I have often read fictions about aircraft, where a pilot does an action that I cannot explain:

During a takeoff with an heavy aircraft (four motors for example), the pilot rolls to the end of the airstrip, and starts to climb. Then he says: I am too heavy to climb enough in order to avoid the mountain/trees/whatever obstacle in front of us. So the pilot starts to dive to the ground in order to gain more speed, and then he climbs again and is able to have enough altitude.

How do you explain this action?

Considering the drag, why does the dive gain enough energy to go to an altitude above the one reach by the take off?

I might not be clear, please don't hesitate to ask details. Here is a schema:

    Not enough         /Climb enough
        /  \          /
strip  /    \        / 
  • $\begingroup$ I don´t think you are using the word glide for what it is usually understood to mean: fly without power exchanging gravitational potential energy to maintain airspeed. Maybe you mean flying in ground effect to accelerate to a safer speed before maneuvering? $\endgroup$ May 19, 2019 at 15:00
  • $\begingroup$ Hi I am not a native English speaker. I mean by gliding: flying with a negative AoA, the nose of the airplane pointing to the ground, with motors powered on. Thus the airplane is exchanging gravitational potential energy to increase his airspeed $\endgroup$ May 19, 2019 at 15:14
  • $\begingroup$ @totalMongot What you describe is diving, not gliding. Gliding is a power off maneuver. $\endgroup$ May 19, 2019 at 17:33
  • $\begingroup$ @Skip Miller Thank you for the clarification, and sorry for the mistake $\endgroup$ May 19, 2019 at 18:41
  • $\begingroup$ Flying with a negative angle-of-attack near the ground will definitely lead to disaster. It never fails. Unless the aircraft is inverted. $\endgroup$ May 20, 2019 at 19:06

4 Answers 4


When an aircraft can barely climb after take-off, it may be flying on what is called 'the backside' of the power curve.

Flying at very low speed:

  1. costs a lot of power because of high induced drag
  2. means that the engines are inefficient because they are optimised for higher speeds.

As a result, the excess power available, that is the difference between the required power and the maximum available power, is very low when the aircraft flies close to the minimum flight speed. And the excess power is what is needed to climb the aircraft.

power curve

When the aircraft is flying below the speed at which the best angle of climb is achieved (Vx), it will be able to improve its climb angle by accelerating. To do so, the aircraft needs to use available power to accelerate, instead of using the power to climb.

Stopping the climb or even descending a bit will convert the available excess power and optionally height (potential energy) into speed. While this initially does not give extra clearance above the obstacle, the available excess power increases drastically. This in turn gives the ability to climb at a much steeper angle, allowing the aircraft to clear the obstacle.

dive to climb

  • 3
    $\begingroup$ Excellent answer! $\endgroup$
    – Terry
    May 21, 2019 at 20:33

Bear in mind that fiction often (usually?) takes liberties with the truth of a technical situation. That said, let's say your 4-engine aircraft is a Boeing 747-200 with JT9D-7Q engines (because that's what I still have a QRH for).

And let's say the paperwork says you weigh 620,000 lbs—it's an empty freighter you're ferrying from Santiago, Chile to Buenos Aires. You look in the performance tables and come up with the following:

  • V1 is 130 knots
  • VR is 140 knots
  • V2 is 156 knots
  • target rotation attitude is 16° nose up

Unbeknownst to you, the ground handling company decided to pick up a little money on the side and put 180,000 lbs of cargo on board for an equally corrupt operation in B.A.

So, you really weigh 800,000 lbs, the numbers for which are:

  • V1 is 160 knots
  • VR is 173 knots
  • V2 is 183 knots
  • target rotation attitude is 13° nose up

You stagger off the ground—God help you if you get a bad gust—with your reference speed 27 knots slower than it should be and your nose 3° higher than it should be. My guess is that you would be on the backside of the power curve, but whatever, your performance is going to be less than it should be.

Some miles ahead you have one of the highest MEAs in the world, 24,000 feet as I remember. Mt. Aconcagua at 22,841 feet (highest peak in the Andes) will be just off to your left. You often have difficulties getting that high that soon. You wouldn't make it today.

The solution: lower the nose and speed up.

And by the way, you're probably going to have to land short of B.A. for fuel. Personally, I'd go to Mendoza, Argentina.

