My aviation teacher told us to memorize some speed values among other things before our flight training:

$V_R$ = 40 kt

$V_S$ = 39 kt

According to wikipedia:

Vr: The speed at which the pilot begins to apply control inputs to cause the aircraft nose to pitch up, after which it will leave the ground.

But I'm somehow confused with the values, since the takeoff speed (according to what I was taught) should be 10-25% greater than Vs, in this case around 43 and 49 kt.

If Vr is the definition mentioned above, which should be takeoff speed? does it have a definition within V speeds? Isn't takeoff speed the speed at which the aircraft leaves the ground?

Thank you


Based on the speeds that you are talking about, it isn't a turbojet definition that you are looking for. The definition for $V_R$ that you gave is for turbojets.

In small GA airplanes, I teach my students to rotate around $1.3V_{S0}$ which really means to slowly bring the nose up to the takeoff pitch attitude. Under no circumstances do I want them to yank the airplane off the ground. If done correctly, the airplane will naturally lift off when it is ready to.


You need to add some time for the rotation to gather enough angle of attack change so the aircraft can lift off. While the pilot pulls to lower the tail and rotates around the main gear, the aircraft continues to accelerate. Also, the 1.3 factor is only required to be reached once the aircraft climbs through 35 ft (or 50 ft, depending on the certification). What happens is that the pilot starts pulling right around stall speed, the aircraft starts to produce more downforce at the tail until the nose gear lifts off, then rotates until the angle of attack has grown to maybe 10° so the wings create enough lift for take-off. At this time, the aircraft will have accelerated to around 1.15 times stall speed and starts to climb, which requires still more pitch change. Now drag will increase by the amount of induced drag produced by the added lift so the rate of acceleration slows down. Still, until the 35 ft are reached the aircraft will accelerate further to 1.3 times stall speed at this point.

That is how the take-off distance is flown for certification to achieve the minimum distance. Depending on the rotation rate (typically 3° to 5° per second) and excess engine thrust, the optimum take-off might require to start the rotation process at less than stall speed. For added safety, it is advisable to stay a little longer on the ground if the available runway permits it.

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    $\begingroup$ Given a target nose up pitch of 13 degrees, the first 747 carrier I flew for taught raising the nose to 10 degrees until liftoff and then going to 13. Their thinking was that that protected against a Vr bug erroneously set too low or a pilot rotating too quickly. $\endgroup$ – Terry Oct 21 '16 at 20:02
  • $\begingroup$ @Terry: This is the best way to avoid a tail strike. Takeoff run will be longer, though. If the runway allows for it, your old carrier's teaching is sensible. $\endgroup$ – Peter Kämpf Oct 22 '16 at 7:28

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