Or is the thrust : weight ratio larger than one so they can fly vertically without losing speed?

  • $\begingroup$ Question needs clarification-- "15 degrees" of what? Do you mean angle-of-attack of the wing, or something else? $\endgroup$ Commented Apr 9, 2023 at 13:33
  • $\begingroup$ Related -- aviation.stackexchange.com/a/56476/34686 $\endgroup$ Commented Apr 9, 2023 at 13:33
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    $\begingroup$ Because they can? P.S. The max performance demos you see at air shows are not done on every takeoff. $\endgroup$ Commented Apr 9, 2023 at 15:08
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    $\begingroup$ Because for a jetfighter is more important "how fast" you reach a target than "how efficient". If flying straight up like a rocket is the fastest way then you fly that way. $\endgroup$
    – sophit
    Commented Apr 9, 2023 at 16:04
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    $\begingroup$ Where does the statement "lift is optimum around 15 degrees" come from? I'm not personally familiar with that generalization. $\endgroup$
    – Ralph J
    Commented Apr 9, 2023 at 21:55

3 Answers 3


There is something called a Rutkowski Profile. Fighters are more interested in the shortest time to get to altitude, not the most efficient. The Rutkowski Profile is flown by following the path through the energy maneuverability envelope where the aircraft has the most excess specific power. This can be calculated by drawing a line that crosses (intersects) all of the specific excess power lines (Ps lines), on the envelope diagram as close to 90 degrees as possible.

Rutkowski Profile

enter image description here

In English,

  1. Accelerate to get to the best (highest) Ps. In a jet, because Massflow through the engine increases thrust, this is always just below the transsonic region, about Mach 0.92, where transonic drag starts to increase.
  2. Climb, holding airspeed, (Mach number) at just below the transonic drag point.
  3. Reaching a certain altitude, (As I recall, in the F-4 it was around FL280-FL300), you nose over and acccelerate through the sound barrier to a pre-specified kias (again, as I recall, in the F-4 it was 610kias)
    (The FL280-FL300 altitude was probably determined by the point where, as you climbed at constant Mach, the indicated airspeed slowed to L/D max.)
  4. Then, climb, supersonic, at 610 kias until you reach maximum altitude/Mach number.

Fighter jets are blessed with a much higher thrust to weight ratio than many other aircraft due to their performance requirements for speed and acceleration.

But they follow the same rules for climbing that all other aircraft follow: excess thrust is required to climb.

The amount of thrust required to climb is determined by the weight of the aircraft weight × sine angle of climb. There is a decrease in drag associated with reduced wing lift requirement.

Modern fighters, with a thrust to weight ratio greater than one, can literally climb straight up like rockets. General aviation aircraft have thrust to weight ratios more in the range of 0.2 - 0.25, so they must climb at a much shallower angle or risk losing airspeed.

why climb so steeply?

Engines have a most efficient speed (RPM) too.

pitch controls speed

The most efficient way to climb is when the engine and airframe are most efficient: Vy.

So, a really hot airplane will climb at a steeper angle to control its airspeed. The same plane could climb at the same speed by choosing a lower climb angle and a lower power setting, holding full power (and afterburners) in reserve, helping extend engine life.


It's faster, and often more efficient, for a high-powered aircraft to climb at a higher climb angle than the one that provides for the highest momentary L/D.

The simple reason is that jet fighters, while capable of supersonic flight, are normally limited to Mach 0.75-0.85 to avoid the transonic region and need for wet thrust. As long as there's thrust to spare, true airspeed will be ~same at whatever angle you climb. This is different from non-aerobatic aircraft that have to choose between climb angle and airspeed.

In ballistics, the longest range for a low-altitude projectile at given energy is achieved at a 45-degree launch angle. This holds true whether it is a bullet or a rocket. At high altitudes, the optimum angle is even higher, since it's more efficient to reach less-dense air sooner.

Winged aircraft can convert thrust to lift more efficiently than pure ballistic projectiles. However, they still do so better at high altitude. Full dry thrust aka MIL is an efficient setting for the engine, wet thrust is inefficient.

As for the efficiency of a steep climb, some basic trig. Say you're an F-16 or a Su-35 with a TWR of 1.0 wet or 0.7 dry, a L/D of ~10, and a ~300 m/s subsonic limit. If you climb at 15 degrees, you will climb at sin(15)*V ~= 75 m/s, no more. At 90 degrees, you can't gain airspeed, and will lose it to parasitic drag.

At 45 degrees, 71% of your thrust is lateral and 71% is vertical. At full dry thrust (70%), you're holding 50% of your weight with vertical thrust. As long as your L/D is at least 2, the rest is supported by lift. You can climb at 0.707*300 ~= 212 m/s, about 2.8 times faster. Your groundspeed loss, from 287 to 212 m/s, is only 26% at -75 m/s.

At L/D=10, each second of 45-degree climb will convert to (212-75)*10 = 1,380 m of range. Whether a slow climb is more efficient depends on how much a difference in L/D a shallower angle makes. To even things out, your fuel burn in a 15-degree climb would have to be ~24%, only 1/3 of the 70% setting for 45 degrees. Both heuristics and simulator experiments suggest it's not. This is before considering the effects of reduced parasitic drag the higher you fly.

In short: It's practical to climb as fast as you can, while keeping your engine and your airfoils in their efficient range.

  • $\begingroup$ Re first sentence-- did you really mean to say a "higher" angle-of-attack than best L/D? "Lower" would seem to make more sense. Comment informed by second set of diagrams here, and three paragraphs of text following: aviation.stackexchange.com/a/56476/34686 $\endgroup$ Commented Apr 9, 2023 at 22:06
  • $\begingroup$ @quietflyer Gets complex in there. Generally you would use a higher than lowest-drag AoA to initiate an unrestricted climb, but at a high enough angle the AoA itself does drop. Edited, it's really climb angle that matters. $\endgroup$
    – Therac
    Commented Apr 9, 2023 at 22:22

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