  • $\begingroup$ I don't understand. You lower the node and trade potential energy for kinetic. Then you need to raise your nose again and ... Convert all your gains back to potential. Without counting drag, this sounds like a 0 sum operation $\endgroup$
    – Antzi
    May 20, 2019 at 1:27
  • 2
    $\begingroup$ @Antzi No, the goal is not to trade potential energy for kinetic. The goal is to increase your speed to that which will enable the wing to perform better. Now, if you're on the backside of the power curve and are already at max power, the only way you'll be able to increase your speed is to sacrifice some altitude. Then, when your speed is up to the right speed you'll be on the front side of the power curve and have performance capability you didn't have at the same power setting on the back size. I suggest doing a search on "back side of the power curve." $\endgroup$
    – Terry
    May 20, 2019 at 2:51
  • 3
    $\begingroup$ I'm hoping the sentence Unbeknownst to you, the ground handling company decided to pick up a little money on the side and put 180,000 lbs of cargo on board for an equally corrupt operation in B.A. isn't from bitter experience :| $\endgroup$
    – Jamiec
    May 20, 2019 at 7:22
  • $\begingroup$ @Jamiec Well, it's from experience but never that turned bitter and never anywhere near 180,000 lbs. Back in the 1990s in freight runs out of B.A. and Santiago, some overloading was to be expected for whatever reason. The two worst overloads I experienced, though, were out of Miami and Chennai (then called Madras). The company did some research on the Miami to Trieste incident because it caused an unplanned fuel stop in Paris. We were 30,000 lbs over due to cargo that had been loaded in the lower holds but not shown on the paperwork. Madras was also about 30,000 lbs but wasn't intentional. $\endgroup$
    – Terry
    May 20, 2019 at 18:51
  • 1
    $\begingroup$ @Antzi don't forget you're pouring 100% power in the whole time. Getting the speed right to climb is what you are after. $\endgroup$ May 20, 2019 at 19:00

The Pilot's Operating Handbook (POH), which is specific to an individual airplane, lists two speeds of interest: Vx and Vy. Vx is the obstacle clearance speed, and Vy is the maximum rate of climb. Vx is less than Vy.

Flying at Vy, the plane will climb more quickly (e.g. how long will it take to reach say 1000 feet above the airport) but because you are moving faster over the ground at Vy, you still might hit that mountain because you get to it sooner.

Flying at Vx sacrifices airspeed for altitude. At Vx you are flying more slowly and are climbing as fast as the plane can. It maximizes your vertical speed, not your horizontal speed.

As altitude increases, Vx and Vy converge and when they do, you are at the absolute altitude limit for your plane.

In your example, it sounds like the plane took off at Vy, noticed the lack of sufficient climb performance, leveled off or descended to gain speed, and continued the climb at Vx.

I have omitted a discussion of ground effect flying as it would be highly unusual for your four engine large airplane to remain in ground effect for long.

  • 1
    $\begingroup$ Perhaps I'm misinterpreting what you're saying. If so I'll remove this comment, but consider rewording your third paragraph. " At Vx you are flying more slowly and are climbing as fast as the plane can." But you said in para 1 that Vy is the maximum rate of climb (and I agree), so how can both statements be true. "It maximizes your vertical speed, not your horizontal speed." Not really, it maximizes the altitude gained for the amount of horizontal distance covered. P.S. How old is "older than dirt" (from your profile)? I'm about to turn 80. $\endgroup$
    – Terry
    May 19, 2019 at 19:55
  • 1
    $\begingroup$ The above answer sounds like it is suggesting that Vx is faster than Vy which is never true -- "it sounds like the plane took off at Vy, noticed the lack of sufficient climb performance, leveled off or descended to gain speed, and continued the climb at Vx." $\endgroup$ May 20, 2019 at 19:12

The only time it will work is if you are below Vx, and you do not need to dive to pick up speed under power, reducing pitch angle to level should be enough. Actually did this once with a model that was too rich (prop RPM slower) and had gained considerable weight from numerous repairs.

Essentially, you climb in stepwise fashion. If airspeed bleeds off below Vx, reduce pitch and keep climbing. But doing a rotation at proper speed and climbing at Vx is the best chance of clearing an obstacle.

What you have depicted is more like a zoom maneuver, which might help insure clearing one row of trees if you are slower than Vx, but a mountain 10 miles distant needs Vx right after liftoff.

  • $\begingroup$ Thank you. The examples I read off in fiction are in fact about clearing one row of trees a few hundreds meters ahead of an air strip with an heavy airplane $\endgroup$ May 20, 2019 at 19:00
  • $\begingroup$ @totalMongot a little fast is far better, then pitch to Vx speed. If you do not see trees and your airspeed holds, you got it. If trees or terrain are still directly ahead, time for plan B. $\endgroup$ May 20, 2019 at 20:12

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