Which airplane will cover more ground distance?
 

Two aircraft are travelling at the same TAS at 1000 feet and 40000 feet respectively, for the same amount of time, which aircraft will cover more ground distance? (Assume nil winds)

RTFQ ..................!

Supposing we consider two concentric circles of 1000 ft and 40000 ft, and travel along them at a speed of x kts or km/h or m/s (any speed unit).
Surely the arc length covered by them in equal time would be equal, but if we look down vertically from 40000 ft, at the end of 1 hr, we would see that the lower plane is ahead, right? I'm really confused.

Is the earth flat? When did you start the stopwatch?
 

Of course the earth is flat and vehicles are either stationary or going flat-out Charlie. Obviously there is no lapse-rate or you would have mentioned it. So the temperature is everywhere invariant unlike a round earth for which constants aren't and variables won't. So on a flat earth with everything under the flat sun constant speed is everywhere constant therefore distance is defined by time. Being at 1000ft I saw the pub open with my eagle eye and instantly flicked my stop watch on. You on the other hand left your glasses at home, are blind as a bat and did not hear the click. So you did not synchronise watches. Therefore you turned up late and its your round. Mine is a Tooheys and I'm drier than a pommie's towel.

If the two aircraft are experiencing the same TAS, and the same wind, and travelling in the same direction, they will cover the same distance over the ground in the same time.

TAS + Wind = Groundspeed
These are all vectors, of course.

That said, the indicated (and calibrated etc) airspeed of the aircraft at high altitude will be significantly less than that of the lower aircraft. This is a measure of the dynamic pressure the aircraft experiences as it passes through the air, which in turn is a function of the density of said air, and the square of the velocity. The density of air at 40000 feet is significantly less than density lower down, hence the difference. This dynamic pressure is also an indication of the sort of energy state the aircraft is in, and the sort of flight regime that would be encountered. The lower IAS (or "Q") means lesser control response, higher angle of attack, and less power available, for example.

In reality, the reduction in the speed of sound with temperature means transonic effects complicate the matter somewhat at high altitudes. Also, in reality, if we are considering identical aircraft, I would imagine it would be pretty difficult to achieve the same TAS at 1000ft as at 40000ft without:

a) Ripping the wings off at 1000ft because of the excessive IAS/Q as mentioned, or
b) Being unable to maintain altitude at 40000ft because of the insufficient lift because of the low IAS (or, in seeking sufficient lift, exceeding the critical angle of attack and stalling).

Sadly it is too early in my career to have experienced flight above 10000ft, however its not very long ago I was in the classroom getting my head around this stuff, and this is how I would explain it to myself.

Objections/comments/criticism welcomed!

seems pretty obvious. the aircraft at 1,000ft spends less time in the climb and more in straight and level cruise so must cover the greatest ground.

circumference is pi times the diameter
so a circle at 1,000ft has a circumference of 12,742,304.8m x pi = 40,031,131m
and a circle at 40,000ft has a circumference of 12,754,192m x pi = 40,068,476m
not that much difference.

Leaving aside conveyor belts, RTFQ and the like, the answer becomes obvious if the second 'vehicle' was a satellite orbiting at 22,236 miles. Now which one covers the greatest ground distance?

Mine's a 4X, enicalyth:)

For all practical purposes, yes the ground distance covered by both will be the same. But I still think actually or rather theoretically, the higher airplane will cover lesser corresponding ground distance than the lower airplane. The air distance though will be exactly the same.

Just like BOAC said. A geostationary satellite rotates at a speed of 3.07 km/s to keep a stationary position w.r.t the earth. But the speed of earths rotation is 460 m/s. Which implies if both the sattelite and earth were moving at the same speed (say 460 m/s) the satellite would lag behind the earth in the same given time.

Here's some calculations (albeit late night ones so excuse any nonsense).

These are calculated assuming no wind, no temperature variations, no losses and starting and ending at a constant speed (ie. no acceleration).

Google tells me the mean radius of the earth is 6,371.009km, so to travel half way around the world (half the circumference) would be approx 20015.09km at sea level. To do the same at sea level +40000ft would mean a radius of approx 6283km and therefore an A to B distance of 20052.79km.

