During the Second World War the RAF had a 10,000 kg bomb, the "Grand Slam", which was dropped from a Lancaster bomber.

I assume that when this was dropped the plane would lurch upwards.

Is there any way of knowing just how big and violent a movement this would have been?

According to Wikipedia, a 'loaded Lancaster' weighed 25,000 kg. I'm not sure if this includes a standard bomb load (approx 6,350 kg) or not, but either way it seems that when the "Grand Slam" was dropped the aircraft would, in an instant, reduce to about half or two-thirds of its previous weight.

Is there any way to estimate how big an upward jump this would cause?

  • 5
    $\begingroup$ Bombers open spoilers as they drop bombs. It's done exactly to prevent that upward jump, mostly for precision so the next bomb can be dropped without need to re-align the plane. I don't have a link now, but there was a cool video with visible row of small spoilers on the wing that are opening one by one, as the plane loses weight in carpet bombing. $\endgroup$
    – Agent_L
    Nov 9, 2016 at 11:29
  • $\begingroup$ If not balanced properly, you could wind up with something a lot like this (cargo broke loose and slid to the back): youtube.com/watch?v=4Z7GkV5DqOI $\endgroup$
    – SnakeDoc
    Jan 13, 2017 at 19:55
  • $\begingroup$ The center of gravity changes too. So there is a need to re-trim the airplane. $\endgroup$
    – jjack
    Nov 3, 2018 at 17:47

4 Answers 4


In steady flight, the lift $L$ generated by the wings equals the loaded weight so that there is no vertical acceleration. Thus, the lift just before the bomb is dropped is $$ L = m_1g = 245\,250 \;\textrm{N} $$ where $m_1=25\,000$ kg is the loaded mass and $g=9.81$ m/s$^2$ is the acceleration due to gravity.

In the time interval just after the $10\,000$ kg bomb is dropped, the lift remains the same until the lift is reduced to again balance the new mass of $m_2=15\,000$ kg. During that transient period, the Lancaster will accelerate upwards at $$ a = \frac{L-m_2g}{m_2} = \left(\frac{m_1-m_2}{m_2}\right)g $$ which works out as an upward acceleration of two-thirds of a "gee": $$ \frac{a}{g} = \frac{m_1-m_2}{m_2} = \frac{2}{3} $$

As @sanchises pointed out, the steady increase in the upward velocity results in a decrease in the angle of attack, so that the aircraft will again settle in a steady climb.

  • 2
    $\begingroup$ There is no upward jump! The discontinuities are with the vertical forces and the vertical acceleration. The vertical speed and the altitude remain continuous when the drop occurs. $\endgroup$
    – bogl
    Nov 9, 2016 at 9:38
  • $\begingroup$ Yes, @bogl. There is no sudden "jump", but a smooth upward motion. $\endgroup$
    – Christo
    Nov 9, 2016 at 9:42
  • 8
    $\begingroup$ @bogl There is no 'jump' but there is definitely a 'jolt' - or more formally, an extremely high jerk for a very short time as the vertical acceleration goes from zero to $\frac{2}{3}g$. $\endgroup$
    – Sanchises
    Nov 9, 2016 at 9:50
  • $\begingroup$ @sanchises Completely agree. That should be pretty uncomfortable. $\endgroup$
    – bogl
    Nov 9, 2016 at 9:53
  • 2
    $\begingroup$ Either way, welcome to the community! $\endgroup$
    – Sanchises
    Nov 9, 2016 at 9:55

The A7E Corsair had an empty weight of around 19,000 pounds. Its fuel load was 10,500 pounds, and with six armed 2,000 pound MK84 general purpose bombs hung under the wing it had a total weight of around 42,000 pounds. This was maximum takeoff weight, and coming off the cat shot the aircraft response felt lazy and sloppy as you did your clearing turn.

By the time we got to the target, which was a rock in the Aegean called Avgo-Nisi, we were at around 7,000 pounds fuel remaining. That put the aircraft at around 26,000 pounds and a 12,000 pound bomb load still under the wings. We cleared the target with a low pass before climbing up to the pattern altitude burning probably a 1,000 pounds. It was a popular fishing spot and so we wanted to make sure the area was safe before our delivery.

We were in combat formation when we rolled in to our 45 degree dive, arming the pylons. I was on the wing of my lead in a loose formation around 400 to 800 feet away. The formation was a tactical position, and if either one of us got hit during combat operations the other was far enough away to survive. A 45 degree dive looks very steep from the cockpit and we were fast at the release point. Minimum altitude for this kind of run was 2,500 feet AGL to keep you out of the blast radius. We were at 1 G when we released 24,000 pounds of bombs on the rock, 12,000 pounds apiece.

I watched the bombs release from the lead's aircraft as I saw mine drop away in the mirrors. I just shed around half of my total weight. I immediately felt the aircraft "jump up." This is the way I characterize it, and someone else might call it a jerky motion. It was definitely stronger than the nose tuck you experience in the A7E breaking the sound barrier. The jump wasn't disconcerting, but definitely noticeable. In fact, it was a welcome change to what felt like a lumbering, gas eating configuration.

