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In my Uni course I've learnt how a piston-engine produces a constant power, as opposed to a constant thrust like in a jet engine, which can be estimated as Thrust = Power / Velocity. However, using this formula I calculated that a Spitfire should be able to go from idle to take-off in under 1.5s (as opposed to 9.5s irl). So my question is this: what else is going on that makes the true take-off time so much longer than my calculated take-off time?

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    $\begingroup$ What does your formula give as thrust a 0 velocity? Or near zero... is that a realistic value? $\endgroup$
    – Ralph J
    Commented Oct 7, 2022 at 16:22

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I propose the following method: the rate of change of momentum has the units of force. When you are standing still on the brakes with full power applied, the prop is taking air which is essentially standing still (zero velocity) and applying a force to its mass which accelerates it to some exit velocity. That force is opposed by the brakes and is equal to the thrust being applied to the airframe by the bearings in the engine case as the prop tries to pull the engine forward.

This means if you know the air mass flow rate through the propeller disc, you should be able to solve for the momentum change and get the thrust that way.

We eagerly await Peter Kaempf's (yes I know it's spelled wrong) opinion on this method!

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  • $\begingroup$ The spelling is fine for someone on a US keyboard. $\endgroup$ Commented Dec 17, 2022 at 12:56
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Thrust = Power/Velocity

Refers to the energy state of the aircraft. Units are mv$^2$/s for Power.

People often confuse this "power" with the power output of the engine, which in its most basic form is fuel burned/second multiplied by efficiency factors giving horsepower output, described as torque at a given RPM.

It doesn't end there. Now we have to further consider the efficiency of the prop at a given airspeed. This is why it is very rare to see prop driven aircraft listed as having a given "thrust".

In reality, the horsepower is consumed by the drag of the propeller, or more properly, the drag torque of the propeller, which is drag force summed from a center of resistance not unlike the "sweet spot" of a baseball bat. Props can also be geared so engine rpm and prop rpm are better matched.

The thrust output is the "lift" the prop airfoil creates.

That is the Force pushing against the inertia, tire friction, and aerodynamic drag of your Spitfire as it accelerates down the runway.

As with the other airfoils on your plane, the wings, quantifying prop "lift" in units of force can be approximated by determining mass flow of air created by the spinning prop as action/reaction forces. Also, thrust at V 0 can be measured (as with model aircraft) "on the bench" by attaching a scale to the back of the engine/prop mount and applying full throttle.

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So my question is this: what else is going on that makes the true take-off time so much longer than my calculated take-off time?


Findings: If evaluated considering propeller thrust, resisting drag, and weight of the aircraft, an accurate and realistic time and distance for reaching take-off velocity can be determined. An adequately developed computational procedure must be used, however. Aspects considered include propeller diameter and blade dimensions for the estimation of thrust, and influence of the propeller slip-stream on lift. Results for two Spitfire aircraft, for which adequate factual information was available, show computed times and distances to takeoff that are comparable, with little difference, to reported field measurements. For aircraft K.9793, a well documented Mark 1 machine, the computed take-off-run time and distance was about 12.39s at 321 yds, while the field reported time and distance estimate was 12.5s at 320 yds, given no wind and standard atmosphere conditions. An incidental investigation regarding computational adjustments to determine take-off time and distance in wind provided similarly comparable results. A brief overview of the basic computational process for take-off time and distance without wind is provided. $^*$A short end-note gives a perspective of the time required to answer this question.


Your very interesting question involves somewhat more than meets the eye, particularly when trying to make an assessment of a plane of this type. A direct solution can be determined only if a number of variously different factors are considered. You see, in order to obtain realistic results to determine the take-off time and distance for the Spitfire, one must, in a literal sense, fly the airplane on paper. This means accounting for various factors of significance that affect the outcome of the determined result.

Three basic factors are at play, namely, thrust, drag, and weight. From the outset, we would have to consider these three factors to determine the required take-off time and distance of a Spitfire. Specific issues in accelerating an aircraft to take-off velocity primarily regard propeller thrust, drag due to rolling resistance during the take-off run, aerodynamic drag of the fuselage and lift-induced drag of the wing, and ultimately, weight of the aircraft and consequent lift produced by the wing during the take-off run. In determining our answer, by assuming the process-forces are linear with respect to the square of the velocity of the aircraft, we only need to consider the mass of the plane, velocity at takeoff, and mean force of acceleration thru the take-off run.

