Suppose I have an airplane with a certain weight M, which uses a certain amount of power P, to fly at a certain speed V.

If I use the same airplane, but quadruple the weight (4xM) and double the speed (2xV), what is the resulting need in power?

I would like to understand the physics of this, for an ideal system. We can ignore how efficiencies change with velocities, etc.

Here is my reasoning: the airplane needs four times the lift to fly in the same mode, which will be provided since the speed has doubled. The only force the engine needs to overcome is the drag, which is four times higher, i.e. the required power is 4xP, four times higher.


Short answer: No. $ P = F \cdot v$ and since $ D = \frac{1}{2}C_D\rho v^2S $ it follows that $P\propto v^3$

Let's take a more simple case: a block moves over a rough, horizontal surface in a vacuum and a force is applied such that its velocity $v$ stays constant. This force is always the same, since the "drag force" is given by the friction coefficient times the normal force. Here, the kind of relationship that you're thinking of would hold: if we're moving at $2v$, we need twice as much power ($P=F\cdot v$).

Look at a fixed distance along that trajectory: we do the same work in both cases ($v$ and $2v$), so we put in the same energy which dissipates through friction, i.e. heat. However, we now need to do the same work in half the time, so the power is double. In a certain way, it's "more difficult".

Now for an aircraft, the drag force itself is proportional to the square of the velocity. So, in conclusion, there are "two effects" which "add up": a quadrupling related to the increased force and a doubling due to the increase in velocity. Hence, the power is multiplied by 8.

Edit (in response to comments below):

Ok, let's do another thought experiment: we place a jet engine in a wind tunnel and keep it at constant thrust and constant mass flow (choked nozzle), which results in a constant velocity increment across the engine. The power contained in the jet (the airflow exiting the engine at higher speed $v_j$) is $\frac{1}{2} \dot m v_j^2$ and the power contained in the incoming flow is $\frac{1}{2} \dot m v_0^2$. The difference between the two is called the "propulsive jet power"; it's the power which is "available" from the acceleration of the flow. Now, you can see that because both terms are squared, it is not just the velocity increment which counts (as it was the case for the thrust), but also the magnitude of the two terms.

So alternatively to the "thrust power" given by $$P_{thrust} = \dot m (v_j-v_0) v_0 = T v_0 = D v_0$$

we can also consider the "propulsive jet power"

$$P_{prop} = \frac{1}{2} \dot m (v_j^2-v_0^2)$$

where the proportionality to $v_0$ at constant thrust is maybe more clear. The difference between these two quantities is called the power loss ($P_{loss}=\frac{1}{2} \dot m (v_j-v_0)^2$) and is a measure of how much of the orginal chemical energy contained in the fuel is lost to kinetic energy in the jet.

You can see that if the velocity differential is constant, the power loss is constant. So if we have shown that the propulsive jet power is proportional to $v_0$ at constant thrust, the thrust power (which is the power that "actually moves you") must also be proportional in this way. And this proportionality is related to the fact that the kinetic energy of a mass is proportional to the square of its velocity.

  • $\begingroup$ Thanks Daniel! The formulas in your short answer makes total sense (and Tanner Swett has already explained it to me under a different topic). However, I have a hard time understanding it intuitively. Imagine that the aircraft is not traveling at all relative to the earth. Instead it fights wind speed (as in a wind tunnel), and we do the same thought experiment. (... more to follow) $\endgroup$ – Björn Morén Jan 2 at 15:55
  • $\begingroup$ To overcome the drag component the engine needs four times the power. The wind speed is double, but oncoming wind speed is irrelevant to how much thrust/force an ideal engine can give. Speed difference between incoming and outgoing air of the propeller is what creates thrust. So in my view there is only one component to overcome; the drag. Take away drag (vacuum) and you are in a space ship, and thrust is constant regardless of speed. $\endgroup$ – Björn Morén Jan 2 at 15:55
  • $\begingroup$ A related question: will an ideal engine/propeller create the same thrust force at the same power, regardless of incoming air speed? As far as I understand the subject, it will. Please correct me if I'm wrong! $\endgroup$ – Björn Morén Jan 2 at 16:01
  • $\begingroup$ I've added an alternative way to look at it above. Some other remarks: be careful not to mix up jet engines and propellers driven by piston engines, these work completely differently. The power of a piston engine is of course independent of the air speed, but the thrust that the propeller then produces is highly dependent on the air speed. The latter is of course also true for a jet engine as a whole. $\endgroup$ – Daniel Jan 2 at 20:44
  • 1
    $\begingroup$ Another analogy along the same lines is: it's harder to climb a ladder quickly than it is to climb a ladder slowly. (As an aside, once you get into going down a ladder, the analogy breaks down. A well-designed ladder-descending machine would gain energy from descending a ladder; but since human beings are not well-designed ladder-descending machines, we have to expend effort to descend ladders.) $\endgroup$ – Tanner Swett Jan 4 at 3:45

Let the initial state have the index 1. If drag is $D$, then $P_1$ is: $$P_1 = D_1\cdot v\;\; \text{or} \;\;\frac{\rho}{2}\cdot v^3\cdot S_{ref}\cdot\left(c_{D0}+\frac{c_L^2}{\pi\cdot AR\cdot\epsilon}\right)$$

Now, state two is: $$M_2 = 4\cdot M_1 \;\;\text{and}\;\;v_2 = 2\cdot v_1$$ At twice the speed dynamic pressure is four times higher, so the lift coefficient $c_L$ stays unchanged. So do aspect ratio $AR$ and zero-lift drag coefficient $c_{D0}$. In this special case the term in the brackets in the first equation has not changed (if we neglect the higher Reynolds number which lets $c_{D0}$ shrink).

