What is the formula for engine thrust as a function function of engine pressure ratio (EPR)?

If the EPR is lower than 1, is the engine producing thrust? For example during descent, when the engines are in idle power.

  • $\begingroup$ If the engine is running it is producing thrust. Even at idle power, the engines are producing thrust, how much depends on what engine. $\endgroup$
    – Ron Beyer
    May 19, 2017 at 11:09
  • $\begingroup$ @Ron Beyer It At 'idle power', with little or no torque applied, a propeller will absorb power from the relative wind, producing drag, that might be seen as a 'negative thrust'. The same will happen with the fan of a turbofan or –in general– at any 'actuator disk'... $\endgroup$
    – xxavier
    May 19, 2017 at 12:23
  • $\begingroup$ I suppose if the "relative wind" was equal to the thrust of the fan or propeller, it could be zero thrust. A B777 engine which produces about 115,000 lbs. of thrust at full power, still produces about 4,000 lbs. of thrust at idle. $\endgroup$ May 19, 2017 at 13:11
  • 1
    $\begingroup$ @xxavier The thrust can't be negative, otherwise the aircraft would move backwards at idle sitting on the tarmac. Quite a few aircraft, props and jets included, produce enough thrust at ground idle to move the aircraft forward at a good clip. The drag in flight is noticeable because it is a difference in velocity, most aircraft can't fly (level) at idle thrust, but there is a positive thrust there, it just isn't enough to maintain level flight. Same as if you just throttled back to 25% or some other low thrust value. $\endgroup$
    – Ron Beyer
    May 19, 2017 at 13:28
  • $\begingroup$ @With enough headwind, a prop aircraft at idle will roll backwards on the tarmac... Please see this: imgur.com/TlztzL4 In both cases, the propeller has the same headwind, but in the upper figure the revs are high enough to give a positive thrust, while in the lower sketch the revs have dropped, and the thrust becomes negative... $\endgroup$
    – xxavier
    May 19, 2017 at 13:32

2 Answers 2


I'm heavily cribbing from Boeing's "Jet Transport Performance Methods" here, so hold on to your seat...

F = m· (V2 – V1)

where is the gas flow rate, mass per unit time

V1 is the initial velocity of the gas

V2 is the final velocity of the gas

Let’s re-write equation 3 in a more complete form:

Fnet = [(m·core air + m·fuel) × Vcore exhaust] + (m·fan air × Vfan exhaust) - (m·air × Vairplane) Note that the force is now called Fnet, for net thrust. What are those terms in parentheses? The term (m·air × Vairplane)is the momentum per unit of time of the air being taken into the engine. The term [(m·core air + m·fuel) × Vjet exhaust] is the momentum per unit of time of the gas leaving the core of the engine. Notice that we are now accounting for the fact that the mass flow of the exhaust is greater than the mass flow of the inlet, because fuel has been added to the intake air. Finally, the term (m·fan air × Vfan exhaust)is the momentum per unit of time of the air leaving the fan stage of the engine. The net force, which is the thrust produced by the engine, is the algebraic sum of these three terms, which is the rate of change of momentum of the gas as it passes through the engine. If we considered only the first two terms, the exhaust flow, we would see only the force created at the fan and the exhaust nozzle. But this is only half of the story, because we must remember that the flow of the air entering the engine also has momentum and the useful work that the engine is doing is the difference between the two.

Note that is "m-dot." The dot should be over the letter "m" but didn't copy well from the original pdf. I bold-texted the part about force created at the fan and exhaust nozzle, because I think that's what you're measuring with EPR. Again, from the JTPM textbook:

This stands for engine pressure ratio. It’s the ratio of the gas flow total pressures measured at two specific points: at the entrance to the exhaust nozzle (station 7) and at the inlet to the fan and compressor (station 2). EPR is defined simply as:

EPR = Pt7 / Pt2

where Pt2 is the total pressure measured at station 2 Pt7 is the total pressure measured at station 7

Hope this helps. Just a little math, but I highly recommend the whole book.

  • $\begingroup$ We have formatting here that can help with the equations but this doesn't seem answer the question. $\endgroup$
    – fooot
    May 19, 2017 at 19:33

NASA has a very website with a turbojet simulator that allows you to play around with different settings and see how the engine responds: Engine Sim. If you go to this site and play around with the app, you can easily see that thrust is more or less proportional to EPR. By "proportional" I mean if you double EPR, you double thrust. Very close to idle it deviates from this behavior, but above idle it is close. You'll also see that EPR is just one of many parameters needed to calculate thrust. i.e. at EPR=2 at sea level static is a very different thrust than at 40,000 ft and Mach 0.8. But, all else being equal, thrust is linear with EPR.

Now, why should that be... well, there is a lot of math involved. On the same NASA site, you'll find a tutorial on calculating thrust (Turbojet thrust tutorial). They don't actually show the derivative of thrust with EPR, but a lot of the basic math is there.

Now, all of the above is valid for a single spool turbojet engine, which basically nobody uses anymore. Everything is a turbofan. For a turbofan, there are now multiple different exit pressures to think about (i.e. both the fan exit pressure and the core exit pressure). Rolls-Royce uses a modified EPR for their turbofans, which takes both fan exit pressure and core exit pressure into account.

This article also has some more details about EPR that you may be interested in: Cockpit Design: EPR v/s N1 indicationhttp://theflyingengineer.com/flightdeck/cockpit-design-epr-vs-n1-indication/

  • $\begingroup$ How do you run EngineSim? The page just displays text for me. $\endgroup$ Jun 19, 2017 at 22:20
  • $\begingroup$ It is a Java applet. you may not have Java installed or there may be a problem with your Java setup. $\endgroup$
    – Daniel K
    Jun 20, 2017 at 11:20
  • $\begingroup$ Oh, apparently Chrome dropped Java support in 2015. $\endgroup$ Jun 20, 2017 at 22:34
  • $\begingroup$ For anyone else having trouble, I made it work using Firefox ESR (32 bit): mozilla.org/en-US/firefox/organizations $\endgroup$ Aug 19, 2017 at 17:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .