In his upcoming movie, Captain America stops a helicopter from taking of, by grabbing the helicopters skid and a piece of railing.

Is this feasible, and how much strength would be required to ground a helicopter like this?

enter image description here

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    $\begingroup$ Could you describe that scene, so those of us not having seen it can understand what you mean? "Hold a helicopter" could refer to quite a lot of different scenarios. $\endgroup$ – celtschk Dec 29 '15 at 10:32
  • $\begingroup$ Are you referring to the Captain America Civil War Trailer, at about 1:55. $\endgroup$ – Gary Walker Dec 29 '15 at 12:13
  • $\begingroup$ yes Sir...that one $\endgroup$ – aswal94 Dec 29 '15 at 12:15
  • $\begingroup$ Well, the answer is not power at all. The appropriate unit is force in newtons or pounds. Not a helicopter person so don't know the model, but if you look up the net lift aka cargo capacity for a given model, you will have your answer (or at least an upper bound to the answer). If you watch the video, you see much of the helicopter in overhead view for a few seconds. $\endgroup$ – Gary Walker Dec 29 '15 at 12:26
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    $\begingroup$ Personally I'm pretty impressed with that railing. $\endgroup$ – AndreiROM Dec 29 '15 at 20:01

In this exact way? Less than it seems, to be honest.

I'll explain with minimal physics, and overly generic simplifications, just for some basic reasoning. This is an answer for a layman.

A helicopter works by generating lift. This lift kinda cancels out the gravity effects, and thus keeps the vehicle aloft. The more lift a helicopter generates, further it raises from the ground. That's easy.

So, to hold a helicopter directly to the ground, you need to pull it down with equivalent forces.


Captain is holding the Helicopter by the side of the landing gear. This means that, should the pilot speed up the Helicopter, it would start tilting to the side, probably making the vehicle crash. This is called a Dynamic Rollover, and can be caused by pretty much a single steel cable still attached to the vehicle. Heck, it can be caused by anything at all, if the pilot is not careful or skilled enough. Helicopters are finicky machines, and prone to disasters if you don't take care.

So, the pilot here is not really pushing Cap - if he does, and Cap happens to be strong as a steel cable, the helicopter would probably start rolling over, crash, and kill everyone on board.

That said, this scene can't really be used to estimate how strong Cap is, since we don't know how much force he is putting here. A steel tie-down cable can withstand up to 4kN (around my region) and is enough to cause a rollover. An Olympic Grade gymnast withstand around 3kN during competitions, so it's not a stretch to assume that Cap is stronger than that.


I think this is a Bell 206, you can't see all of the copter, and there may be another model that looks about the same.

If it is a Bell 206, the dry weight is 2331 pounds and the max. takeoff weight is 3200 pounds. If you allow 250 pounds for pilot, fuel and misc. you have a max. net thrust of about 620 pounds. The amount Captain America is using is almost certainly less than that, as he is not vertically "stretched" the side thrust is quite limited as the copter is not obviously tilted at steep angle. So in this shot it is likely considerably less than 200 pounds. I.e., you could actually make this shot without having the strength of Captain America, just a normal stunt double, or even Chris Evans himself - the shot does not look particularly dangerous if wearing a wire to catch him in case of a fall.

For a movie shot, the studio would very likely select a helicopter with limited power as a simply safety measure (as well as being a cheaper model that the more powerful ones). What is important to the studio is that a shot looks impressive, not that it is actually impressive.

  • $\begingroup$ very good observation, i was looking for the model number as well.. $\endgroup$ – RicoRicochet Dec 29 '15 at 13:22
  • $\begingroup$ There must be more to it than that. At the very least I'd think momentum would be a serious issue. For instance, say the Chopper briefly jogs back towards Captain America, slackening his grip. It can then reverse direction, and even if it only attains a very modest speed before he's taut again its momentum will exert thousands of pounds of force (even laterally). Easily enough to dislocate a shoulder, if not sever an arm. In other words, he's fighting not just the static thrust of the chopper, but its accumulated momentum as well. $\endgroup$ – aroth Dec 29 '15 at 13:53
  • $\begingroup$ @aroth - using the maneuver you describe, the copter would be not be able to generate much horizontal velocity in a limited distance.Also the copter motion would not be ignored by CapAm, i.e., he would resist the horizontal acceleration of the craft because he does not want it to build up momentum since he would like to keep his arms and shoulders intact and in working condition. Analyzing movies often leads to disappointing answers. $\endgroup$ – Gary Walker Dec 29 '15 at 14:01
  • $\begingroup$ I always find it amazing how much rag doll physics (assuming Captain America won't responds to the helecopter's changing motion) is intuitive, and just how insanely strong or fast one can appear simply by breaking that assumption! $\endgroup$ – Cort Ammon Dec 29 '15 at 14:41
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    $\begingroup$ @aroth - you are simply incorrect. The combination of velocity and mass is momentum (kg m/s), not force (kg m/s^2). Stopping a 3000 pound (the Bell 206 is a little over 1500 lbs) object moving at 0.75 mph using a conservative 2 feet stopping distance requires a force of only 62 pounds. I answered the question as asked re: the movie, not some otherquestion. $\endgroup$ – Gary Walker Dec 29 '15 at 15:31

You need to apply a force equal to the excess force applied by the rotors above what's needed to keep the helicopter hovering plus the weight of the person holding it. I'll leave someone else to put some numbers to that, but it doesn't strictly need to be very much, just enough to destabilise the chopper should it try to move away.

  • $\begingroup$ number is what i seek for.....and what that amount of force can do in a different scenario. $\endgroup$ – aswal94 Dec 29 '15 at 12:37

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