Water as weight rather than lead?

[cool_hardware52]

Originally Posted by Charlie99
If I replace my 10 pounds of lead with gold, how much less weight do I have to carry on the weightbelt?

On today's market? About $63,040.00.

I think it's your wallet that will be lighter....



Tobin

 

[donacheson]

the question:

can you use water as weight, instead of lead weight?

For example, say that you take containers of water instead of lead. Yes, water weights six pounds per gallon, so you would have to take two gallons of water to equal 12 pounds, but...

Say you tie two milk gallons full of water to your waist...

Wouldn’t it be the same as wearing lead weights?

We were having this discussion, and someone was saying that wouldn’t work, because the water would “float on water."

is this correct?

Yes, you'd increase your weight by 12 pounds. Unfortunately, the water "weights" would also displace 12 pounds of water, so by Archimede's principle, you'd also be 12 pounds more bouyant.

 

[donacheson]

Ah, that's the tricky thing about water. It's actually densest right before it freezes, then it becomes less dense. That's why ponds freeze from the top down.

Actually, water is most dense at 3.98C, about 39.2F.

 

[widebody50]

Let's not be too hasty here. I think the original poster may have been referring to using "Heavy water" as a substitute for lead weights. In that case, it certainly would be heavier than water. Approximately 10% heavier, in fact. Added benefit is that if you are diving near a nuclear plant you are automatically protected from radiation, since "heavy water" is also a moderator of nuclear fission. The price, however, might be a tad higher than lead weights. Hydrogen sulfide separators are kinda pricey.

Mr. Science.

Actually, Mr. Science, not ot be too picky, but...most commercial nuclear facilities in the U.S. use light water as a moderator. While it is true that heavy water is also a good moderator for neutrons, the last part of your statement is why most U.S. nukes are designed for light water reactors. Additionally, just that fact that you are diving in water, and would be at least several hundred yards from the operating reactor of any facility (or else they would be shooting at you), you would have more than enough shielding from any radiation source that could possibly pose a problem for said diver. Water is an excellent source of radiation shielding. Anybody got any donuts? (mmmmmmmmmmmmmmmm donuts.........)

And, while I'm on a roll...the original poster stated that a gallon of water weighs about 6 pounds. It is actually more like 8.5 pounds per gallon. The old saying still holds true: a pint's a pound the world around.

Mr. Know-it-all (or so I think)

 

[Don Burke]

If I replace my 10 pounds of lead with gold, how much less weight do I have to carry on the weightbelt?

On today's market? About $63,040.00.

Heck, you can get a whole gram of osmium for less than that.

 

[widebody50]

I've often thought of depleted uranium for weights. I've dealt with that stuff and GEEEEZ is it heavy (dense). The problem is cutting the stuff. There really isn't a radioactivity problem, hence "depleted". A couple of pieces about the size of 2 "D" cell batteries would give you about 10 pounds of weight.

Depleted uranium has about twice the density of lead.

Welllllllllllllll, not exactly. "Depleted" means that the Uranium nuclide of choice (ie, Uranium-235, which is used in most commercial and military reactors in the U.S.) has been all but used up, but what is left behind is some pretty "hot" stuff. And, it's going to be that way for a VERY long time! See, the Uranium is "enriched" so that the desired Uranium nuclide is at a higher percentage than naturally occuring Uranium. Once that nuclide is used up (depleted) then the fuel is put into long term storage. Maybe the Uranium you have dealt with is what is left after the enrichment process. If it hasn't been subject to a neutron flux then it wouldn't be any more or less radioactive than what is in the ground. In that case, I guess you may be right.

 

[android]

No, depleted barely radiates. It was used for counterweights on some airplanes I worked on in the Navy. The A-4 stabilizers if I remember right. You did have to fill out some special forms if you had to remove/replace it though.

 

[No Fish]

Say I have a BCD that consists of a cooling unit.,,,say.... for the sake of this discussion, when I have it on, I am neutrally bouyant.

and I am in 70 degree water,

Now thowing the switch, (zzzzzzzppp...whrrr...whrrrrr....hummmmmmmm) I cool off the water surrounding me...... estimate a 3-foot distance all of the way around.

Now what happens, I think I know.......

 

[Charlie99]

and I am in 70 degree water,

Now thowing the switch, (zzzzzzzppp...whrrr...whrrrrr....hummmmmmmm) I cool off the water surrounding me...... estimate a 3-foot distance all of the way around.

Now what happens, I think I know.......

Water at 4C is 1.0g/cm3. Water at 70F (about 21C) is 0.998g/cm3.

IF your body doesn't change density, then you would gain buoyancy in the colder water. Only about 2 parts per 1000, or about 1/2 pound for diver + gear of 250 pounds total.

OTOH, if various bits of your body shrink in cold water, you might be denser too. :banana:

Come to think of it, somebody who purposefully dives in 4C water has got to be really dense.

 

[WJL]

Water at 4C is 1.0g/cm3. Water at 70F (about 21C) is 0.998g/cm3.

IF your body doesn't change density, then you would gain buoyancy in the colder water. Only about 2 parts per 1000, or about 1/2 pound for diver + gear of 250 pounds total.

OTOH, if various bits of your body shrink in cold water, you might be denser too. :banana:

Except this theoretical cooling unit has to put that heat it took out of the water somewhere - the only place handy being No Fish's body. So, if anything, he's going to get a LOT hotter.