Understanding all this has been a work in progress, but now from start to finish we have:

Fuel/second ---> shaft Horsepower---> RPM ---> Thrust ---> Drag

Drag Force = $\rho$ × Area × Cdrag × V$^2$

Velocity is proportional to $\sqrt{Drag}$.

Transmission efficiency of a properly pitched propeller can be considered (as 1-sin $\theta$), or around 0.8. In other words transmission losses are only slightly worse than automobiles, which have direct force transmission to the ground.

All well and good. Power, as % BHP in the POH, is the throttle, or fuel flow rate. RPM is proportional to the square of fuel flow rate. But the propeller "steals back" efficiency (by V$^2$ in its equation) so thrust is proportional to fuel flow rate. Velocity is derived from Thrust = Drag as $\sqrt{Drag}$.

Another, much simpler example: parachute and weight. Double weight and velocity increases by the $\sqrt{}$ of increase of force, or 41% No need to go up to "Power terms". Thrust = Drag!

So, what about multiplying Thrust or Drag by velocity to compare with another Force × Velocity. For example, we know that Drag will be 9× more at 3× speed. Yet if we use, for example, 120 kg of prop drag at 450 km/hr, is that the same or comparable to 40 kg of airplane Drag at 150 km/hr?

450 × 40 = 150 × 120 = 1 : 1. No

Only when velocity is squared and the correct Drag value is plugged in does the equation work as:

V1$^2$ × Drag 2 = V2$^2$ × Drag 1

The correct value for Drag at 150 km/hour is 13.3.

Not sure about the value of converting Thrust or Drag into "Power" when "steady state" seems to be a balance of forces at the same Velocity.

Data for this question is taken from the Cessna 172 POH as follows at 2500 feet:

RPM....% BHP.....True Airspeed....GPH.....Range
2700.......87 %.............139...............9.6.......545..
2600.......78 %.............133...............8.6.......590..
2500.......70 %.............128...............7.7.......630..
2400.......63 %.............122...............7.1.......655..
2300.......57 %.............116...............6.6.......665..
2200.......51 %.............109...............6.2.......665..

In question is the relationship between GPH and RPM, which suggests that fuel is converted into piston Force (and Heat), and $\Delta$ RPM is to the square$^1$ of that Force. So is % BHP! And airspeed. Mechanisticly, this suggests as follows:

Fuel potential energy --> Force/Heat --> Force$^2$ = Prop $\Delta$V

Prop "lift" equation $\Delta$V squared --> Thrust = Drag

Resultant Velocity proportional to $\sqrt{Drag}$. The chart data seems to support this.

$^1$ not cubed

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    – Ralph J
    Commented Feb 5, 2023 at 0:47

4 Answers 4


"Velocity is $\sqrt{Drag}$" is not a meaningful statement. It confuses causality and is simply wrong.

Drag (at fixed drag coefficient) is proportional to Velocity squared. $D=0.5 \rho\, C_D\, V^2\, S$. However, when you change velocity, you must also change your lift coefficient -- which changes the drag coefficient. The parasite term of drag is proportional to $V^2$, but the induced term ends up proportional to $1/V^2$.

At cruse, when thrust is equal to drag, we sometimes call the power associated with thrust the "Thrust power". It is equal to the product of thrust and velocity. Full stop. $P=T\, V$. Since drag is equal to thrust, we can also think of this as the "Drag Power" $P=D\, V$. It can be thought of as the rate of energy dissipated into the air due to drag.

If you want the power associated with flight, this is it. You're done here. However, you also want to see the connection from here through to fuel flow.

The propeller efficiency (say $\eta_P=0.8$) means that the shaft power is larger than the thrust power $P_{shaft}=P/\eta_P$.

The energy contained in the fuel is characterized by the lower heating value ($\mathrm{LHV}$). We think about fuel consumption for a piston engine in terms of brake specific fuel consumption ($\mathrm{BSFC}$).

