# Aircraft - static take off - how is this possible?

From this Aviation SE answer, and personal experience, aircraft routinely are held on their brakes while the engines are run up to full power, prior to take-off. Intuitively, to me at least, this is hard to explain. Given the formidable power of jet engines, especially military aircraft, why is this possible? Why doesn't something break, or why doesn't the aircraft just skid down the runway, wheels locked, tyres smoking? For instance, it's not that hard to wheel-spin a stationary car, or vice versa slam on the brakes in a moving car and break the adhesion between the tyre and road surface.

• Because they are designed for that scenario? Commented Jan 9, 2016 at 1:17
• The heading is not a good fit to the question. Power is related to force and velocity. The engines are delivering ZERO power to the aircraft while it is stationary. The force delivered is close to the same as that delivered when they are at cruise speed. The air flow past or through the engines is similar to that at cruise speed. ie the main difference is that the force is not being used to change velocoty or oppose airflow but to oppose braking forces. Calculate force from engine at cruise speed and you'll see the force is significant but not 'unopposably immense' at zero velocity. Commented Jan 9, 2016 at 1:25
• I don't think jet engines actually deliver full power at standstill. The air intake would be far smaller than at high air velocity. Apart from that Jon Custer got it right. If you can do a 15g turn in that machine, revving the engines won't do squat. It's not even close to stressing the structure even a little. Commented Jan 9, 2016 at 1:38
• "For instance, it's not that hard to wheel-spin a stationary car" - for an aircraft, unlike a car, the wheels are not directly driven. Commented Jan 9, 2016 at 2:09
• @Jon, Come on, that's just re-stating the question in different words - what is it about the design that allows this? Commented Jan 9, 2016 at 20:58

Why doesn't something break, or why doesn't the aircraft just skid down the runway, wheels locked, tyres smoking?

Basically, the aircraft you're accustomed to simply don't have all that much thrust, particularly compared to the force required for hard braking.

An example: a commercial airliner will typically have a rollout time of about 30 to 35 seconds, and a liftoff speed of 120 to 140 knots (70 m/sec). Then acceleration a is given by $$a = \frac{\Delta v}{\Delta t} = \frac{70}{35} = 2 \text{m/sec}^2 = 0.2 \text{ gs}$$ While this is respectable, it implies that, for a static takeoff, the brake coefficient of friction must be greater than 0.2, and that is not hard to do at all.

Military aircraft (fighters, especially) need greater performance, of course, and a thrust-to-weight ratio greater than one is possible. A static takeoff for such frisky aircraft would indeed be problematic, requiring tires with friction coefficients greater than one. However, land-based fighters don't normally do static takeoffs, while carrier-based fighters do. But carrier aircraft are hooked to catapults, and the brakes are not the dominant factor.

• Nice idea to use the approximate time to take off to estimate thrust! Commented Jan 9, 2016 at 4:03
• IIRC, in the case of carrier catapult launched aircraft, something does break. While the aircraft is "locked in", the pilot powers up to full throttle; then steam pressure is applied to the catapult piston, but this is not instantaneous, so the mechanism is restrained by a component that breaks when the required "launch" force is attained, and that's when the shuttle (and attached aircraft) start moving. Commented Mar 9, 2019 at 1:11

If you apply the front brakes, and then increase thrust along a line that is some distance above the ground, you increase the downward force on the nose wheel. This helps to prevent skidding - although it would not be sufficient if the thrust was sufficiently large.

You can compute this with a simple diagram:

I drew the limiting case where the thrust of the engine is big enough to cause the plane to "tip over" the nose wheel - note that I made the wheels quite high to get a clearer view of angles. To reach this point, the engine would have to have a thrust that is greater than the weight of the plane by a ratio given by $\frac{d}{h}$ where $d$ is the horizontal distance from the center of mass of the plane to the point of contact of the nose wheel.

Now the plane may start skidding before that - but assuming that we have tires with a nice high coefficient of friction (around 0.7 seems reasonable for a dry surface) then you can support a good fraction of the weight of the plane in thrust before it will skid. And when a plane is stationary, it may not develop the full thrust that the engine is capable of in flight, since you are limited in the amount of air you can take in (even if you were able to create a perfect vacuum at the intake of the engine, you would be limited in air flow by the fact that atmospheric pressure has to push air towards the engine. Once you have picked up a lot of speed, air is coming towards you at high velocity and therefore there's more air to move, and you can develop greater thrust. On the other hand, you need to accelerate the air to greater speed in order to get a net increase in thrust, and that offsets things a bit - so "it depends" on details of the design. See for example this article on jet engine performance).

At any rate, you can assume that the engine, while revved up, is probably not operating at maximum thrust; but it can be important to get it up to temperature before demanding maximum output, since the clearances between turbine blades and engine wall are tightly controlled and optimized for when the engine is "at temperature". By slowly revving the engine up, you prevent massive thermal shock and prolong the life of the engine - even if it's not running at 100% thrust.

So although the engine may be screaming, you're not going to skid, topple, or break the plane.