# Which one needs longer runway of an airplane between take off and landing for a constant weight?

Consider we have an airplane that the weight will be constant when take off and landing (in fact will be no such common airplane right now as the fuel will be less due to engine consumption. But in an electric airplane, that will be possible, I guess). What I am asking is, which one is required longer runnway from an airplane starts taking off in the edge of the runway until it just leaving the runway (ground), compared to when that airplane is landing, from it start touching the runway until it can stop safely in the runway. How do we calculate the required runway for landing?

• Unclear what you're asking. I very much doubt that an anvil would be at all helpful in takeoffs or landings: merriam-webster.com/dictionary/anvil – jamesqf Feb 1 '19 at 4:21
• Which one is need longer runway of an airplane between take off and landing for an airplane with the same weight? – AirCraft Lover Feb 1 '19 at 4:24
• Apologize if it was not clear. I have changed it to runway. Hope it clear. – AirCraft Lover Feb 1 '19 at 4:29
• Are you interested in a specific airplane? The reason I ask is because it will be easier to answer based on published data; the manufacturers draw from experience and flight testing, because calculating it for practical purposes from a single equation is not possible. – ymb1 Feb 1 '19 at 4:41
• As far as my understanding, there is equation to bind Lift force (L) to the air density (rho), the airspeed hit the wing (v), the wing area wide (A), and the coefficient of lift of the wing of a specific airplane (Cl). L=0.5*rhov^2*ACl. The weight of course we can get. Lifting force is the opposed force to the gravity+drags. When the Lift is same with the gravity and drags, then it is the time airplane to leave the ground. That what my understanding. And the requires runway we may calculate based on the required speed and the acceleration. But so far I don't have such a formula for landing. – AirCraft Lover Feb 1 '19 at 5:19

Generally, aircraft require longer distance for taking off than landing roll.

The standard definition for take-off phase is from the start of acceleration (stopped aircraft) to the point where aircraft reaches 35 ft above ground. Similarly, the landing phase is from 35 ft to the final stop. What you mentioned is a sub-phase, which is called "ground roll".

When the aircraft is accelerating for the take-off, the engins' thrust must overcome the aerodynamic drag and the ground friction, and the resultant will contribute to the acceleration. But when the aircraft is landing the reversed thrust in addition to an increased aerodynamic drag (due to spoilers and increased flap deflection) and increases ground friction (due to application of brakes) are decelerating the aircraft. This is physical reason.

While there are some engineering methods (such as equations or charts, usually presented in aircraft design books i.e. Roskam or in aircraft performance books), the most accurate method is to writing the force balance acting on the aircraft and computing the acceleration at any time step, and solving the position differential equation for the distance. But this method requires details about aerodynamic and thrust characteristics. This method somewhere was called time marching.

• Don't you have something like formula how to calculate or to estimate the runway requirement for landing? Appreciated if you have. – AirCraft Lover Feb 1 '19 at 16:35

Landing distance will always be shorter, due to higher deceleration values but mainly due to the definition of TO length, which takes into account obstacle clearance and rejected TO distance after engine failure.

Torenbeek, Synthesis Of Subsonic Airplane Design, section 5.4.6 gives the runway length from touchdown to standstill as:

$$S_{run} = \frac{{V_{td}}^2}{2 a} + V_{td} \cdot \Delta t$$

With

• $$a$$ = max deceleration.
• $$V_{td} \cdot \Delta t$$ = mean inertia time to reach max deceleration, approx. 1.5 to 2 seconds (from Torenbeek Appendix K)

Values for $$a$$/g on dry concrete are given as:

• 0.30 - 0.35 light aircraft, simple brakes
• 0.35 - 0.45 turboprop without reverse thrust
• 0.40 - 0.50 jets with ground spoilers, anti-skid devices, speed brakes.
• 0.50 - 0.60 as above with nose wheel braking.

For wet concrete, as in this answer