BSFC is commonly used in the terrible units of lbm/(hp⋅h). I.e. pounds of fuel consumed per shaft horsepower each hour. So, the fuel flow rate looks something like:

$\dot{m}_f= \mathrm{BSFC}\, P_{shaft}$

The $\mathrm{BSFC}$ is related to the $\mathrm{LHV}$ by the cycle efficiency of the engine $\eta_c$. Obviously a combustion engine is not perfectly efficient -- it gives off substantial waste heat to the air (whether air cooled or via the radiator if liquid cooled) as well as waste heat in the exhaust. All that waste heat is lost energy that was originally contained in the fuel. The cycle efficiency of a piston engine is very low -- typically less than $\eta_c < 0.3$.

$BSFC=\frac{1}{\eta_c LHV}$

We can combine these equations together...

$\dot{m}_f= \frac{D\, V}{\eta_P \eta_c LHV}$

Or, another step further...

$\dot{m}_f= \frac{0.5 \rho\, C_D\, V^3\, S}{\eta_P \eta_c LHV}$

If we assume a simple parabolic drag polar $C_D=C_{D,0}+K\,{C_L}^2$, then we get...

$\dot{m}_f= \left(0.5 \rho\, C_{D,0}\, V^3\, S + \frac{K\ W^2}{0.5 \rho\, V\, S}\right) \frac{1}{\eta_P \eta_c LHV}$

This gives a parasite power term that is proportional to $V^3$ and an induced power term that is proportional to $1/V$.

IMPORTANTLY, this is not the whole story. Buried in this are the facts that propeller and cycle efficiency are not constant. Propeller efficiency $\eta_P$ will vary with speed and throttle setting (power). Cycle efficiency $\eta_c$ will vary with throttle setting (power). In particular, most engines are most efficient operating at near maximum power -- and are very inefficient operating near idle.

If you look at a given aircraft's POH and consider the fuel flow values, you will not see perfectly $V^3$ behavior because of many things -- there is also the $1/V$ term -- and the variation of $\eta_P \eta_c$ is complex and nonlinear.

Edit Below

If we divide the fuel flow by $V$, we get the specific range -- a fuel economy measure like MPG for a car.

$SR= \left(0.5 \rho\, C_{D,0}\, V^2\, S + \frac{K\ W^2}{0.5 \rho\, V^2\, S}\right) \frac{1}{\eta_P \eta_c LHV}$

Which will work out in ft/slug or m/kg, but can be thought of (converted to) nmi/lbm or mpg.

You repeatedly assert that observation of POH data leads you to conclude that various behavior is linear. This is an erroneous conclusion. Taylor's series for any smooth function is:

$f(x_0+\Delta x)=f(x_0)+f'(x_0)\Delta x+\frac{f''(x_0)}{2}(\Delta x)^2+\frac{f'''(x_0)}{6}(\Delta x)^3+...$

When $\Delta x$ is small, then $(\Delta x)^2$ and other higher powers will be smaller and smaller such that they can be ignored. This leaves us with a linear function:

$f(x_0+\Delta x)=f(x_0)+f'(x_0)\Delta x+...$

I.e. everything is linear on a small enough scale.

And on a slightly larger scale, you can keep the quadratic terms and drop everything higher...

$f(x_0+\Delta x)=f(x_0)+f'(x_0)\Delta x+\frac{f''(x_0)}{2}(\Delta x)^2+...$

I.e. everything is quadratic on a slightly larger scale.

If you are looking at POH data, the manufacturer will supply data points at a resolution such that a pilot can linearly interpolate between any given points (or possibly just use the nearest point) to get a reasonable answer. As you point out, we know that fuel economy is bad at low speed -- so the POH does not give data there. It is not useful to a pilot.

However, if we are looking to understand the physics of flight, then we prefer to look at a wider range of data. A range sufficient to show the higher order terms and the more complex interactions.

Edit 2 Below:

The claim was made that thrust can not increase faster than V^2. Below is a plot of thrust available and thrust required (drag) for an F-5E at 36,089 ft altitude.

enter image description here

If this is insufficient to show that drag can increase faster than $V^2$, we can overlay some analysis...

Here I take the lowest drag point -- since this line corresponds to L=W, it is the point of best L/D. If the drag polar were of simple parabolic form, induced and parasite drag would be equal there. I then plot lines proportional to $1/V^2$ and $V^2$ that go through the half-minimum-drag point. I also plot their sum.

enter image description here

The gold line does increase slower than parasite drag's $V^2$ -- because of the $1/V^2$ term of induced drag.

