In descent, an aircraft must lose both potential energy (in order to get down to the ground) and kinetic energy (in order to land at a safe speed).

Weight and drag can take care of that. An unpowered plane will eventually slow down and sink. However, sometimes it won't happen fast enough without additional control input, and the pilot must act to make the plane less aerodynamically efficient, and dispose of some surplus energy that would otherwise be retained longer.

If a plane completed its descent from top to bottom without using the control surfaces to:

  • dump excess lift
  • bleed off excess speed along the way

until actually landing, that would mean no waste of this energy in descent: all of it was used productively to get the aircraft to its destination, and none of it needed to be thrown away to get the plane low enough and slow enough fast enough.

In a particular descent, would it be possible to calculate or estimate how much of that energy in practice does in fact get "wasted", i.e. needs to be deliberately thrown away?

  • $\begingroup$ I think the question is inherently somewhat problematic I guess, in this conception that certain part of the potential energy loss is "productively used" and a certain part of it is not. See comments below my answer. $\endgroup$ May 6 '19 at 23:20
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    $\begingroup$ The plane has an altitude and some hundreds of km/hr (potential and kinetic energy) but the goal is to stop the plane so that passengers can come out safely. You want to dump all this energy. I agree that the question is problematic... You probably want to pose the question differently.. $\endgroup$
    – ares
    May 6 '19 at 23:39
  • $\begingroup$ To calculate this energy you simply need to know the mass, the altitude and the velocity of the plane. If you wanted to calculate how far the plane would 'glide' (converting altitude into kin. energy without airbreaks) you need the aero coefficients and simple flight mechanics equations. So for a specific answer you need to specify the type of plane. Otherwise the question is overly general. $\endgroup$
    – ares
    May 6 '19 at 23:44
  • $\begingroup$ I guess "deliberately thrown away" could mean any energy lost due to the deliberate increase in drag caused by deploying the flaps and spoilers. Keeping in mind though that as we slow down below high-speed cruise speed we reap some benefits in efficiency. So compare the actual energy change, to the energy change that we'd see if we maintained the cleanest configuration for the same time period, and brought the plane to the minimum sink rate airspeed as efficiently as possible and then held that airspeed for the same period of time? Something like that. There could be other variations. $\endgroup$ May 6 '19 at 23:44
  • $\begingroup$ @quietflyer Exactly, as the question says, "using the control surfaces". $\endgroup$ May 7 '19 at 6:13

A meaningful numeric value is going to be hard to calculate, because so much depends on the aircraft you're working with. In practice, if I can pull off power at the calculated Top of Descent point, descend, meet any ATC crossing restrictions, fly the approach, configure as late as possible, and push up the power right at the point where I'm required to be stabilized for the approach, then I've flown an efficient approach & wasted none of the energy that I started with.

If I have to add power sooner than that, I started down too soon (or wasted energy in some way); if I have to deploy drag (which includes configuring early), then I started down too late & so I have excess energy that I need to dissipate.

I suppose "amount of time speedbrakes were up" would be one way to score how much energy had to be wasted; add to that how long flaps & gear were out before they were "needed" (given the requirement to be stabilized & configured at a given point in the approach).

Here's the complication, though. If I see early on that I have too much energy, I can push the nose down & gain speed, which increases the parasite drag of the aircraft -- thus dissipating energy. If you have enough time at the higher speed (not just push the nose down, gain speed, pull the nose up, lose speed -- that doesn't accomplish very much), then you can get rid of excess energy without having to deploy speedbrakes or configure early. And the reverse, if my standard descent is at 280 knots (above 10,000', obviously) but I slow up to 250 knots early enough, I can "gain" some energy, since I'm losing less with less parasite drag. How you could account for all of that in a "score" is beyond me.

In the ideal case, I can descend to glideslope intercept altitude, level off there, let the speed decay, then at the point when I need flaps to fly any slower, I extend the first increment of flaps, let the speed keep dropping off, start down the glideslope (speed decreases more slowly or if the aircraft is light enough, not at all), at about 500' above the "stabilized approach" gate I'll lower the landing gear & extend more flaps, so that as I approach that gate I'm extending the last increment of flaps, my speed reaches VTarget, and I add power. No extra time with power above idle, a fuel-efficient approach.

ATC may not let me do that, though; they may keep me fast until close to the runway, in which case I'm extending flaps & gear closer to their maximum speeds so as to dissipate more energy more quickly and still be configured & stabilized at the appropriate point. (The 737 dissipates speed pretty quickly fully configured at idle power, even descending on the glideslope.) So did I "waste" some energy in that scenario? Yes. But keeping ATC happy so they can get efficient runway utilization is a worthy goal as well.

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    $\begingroup$ I think this answer touches on an important point: while it's possible that you need to waste some energy because you're high or fast, the alternative may be that you need to expend some energy if you're low or slow by adding power at low altitude (which is much less efficient). And if you take into account that lower runway utilisation means more airplanes have to take a detour or hold, it seems that "energy wasted" is a function of the airport environment rather than a single number. $\endgroup$
    – Sanchises
    May 7 '19 at 8:39

It has been studied by Eurocontrol for its Continuous Descent Operations (CDO).

