Nose weight, now Tail Helium

[Bob Richards]

I guess I was tending to approach the problem in terms
of metrology rather than physics. In the work I do, I
trust my own measurements more than taking someone
else's word. Even then, I try to think of more than
one way to measure something just to make sure I have
correlation (is the equipment working properly!).

I guess I should also say:

ASSUMPTIONS: densities, found on the internet, are
correct. :-)

Bob R.

--- Rcmaster199 at aol.com wrote:
>  
> BRAVO!! 
>  
> The whole point was that it was a simple comparison
> of densities.  Eventhough 
> it sounded alot more difficult than that.
>  
> MattK
>  
> In a message dated 3/18/2005 12:29:05 AM Eastern
> Standard Time,  
> bob at toprudder.com writes:
> 
> 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.
> 
> 
>  
> 
=================================================
To access the email archives for this list, go to
http://lists.f3a.us/pipermail/nsrca-discussion/
To be removed from this list, go to http://www.nsrca.org/discussionA.htm
and follow the instructions.