Re: gravitational force -airplane motion and earths rotation

From: Richard Saam (rdsaam_at_att.net)
Date: 08/26/04 

Date: Thu, 26 Aug 2004 09:30:30 +0000 (UTC)

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John T Lowry wrote:

>
>Here is a snip from an article I wrote on this subject. The Physics
>Teacher rejected it because it was too complicated, suggesting I send it
>to the American Journal of Physics. AMJ rejected it because it was too
>simple! Oh well....
>
>The airplanes (one headed straight east, the other straight west, were
>taken over the equator, no wind, at altitude 2000 m, speed 200 m/s (389
>KTAS).
>***
>Because centrifugal terms for the eastbound and westbound airplanes are
>identical (since that term depends only on the rotation of the
>non-inertial frame and the common location of the two objects), it might
>appear there is no dynamical difference between the airplanes. But
>("just to be on the safe side") compute the Coriolis terms. On the
>eastbound airplane E, the Coriolis acceleration is 2SV outwards (to
>airplane E, upwards). On the westbound airplane W, 2SV inwards
>(downwards to airplane W). Unconstrained airplane W would be forced down
>by this Coriolis force and so to counteract W's higher effective
>acceleration of gravity - higher than E's by 4SV = 0.0583 m sec!2 -
>airplane W requires more lift. Higher lift is accomplished by W's taking
>a slightly higher angle of attack, the angle with which its wings meet
>the relative wind. But more lift means more induced drag (drag due to
>lift), hence - everything else being equal, as we have specified - a
>slightly higher fuel consumption rate for the westbound airplane. Since
>the acceleration of gravity near the equator is 9.780 m sec!2, there is
>a lifting acceleration difference of 0.0060 g. For a 10,000 lbf airplane
>(the scales having read somewhat light because of the centrifugal
>pseudoforce), perhaps at this air speed a World War II fighter, that is
>an effective weight reduction of 60 lbf just for having turned from
>westbound to eastbound. In practice, however, negligible. We apologize
>that we don't have space here for a meaningful discussion of induced
>drag.
>
>***
>
>
>
>John T. Lowry, PhD
>
>Flight Physics
>
>
>
On the subject of Coriolis Correction, the inside back cover of 'AIR
ALMANAC' published by The United States Naval Observatory indicates a
table to correct aircraft celestial bubble sextant fixes for coriolis
effect. The table heading states "To be applied by moving the position
line a distance Z to starboard (right) of the track in northern
latitudes and to port (left) in southern latitudes". Corrections in
nautical miles are in terms of Latitude and Ground Speed in Knots. The
correction is zero at the equator for all Ground Speeds proceeding
towards maximum at the poles. At 45 degree latitude and 500 knots the
correction is 9 nautical miles.

Assuming an aircraft moving east at 45 degree north latitude, the 9
nautical mile correction would be applied to the south.

Assuming an aircraft moving west at 45 degree north latitude, the 9
nautical mile correction would be applied to the north.

Logic would indicate that
 (cos(45 + 9/60) - cos(45 - 9/60)) / cos(45)
be representative of difference in pseudo distance traveled by aircraft
due to coriolus effect.

The value is .00524 (.524%)

So for a long trip of say 4000 nautical miles there would be pseudo
distance difference of 21 nautical miles.

There should be no effect on opposing great circle routes.

Richard

>
>

The airplanes (one headed straight east, the other straight west, were
taken over the equator, no wind, at altitude 2000 m, speed 200 m/s (389
KTAS).
***
Because centrifugal terms for the eastbound and westbound airplanes are
identical (since that term depends only on the rotation of the
non-inertial frame and the common location of the two objects), it might
appear there is no dynamical difference between the airplanes. But
("just to be on the safe side") compute the Coriolis terms. On the
eastbound airplane E, the Coriolis acceleration is 2SV outwards (to
airplane E, upwards). On the westbound airplane W, 2SV inwards
(downwards to airplane W). Unconstrained airplane W would be forced down
by this Coriolis force and so to counteract W's higher effective
acceleration of gravity - higher than E's by 4SV = 0.0583 m sec!2 -
airplane W requires more lift. Higher lift is accomplished by W's taking
a slightly higher angle of attack, the angle with which its wings meet
the relative wind. But more lift means more induced drag (drag due to
lift), hence - everything else being equal, as we have specified - a
slightly higher fuel consumption rate for the westbound airplane. Since
the acceleration of gravity near the equator is 9.780 m sec!2, there is
a lifting acceleration difference of 0.0060 g. For a 10,000 lbf airplane
(the scales having read somewhat light because of the centrifugal
pseudoforce), perhaps at this air speed a World War II fighter, that is
an effective weight reduction of 60 lbf just for having turned from
westbound to eastbound. In practice, however, negligible. We apologize
that we don't have space here for a meaningful discussion of induced
drag.

***

John T. Lowry, PhD

Flight Physics

Relevant Pages

  • Re: gravitational force -airplane motion and earths rotation
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