Let’s take a closer look. During takeoff, aerodynamic drag, lift-induced drag, and rolling resistance comprise forces opposing the accelerating thrust of the propeller. During the take-off run, lift of the wing reduces rolling resistance by taking weight off the landing gear. However, as the aircraft is accelerated, lift from the wing also creates drag and the force of this drag increases as speed of the aircraft, and hence force of lift, increases. Included in resisting forces is aerodynamic drag of the aircraft itself (i.e landing gear, canopy, antenna mast), as well as drag caused by portions of the aircraft in the slip-stream flow of the propeller. So, for any plane to take off, especially in still air, the force of acceleration must, at the start, move the aircraft against resisting forces that are due to the initiation of lift and dead weight (mass inertia) of the aircraft itself, drag caused by resistance of wheel bearings and landing-gear wheels rolling on the ground, aerodynamic drag caused by the aircraft fuselage and other appurtenances such as cowl vents, induction intake, and engine-cooling inlet. All of these opposing forces must be overcome by thrust of the propeller. That thrust is developed through rotation of the propeller by the aircraft engine.

There are also several unknown factors. For instance, given the presumed surface condition of the runway, and still air, we could assume minimal rolling resistance and optimum conditions for aircraft performance upon which to base an analysis. With little information regarding piloting practice with this airplane, we have to presume the most conservative practice in operation of the aircraft, and use that as a starting point. And we need additional information about the airplane and its propeller. Some historical data and relevant information are available, however. Most of the field test data for Spitfire take-off measurements, among other things, were measured at grass airfields near sea-level. Data during these tests regarding wind, temperature, and pressure, were noted. A host of specific information about the aircraft being tested was carried in the test reports. These data include engine type, propeller blade dimensions, and all-up weight of the aircraft tested. Exactly the information needed upon which to make an assessment of aircraft take-off performance. With this information, the propulsive force during takeoff can be estimated, as can the time and distance required for takeoff. These estimates can be compared with field measurements.

But there are additional, basic primary considerations: This aircraft has a tail wheel. Consequently, as this aircraft is accelerated to take off, the tail is raised while keeping the nose slightly elevated. In attempting to determine the most reasonably short time and distance to takeoff, one would presume the aircraft would leave the ground at airspeeds just above stall velocity. For instance, a rather basic standard procedure in computing take-off airspeed for an aircraft of this type, is to a) assume a representative take-off lift coefficient at stall and determine the consequent stall velocity, then, b) multiply that velocity by a factor of 1.2 to assure sufficient airspeed to be in the positive command region of the power curve as the aircraft achieves flight velocity. This assures that as the pilot folds the landing gear and the aircraft leaves ground-effect, sufficient airspeed is available to maintain control. However, on takeoff, as the aircraft is near its lowest possible flight velocity and throttled at full power, the propeller slip-stream will have a significant influence on lift. Consequently, to examine the Spitfire time-to-takeoff problem given a general approach and relevant factors, as just previously discussed, we must also consider the propeller slip stream and its influence on lift. And finally there is aircraft pitch. In addition to the propeller slip stream, we must account for the attitude of the wing in generating lift at takeoff. By adjusting pitch, take-off velocity can be closely determined so that calculated take-off time and distance can be adequately compared with actual field measurements. Essentially, this total process gives us a reality-check on our results.

The propeller is, perhaps, the most significant element in aircraft performance regarding this airplane. Spitfire aircraft were operated on takeoff by advancing the throttle to rated take-off power at full boost. As this occurs, the propeller generates tremendous thrust by inducing airflow through the propeller disk. As aircraft motion is initiated at the start of the take-off run, the force of this thrust can well exceed 2400 lbs. As this flow passes behind the propeller, the diameter of the induced flow is contracted and the flow velocity is increased by a factor of about two. This accelerated flow is the propeller slip stream, which then flows along and over the fuselage, and consequently over the near-fuselage, inner portion of the wing. A resulting favorable consequence is that this higher-velocity slip-stream flow augments lift of the wing.

