How efficient are airplanes at reversing direction?

Factual Questions
1

I may be phrasing the question poorly, so please bear with me. I always had the assumption that it’s harder to turn around and reverse direction with a rocket in a vacuum than an airplane in the atmosphere, because with a rocket you have to spend fuel to cancel your momentum, and with a plane you can use the air to do that.

Now that I think about it, I don’t really know the mechanics of how the plane does that, or how efficient it is, compared to a straight reaction thruster. I assume it’s much more efficient at least from a fuel consumption point of view, but with regards to the physics, is it just a matter of deflection of air, or is there some more complex process going on here?

In other words, how much energy does it take to turn an air plane 180 degrees and travel at the same speed compared to a rocket?

2

If I understand your question right, and I’m not sure I do, the answer is that it only takes slightly more energy to turn an airplane around than it takes to continue flying in a straight line. Airplanes are fairly efficient at turning.

Planes turn by increasing their bank angle (having one wingtip lower than the other). This creates a sideways component of lift that gradually changes the direction of flight.

To maintain altitude during this process, the pilot has to add a bit of up nose-up attitude to compensate for the reduction in upward lift. This increase of angle of attack creates an increase in drag, which means that to maintain both speed and altitude during a 180 the pilot has to increase engine power, thus burning slightly more fuel.

So it does take more energy to turn, but it’s incremental – unlike the rocket scenario you mentioned in which you’d have to completely arrest forward momentum, turn around, and accelerate back to speed in the other direction.

Does that help?

3

Edit: I mixed two scenarios with and without gravity, I feel stupid.

4

You can also reverse by pulling into a climb and looping over the top, rolling the right way up again and descending to your previous height, which wastes even less energy (but is not really comfortable enough for civilian transport :smiley: ). And as Figaro says, this is a darn sight more efficient than basically throwing away all the energy it took to get up to speed and then spending the same yet again to go the other way.

If you really want to change direction drastically in space, it helps if you can take advantage of a nearby gravity well. When Apollo 13 blew, it had next to no delta-vee left and a largish velocity away from Earth. It was only thanks to the proximity of the Earth itself and the Moon that it was able to return.

Unfortunately, this is completely wrong, except for the “and to fly at all” part, which is the right answer to a different question.

5

Yup. I realized that much :smack:

As to the OP, you can’t really compare efficiency of rocket in no gravity/no atmosphere to plane with gravity/atmosphere. Plane does a lot of it’s acceleration at the take-off, and then just constantly spend energy to keep flying. Rocket needs only initial impulse, and then spend nothing to keep going. But to change movement vector by 180 degrees and keep it’s velocity rocket needs to use thrust equal to 200% of initial thrust, but in opposite direction (weight of fuel omitted for simplicity). For plane to change direction by 180 degrees you need only to slightly increase your energy expenditure (either by losing altitude or increasing throttle). Anybody care to do some calculations including time/speed/fuel consumption/distance/etc.?

6

I guess my question really is about how an airplane turns. I understand on a general level what Figaro and Malacandra are saying about turning, but some part of me still thinks it’s odd that a plane can turn easier than a rocket. I guess I did forget that you need to constantly provide thrust to fly a plane in the first place, like puppygod said, but my instincts say, if you have a plane and a rocket with the same mass, it ought to take more energy to reverse the direction of a plane, since not only do you have to cancel out its momentum, you also have to deal with drag and maintaining lift and all that. This is clearly not the case though, at least with respect to the amount of fuel they would need. Part of that is probably because a plane can use ambient air as reaction mass, where as the rocket has to provide it. But the part I’m having trouble is, what is going on that allows the plane to turn. How do the forces balance, and how is momentum being conserved? With a rocket I can picture this pretty easily. With an airplane I’m not so sure.

7

This may not be directly germane to the OP, but (IANAP) my understanding is that there is something called a coordinated turn that requires control from the tail as well as banking. You can turn without banking and you can bank without turning, although those are not textbook methods.

8

Hmm… if the plane is using a certain amount of energy anyway to keep aloft, it’s a bit like it’s building its own “road” under it. A road of air. That road becomes banked in a turn, much like the surface of a velodrome. So a plane keeps its momentum in a turn in pretty much the same way a cyclist in a velodrome does.

