PPRuNe Forums - View Single Post - Lift due curvature of the earth?
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Old 3rd Apr 2013, 10:30
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ATCast
 
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Depends on how fast you are flying and in which direction. Concorde performance brochures were calculated for flying north/south so no effect
There is an effect, even when flying North - South. But the North - South effect is the same as the South - North effect.

The East - West effect is different from the West- East effect.

For simplicity I omitted the effect of earth rotation. You can include the rotation but it becomes a bit more complex. Let's not include any effects of the earth rotating around the sun, effects of the moon; these are negligible for these purposes but make it far more complex.


Flying eastward around the earth at the equator would yield 2.67 rotations around the earth centre when the circumvention takes 40 hours. One rotation for you flying around, 1.67 rotations the earth does by itself in the same time. The resulting rotational speed is one rotation every 15 hours.

Flying westward at the same speed would yield 0.67 rotations around the earth centre, still in the same directions as when one would fly eastward. The rotational speed is one rotation every 60 hours

In the latter case contribution to the lift would be negative because the rotation around the earth's centre would be slower than normal.

If you are standing still on the equator you are moving around the earth centre with a speed of approximately 462 m/s. That means that the "lift" you get from standing still is approx 0.34% of g.

Moving east at an additional 280 m/s would give a total of 740 m/s, resulting in 0.88% g centripetal force, which is a difference of 0.54% g from standing still. So in effect you need to produce 0.54% less lift than your weight.

Moving west at 280 m/s would give a total of 180 m/s, resulting in 0.05% g. So in effect you need to produce 0.29% more lift than your weight.

The difference in lift to generate is 0.83% between flying eastwards or westwards at the equator.


Moving north at the equator at 280 m/s gives a resulting speed of
sqrt(280^2 + 462^2 ) =~ 540 m/s
This results in a centripetal acceleration of 0.46% g, which is about 0.12% g more than when standing still at the equator. So flying north-south or south-north over the equator with 280 m/s requires 0.12% less lift than your weight at the equator.

Moving on the South-North-South track over the North pole also benefits from 0.12% centripetal acceleration when flying at 280 m/s. To counteract the Coriolis effect a bank angle of approximately 0.23 degrees to the left is needed, but this has no significant impact on the required lift.
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