So what happens as the aircraft begins to move? With full power, thrust of the propeller, the accelerating force moving the airplane, is at maximum. This thrust, at zero forward velocity, is called static thrust. Just as the plane begins to move, thrust becomes no longer static yet propeller efficiency increases. Once in motion, available thrust from the propeller is consumed by opposing resisting forces due to rolling resistance of the aircraft as the run begins, and ultimately just aerodynamic drag, induced drag, and lift remain as the aircraft achieves flight. The sum of these resisting forces over the course of the take-off run is taken to reduce available thrust of the propeller in linear relation to the square of the velocity of the aircraft. Consequently, one may calculate the accelerating force available from the propeller as a linear function of the square of the velocity of the aircraft. The aircraft, of course, will achieve flight at the moment that lift is equal to weight of the aircraft. Nevertheless, in order to determine the time and distance this will incur, we must know the aircraft weight (of course) and landing gear rolling resistance at the start of the run, area of the wing, slip-stream velocity due to the propeller, the resulting effective lift coefficient being developed by the wing, and ultimately, velocity and drag of the aircraft at takeoff.

And now a word from our sponsor –


Suppose we know the velocity at which an aircraft will take off, and the thrust – the force of acceleration – necessary to accomplish this, then what is the time and distance necessary for the aircraft to achieve this velocity? $ $ Let’s look at the math…

We work with variables $m$ = mass, $F$ = force, $t$ = time, $V$ = velocity, and $S$ = distance. In this case, we take $F$ to be the mean force thru the path traveled. If we already know $m$, and forces opposing $F$ are linear with regard to $V^2$, then these aspects follow:

$$F = m \cdot a$$ $$a = \frac{F}{m}$$ $$a \cdot t = V$$ $$V \cdot \frac{t}{2} = S$$ $$so \ \ V = \frac{2S}{t}$$ $$but \ \ t = \frac{V}{a}$$ $$therefore \ \ t = \frac {m \cdot V}{F} .$$ Consequently, with already given mass $m$, if we know the take-off velocity $V$, and mean accelerating force $F$, then we can determine the time to takeoff. Now, however, $$given \ \ a = \frac{F}{m}$$ $$then \ \ V^2 = 2\ \ \frac{F}{m}\ S $$ $$reworking, we find \ \ S = \frac{m V^2}{2 F} .$$ Consequently, again, with given mass $m$, by just knowing the take-off velocity $V$ and mean accelerating force $F$, we can determine also the distance required for the take-off run.


Given what we need to know, how was Spitfire take-off time and distance calculated? In the abstract, one must calculate the take-off velocity first, and then determine the time and distance needed for available thrust to achieve that velocity, presuming available thrust, opposing drag, and consequent acceleration, are linear in relation to the square of velocity. Available thrust is estimated as being equal to static thrust less opposing drag. For the question at hand, the calculations were based on an empirically based understanding of propeller thrust, an accounting of the slip-stream influence on lift just as the aircraft achieves flight velocity, and conceptually direct aspects regarding drag, acceleration, velocity, and distance. Resulting time and distance are dependent on a determination of the mean force of acceleration thru the takeoff run. In short, all we need to know to determine take-off time and distance, is the velocity of the aircraft at takeoff, and mean thrust (force of acceleration) during the takeoff run. That mean force is determined in relation to the thrust at the start of the take-off run, and thrust as the aircraft achieves flight.

Let’s take a quick descriptive overview of the computational accounting of thrust, drag resistance, lift, and velocity at takeoff. Remember, the aircraft is very near or at its stalling velocity at takeoff save for the higher-velocity, propeller-accelerated slip-stream flow over the wing. The significance of the results will be summarized. Here is the process –

