Lift due curvature of the earth?
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Lift due curvature of the earth?
Hi,
There is a lot jokes that Russian airplanes use this to get airborne but how does it really effects lift? According my calculations: Schuler period is 84min, lets say jet airplane can go around the world in 40 hours. 1.4/40*100 we get around 3.5% of the lift generated by the curvature of the earth. Is that true? If so is it taken into account calculating lift of the airplanes in cruise?
There is a lot jokes that Russian airplanes use this to get airborne but how does it really effects lift? According my calculations: Schuler period is 84min, lets say jet airplane can go around the world in 40 hours. 1.4/40*100 we get around 3.5% of the lift generated by the curvature of the earth. Is that true? If so is it taken into account calculating lift of the airplanes in cruise?
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Is that true?- if it wasn't, satellites would not stay in orbit and aeroplanes would fall out of the sky if they flew too slowly - wait a minute...........................
If so is it taken into account calculating lift of the airplanes in cruise? - I very much doubt it
If so is it taken into account calculating lift of the airplanes in cruise? - I very much doubt it
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The ratio of the Schuler time to the time that it takes to travel around the world is not the percentage of lift that you get for free.
The centripetal force is given by F = m * V^2 / r
Where:
m - aircraft mass
V - aircraft speed
r - radius of the earth
Lift is given by L = m *g
Where:
g - gravitational acceleration
The percentage of Lift "generated by curvature of the eart"
V^2 / (g*r) * 100
Substituting:
r =~ 6400000 m
V =~ 280 m/s
g =~ 9.81 m/s
Gives 0.12%
My guess is that is within the error margin of most measurements and calculations and therefore it is probably not used.
For validation of the above you can calculate the "Schuler speed" and substitute that:
V =~ 40000000 / (84*60) =~ 7936 m/s
Gives 100.31% ~ 100% Q.E.D.
The centripetal force is given by F = m * V^2 / r
Where:
m - aircraft mass
V - aircraft speed
r - radius of the earth
Lift is given by L = m *g
Where:
g - gravitational acceleration
The percentage of Lift "generated by curvature of the eart"
V^2 / (g*r) * 100
Substituting:
r =~ 6400000 m
V =~ 280 m/s
g =~ 9.81 m/s
Gives 0.12%
My guess is that is within the error margin of most measurements and calculations and therefore it is probably not used.
For validation of the above you can calculate the "Schuler speed" and substitute that:
V =~ 40000000 / (84*60) =~ 7936 m/s
Gives 100.31% ~ 100% Q.E.D.
Last edited by ATCast; 3rd Apr 2013 at 08:00.
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If so is it taken into account calculating lift of the airplanes in cruise?
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The effect of rotation is greater than you might think; any artilleryman will tell you that when precision shooting a correction must be applied to take it into account. Shooting North-South and vv the correction is entirely to the azimuth, and East-West and vv it's to the range. As you might guess, on other bearings it's a mixture.
I forget the values of a typical correction; could it have been 10m on a 35 second time of flight when shooting, say, due North?
I forget the values of a typical correction; could it have been 10m on a 35 second time of flight when shooting, say, due North?
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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
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.
To counteract the Coriolis effect a bank angle of approximately 0.23 degrees to the left is needed
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Earth's rotation is one of the considerations airport designers consider when locating park spots. West facing is safer.
That in turn would have to be countered by a bit of rudder trim, would it not?
Isn't there another thread on that subject?
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Flight Controls and Earth Rate
A related issue that does require more consideration than lift is the angular rate impact of Earth's rotation. Earth rate is ~15 deg/hour (slightly less than 360 degrees in 24 hours). Depending on lattitude, heading, and speed an airplane will see quite a range of steady angular rates while at cruise. At the equator heading West pitch rate will be nose up. Turn to the East and pitch rate will be nose down and quite a bit larger. Head north and you will find right wing down roll rate to maintain wings level. Head south and left wing down roll rate is needed. There is a noticable difference between maintaining constant pitch and roll attitude and having the associated angular rates be zero.
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Head north and you will find right wing down roll rate to maintain wings level. Head south and left wing down roll rate is needed. There is a noticable difference between maintaining constant pitch and roll attitude and having the associated angular rates be zero.
In my book if something is constant, that the associated rate is zero. Per definition...
Don't you mean the associated roll/pitch moment?
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I thought this was going to be a thread about early model A340's
Last edited by ATCast; 3rd Apr 2013 at 17:36.
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You confuse me.
In my book if something is constant, that the associated rate is zero. Per definition...
In my book if something is constant, that the associated rate is zero. Per definition...
So - how does a 340 manage to fly, anyway?
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That's why we learned about Earth Rate Compensation and Transport Rate Compensation during those hours of ATPL theory. Yes, Inertial Systems are stable relative to space coordinate systems, indeed need to be perturbed to keep them stable relative to the Earth coordinate system that, itself, moves and tumbles through space.
