Nose weight

[Bob Richards]

Ok,

Air density(NTP) - helium density(NTP) = 
0.001205 g/cm^3 - 0.000166 g/cm^3 = 0.001039 g/cm^3

So, replacing a volume of air inside the airplane will
only save approx 1 gram per liter, at normal
atmospheric pressure.

But, your original question involved stuffing 15 grams
(mass) of helium into 1000cm^3. At normal pressure,
you would have only .166g, but you would have to
pressurize it to stuff the 15g into the stab, and
would end up with 15g of helium instead of the normal
12g of air. So, to answer your question, 15g of helium
in the stab would hurt, not help.

Bob R.


> --- Rcmaster199 at aol.com wrote:
> >  
> > In a message dated 3/15/2005 10:58:39 PM Eastern
> > Standard Time,  
> > jivey61 at bellsouth.net writes:
> > 
> > > I  guess the only real way to weigh it would be
> to
> > > completely evacuate a  tank (perfect vacuum?),
> > weigh
> > > that, then fill the tank (atmospheric  pressure
> > only)
> > > with helium and weigh that.
> > > 
> > > Talk about  subject drift!!!
> > > 
> > > Bob R.
> > 
> > 
> > I am surprised at you. There's a far simpler way
> > 
> 

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