DM Physics

[amascuba]

I'm not the best with metric, but the basic math should be the same.

SAC = Surface Air Consumption

SAC = (BAR / Time) / ATA

33m = 3.3ATA (of water) + 1ATA (atmosphere) = 4.3ATA

(7 / 1) / 4.3 = 1.63

25m = 2.5m + 1 = 3.5

1.63 * 1 * 3.5 = 5.71

That's not correct is it? I believe I got volume and pressure mixed up here.

 

[vondo]

That's not correct is it? I believe I got volume and pressure mixed up here.

Number is right, units got messed up along the way and no units on the final answer. 6/10 points.

 

[Rick Murchison]

I'm coming up with the same answer as Mike. Negative displacement of 106.8lsw (110kg/1.03 kg). Add 106.8lsw plus 38.8lsw (40kg/1.03kg) = 145.6lsw.

1.03 is actually the high end of the sea water density scale, which is "normally" defined as about 1.018 to 1.029... so the "correct" answer is anywhere from about 145 to 150 liters additional displacement required; if you include the Med (some parts of the Med get as high as about 1.038!) then the low end drops to a bit under 144 liters.
I'm curious as to the "book" answer - if they use 1.03 they're using a density higher than the Atlantic (including the Caribbean), Pacific and Indian oceans...
But somehow that wouldn't surprise me
Rick