I see it a lot in movies, curious how say...an Robinson R44 would respond to a 200-pound man jumping and grabbing onto the skids while it's ~10ft off the ground and in the process of taking off??

Edit: more specifically, how dramatic would the effect be?
with Hollywood physics, the chopper would stumble as if the hand of god lightly slapped its side?

  • $\begingroup$ Well, if it was in a hover, then lift=weight and any addition of weight would cause a descent. In the case of taking off, lift>weight, so as long as that ratio remains true, the helicopter will continue to take-off (or hover). It works in the opposite way when somebody jumps out of a helicopter, which happens in the real world quite often. $\endgroup$
    – Ron Beyer
    Nov 25, 2016 at 3:32
  • $\begingroup$ Oh sorry, I should have clarified more. I meant to ask just how dramatic the effect would be. In hollywood physics, the heli's would dramatically stumbles as if the hand of god lightly slapped one side of it. $\endgroup$
    – House3272
    Nov 25, 2016 at 6:26
  • 2
    $\begingroup$ @House3272 You should include that in the question so people can see without reading the comments. $\endgroup$
    – Xylius
    Nov 25, 2016 at 10:27
  • $\begingroup$ There's some pretty dramatic footage of a news helicopter that helped rescue people from the Potomac after the crash of Air Florida flight 90. They picked people up with ropes and dragged them onto the skids. The pilot was expecting it, but it didn't seem to budge at all. And it's not a helicopter designed for rescue $\endgroup$
    – TomMcW
    Nov 25, 2016 at 20:02

2 Answers 2


The total mass of a light helicopter is about 1000 kg (depending on type and actual payload, of course). The 200 pound guy adds about 10% to the total mass (and weight). The skids are relatively close to the centre of gravity (COG).

Immediate impact

  1. Kinetic energy

In terms of mechanics, the jumping man will have a mostly inelastic collision with the helicopter, and he will transfer both, kinetic energy and impulse accordingly. One may simply choose the helicopter as reference coordinate system to calculate the resulting change of (vertical) speed for the helicopter: $$\Delta v_\rm{H} = \frac{m_\rm{G} \Delta v_\rm{G}}{m_\rm{H}+m_\rm{G}} = \frac{1}{11} \Delta v_\rm{G} \ \ \ \ \ \ (m_\rm{H}=1000 \rm{\,kg},\ m_\rm{G}=100 \rm{\,kg})$$ In the movie, the jumping man will just make it. The relative speed will be close to zero, and divided by 11 will be even closer to zero. If he had a significant speed relative to the skid, he would not be able to hold himself.

The transfer of kinetic energy is negligible for the stability of the helicopter.

  1. Added weight

The total weight of the aircraft increases by 10%. This sudden change creates a jolt which is clearly perceptible inside the helicopter, also suddenly changing the vertical acceleration. The (vertical) speed is the mathematical integral of the (vertical) acceleration. The speed will steadily change. The same holds true for the (vertical) position of the aircraft, which is the mathematical integral of the (vertical) speed.

In other words, the inertia of the bigger mass will prevent sudden change of vertical speed. For the pilot, this 10% disturbance will be very obvious, and he needs to counteract.

(Note that there was a related question recently: What would happen if a plane lost half of its weight instantly?)

  1. Imbalance

The imbalance caused by the guy hanging somewhere else than under the COG creates a torque. This torque creates an angular acceleration. This angular acceleration integrates to angular speed, which integrates to roll and pitch respectively. While the angular acceleration starts suddenly, the angular speed can only change steadily.

The helicopter will not loose its attitude in a fraction of a second, like it happens in Hollywood movies.


The pilot will use the Collective control to counter the added weight, and the Cyclic control to counter roll and pitch. Provided that the pilot reacts timely enough, it should be possible to stabilize the helicopter. 100 kg additional mass should not prevent the helicopter from climbing, unless it was already overloaded and without safety margins before the 200 pound guy arrived.

More critical is the torque created by the offset load and the resulting COG shift. I am not aware of the roll and pitch performance of small helicopters, but I believe that they are able to fly with somebody standing or hanging on a skid. Asymmetric load of this magnitude will likely impact stability and manoeuvrability, but will not necessarily cause a crash.

If the fearless 200 pound guy instead would jump and hold to the tail skid, the additional weight would be much further from the COG, and the outcome probably be disastrous.


This does not mean that it is safe to continue the flight with the 200 pound guy hanging anywhere on one skid. Pilots always need to verify if the amount and distribution of the load is within the performance limits of their aircraft. See "Helicopter - Weight & Balance, Performance" for some explanation and example (note Figures 2-5, 2-6, and 2-7). Depending on the helicopter performance and the load distribution before the guy jumps, the new distribution may be within the specified safe longitudinal and lateral COG envelopes, or not.

Unbalanced load on helicopter skid

  • 4
    $\begingroup$ I assume all super-villain helicopters come with a manufacturer COG table for mass with Hollywood leads hanging from a skid, for easy reference? :-) $\endgroup$ Nov 25, 2016 at 21:32
  • $\begingroup$ @DanSheppard Indexed by hair color, because I'm pretty sure I haven't seen a superhero or a supervillain wearing a hard hat while pulling that stunt. $\endgroup$
    – user
    Nov 29, 2016 at 14:29

$200\,\mathrm{lb}$ is about $90\,\mathrm{kg}$. As long as the helicopter was generating a lift force at least $90\times 9.81 \approx 880\,\mathrm{N}$ in excess of its weight, then that lift force is enough to continue to accelerate plus helicopter upwards. Wikipedia says that the loaded weight of an R44 is about $1\,134\,\mathrm{kg}$ ($2\,500\,\mathrm{lb}$) so a force of $880\,\mathrm{N}$ would be accelerating the helicopter alone at about $0.78\,\mathrm{ms}^{-2}$ ($2.5\,\mathrm{ft\,s}^{-2}$). So, as long as the helicopter was already accelerating at least that fast, it will continue to accelerate upwards.

I'll let other people deal with the effects of the unbalanced load.


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