At least that’s how I visualise it.

9

Bolding mine. This is where you’re going wrong. You don’t have to cancel out a plane’s momentum to turn it. You just need to adjust the direction of lift.

10

Yes, absolutely - banking the wing changes your direction but not your facing, kicking the tail round with the rudder does the opposite. There are generally self-correcting mechanisms that see to it that the plane will end up facing the way it’s going anyway. For instance, dihedral (both wings slant upwards from the horizontal, as seen from in front) will automatically bank the plane into a rudder-only turn, which is used a fair bit in simple model aircraft. Similarly, the tail of a plane that’s sideslipping will be kicked around by the fin even if you don’t touch the rudder. But coordinating turns is safer and more controlled, especially if the plane’s designed for agility rather than stability.

11

Exactly. What you’re doing is using part of the plane’s momentum to provide your lateral thrust vector. Keep that up for long enough, and your 180 turn will happen. It costs you some kinetic energy to do that, but by no means all.

12

The coordination of rudder is to prevent skidding of the tail in a turn. It increases the efficiency and comfort of the turn, but it is not responsible for the change of direction. You cannot, practically speaking, turn an airplane without banking, but you can bank without turning. Landing in a crosswind requires banking into the wind with an opposite application of rudder to maintain a straight direction of flight, for instance.

13

Let’s reconsider this. I’m using a rusty recall of physics, so please jump in and point out the errors.

The solution is based on:

  1. Conservation of linear momentum
  2. Conservation of mechanical energy

In the plane case, the plane is traveling north (pos. velocity) through some huge mass of air. We have:
momentum = m(plane)*v(plane) + m(air)0
energy = 0.5m(plane)v(plane)^2+0.5m(air)*0^2+plane’s fuel

After it reverses course to the S by banking and returning to the same altitude and speed we have:
momentum = m(plane)*-v(plane) + m(air)v(air)
energy = 0.5m(plane)v(plane)^2+0.5m(air)*v(air)^2+plane’s new fuel level

The momentum of the plane does have to be ‘canceled out’ in a sense, by accelerating the body of air it banks through. The delta in fuel levels (considering fuel just as some form of potential energy) is
0.5*m(air)*v(air)^2, so the energy expended depends on the mass of air moved by the plane turning.

In the rocket example
momentum = m(rocket)*v(rocket) + m(rocket ballast)v(rocket)
energy = 0.5m(rocket)v(rocket)^2 + 0.5m(rocket ballast)*v(rocket)^2 + rocket fuel

After reversing thrust and returning to initial speed, reversed heading we have:
momentum = m(rocket)*-v(rocket) + m(rocket ballast)v(rocket ballast)
energy = 0.5m(rocket)*v(rocket)^2 + 0.5&m(rocket ballast)*v(rocket ballast)^2 + new rocket fuel level

The change in rocket fuel energy is:
m(rocket ballast)*v(rocket ballast)^2 - m(rocket ballast)*v(rocket)^2

Again, the amount of energy used depends on the mass of ballast used to reverse course.

For each of these scenarios we can artificially select values for planes and rockets of equal mass and velocity in which either the plane or rocket expend more energy.

In reality, I suspect that the plane effects an enormous mass of air, especially when compared to a rocket’s ballast (you have to launch the ballast with the rocket after all), which is why it expends far less energy.

*I completely neglect friction, which is a small advantage in favor of the rocket.

14

You can turn without banking though it is not useful, not particularly comfortable, and requires opposite aileron to stop the wings from banking into the turn.

Also, it is not necessary to increase power in most turns, normally you’d just accept a slight drop in airspeed.

15

I don’t mean cancel out the momentum as in having to stop the plane completely in mid air, just that the vector component in its previous direction of travel has to be reversed.

The analysis in hdc_bst’s post is what I was looking for.

The main thing I have trouble grasping is, how does the plane affect that enormous mass of air? Seems like it would have to be an amount close to the mass of the plane itself. Is it purely through deflection, or are there some more subtle aerodynamics going on?