We also see that neither simple curve keeps up with the real drag. This is because of the changes in $C_{D,0}$ and $K$ with Mach number and the non-parabolic parts of the drag polar at high $C_L$ (for the low velocity region of the curve approaching stall).

In the real world, these non-ideal behaviors can drown out the idealized trends. In this case, in the transonic region, $C_{D,0}$ more than doubles between M 0.9 and 1.1 and then remains nearly constant supersonically.

For a piston prop, you won't experience this -- and admittedly, it is one of the most drastic changes in the physical world. They called it the sound barrier for a reason. Unfortunately, I don't immediately have my hands on a similar original chart for a piston-prop aircraft.

Note that the textbook approximations for jet aircraft performance would also tell you that thrust available from a jet is constant with Mach number. It clearly is not.

  • $\begingroup$ $\eta$ is not that complex approaching it from the car point of view. Up at cruise AoA is changing little. Prop efficiency is near 1. But you are absolutely right when things get slow, AoA increases, lift vector tilts back, airflow becomes more turbulent. That's why it is fascinating that Vmin power is at the slow end. $\endgroup$ Commented Feb 4, 2023 at 10:51
  • $\begingroup$ I wasn't implying drastic behavior in $\eta_P\eta_c$. I was implying behavior that is not easily put into an analytical form. I should have mentioned a few more, $C_{D,0}$ will change with Mach and Reynolds number, and $K$ will change with Mach number. The common textbook assertion (for example) that best range is at best $C_L/C_D$ for a piston prop aircraft is based on constant $\mathrm{BSFC}$, $C_{D,0}$, $K$, $\eta_P$ -- none of which are true. Most books do not make it clear that these are approximations -- not gospel. $\endgroup$ Commented Feb 4, 2023 at 20:06
  • $\begingroup$ So, as mentioned with Aditya, steady state is a maintenance of T = D. Power term V falls out because V is equal, in this case T and D are on the same prop. If one changes V, yes, additional force is required to overcome inertia, but the drag at V2 =Thrust at V2. But drag and thrust are not linear to V. My search has yielded that thrust must increase by the square of Velocity increase, but not more (theoretically). However factors you mentioned such as AoA, Mach numbers, etc. certainly play a role. $\endgroup$ Commented Feb 5, 2023 at 3:46
  • $\begingroup$ T=D is a statement about conservation of momentum. Thrust power equal to Drag power is a statement about conservation of energy. There are certainly times when thrust can increase faster than V^2. I'll post an image as another update. $\endgroup$ Commented Feb 5, 2023 at 5:31

First of all, velocity is not $\small \sqrt{drag}$ ; velocity is proportional to $\small \sqrt{drag}$. Additionally, as Rob McDonald said, this is true only for a constant drag coefficient, which is not the case in practice. And as Sophit said, at low speeds, total drag reduces with increase in speed. This is because at those speeds, the effect of reducing $C_{di}$ is greater than that of increasing dynamic pressure.

Fuel/second ---> shaft Horsepower---> RPM ---> Thrust ---> Drag

I would disagree. You start with an expression for Power (fuel/second), proceed with an expression for Speed (RPM), and finally end with an expression for Force (drag).

I my opinion, it would be more appropriate to end with the quality that we started with. If we start with power, we must end with power. The order should look like:

fuel power ---> shaft power ---> thrust power ---> rate of doing work

$fuel\ power = \large \frac{fuel\ energy}{time}$

$shaft\ power = rotation\ speed \cdot torque$

$thrust\ power = speed \cdot thrust$

Rate of doing work

The thrust power does work on:

  • The aircraft

  • The air particles

Energy does work, and so rate of doing work is same as rate of change of energy. So let's look at rate of change of energy of aircraft and air particles.

Aircraft: It possesses energy in two forms, kinetic and gravitational potential. Therefore, the rate of change of energy of the aircraft is:

$$\left( ½ \cdot m \cdot \frac{dv²}{dt} \right) + \left( m \cdot g \cdot \frac{dh}{dt} \right) $$

Air: The aircraft does work on air particles by imparting kinetic energy into them, and this manifests itself as drag. The work done on air is:

$$½ \cdot ρ \cdot V³ \cdot C_d \cdot S$$

this is known as drag power $-$ the power required to overcome drag.

