Misconceptions and Fallacies

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del_mo:

I don't know if everybody else missed it, but they had more than one-

>>Neoprene works by creating and maintaining a thin layer of body-temperature water next to your skin.<<

Wow, how does the neoprene know to gather only the body-temperature water, and keep the *cold* water out?

>>Drysuits have been tried for caving, but they are even less breathable than neoprene and are prone to punctures in caves. Then they become water bags with no insulation and negative buoyancy<<

Here is the suggestion, again, of negative bouyancy caused by the addition of *water* underwater.

>>You need a much looser suit than a diver would wear. Otherwise, you will be miserable.<<

Hmm, how is a loose wetsuit going to hold in all that *warm water*?

>>It is also very easy to undo the beaver tail to relieve oneself, and it is easy to take the whole jacket off to put a shirt on underneath.<<

Wow, doing the Warhammer in a cave? It's nasty enough in the ocean, but in an enclosed environment? And how is a shirt going to help?

I can't keep reading this site- how do cavers keep from killing themselves?
gomi_
 
charlesml3:
That is just absurd. The air is shot in at high pressure because you're pushing it into a pressure vessel. Oxygen and Nitrogen do not have sufficiently different weights such as to stratify in the tank. Fill it slow. Fill it fast. Roll the tanks. It makes NO difference.

I cannot understand why people try to make more of this than it is. Nitrox is AIR with a different ratio of Oxygen and Nitrogen. Air will not stratify in a tank (or anywhere else, for that matter). Nitrox won't either. Why would it?

-Charles

I used to think like you.
The gases will not stratify, Dalton sees to that, but if they are put in separately they take some time to diffuse. They are quite viscous at 200bar.
There have been several threads where various posters have worked through the calculations and the diffusion velocities are very slow and rolling the tanks will help speed up the process.
 
Blackwood:
That reminds me of a good one:

Myth: When you are neutrally buoyant, you are weightless.

I agree with the poster you quoted; if you are neutrally buoyant in the water you are weightless. But as he stated you are not mass less ie without mass.

Mass and weight are two different things. The terms are not interchangeable.
 
miketsp:
I agree with the poster you quoted; if you are neutrally buoyant in the water you are weightless. But as he stated you are not mass less ie without mass.

Mass and weight are two different things. The terms are not interchangeable.

I agree that they are not interchangeable. If you have mass, you necessarily have weight (unless, of course, you manage to get yourself an infinite distance away from any other masses).

Neutral buoyancy means buoyant = weight (i.e. the sum of all forces in the vertical direction = 0), no? One can't be zero unless the other is also.
 
Blackwood:
If you have mass, you necessarily have weight....

That's just incorrect. Take it from a physicist. "Weight" is defined as the net force acting on you. If you are in freefall (such as in orbit) or if you a neutrally bouyant, that force is zero. So, you are by definition weightless.
 
vondo:
That's just incorrect. Take it from a physicist. "Weight" is defined as the net force acting on you. If you are in freefall (such as in orbit) or if you a neutrally bouyant, that force is zero. So, you are by definition weightless.

Hmm... I seem to remember weight being defined as the gravitational force attracting masses together.

So you are saying that if I go stand on a scale it should read zero lbf? After all, the net force acting upon me would be zero...



According to the dictionary, "weightless" is a "condition being without apparent weight, as a freely falling body or a body acted upon by a force that neutralizes gravitation." So I guess when you are neutral you are weightless by that (odd) definition. But I'd love to see a physics text that defines weight as the net force acting on a body.
 
Quite simply, Newtons Second Law covers it.

Force = Mass * Acceleration

In the specific case of weight its only acceleration in the vertical direction. So, a neutrally buoyant diver has no acceleration in the vertical direction (positive or negative).

Thus:

Weight = Mass * 0m/s^2. Plug in any mass and you will have 0 weight.
 
To me, it seems that weightless should mean "without weight." But there is a definition for weightless that makes my post wrong.

This conversation has taken a turn for the useless, but I continue...

vondo:
Underwater? It sure will.

Anywhere.

If weight = the net force acting on you, you never have weight when you are in equilibrium. The cars in my garage are in equilibrium, thus they weigh nothing. The garage, the house itself, the stores down the street, the street... everything has no weight.

Wrong.

If I have mass, I have weight. Period.

However, if other forces are equally opposing that of gravity, I am "weightless" according to Webster.


JimC:
Quite simply, Newtons Second Law covers it.

Force = Mass * Acceleration

In the specific case of weight its only acceleration in the vertical direction. So, a neutrally buoyant diver has no acceleration in the vertical direction (positive or negative).

Thus:

Weight = Mass * 0m/s^2. Plug in any mass and you will have 0 weight.

The acceleration in question is that of gravity, or roughly 9.8 m/s^2 near the surface of the earth. Otherwise, as I sit un-accelerating in my chair or stand un-accelerating on a scale my weight is nill. That is clearly not the case.
 
How about "A mass that has weight, but due to it's current state, acts weightless"?

-----

Mike.
 

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