Calculating it will be aircraft type dependent; however, using the study figures it can be estimated as requested.

For those flights currently flying non-CDO profiles, the average time in level flight from the ToD was 217 seconds with per-flight savings estimated at 46kg [of] fuel.

Using the jet fuel specific energy, an estimate can be 9 MJ of energy for every 1 second of not being on the CDO profile.


In a particular descent, would it be possible to calculate or estimate how much of that energy in practice does in fact get "wasted", i.e. needs to be deliberately thrown away?

The same question, but from a different perspective, is "how much energy margin is typically retained during an approach?" Having flown gliders, this is a near-and-dear question to my heart.

Assume you are going to deadstick the landing. From a point in the sky at a certain height (potential energy) you can come up with an altitude versus distance chart to compute how far away an airport can be and basically roll out at the end of the arrival runway having not used brakes and not moving. If assuming standard atmosphere, you can derived a closed form equation. The potential energy at the start is the bare minimum energy needed to get to that point. (There is some kinetic at the start too, but gets dwarfed by the potential.)


Your speed to fly would be L/D max speed.

To figure out the excess energy, just look at your start altitude, and then the finish altitude for the actual distance to fly. The energy at that finish altitude is the excess energy "wasted" or to be considered as margin.

I used to do this. On an OCF (out of control) flight, we'd start our last inverted spin at 27,000 ft, in our working area in South Texas. My goal was to pull the power to idle at the start of the spin, and if possible not touch it again all the way to parked in the chocks (very runway dependent). The working area was about 25 miles away. We'd come out of the spin, orient, and fly at max range-ish profile to the field, get a sense for the pattern, and if open, basically dive bomb to a tight initial, enter the break at 400+ kts, break at the numbers, fly the circling approach, and if on the right runway, the turnout at the end was right by the parking, and if they were almost full, the last spots were by the turnout, with a wee bit of downhill to the chocks. All at idle throttles. Woot!


First one may consider a glider at Vbg. This is the most efficient use of its potential energy. "Wastefullness" would be any departure from Vbg and could be quantitatively determined from flight data and drag curve graphs.

For powered aircraft, the same determination could be made from departure from both aerodynamic and engine/prop peak efficiencies.

Also, consideration of departure from direct flight due to holding patterns or other traffic control could play a role in long term efficiencies for a large fleet of aircraft.

However, other factors, such as safety or expediency limit the applications of the concept. A flight from New York to Los Angeles will not slow down to save fuel. New York to Atlanta, may not be a bad idea.

  • $\begingroup$ I'm only concerned about the descent phase of flight in this question. $\endgroup$ May 7 '19 at 6:15
  • $\begingroup$ @Daniele Procida Would that be wasted as far as diversions or holding, or would it be related to the way the aircraft was flown? There are also speed "limits" based on altitude for airliners, and changes in drag when they re-configure. If you can be more specific about the application, it might help. $\endgroup$ May 7 '19 at 7:44
  • $\begingroup$ I'm just interested in the aeroplane as a flying object subject to physical laws. $\endgroup$ May 7 '19 at 13:52
  • $\begingroup$ @Daniele Procida This is a very good subject as they really try to BUILD aircraft to be fuel efficient, yet do not FLY them that way. Why not include climb and cruise as well? $\endgroup$ May 7 '19 at 17:01
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    $\begingroup$ I agree that the question of fuel efficiency is important, it's just not what I'm asking about. It's a question about physics in aviation. $\endgroup$ May 7 '19 at 18:44


Let's take the simple case where mass is assumed to stay constant during the descent.

The formula for kinetic energy is KE=(1/2) * m * (v^2), where m is mass and v is velocity.

Compare the kinetic energy at start and end of descent, using True airspeed. If True airspeed is constant, Kinetic energy is constant.

Now compare the (gravitational) potential energy at start and end of descent. Potential energy = mgh, where h is elevation and g is the gravitational constant and m is mass.

The energy lost is the decrease in potential energy, minus any gain in kinetic energy. Or the decrease in potential energy, plus any loss of kinetic energy.

If airspeed is constant throughout the descent, we are losing ALL of the potential energy we had at the start of the descent-- NONE of it is being converted to kinetic energy, or to any form of stored energy.

Too bad we aren't coming down by using a prop as a windmill to power up a battery as well as create drag, rather than by opening spoilers and such.

Alright, expanding this answer to consider comparing the energy lost in the actual descent to touchdown, to the energy lost in in some representative "ideal" maneuver.

What is the "ideal" maneuver?