Therefore, the total work done by thrust power (in a unit time) is:

$$thrust\ power = \left( ½ \cdot m \cdot \frac{dv²}{dt} \right) + \left( m \cdot g \cdot \frac{dh}{dt} \right) + \left( ½ \cdot ρ \cdot V³ \cdot C_d \cdot S \right) $$

Note here that if thrust power is less than drag power, then either $\frac{dv²}{dt}$ or $\frac{dh}{dt}$ (or both) will be less than zero. This means that the aircraft will either lose gravitational potential energy, kinetic energy or both. Likewise, if thrust power is greater than drag power, $\frac{dv²}{dt}$ or $\frac{dh}{dt}$ (or both) will be greater than zero.

  • $\begingroup$ is kinetic energy ever "transfered" in propulsion? Or is momentum transfered?. With drag increasing V "strikes twice as many molecules twice as hard", which is 2 × 2mv. $\endgroup$ Commented Feb 4, 2023 at 9:52
  • $\begingroup$ @RobertDiGiovanni Both KE and momentum are imparted. Assuming constant $C_d$, If speed increases by a factor of 2, momentum imparted will increase by a factor of 4 and kinetic energy by a factor of 9. $\endgroup$ Commented Feb 4, 2023 at 10:26
  • $\begingroup$ I think we are essentially in agreement with thrust = 1/2m dv$^2$/dt. Prop RPM is proportional to fuel burn rate$^2$ because the cylinder Force as Average psi/cylinder Area must work to equal prop drag. Essentially, fuel potential energy is "downrated" to mv in propulsion of the pistons. Therefor it takes the square of fuel flow to increase RPM (see POH). But thrust from the prop is a function of V$^2$. I'm fairly sure that this is why early jets used a lot more fuel than fan jets or turboprops. (afterburners too). The propeller greatly improves efficiency (for a Cessna 172 speed range). $\endgroup$ Commented Feb 4, 2023 at 14:19
  • $\begingroup$ @RobertDiGiovanni: $\left( ½ \cdot m \cdot \frac{dv²}{dt} \right)$ is the time derivative of KE. KE is energy, and its time derivative $ \left( \frac{d(energy)}{dt} \right) $ is power. In the LHS, you have thrust, where you should instead have power (so thrust must be multiplied with speed). Also, this is true only when $C_d$ (drag) is zero and $∆h$ (instantaneous change in height) is also zero. $\endgroup$ Commented Feb 4, 2023 at 21:31
  • $\begingroup$ And the reason RPM is proportional to (Fuel Flow)² instead of (Fuel Flow)³ might be that the engine is quite inefficient at low RPMs. If the prop was driven by a constant efficiency engine, I'd expect RPM to be proportional to (Fuel Flow)³. $\endgroup$ Commented Feb 4, 2023 at 21:36

This answer is provided in response to concerns expressed by the moderator regarding extended comments. Elements of the question are very interesting, although the general thread of the question is difficult to follow. The answer Mr DiGiovanni desires is one that apparently shows the relationship between engine torque (or propeller thrust) and various parameters regarding engine performance, including fuel flow and thermodynamic processes in combustion.

Based on the information that was provided in the question, a considerable amount of additional information can be developed about your plane. However, to answer your question, the following are necessary –

  1. The question must be stated clearly regarding the subject of the question. This means clearly abstracting the issue at hand, and stating that issue along with relevant information. Understandably, that is not always easy, and in some cases may take a considerable amount of time and thought. Keep in mind, a lot of time is spent in trying to answer your question by thinking through and evaluating what it is you are trying to find out.

  2. In the process of providing an answer, data that are provided are usually validated by the person answering your question. This is usually done by referencing these data directly to a source that shows the same data, or allows the same data to be developed in context. However, such is not the case with this question. A desirable solution to this issue is to provide an image of the source of these data that clearly references the document at hand, or provision of a link where these data can be examined directly. Consequently, for information provided in your question, there is a concern regarding the validity of the data. Having credible data with unquestionable validity is helpful.

