How does the thrust to weight ratio work in terms of aeronautics and rocketry?
I have been reading about fighter jets recently, and came across the thrust/weight ratio. Using the F-22 Raptor as an example, it has two engines that put out 104kN of thrust (without afterburners). This gives a combined thrust of 208kN. Its unloaded mass is 19,700kg, which is equivalent to a weight of 193kN. This gives it a thrust to weight ratio of 1.078:1.
All this makes sense to me. Where I have trouble though, is the implications of this figure on climb rates. Modern fighter jets can famously accelerate perfectly vertically, with some reaching Mach 1 within a short period of time. However, given a thrust/weight ratio of 1.08:1, it seems that the thrust of the engines is only just greater than the weight of the plane. The pilot should experience 1.08g of acceleration, and hence climb at 0.78m/s/s. This is obviously not the case, as evidenced by any number of aerobatics videos.
So where have I gone wrong? How does the thrust/weight ratio convert into climb rates?
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This is probably wrong, but since your original figures were in kiloNewtons, maybe your calculated accelerations should be in km/s/s?
Mariah is right, you should measure in Newtons then you can being calcuating the accerlerations by m/s/s or use Mariah’s way. Both ways are more sense. The calucating in m/s/s should be 780m/s/s?
Note that the pilot will initiate the climb after picking up at least some horizontal airspeed first (even if this is done very quickly after takeoff), the pitch up maneuver will translate most of this into vertical airspeed, so you aren’t starting off from zero.
@MasterAir16 and @Mariah Thanks, but both the kN figures cancel out when I divide thrust by weight (208,000 divided by 193,000 is still 1.078:1). The thrust to weight ratio has no units, and that is the figure I used to calculate the acceleration.
@hiphiphopflipflapflop That is true, but it should still take a long time to climb to any great speed. If the plane is travelling at take-off speed, which is about 250–300km/h, at the calculated rate it should take about four minutes of continuous vertical motion to reach Mach 1. In this time, it would reach around 42km altitude, which is more than double the service ceiling for the Raptor. I’m certain my calculations are wrong somewhere between thrust/weight and vertical acceleration, but I cannot find the error.
You need to know the speed during each phase of the maneuver to figure this all out. It may very well be that the aircraft is already above Mach 1 at the start of the climb.
@RocketGuy True, I don’t have exact figures for how it achieves this. The Wikipedia page for the F-15 Eagle says “the thrust output of the dual engines is greater than the aircraft’s weight, thus giving it the ability to accelerate in a vertical climb.” Apparently this is not a common ability, so it seems my calculations are correct, and vertical acceleration is minimal in the few cases where it is possible.
But the F-15 Wikipedia page also says “Rate of climb: >50,000 ft/min (254 m/s)”
So 568 mph.
That’s not Mach 1, but Mach 0.74 isn’t slow.
@jaytkay Maybe the maximum rate of climb is not vertical. If it were on a 45 degree gradient, the wings would take some of the weight, allowing the thrust to give more upward acceleration.
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