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<DIV>BRAVO!! </DIV>
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<DIV>The whole point was that it was a simple comparison of densities.
Eventhough it sounded alot more difficult than that.</DIV>
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<DIV>MattK</DIV>
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<DIV>In a message dated 3/18/2005 12:29:05 AM Eastern Standard Time,
bob@toprudder.com writes:</DIV>
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size=2>Ok,<BR><BR>Air density(NTP) - helium density(NTP) = <BR>0.001205 g/cm^3
- 0.000166 g/cm^3 = 0.001039 g/cm^3<BR><BR>So, replacing a volume of air
inside the airplane will<BR>only save approx 1 gram per liter, at
normal<BR>atmospheric pressure.<BR><BR>But, your original question involved
stuffing 15 grams<BR>(mass) of helium into 1000cm^3. At normal
pressure,<BR>you would have only .166g, but you would have to<BR>pressurize it
to stuff the 15g into the stab, and<BR>would end up with 15g of helium instead
of the normal<BR>12g of air. So, to answer your question, 15g of helium<BR>in
the stab would hurt, not help.<BR><BR>Bob R.</FONT></BLOCKQUOTE></DIV>
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