We pulled out of the dive hard and had our noses high in a turn when I felt the blast wave go through my internal organs. It was a weird feeling, like someone rearranging the position of my stomach. I must have been at 4,000 to 6,000 feet above the detonations and could hear the explosions over all the noise in the cockpit. It was massive.

  • 3
    $\begingroup$ I like this answer even though some readers may consider it to be too much information. - It is a very clear demonstration that yes, an aircraft does indeed bounce upward upon shedding a large fraction of its weight, but that is far from the most exciting thing that happens on a typical day when using an aircraft in this manner. $\endgroup$ Jan 13, 2017 at 4:53
  • 1
    $\begingroup$ I appreciate your comment. Thank you. Using day-to-day experiences of carrier aviation brings home the nature of what we do. I learned best from stories told in briefs, at safety standowns, during NATOPS, or even in the dirty shirt mess. Also, I think there is a common misconception among some, that the bombers today carry less payload than those of old. Lastly, I think valuable information can also be engaging as well, but that does take space. My intention certainly isn't to write for everyone. I, personally, like scanning an answer and if so inclined will read it. Again, thank you. $\endgroup$
    – Aaron
    Jan 13, 2017 at 13:46
  • $\begingroup$ @Aaron, great to hear from someone who did it and lived to tell the tale. Thanks. $\endgroup$ Nov 3, 2018 at 10:35
  • $\begingroup$ While 12.000 lb is a pretty sizeable bomb load, it's less than the 10.000 kg bomb load in the question - The Mk84 is 925 kg, so x6 that's merely 5550 kg. Then again, the A7 was carrier-based. A B1-B will carry 24, so a bit over 22 tons. Relatively, the Lancaster dropped 40% of its take-off weight, versus 30% for this A7, so in that sense there's been some regression. $\endgroup$
    – MSalters
    Nov 5, 2018 at 14:14

(A new question was recently asked, then closed as a duplicate, linked to this older question.)

Answer: You would have a pitch "phugoid"

Until airspeed decreases, lift is greater than weight

Flight path curves (accelerates) skyward

There is a very temporary reduction in angle-of-attack, until the aircraft's pitch stability dynamics overcome pitch rotational inertia and pitch the nose up into line with the new direction of the flight path, restoring the original angle-of-attack. This effect may be negligible.

The changing direction of the flight path causes the weight vector to have an aftward component relative to the aircraft's flight path.

Thrust, drag, and the fore-and-aft component of weight are no longer in balance, and airspeed starts to decrease.

Eventually the aircraft will come back into equilibrium in a stabilized climb at a constant airspeed, lower than the original airspeed. But we might see the aircraft's pitch attitude "overshoot" and "undershoot" the final steady-state pitch attitude several times, as the flight varies between steeper and shallower climb angles, before this happens. Likewise we might see the aircraft's airspeed "undershoot" and "overshoot" the final steady-state airspeed several times before settling into equilibrium. Basically what is going on here is that a change in the climb angle drives a change in airspeed which drives a change in the magnitude of the lift vector which drives a change in the climb angle, and round and round it goes till everything settles into balance.

All this is assuming the pilot is making no corrections-- such as to hold a constant pitch attitude-- and the aircraft's pitch stability dynamics are acting to try to hold some given trimmed angle-of-attack, and the aircraft's dynamic pitch stability characteristics are such that a pitch "phugoid" tends to eventually damp out.


Another way to look at this is what would happen if IAS increased 100 mph, which could happen entering a very strong microburst. This is where building models and gaining experience with tail design and CG/trim condition can provide some key insights.

It is very desirable NOT to have "phugoid" behavior when lift/weight vector comes out of balance. There is a relationship in wing and tail design that is very fundamentally important as the horizontal stabilizer has (with the Lancaster) a long torque arm and end plates, giving it directional stability. We also must remember pitch change will cause wing center of lift to move forward as well. Trim condition/CG placement is also very important.

So what will happen? With the lift/weight vector suddenly out of balance, the first move is straight up. Vertical acceleration. Downward pressure on the H stab will begin to pitch the nose up, increasing AOA. Any momentary DECREASE in AOA is trivial, the tail downforce will be pushing the nose up. The increase in AoA will move center of lift FORWARD in the wing, further pitching nose up. With no pilot control inputs, the airspeed will decrease, the plane reaches the top if its ascent, and will then pitch down and recover as it "becomes one" with the air mass and is no longer being accelerated by it.

Trim condition is also very important here. One of the holy grails in free flight is to have the plane rise and fall with a gust without excessive pitching. This is accomplished by avoiding CG placement too far forward, which requires much more elevator down trim. This is one case where having CG further back and less elevator trim is beneficial. As long as the aircraft is directionally stable, there is no need for having CG forward by rote!

So, suddenly reducing weight will lead to a rise/ pitch up/ airspeed reduction/ pitch down/ recovery cycle in a properly designed and trimmed aircraft, along with considerable relief for those carrying out the task.

  • $\begingroup$ After pitch stabilizes, aircraft will climb unless throttle is reduced. $\endgroup$ Nov 3, 2018 at 12:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.