change occur in the top 10 ft? To some extent it depends on what exactly you mean. the pressure at the surface is 1 bar (1 Atm give or take). The pressure at 10 ft is 1.25 Bar so in descending the first 10 ft the pressure has increased by 25%. If you are at 75 ft the pressure is 3 bar, descend 10 ft and it is 3.25 bar so it has increased by 8% much less than the % increase in the first 10 ft.
I am aware my figures are near estimates, not absolutely exact.

I think that should have read 10 meters. Not feet.

 

[Degenerate]

I think that should have read 10 meters. Not feet.

Don't think so, because those numbers would be way off then.
Pressure at 10m is 2 bar, not 1,25.
Calculating pressure is so easy with metric, just take depth in meters, divide by ten and then add 1.
27m is 3,7 bar
42m is 5,2 bar
And so on.

Although, 10ft is 3.04 meters which means pressure at 10ft is 1.304 bar, and not 1,25.

Then again I could have completely missed what you meant, haven't read the whole thread

 

[Storker]

Calculating pressure is so easy with metric

 

[tridacna]

@Degenerate Sigh...read the question. Maximal % pressure increase.

Pressure at surface is 1 bar

Pressure at 10m is 2 bar. Just doubled.

Pressure at 20m is 3 bar. Did not double from 10m. How much (%) did it increase? You can do this.

Pressure at 30m is 4 bar. Less than double from 20m. Can anyone calculate % change? Not too hard. Hint-% pressure change is not linear.

.....

When did maximum pressure change happen? This is not rocket science.

 

[Degenerate]

snip

Yeah I just re-read the posts and realize I put my foot in my mouth there.
I know the biggest percentage pressure increase takes place at the first 10m, I just blanked out on the percentage part when posting my last reply and focused solely on the pressure increase in bar, sorry

@Storker While I do enjoy a good can of worms, that is one I definitely don't want to open.

 

[BlueTrin]

I think that should have read 10 meters. Not feet.

So you are both right, the rate of (proportional) change is always greater in the last bit, this is true regardless of the unit. i.e. the rate of change will be greater between 0 to 10m than between 10 to 20m. It is also greater between 0 to 10ft than between 10 to 20ft.

Actually to be exact, he pressure change per depth change is a linear constant.

• d(bar) / d(depth in m) = 0.1
• d(bar) / d(depth in ft) = 0.03048

i.e. the bar pressure is b(x) = 1+ 0.1 * x (x is the depth in meters)


What is not constant is the rate of change of the proportional pressure, which is the rate of change of volume of the balloon.

so if you write down the proportional rate of pressure change as:

• c(x) = b(x+dx) / b(x)

for example if we compute this between 0 and 10m:

• x = 0m
• dx = 10m
• c(x) = 2 / 1 = 200%, i.e. the balloon at 10m will double in size when it reaches 0m

so the rate of this change in regards to the depth is:

• d c(x) / d x = d/dx( b(x+dx) / b(x) )
• = d/dx( [1+0.1*(x+dx)] / [1+0.1*x])
• = d/dx( 1 + dx/(0.1 + x) )
• d c(x) / d x = 1 / (0.1 + x)

So the rate of change, the formula above, decreases when x (the depth) is increasing.
(plus/minus all mistakes that I usually do)