Ideal Option 1) -- at the same point in time that the descent would be normally be initiated, we instead smoothly pull back on the yoke to transition to the angle-of-attack that gives the max ratio of lift/ drag (unless this would arc the flight path up beyond vertical unto the second quarter of a loop, then we have to use a slower rate of increase of angle-of-attack to pull less "G".) Anyway we end up doing a (possibly rather steep) "zoom" climb to gain altitude and lose airspeed and then we let the flight path arc back down into a steady-state glide at the angle-of-attack (and airspeed) that gives the max ratio of lift/ drag. As we overfly the point where we'd touch down with the real-world method, we note our altitude and airspeed; this gives us a basis to compare the actual energy lost in the real-world method with the energy lost by the "ideal" maneuver.

2) Ideal Option 2-- we roll back time to the point where we can glide all the way to touchdown with the plane in a clean configuration, at the speed that gives the flattest glide, after initiating a highly efficient transition to the best-glide speed such as that described in Ideal Option 1. At the moment of touchdown, compare the fuel in the tanks at the moment of touchdown with this Ideal Option, to the fuel in the tanks with the standard method of a longer cruise period followed by a steeper shorter descent. Also consider the difference in airspeed at the instant of touchdown. Now we have a basis to say how much energy is "wasted" with the standard method (but how do we evaluate the energy stored in the fuel? Do we consider ALL the energy stored in the chemical bonds, or just the energy that the engine would actually be able to extract?)

3) Ideal Option 3-- instead of descending we just maintain cruise speed and altitude so we lose neither kinetic energy nor potential energy. So no energy at all is expended with this ideal option. Well, that's a little silly and shows that we really do need to take into account the potential chemical energy we're taking out of the fuel tanks, both with the actual method and with the ideal method.

Since we're now considering the energy stored in the fuel, we now ought to revisit the ideal option 1. How much power should we carry? Just enough that the engines create zero thrust-- keep the fans turning under power so we're not using them as giant windmills? Maybe just a little less than that? Surely we don't want to chop the fuel flow to zero do we? Then the fans would make way too much drag.

On the other hand if we are considering, say, the energy that could be released by the fuel in a nuclear fusion reaction, then we better leave it all in the tanks for the "ideal" descent option, as it is too precious to waste any.

While we're at it we should really note that with the real method-- and arguably with the ideal methods too -- all the kinetic energy that we're carrying at touchdown is wasted-- for the real-world method at least I guess we really need to measure all the way to the point where the plane comes to a stop with zero energy. Also consider all the fuel wasted as we apply reverse thrust. None of this comes into play for the "ideal" methods; we get to use the most efficient method of flight possible to keep the plane moving through the air till we are overflying -- the touchdown point? The point on the runway where the plane would come to a stop in the real world? The gate at the terminal? With the no-landing ideal methods such as #1 and #3, as we're about to overfly the gate at the terminal, high above it, do we get to ease the yoke further back and bleed off airspeed all the way to stall speed, helping us to maintain altitude for a while without burning (much) fuel, or even to gain some altitude? Sure, why not.

Lots of different ways to set up this problem.

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    $\begingroup$ Kinetic & potential energy for aircraft don't work the same as KE & PE in marbles rolling down ramps and across tables in the Physics lab. A ball rolling across a level table can be said to "lose" no KE or PE; an aircraft can't do that. Some sort of energy (chemical, as thrust, or KE as slowing down, or PE as drifting down) MUST be expended to overcome induced & parasite drag and keep flying. Doesn't mean your PE is "wasted"; it's being put to use. If you deploy lots of drag in order to descend quickly, then you're wasting your PE, and that is what the OP is asking about. $\endgroup$
    – Ralph J
    May 6 '19 at 22:55
  • $\begingroup$ Hmmm, I don't think I agree. A glider with a very high L/D ratio loses very little potential energy per mile travelled. An airliner coming down w/ flaps full down and spoilers full open and those big fanjets hanging out in the wind has a much poorer L/D ratio and a much steeper glide angle and loses a great deal of potential energy per mile travelled. Guess I'm not seeing the problem with my answer. PE lost is PE lost. If it's not being converted to KE or stored energy than it's lost. You seem to be using "wasted" in some kind of subjective sense. $\endgroup$ May 6 '19 at 23:04
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    $\begingroup$ Your answer is right in the physics classroom, but look at how the OP defines wasted, or "no wasted energy." That's where I think this answer doesn't address the question as stated. $\endgroup$
    – Ralph J
    May 6 '19 at 23:14
  • $\begingroup$ @RalphJ maybe you want to define "wasted" related to the total energy ACTUALLY lost in the descent compared to the total energy lost in the MOST EFFICIENT POSSIBLE descent for that particular aircraft? I don't think that's implicit in the question though. And then we have to decide if we want to define "most efficient" as minimum energy lost per unit horizontal distance travelled, or per unit time, or per unit vertical distance descended, or what-- anyway I don't see that as being implicit in the original question. $\endgroup$ May 6 '19 at 23:16
  • $\begingroup$ OK I see he did put some constraints like that in the question-- I think the question is inherently somewhat problematic I guess, in this conception that certain part of the potential energy loss is used "productively" and a certain part of it is not. $\endgroup$ May 6 '19 at 23:18

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