  1. Calculate the estimated stalling velocity at takeoff using a simple estimate of maximum $C_L$ at stall. Adjust $V_{stall}$ by a factor of 1.2 to obtain estimated take-off velocity, $V_{t.o.}$. Derive the estimated $C_L$ at takeoff from the estimated $V_{t.o.}$.
  2. Calculate propeller static thrust and slip-stream velocity. Then, for estimated dimensions of the fuselage frontal area, determine the slipstream diameter and region of the slipstream flowing over each wing.
  3. The revised $C_L$ allows us to calculate the effective $C_L$ due to the slip stream. This requires knowing the specific wing chord and span dimensions, and consequent free-stream velocities relative to the wing, both interior to and exterior of the slip stream. We will also need to know the aircraft pitch and resulting angle of attack of the wing for each of the wing areas relative to the slip stream. For the total wing, the effective $C_L$ and $V_{t.o.}$ is now determined.
  4. Using the aircraft polar, calculate the force of drag on the aircraft at takeoff velocity. Also, determine the rolling resistance force of the aircraft at the moment the take-off roll begins. The mean thrust of acceleration due to the propeller is determined at the start of the run, and end of the run, using these values. The mean force of acceleration is determined as $$F = \frac{F_s – F_e}{ln (\frac{F_s} {F_e})}$$ where $F_s$ is static thrust less rolling resistance starting the run, and $F_e$ is static thrust less total aerodynamic drag just as flight velocity $V_{t.o.}$ is achieved.
  5. Determine consequent time and distance for takeoff using the mean force of acceleration during the run, and the determined velocity at takeoff. Note: The mean force of acceleration is naturally biased toward the end of the take-off run. Consequently, within limits, errors in the initial determination of propeller static thrust have a minimal consequence to our final answer.

The actual field assessment reports of performance of these WWII UK Spitfire aircraft can be reviewed at this site. Calculated performance of two aircraft, K.9793, and P.7280, was compared with field tests. These two aircraft had published results estimated for zero wind and standard atmosphere given time-at-test conditions at, or very near, sea level. Calculated results, herein, are taken at sea level. Note: In order to be comparable with field measurements, without error, all results were calculated in English-system units, and are so presented, herein.

For aircraft K.9793, a Mark 1 machine, the calculations were for the aircraft as equipped, and the salient following were ascertained: Propeller diameter, 10 ft 09 in. Blade chord at standard 0.7r position, 8.72 in. Rated power at takeoff, 1030 hp @ 3000 rpm. Reducer, 0.477:1. Static thrust estimated at takeoff, 2413 lb. Mean thrust thru the take-off run, 2316 lb. Aircraft take-off weight and pitch, 5,935 lb @ +3.4deg. Take-off run time at distance, 12.39s at 321 yds. Documented time and distance, estimated for no wind - standard atmosphere: 12.5s at 320 yds. Propeller fitted: dual pitch, 3-blade de Havilland with forged and machine-finished duralumin blades.

For P.7280, a Mk-II machine, the standardized aircraft was as follows: Propeller diameter, 10 ft 10 in. Blade chord at standard 0.7r position, 10.5 in. Rated power at takeoff, 1175 hp @ 2850 rpm. Reducer, 0.477:1. Static thrust available at takeoff, 2859 lb. Mean thrust thru the take-off run, 2750 lb. Aircraft take-off weight and pitch, 6172 lb @ +6.0 deg. Take-off run time and distance, 9.48s at 215 yds. Documented time and distance, given for no wind - standard atmosphere: 9.5s at 230 yds. Propeller fitted: dual pitch, 3-blade Rotol of machine-finished hygdulignum (birch/phenolic-resin laminate) blades.


Summary: So yes, we can calculate realistic numbers for take-off time and distance for the Spitfire. Several aspects must be considered. As previously noted, the Spitfire evaluation was accomplished by initial calculation of take-off velocity given the aircraft at its full performance with fitted propeller and Merlin engine. Initially calculated factors include propeller static thrust and slip-stream velocity for the propeller being evaluated, slip stream influence on lift, and determination of effective lift through the take-off run. Known are aircraft weight and wing area. We will know propeller thrust and rolling resistance at the start of the take-off run, total aerodynamic drag and velocity at takeoff, and remaining propeller thrust at the end of the run. We only need to sum thrust and drag and determine take-off velocity to determine distance and time essentially by simple integration. Lift at takeoff is equal to the weight. The slip stream from propeller thrust augments this lift. Time and distance are derived from the mean of the force of acceleration at run initiation and as the aircraft achieved flight, given the initially determined velocity of the aircraft at the moment of flight. In setting up the calculation process, we need aircraft pitch to be varied and consequent results noted. Other than given data regarding the aircraft, we only need, essentially, two resulting numbers in order to determine time and distance to takeoff. These two resulting numbers must be calculated based upon the developed information. Those two numbers are velocity of the aircraft at takeoff, and the mean force of acceleration thru the take-off run.


$^*$As a brief trailing end note: Problem assessment, conceptual evaluation, background research to obtain adequate information upon which to base an analysis, mathematical development and computational procedure assembly, results assessment and validation, as well as development of the text for posting of this answer, required approximately 2.3 months from start to completion.

