What factors affect take-off velocity and runway length for a small jet craft?

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To calculate the speed of a small jet craft during take-off, the thrust of 2,300 lbs (10,231 Newtons) and weight of 22,000 lbs (9,979 kg) yield an acceleration of approximately 1.025 m/s². Using the formulas for velocity and time, the speed at a distance of 400 meters is calculated to be 28.6 m/s (64.1 mph), with a time of 27.9 seconds to reach that distance. Factors such as drag, rotation velocity, and take-off roll significantly influence performance, and adjustments may be needed based on conditions like altitude and flap position. For precise calculations, consulting the aircraft's information manual or pilot's operating handbook is recommended. Understanding these dynamics is crucial for accurate speed assessments during take-off.
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I'm trying to establish the speed of a small jet craft as it accelerates down the runway. What I know is that the jet engine generates 2,300 lbs of thrust, and that the total weight of the craft at take-off is 22,000 lbs. I don't have the coefficient of drag but it's a pretty advanced design. How difficult is it calculate the speed for any given distance and the length of time it took to get there? Does anyone have a formula?
TIA
 
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Best to convert to SI units first.

2,300 lbs thrust = 10231 Newtons

22,000 lbs weight = 9979 kg

Acceleration, a = force / mass = 10231 / 9979 = 1.025 ms^-2

The following three formulae relate initial velocity, u; final velocity, v; time, t; acceleration, a; and distance, s

v = u + at

v^2 = u^2 + 2as

s = ut + \frac{1}{2}at^2

assuming a starting velocity, u, of zero, then the speed at a given distance, for this aircraft is:

v = \sqrt{2as} = \sqrt{2.05 \times distance}

and the time it took to get there is:

t = \sqrt{\frac{2s}{a}} = \sqrt{1.95 \times distance}

example: after 400 metres the speed will be:

v = \sqrt{2.05 \times 400} = 28.6 m/s = 64.1 mph

and the time taken will be:

t = \sqrt{1.95 \times 400} = 27.9 seconds

[edited to correct a = m/F to a = F/m :redface: (which as a is fairly close to unity, didn't affect the answer that much)]
 
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Thanks! That's very helpful.
 
The drag coefficient is indeed very complicated for an airplane. Even the static friction of the wheels rolling along the tarmack changes as the airplane accelerates and begins to shift its weight from its wheels to its wings.

The velocity at which an airplane's nosewheel can be lifted from the ground is called the "rotation velocity," and the length runway used by the take-off is called the "take-off roll." Many factors affect both the rotation velocity and the take-off roll. Changes in the density altitude (local air pressure), for example, affect both the thrust generated by the engines and the lift generated by the wings. The position of the airplane's flaps is extremely important -- flaps generate a lot of lift, but they also create a large amount of drag. Flaps can be used to allow an aircraft to take-off from a short runway, for example.

If you really need precision, I suggest you get a copy of the aircraft's information manual or pilot's operating handbook. It will contain graphs indicating the necessary rotation velocity and take-off roll for all sorts of different conditions.

- Warren
 
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