Date: 03 Mar 98 03:12:49 From: firstname.lastname@example.org Organization: Netcom Online Communications Services (408-241-9760 login: guest) References: 1 2
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In article <airliners.1998.358@ohare.Chicago.COM> email@example.com writes: >Nathan Pusey wrote: >If you were to measure your weight on an aircraft at FL400 and your >weight at sea level, you would find that you would weigh less at >altitude. For example, I weigh roughly 130 pounds at sea level, yet I >would only weigh perhaps 110 pounds at 40k feet. There are formulas to >determine this which can be derived from the relationship: > > F = G*m1*m2/r^2 So... If the Earth is F1 and you are F2, then F1/F2 -> R1^2/R2^2. Given an earth diameter of 6300 kilometers and the fact that at 40,000' you're at 10.1 kilometers, that gives us what, 6300^2/6300^2 = .9968. So I would anticipate that if you weigh 130 pounds at sea level, you'll weigh, oh, 130 pounds at 40,000'. > Since it is much easier to move an object through a very low-density >fluid than it is to move it through a higher density fluid, this means >that an aircraft will require less power in order for it to overcome >frictional forces due to wind resistance (ie: drag). This is a happy cooincidence, given the fact that engine power drops along with density and pressure (your steroid-ridden CFM-56 at 25K lbs at sea level is going to be making 2-3K lbs at altitude), however, it is not a primary factor. The primary factor for flying at high altitude is the thin air. The airplane, being an aerodynamic creature, must have a faster inertial airspeed in order to remain within safe parameters. Thus, assuming that engine costs can be kept in reason, one will simply get to the destination a whole lot faster. See firstname.lastname@example.org's article, <airliners.1998.359@ohare.Chicago.COM>, for details. -- Robert Dorsett Moderator, sci.aeronautics.simulation email@example.com firstname.lastname@example.org "Bother," said Pooh when his engine quit on take-off.