Posted by TDI_2 on July 16, 2002 at 10:14:00:
In Reply to: Re: Weight Adjustment for Diving in Fresh & Salt Water posted by Chuck Tribolet on July 16, 2002 at 08:27:59:
As Eins pointed out, there are several good approximations out there that can get you close, as long as you dont mind "keep the change."
I was amazed by Chris' calculus (summations) approach to the same issue.
If you dive regularly on the Coast, and for a change you would like to go dive at one of the several large freshwater lakes, dividing 1.02564 into your total Ocean weight (diver and all gear, like Chuck emphasized) gives you the freshwater equivalent. Then all you have to do is subtract, to figure out how much weight to pull off your belt.
Last time I went diving at a local lake, we just wore our Ocean weights. Guess we were all just lazy. And very negative too. That is what raised the question in my mind, so I researched it, and then brought it here to this great Forum for confirmation.
Right now the freshwater vis is not much better than in the Ocean. We had 5 to 10 feet of vis depending on depth, due to algae growth in the bright summer sunshine and long summer days. Deep lakes have thermoclines as well, just like the Ocean, as I found out. Its warm for the first 15 feet near the surface, but as cold as our Pacific Ocean below when you go deep, really deep.
I forgot who pointed out that the displacement of the weights themselves is also a factor in converting buoyancy, though negligible for our own purposes. There are some extremely bright people here who dont miss any details.
1.02564 comes from the ratio in similar units (pounds, kilos, coconuts etc) of seawater to freshwater. In pounds per cubic foot, it is 64.0 / 62.4 .
Someone else also pointed out a different twist, that you can also compute your own cubic diplacement by diving your total weight by the weight of a unit of seawater displaced, or of freshwater displaced as well. For example, 331 / 64 = 5.2 cubic feet of seawater, with tech gear on. So a goodly portion of that will be your twin tanks, and your stage bottles.
I doubt that the difference between the salinity of the Red Sea would make the equation vary significantly , but Jason is right, that certain seas have differing salinity and therefore different weights and densities. This is an interesting question, if youre diving in the Near East [Egypt or Israel] or the Mediterranean [Italy or Greece or South France].
Taking the reciprocal should work as well, for adjustments from pool to Ocean. 62.4 / 64 = 0.975 .
THIS is the factor that I remember Ken Kurtis mentioning before, in a post several months or even a year or so ago. And that is why I asked him about going the other way, from Ocean to lake.
However, the key is in what to do with the 0.975 . You have to divide this INTO the weight of the diver and all gear. Or else the pool-to-ocean conversion does not work.
Ocean-to-lake example:
217 lbs diver
100 lbs tech gear
14 lbs lead wts
331 total dry weight
331 / 1.02564 = 323 lbs equivalent freshwater weight.
331 - 323 = 8 lbs to remove
Pool-to-ocean example:
217 lbs diver
100 lbs tech gear
6 lbs lead wts
323 total dry weight
323 / 0.975 = 331 lbs equivalent freshwater weight.
331 - 323 = 8 lbs to add
I doubt it would be diplomatic to ask everyone in a scuba class to stand up on a scale with their wetsuits on, before they get them wet in the pool.
And if you asked them what they weight, they would probably just lie. This is California. Nobody tells the truth about how much they weigh, or how old they are, or how much they earn!
So this is one of those formulae that would probably only work for yourself, and not for your scuba students nor for your girlfriend/wife.
Therefore you may want to go with Eins' approximations instead, after all! And keep the change.