1
$\begingroup$

In the Seneca (PA34-200) AFM, in the ENGINE FAILURE DURING TAKEOFF section, it is noted that if engine failures occurs during takeoff ground roll and 100 MPH has not been attained, the pilot must abort the take off and if attained the pilot must continue. Since Vmc is 80 MPH, why was 100 MPH chosen as the cutoff point for go/abort?

  1. ENGINE FAILURE DURING TAKEOFF
    The single engine minimum control speed for this airplane is 80 mph (CAS) under sea level standard conditions.

    a. If engine failure occurs during takeoff ground roll and 100 mph (CAS) has not been attained, CLOSE BOTH THROTTLES IMMEDIATELY AND STOP STRAIGHT AHEAD. If inadequate runway remains to stop, then:
    (1) Throttles - CLOSED.
    (2) Brakes - apply maximum braking.
    (3) Master switch - OFF.
    (4) Fuel selectors - OFF.
    (5) Continue straight ahead, turning to avoid obstacles as necessary.

    b. If engine failure occurs during take-off ground roll or after lift-off with gear still down and 100 mph (CAS) has been attained:
    (1) If adequate runway remains, CLOSE BOTH THROTTLES IMMEDIATELY, LAND IF AIRBORNE, AND STOP STRAIGHT AHEAD.
    (2) If the runway remaining is inadequate for stopping, the pilot must decide whether to abort the takeoff or to continue. The decision must be based on the pilot's judgement considering loading, density altitude, obstructions, the weather, and the pilot's competence. If the decision is made to continue, then:
    (a) Maintain heading and airspeed.
    (b) Retract landing gear when climb is established.
    (c) Feather inoperative engine (see feathering procedure).

Original image of text.

Improve this question
$\endgroup$
10
  • $\begingroup$ It's specific to this aircraft. The text has explained for you. If you are faster you might not be able to stop within the runway, then keep flying would be safer, relatively. Whenever there is enough runway to stop, stopping is always safer. $\endgroup$
    – user3528438
    Oct 30, 2017 at 22:24
  • $\begingroup$ Probably below 100 mph the aircraft is not controllable with one engine giving full thrust because the rudder does not provide enough control authority at those speeds. $\endgroup$
    – DeltaLima ♦
    Oct 30, 2017 at 22:37
  • 2
    $\begingroup$ @DeltaLima But VMC is 80 so it means we can control the plane above 80mph . we are assuming that we are in sea level . I found no answer for this on internet yet :( $\endgroup$
    – Amin NajibZadeh
    Oct 30, 2017 at 23:09
  • 2
    $\begingroup$ Please avoid posting text inside images, take your time and copy the text in the body of the question here. $\endgroup$
    – Federico
    Oct 31, 2017 at 6:41
  • 3
    $\begingroup$ @tj1000 - VMC is the calibrated airspeed at which, when the critical engine is suddenly made inoperative, it is possible to maintain control of the airplane with that engine still inoperative, and thereafter maintain straight flight at the same speed with an angle of bank of not more than 5 degrees. Are you maybe talking about some other VMC? $\endgroup$
    – Steve V.
    Oct 31, 2017 at 23:10

2 Answers 2

Reset to default
1
$\begingroup$

I finally found out an answer which makes more sense to me . By test pilot , vmca masured with a lot of citation and one of them Is 5 degrees angle of bank towards the live engine . So it means that we can't have that 5 degrees because we are still on the ground. for each 1 degree , our speed decreases 3 knots .So 5 degrees = 15 knots which is equalls to 20mph ! 80+20 = 100 mph ! This was the best answer I have got until now.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review $\endgroup$
    – mins
    Nov 1, 2017 at 18:44
  • 2
    $\begingroup$ @mins this adds an answer in addition to the "thanks." $\endgroup$
    – fooot
    Nov 1, 2017 at 18:55
  • 1
    $\begingroup$ @fooot: Sort of... it is a thank you and a proposal that "Im not sure if this is the exact answer". $\endgroup$
    – mins
    Nov 1, 2017 at 21:42
0
$\begingroup$

Unless one was privy to the certification testing and POH drafting of this aircraft, the exact answer is likely not available. However there are several factors which come into play.

First Ek = 0.5m(v2) so the relative kinetic energy at 80 will be less than 2/3 of the energy at 100 mph. The distance to stop at 80 will be substantially less than at 100. Terminating the takeoff substantially reduces flight risk on one engine, taking off a at a speed very close to VMC with onset drag from the failed engine, and gear extended.

Accelerating to 100 mph will take substantially longer with one engine, and likely it was determined that the stop distance in this situation is less than the remaining distance to takeoff.

Takeoff close to VMC, with the gear down and the onset drag of the failed engine, impair substantially the ability of the aircraft to achieve 100 mph, and risk is reduced by landing the aircraft. Aircraft handling will also be impaired, due to the low speed (and onset drag). If the distance required to accelerate to 100 mph under those conditions exceeds the distance to stop, then the aborting the takeoff reduces the flight safety risk.

14 CFR 25 aircraft require that the takeoff distance and the accelerate-stop distance be less than the available runway length. The Seneca was certificated under Part 23, so this requirement does not apply, yet the concept remains valid. Accordingly, a V1 speed is not defined for the Seneca.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .