1) Is $V_1$ used to determine minimum runway length? How is $V_1$ calculated?

2) If it is used to determine the minimum runway length, and for example we have a runway of 8000 ft and my airplane's accelerate stop distance required (ASDR) is 4000 ft and takeoff distance required (TODR) is 5000 ft, can my $V_1$ be as low as 15 kts? Since (let's say) below 15 kts, of course you will have the 4000 ft required to stop and also have enough distance to takeoff with 1 engine inoperative.

3) Let's say we have a very long runway so an airplane can make a rejected takeoff even at rotation speed ($V_R$). Can $V_1$ be the same as $V_R$?


2 Answers 2


Takeoff Field Length

The “FAR” Takeoff Field Length determined from the FAA Approved Airplane Flight Manual (AFM) considers the most limiting of each of the following three criteria:

1) All-Engine Go Distance:

115% of the actual distance required to accelerate, liftoff and reach a point 35 feet above the runway with all engines operating:

All-Engine Go Distance

2) Engine-Out Accelerate-Go Distance:

The distance required to accelerate with all engines operating, have one engine fail at $V_{EF}$ at least one second before $V_1$, continue the takeoff, liftoff and reach a point 35 feet above the runway surface at $V_2$ speed:

Engine-Out Accelerate-Go Distance

3) Accelerate-Stop Distance:

The distance required to accelerate with all engines operating, have an engine failure or some other event at $V_{EVENT}$ at least one second before $V_1$, recognize the event, reconfigure for stopping and bring the airplane to a stop using maximum wheel braking with the speed brakes extended.

Accelerate-Stop Distance

$V_1$ Speed

$V_1$ is the speed chosen by the manufacturer as "the pilot’s initiation of the first action to stop the airplane during the accelerate-stop tests." Note that this is particularly important in that it is not a takeoff decision speed as some pilots like to think of it. The decision must have been made prior to $V_1$ so that there is time to react by $V_1$.

This speed is chosen by the manufacturer and must be between the engine failure speed1 and rotation speed. Within this range, they could choose any speed, but typically they choose the speed which gives a balanced field length (see next section).

There are also other factors to consider such as the maximum brake energy limit ($V_{MBE}$) which will prevent $V_1$ from being too high.

Balanced Field Length

If we takeoff from a fixed length runway (like, say one that matches the Takeoff Field Length from above), changing the $V_1$ speed will affect the weight that we can successfully continue the takeoff or abort within the available runway.

At a particular weight, a lower $V_1$ will enable us to stop sooner if we abort but will take more distance to continue the takeoff. A higher $V_1$ will require more distance to stop but will take less distance to continue. If we graph these values, there will be a point where the $V_1$ speeds for the abort/continue lines meet, and this is known as the balanced field length. It is the value typically given in the AFM:

Balanced Field Length

This is nice because it gives the pilot one important number to remember instead of different stop/continue $V_1$'s.

If they were to choose an extremely low $V_1$ as in your question, then it would require considerably more runway in order to continue the takeoff than if it were higher, and it would not be balanced. Some AFM's do include separate charts for accelerate-go and accelerate-stop though so you could figure out exactly how much runway that you would need if the engine failed at a lower airspeed.

Rotation Speed

$V_R$ is determined independently from $V_1$, and if it is ever less than $V_1$ then it must be raised to be equal to $V_1$.


The regulatory information for takeoff performance on transport category aircraft is listed in 14 CFR Part 25, Subpart B, Performance, 25.101 - 25.113.

Of particular interest for this question would be 14 CFR 25.107 - Takeoff speeds and 14 CFR 25.113 - Takeoff distance and takeoff run

An easier to read document is the FAA's Pilot Guide to Takeoff Safety which covers this and a lot more. Much of my answer was taken directly from this document, and I would highly recommend anyone that is flying a multi-engine aircraft to read the entire thing.

1 $V_{EF}$ itself must be above the minimum controllable airspeed on the ground plus acceleration while the pilot reacts to the engine failure.

  • 1
    $\begingroup$ Why must $V_R$ be strictly larger than $V_1$? Why not $V_1$ being reduced to $V_R$ (for instance)? $\endgroup$
    – Hugh
    Apr 29, 2014 at 7:13
  • 1
    $\begingroup$ @Hugh It's a regulatory requirement, however within the range of allowable values, the manufacturer chooses $V_1$, so they could lower it if they wanted. If they had to do that, it likely wouldn't be balanced though. $\endgroup$
    – Lnafziger
    Apr 29, 2014 at 12:42

Just to make sure we're on the same page, I'll reiterate what should be obvious: $V_1$ is a speed at which you may stop in the available runway space or continue the takeoff.

In theory, $V_1$ could be any speed that matches this criteria -- it is not necessarily one speed. However, in practice we refer to a few different $V_1$ speeds:

  • Minimum $V_1$
  • Balanced $V_1$ (distance to takeoff is the same as distance to stop)
  • Maximum $V_1$

The reason why $V_1$ cannot be greater than $V_R$ is that you're already or about to be in the air. If you attempt to stop the aircraft rotating, you might be too aggressive and destroy the aircraft by tring to force the airplane down during or after rotation.

You also can't have a $V_1$ of something like 15 knots, as it may not be below $V_{MCG}$. If you did continue with full thrust on one engine at 15 knots, you would not have enough authority from the rudder to prevent the aircraft going off the edge of the runway, which is extremely dangerous. This is even more of an issue if you decide to continue the takeoff at $V_1$, which would almost certainly result in a runway excursion. Typical $V_{MCG}$ values for airliners can be above 100 knots, making your proposed situation impossible.

For all cases, using the balanced $V_1$ provides the minimum runway distance required, though company policy may require using a reduced $V_1$.

  • $\begingroup$ So the minimun value for V1 is Vmcg? Thanks $\endgroup$
    – camilogv
    Apr 24, 2014 at 5:00
  • 1
    $\begingroup$ @camilogv: yes, because if they continue the takeoff past V1 they risk having a runway excursion as they're below Vmcg. $\endgroup$ Apr 24, 2014 at 5:06
  • $\begingroup$ Last question When V1 is balanced, it is said to be BFL? $\endgroup$
    – camilogv
    Apr 24, 2014 at 6:20
  • $\begingroup$ @camilogv: sorry for not getting back to you earlier. The V1 is always balanced if you are using a balanced field length. $\endgroup$ Apr 24, 2014 at 11:42
  • $\begingroup$ @Qantas94Heavy: In that case, in an aircraft with some engines closer to the aircraft centerline than others, like a DC-10 (one engine under each wing, one mounted on the vertical stabiliser) or a 747 (one inboard and one outboard engine under each wing) could min-V1 for a failure of a centerline or closer-to-centerline engine be, at least theoretically, lower than min-V1 for a failure of a further-outboard engine? $\endgroup$
    – Vikki
    May 4, 2019 at 21:16

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