  3. A considerable amount of information can be developed by the data that were provided. However, the analysis cannot be validated. A considerable amount of time was taken to evaluate your plane based not only on the provided data, but by cross comparison with similar aircraft in the literature. Although the results of the analysis are interesting and appear to be along the line of interest posed in your question, little else can be done to afford credibility to the answer because of seeming inconsistencies apparent in data provided with the question. That does not mean those data were in error, just that those data could not be validated and are in question. Take a look at a summary analysis, for what its worth. Perhaps this will be helpful…

enter image description here

  • $\begingroup$ Really don't need the preamble to the Gettysburg address here, but the graphs are great. +1. I spent a week on this, not being a mathematician. But the similarities of HP and drag are amazing. Both have units of weight and velocity. [A][B] = [A][B]. But the trick was realizing fuel is only converted into Force. The expression becomes A$^2$ = [A][B]. If you plot HP$^2$, this gives the prop drag curve, or RPM, which by way of thrust, is proportional to the aircraft drag. Plotting RPM would be great if you have the time. $\endgroup$ Commented Feb 8, 2023 at 21:22
  • $\begingroup$ Special request, large graph. Please divide out V from HP available and replace HP required with GPH. I'm ready for the "real" Vx and Vy graph. PS I think RPMs are going higher as V increases because prop AoA (fixed) will get less and less. More RPM would help offset decreasing prop efficiency. $\endgroup$ Commented Feb 9, 2023 at 4:37
  • $\begingroup$ @RobertDiGiovanni Hey, I think I found your source. You are looking at the K model, but for which year?? I might be able to pin down a bunch of interesting info you would like to see, and adjust those graphs as you requested. Tell me your specific POH source, please.... $\endgroup$ Commented Nov 22, 2023 at 0:15
  • $\begingroup$ not sure there will be significant differences in data for K's (if there were, it would be a new letter). I believe it was 1969-1970. Thanks. $\endgroup$ Commented Nov 22, 2023 at 0:38
  • $\begingroup$ @RobertDiGiovanni Ok... Found the source of your data, and supplemental data, both are for the 172M. We will see how this goes for an update/revision of the answer that was given, including your requested graphs if possible. These data were validated. Data given in your original reference was for airspeed in mph, not kts. All were standardized to kts in the revisions... $\endgroup$ Commented Jan 18 at 21:30

Some definitions. Work, or mechanical energy, equals force times distance. Mechanical power is force times distance per unit of time, or force times velocity.

Let's assume that propeller power output is indeed proportional to engine power, and that engine power is indeed proportional to fuel flow. Thrust is thus proportional to engine power divided by velocity. The other way round, velocity equals power divided by thrust, only works in steady state.

If thrust exceeds drag, we have excess power, which we can put in extra kinetic energy (speed up) or potential energy (altitude). In cruise, all the fuel energy (conventially known as chemical energy, not as potential energy!) is spent on drag, and kinetic energy remains constant. So there's no conversion from fuel energy to kinetic energy.

Drag force is proportional to velocity squared according to experimental physics. So if drag at 450kmh is 120kgf, it is 120/(450/150)² at 150kmh, indeed 13.33kgf. Power is 450x120=54000kgfkmh and 2000kgfkmh respectively, a whopping factor 27 difference (not 1:1 or 3:1)! That is because power is force times velocity, so drag power loss is proportional to velocity to the third power...

No idea where you got that velocity is the square root of drag.

  • $\begingroup$ because the propeller has an efficiency near one, more like cars. For rockets, we are only using momentum as a propulsive force. Like a recoilless rifle. That's proportional to V$^3$. A car with direct contact with the road is V$^2$. Propeller fuel consumption rate is closer to cars with respect to velocity and use of fuel per distance is linear with velocity. I'm working back from POH data. Don't mean to make a fuss over this. $\endgroup$ Commented Feb 4, 2023 at 10:08
  • $\begingroup$ Propeller efficiency is not that near to one. Fuel per distance is not linear with velocity. All things are linear if you consider them in a small enough range. You do not capture the nonlinearity of the world if you only consider small ranges. $\endgroup$ Commented Feb 4, 2023 at 20:10

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