To put a slightly difference spin on it, let's assume we need to travel A > B in 30 hours. At sea level this would require a speed of roughly 667.17km/h and at 40000ft would require a speed of roughly 668.43km/h.

Hope this makes some sense!

Thanks everyone for your replies. I think the last post makes the same point that I was making earlier. On the surface (no pun intended) this question looks very simple, but it's not! :ok:

hi
Its given TAS is const therefore aircraft flying at lower altitude will travel a longer ground dist in a given time.

Don't forget to include the time compression as ruled by Einsteins theory of general relativity - the higher aircraft is subject to a lower gravitational field... :}

The aircraft travel the same distance.

The wind is nil. Wind is measured relative to the ground. Therefore the air at 40000 feet is entirely stationary relative to the ground and the air at 1000' is entirely stationary relative to the ground. The aircraft have the same TAS and therefore cover the same distance relative to the ground.

The points made by other posters about having to travel further when you are higher because of the larger circumference you are traversing is a red herring. The fact that the wind is nil compensates for this effect. (Basically the air at 40,000' is moving faster than the air at 1000' in order to remain stationary reference the ground.)

There is no mention in the question of a climb and there is no reason to assume the question refers to aircraft that start on the ground and climb to altitude. They could just as easily climb to altitude and get stabilised at the correct speed before beginning the experiment. And that is what the question implies.

That's an interesting point. Therefore (and I know it must be infinitesimally small) if you take off and climb to 40,000' you are climbing into a wind gradient in your still air theory.

Yes that would be true.

AerocatS2A, would it then be true to say that the higher aircraft is indeed flying with a faster ground speed than the lower aircraft with the same TAS and nil wind?

No, the only thing moving faster is the air the aeroplane is flying through. It must move faster in order to remain stationary reference the ground and meet the "nil wind" requirement. As long as the nil wind requirement is met, the aeroplane can fly with the same TAS as a lower aeroplane and cover the same ground distance.

Basically the previous posters are correct that something that is higher has more distance to travel because of the increasing circumference around the Earth as you get further from the centre of the Earth. What they missed was that this extra distance is entirely offset by the extra speed* the air has in order to remain stationary over the ground. It then comes back to the basic formula of TAS + wind = ground speed and ground speed / time = distance. With nil wind the TAS = ground speed and both aircraft travel the same distance over the ground.

Of course in real life the wind can't remain completely stagnant like that as you go higher. So the question is purely academic, and I think it's just designed to see whether you understand what TAS is. You're probably not meant to (and don't need to) start considering extra distance travelled the further you get from the centre of the Earth.

*As measured by an outside observer. The speed reference the ground is still zero.

The one at 40,000ft.

AerocatS2A

To test your theory let’s consider two altitude such that the still air at the higher altitude must move twice as fast as the still air at the lower altitude, so that both parcels of air remain stationary relative to the ground. (I know that the atmosphere isn't really thick enough for this, but it is only the geometry that matters)

Now let’s imagine that we have two aircraft, one at each of the two altitudes, hovering in this still air, such that in each case their TAS is zero.

After 1 hour the earth will have rotated through 15 degrees. The two parcels of air and the two aircraft will have moved through the same 15 degrees to remain over the same point on the ground. But the inertial distance moved by the higher aircraft will be twice that moved by the lower aircraft. You theory is good so far.

Now lets set both aircraft to be flying in the direction of earth rotation at a TAS of 10 knots. After one hour the earth and the air will have rotated through 15 degrees. The lower aircraft will have moved through 15 degrees plus whatever angle equates to 10 nautical miles at that altitude. The higher aircraft will also have moved through 15 degrees plus whatever angle equates to 10 nautical miles at that higher altitude.

But the angle equating to 10 nautical miles at the lower altitude will be greater than the angle equating to 10 nautical miles at the higher altitude. In this case it would actually be twice as much. So the lower aircraft will have moved through a greater angle and a greater distance over the ground.

Yes Keith I think you are right. I missed that the air distance is greater for a given ground distance so the the ground distance travelled for a given airspeed will be less.

Must've been having a slow brain day.