Power is the product of drag and speed, and drag in turn is proportional to speed squared, at least in a first order approximation. So in combination power must increase with the cube of speed if mass is quadrupled and speed doubled.

If you use other factors, the result depends where on the drag polar state 1 is, in other words, what is the ratio of zero-lift drag to induced drag. If mass is quadrupled without a change in speed, the lift coefficient will also quadruple and its square will increase by a factor of 16. If zero-lift and induced drag were equal in state 1 (that is the minimum drag polar point), now the term in the bracket grows by a factor of 8.5, and so does your power requirement.

If, however, only speed changes, the lift coefficient in state 2 will only be a quarter of that in state 1 and the drag coefficient in state 2 will be dominated by its zero-lift part. Again, if both parts were equal before, now the drag coefficient in state 2 is only 53% of its value in state 1. But since speed has doubled, the required power will now be 4.25 of that at state 1.

If you want to see how things really change (including Reynolds number effects), take a look at this answer and definitely at this one, too.

  • $\begingroup$ Considering 2x speed change case maintaining same AOA (same lift coefficient), would the lift be 4x? With same AOA do (CDo + coefficient induced drag) - total drag - increase 4x. Did you calculate the speed change case to produce equal lift by lowering AOA? Quadrupling weight with no speed change seemed out of realistic consideration. Observations so far seem to point to lower viscous drag relative to weight (less surface area ratio), and less frontal drag seem to be helping, along with the obvious flying higher. $\endgroup$ – Robert DiGiovanni Jan 6 at 3:14
  • $\begingroup$ Thanks Peter. I need to study this topic more before I can grasp your explanation. $\endgroup$ – Björn Morén Jan 6 at 12:36
  • $\begingroup$ Peter, your speed change only consideration really helps illustrate one point about induced drag - the lift requirement - does not change much with increased velocity and, as you said, most drag increase comes from CDo. The math matches a 4 fold power increase approximation. $\endgroup$ – Robert DiGiovanni Jan 7 at 3:14

It might be easier to break this down into 2 parts.

  1. Thrust required to overcome increased drag at 2x speed D is proportional to Vsquared, so 4x thrust is required, theoreticly yielding 4x lift.

  2. Power discussion is a little more complicated, as it involves Force x distance/time or Force x Velocity of the aircraft and/or the power output of the engine, in terms of fuel burned per given unit of time.

Comparing doubling of speed with respect to distance shows that while the faster aircraft is more "powerful" due to increased speed and weight kg meters/second squared x meters/second = kg meters squared/seconds cubed, the difference in fuel consumed to go the same distance would be 4xthrust x 0.5 time = 2x fuel needed.

So now we need to define "power" for this application. As far as the instantaneous state of the aircraft, it is much easier to think in terms of Forces, and yes, both piston and jet are the same here. Forget about pulleys and gears and just look at thrust. Thrust = Drag. A prop overcomes drag to produce thrust by burning a certain amount of fuel, just like a jet.

The real key is efficiency. So assuming our scale-up to yield 4x thrust produces no change in efficiency, we must burn 4x more fuel, but only for half the time to go the same distance. But the propeller is the issue. One may attempt to keep efficiency close with a variable pitch prop, but in reality, attempting to increase thrust 4x would probably overheat the engine rather quickly even if it were possible. Increasing weight by that magnitude would more likely involve bigger wings and more engines.

Check this out, as ton miles per gallon:

"good" mileage car 2 tons x 30 miles/gallon = 60 ton miles/gallon
60 mph

tractor trailer 40 tons x 7 miles/gallon = 280 ton miles/gallon 😊 60 mph

Cessna 172 1 ton x 12 miles/gallon = 12 ton miles/gallon 110 mph

Boeing B29 65 tons × .5 miles/gallon = 32.5 ton miles/gallon
240 mph

Boeing B52 240 tons x .18 miles/gallon = 43.2 ton miles/gallon
600 mph

Boeing 747 400 tons x .20 miles/gallon = 80 ton miles/gallon
600 mph

Avro Vulcan 85 tons x .28 miles/gallon = 23.9 ton miles/gallon
Mach 1

Concorde 185 tons x .24 miles/gallon = 44.3 ton miles/gallon Mach 2

North American XB 70 270 tons x .16 miles/gallon = 43.4 ton miles/gallon Mach 3

Freight train 500 plus ton miles/gallon 😄

Wandering albatross 12 to 20 lbs up to 60 mph

  • $\begingroup$ Thanks Robert, this really makes a lot of sense from the perspective of amount of fuel (energy). $\endgroup$ – Björn Morén Jan 6 at 12:39
  • $\begingroup$ The XB-70 was the proud achievement of North American Aircraft. What an insult to attribute it to Boeing! $\endgroup$ – Peter Kämpf Jan 6 at 15:33
  • $\begingroup$ Oh sorry, I thought the B stands for Boeing $\endgroup$ – Robert DiGiovanni Jan 6 at 20:54
  • $\begingroup$ Yes, Bjorn It seems V cubed more describes an object that has velocity and is accelerating, what I try to do when I hit a golf ball. Perhaps a description of an energy state. We do not seem to need 8 times more fuel to double our speed (thank goodness). So I wonder. $\endgroup$ – Robert DiGiovanni Jan 6 at 21:02

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