# Water pressure questions

hi, I'm doing a small research project on underwater habitats. Nothing serious just in my spare time, and simply wondered if the formula P = depth * 0.1 is accurate enough to determine the water pressure at different levels or if it's to crude to be used at all. if i may have missed some points or posted in the wrong section please let me know. Thanks in advance.

I can honestly say I've never heard of that formula. I can't even see how it works.

Pressure is force per unit area (or in some cases given as mass per unit area). Your equation ignores the mass of fluid and the area over which it is applied.

All your equation does is reduce the depth by an order of magnitude.

If anyone else knows how your equation works (or yourself) please do explain, otherwise I'd have to say it's useless.

1kg of water applied to 1m2 isn't the same as 10kg of water applied to 1m2. Your equation would make them out to be the same (according to your equation you would get a pressure of 1 for both).

afaik, at zero depth (shoreline) the pressure is 1 atmosphere and at the depth of ten meters the pressure have doubled roughly to 2 so with this formula i've calculated that the pressure at 5 meters the pressure would be roughly 1,5 and at 53 meters the pressure would be 6,3. if i need another formula don't hesitate to correct me. i even missed +1 at the end of it.

P=depth*0,1+1
now i get it to work

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i now noticed that the formula is very crude and not working properly. i think this thread clearly shows my lack of education in the matter. but i still need a working formula. any thoughts?

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You can use the following formula to get pressure at certain depth:

(hmm for some reason I can't get tex working)

p = p0 + rho * g * h

where p0 = initial pressure, in this case 1 bar
rho = density of water = 1000 kg/m3
g = gravity acceleration = 9.81 m/s^2
h = depth which you want the pressure at.
So we plug those in and get
p = 1bar + 9810Pa * h and then we convert 9810Pa into bars
p = 1 bar + 0,0981*h
As you can see the approximation varies depending on which values you use for the g and rho.

ok, thanks. now i have a better understanding on matter :)