This was a very interesting question. Thanks for asking!

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what else is going on that makes the true take-off time so much longer than my calculated take-off time?

I think that the poor propeller doing the whole job in front (on in the back) of the piston engine has been totally forgot! The piston engine is producing power but then it's the job of the propeller to convert this power in a thrust. The easiest way to account for propeller thrust, is just by using basic physics.

Being $P$ the power supplied to the propeller by the engine, the propeller changes this power in a thrust $T$ moving the airplane at a speed $V$. In this process, some 20% of the power (at the design point) is lost in inefficiencies which are incorporated in an efficiency factor $\eta$:

$T=\eta \frac{P}{V}$

$\eta$ depends on the speed, with a typical trend visible in the following plot¹:

propeller efficiency factor

As visible, $\eta$ depends on the blade pitch (as conventionally measured at 75% of the span) and on the advance ratio $J=\frac{V}{nD}$, where $n$ is the rotating speed and $D$ the diameter. The theory behind this plot and therefore the plot itself is valid for a more or less broad range around the peak: at $V=0$ it gives $\eta=0 \Rightarrow T=0$ which is not realistic since also at zero speed the propeller provides thrust. For that part of the thrust (so called static thrust) you can refer to this answer.

Just try to redo the math taking this into account and let's see if you get a more realistic result.


¹from: McCormick B.W. Aerodynamics, Aeronautics & Flight Mechanics. John Wiley & Sons, Inc.

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Prop takeoff calculation is a bit tricky since so much changes between standstill and flight. The most important part is propeller efficiency $\eta_P$ which theoretically would become zero at rest if you use the usual formulas.

The trick is to not take flight speed but the speed in the propeller plane for realistic values. This can be assumed to be half of the eventual speed increase of the air going through the propeller disk $\Delta v$. Then even the efficiency formula for flight can be used throughout the takeoff run. The expression $$\eta_P=\frac{v_∞}{v_∞+\frac{\Delta v}{2}}$$ becomes $$\eta_{P_{at\,rest}}=\frac{\frac{\Delta v}{2}}{\Delta v}=\frac{1}{2}$$ which is consistent with reality as the upper theoretical limit for propeller efficiency at rest. Viscous and induced losses reduce this by about 20% for a constant speed propeller and possibly much more for a fixed pitch propeller and the two-position variable pitch prop of early Spitfires.

The proper form of this equation is $$\eta_P=\frac{v_∞+\frac{\Delta v}{2}}{v_∞+\Delta v}$$ Strictly, this is the correct efficiency equation for the full speed range, but was a bit impractical in the time before computers because using it requires an iterative process in which thrust and $\Delta v$ must be balanced. Therefore, all books of authors who blindly copy what has been published before will show you only the first equation. The error is small once $v_∞$ becomes large, especially since $\Delta v$ drops with increasing flight speed.

But there are more losses. Once the prop is running, the airplane will produce aerodynamic drag even before it has started to move. The propwash flowing over hull, tail and wing root will produce viscous drag which of course only grows once the airplane picks up speed.

A very interesting paper on the efficiency of an installed propeller can be found here. By comparing the gliding qualities of a Luscombe 8E with drag in level flight, the authors measured the drag increase of the airframe caused by the propeller wake. A drag increase of 30% from the propeller effect was measured, resulting in an overall propeller efficiency of only 62% when the propeller alone had an efficiency of 81%.

The smallest bit that is still missing is rolling resistance. This adds maybe 2 - 3% of overall drag and goes away as the wing lifts the airplane up. The price for this is induced drag on the wings, of course.

Since you post on Stack Exchange, it is safe to assume that you have a computer at hand and know how to use it. Write a small program which calculates all forces and integrates them for a small time step, then repeat at the new speed until you reach lift-off speed. The correct take-off time and distance will be a byproduct of this procedure.

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  • $\begingroup$ Please see attached article graphs. $\endgroup$ Commented Dec 17, 2022 at 17:49
  • $\begingroup$ @RobertDiGiovanni The author also makes the mistake of zero prop efficiency at zero speed. This contradicts his own thrust curve which looks quite realistic. $\endgroup$ Commented Dec 17, 2022 at 19:40

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