Edit: fixed a formatting problem

16

The plane continually affects an amount of air equal to its own weight *, to create a lift vector exactly equal and opposite to its weight - that’s what’s keeping it up. If you begin a gentle turn with, say, ten degrees of bank, and want to keep your altitude the same, then you must affect rather more - you have a vector equal to the plane’s weight directly upwards, and a horizontal vector, and the actual lift vector (perpendicular to the wing) is equal to the root of the sum of the squares of these - or weight divided by cos 10 degrees (weight times sec 10 degrees, if you prefer). You have to adjust thrust to compensate for the increased lift you now need, but not by an enormous amount.

You then need to keep this up until your heading has changed by the required amount, but I am definitely going to leave that piece of math as an exercise for the student. :smiley:

Of course, as you’re in a constant gravitational field while all this is going on, you can divide through by g and talk about mass instead of weight, but strictly speaking you’re balancing forces, not masses.

*Hmm, is it, though? It depends on how much air it’s pushing downwards and at what velocity. I may be oversimplifying too much here… :dubious:

17

Let’s simplify this a bit:

You can bank a plane (in air) without applying any additional energy whatsoever. This involves using the gravitational potential energy stored in the plane to transfer momentum from the plane to the atmosphere by banking and giving up altitude for change in momentum. Once you’ve done that, calculate the lift time required to go from the lower altitude to the higher initial altitude, and fuel consumed in doing so; voila!

For a rocket it is even easier; assuming that you’re in free space, calculate the propellant mass expenditure using the Tsiolkovsky rocket equation and the specific fuel consumption from the vehicle specific impulse and/or motor characteristic effective exhaust velocity. Compare SFCs between the two scenarios to get comparative efficiencies.

Note that it is a trivial question to answer in qualitative form: oxidizer-carrying rockets are hugely inefficient (in terms of power consumption) in comparison to air-breathing engines, even without considering the benefits of aerodynamic lift versus direct thrust.

It is, of course, possible to alter and even reverse course of a rocket in space by using a swing-by maneuver to transfer momentum from or to a large moving body (up to twice the relative speed between the body and the vehicle). In this way you can gain huge “efficiencies” over purely ballistic “straight” geodesic courses from one point to another, albeit often by taking a very circuitous and convoluted path. I wrote a Staff Report on this as a guest contributor, but it seems to have been eaten by the recent changeover.

Stranger

18

I think I see it now. I did some reading about lift, and I was thrown off by the old text book explanation of lift being generated by Bernouli’s principle I remembered, which apparently is wrong.

That does bring me to another question though. How are the momentum of the plane and the air balanced during level flight? If the plane is maintaining the same altitude, then it’s got no momentum in the vertical axis, right? If it’s pushing air downwards to provide lift, what’s keeping the momentum conserved? Am I misunderstanding something about the conservation of momentum?

19

All the smart people told me that for all practical purposes, energy can not be created or destroyed.

So we take an plane of X mass and a rocket of that same X mass to the same Velocity V42 and then change the direction 180 degrees.

Would not all the variables, when added up not be equal?

Great big ones and itty bitty ones? All adds up to equal, right?

If you magically put them in different places with no energy penalty, then the thought experiment can be loaded to come out any way you want…

YYMV

20

The momentum of the entire atmosphere is conserved, and the momentum of the airplane is conserved. It is not just a sheet of air that is pushed down. Two vortices are created by the wings. The upward side of the vortex balances the downward side, but the wing only acts on the downward side.

Consider two rows of end-to-end floating logs a foot or so apart. With a lot of talent, it would be possible to run down the two rows, leaving the logs spinning behind you, and using the rotational inertia as well as the flotation to keep yourself from dropping into the water. Now suppose the logs are water-logged, and have neutral buoyancy. You can still exploit the rotational momentum to stay out of the water. Since the logs are neutral bouyancy, they have the same density as water, so if you could somehow spin a cylinder of water, you wouldn’t need the logs at all…this is what a hydrofoil does…it has underwater wings.

So the key is that the aircraft suspends itself using the rotational momentum of two cylinders of air, not the linear momentum of one sheet of air.