I recently re-watched Captain America: The Winter Soldier, in which Steve Rogers spectacularly throws his shield at his enemies to incapacitate them. Action scenes like those left me wondering: would the thrown shield’s impact actually kill those soldiers in real life? In the MCU, Steve throws around his shield like it’s no big deal, hitting villains left, right and center, but it’s always left unclear whether the impact leaves the villains simply unconscious or actually dead. To find out what would happen if one were caught in the crossfire of Cap’s shield, I took to physics.

To calculate the famous vibranium shield’s impact, I needed to first determine the shield’s physical properties, such as its weight. According to the Marvel Database, Captain America’s shield weighs 12 pounds (5.44 kg) and is 2.5 feet in diameter (0.762 meters). Now, I know what you may be thinking, “Captain America’s shield is made of vibranium, so doesn’t that change the values?” But, the problem with taking this fact into account is that we don’t know much about vibranium as an element, so it’s impossible to replicate those results in real life.

The simplest way to account for vibranium is to take into account *exactly* what we see on film. Using the incredibly powerful Open Source Physics software, I took various trials of Steve throwing his shield and tracked it to get various velocities (the gif below shows one of those). I then took all the values and averaged them out (which gave surprisingly similar results) to get a velocity of 18.77 m/s (~42 mph).

So, abiding by the rules that the MCU gives us, we have a 5.44 kg disk of metal flying at someone at 42 mph (18.77 m/s). Sounds pretty dangerous, huh? So, what’s happening to the poor guy that’s getting a full serving of justice? Well, let’s first look into energy. The formula for kinetic energy is \(\frac{1}{2}mv^2\) and using the values we got before we can find the kinetic energy this shield has. This value turns out to be 961.9 J. This is the translational kinetic energy. But when Steve is throwing his shield, it’s not only moving linearly but also spinning. That means we need to calculate the rotational kinetic energy of the system using \(\frac{1}{2}Iw^2\). To calculate this, we would need to approximately figure out the moment of inertia (I) of the shield and the angular velocity (w) of the shield. First, let’s determine moment of inertia.

So, Cap’s shield kind of looks like the top part of a sphere, if you were to chop off a slice of it. We know the radius of this shield and the depth of the shield, so if we can find the radius of the sphere the shield came from we can determine the moment of inertia. Take a look at the picture below for clarification.

To find the radius of the bigger sphere, we would use the Sagitta Theorem. Using this theorem, we take the sagitta (or the length from the top of the shield to the ground if it is lying flat) and the half chord length (the radius of the shield) and use it to find the radius. We do this with the formula \(s=r\pm\sqrt{r^2+l^2}\). This gives us the sphere radius of 0.8609 meters.

Now that we have the radius of the sphere, calculating the moment of inertia is a simple integral. Check out the picture for my work, but instead of boring you with the calculus, I’m going to tell you the value which is 0.645576 \(kgm^2\). Quick note, in this case, we’re assuming the shield is a thin shell as it is probably closer to the rules of vibranium than a shield with thickness.

Now, to find angular velocity. To do this, we need to go back to the videos. All I had to do was count the amount of times Cap’s shield rotated in a certain time frame. This was actually difficult for two main reasons: 1) Cap’s shield is going 18.8 m/s and 2) camera angles, but I managed to get a good estimate by counting the rotations in slow motion and then speeding the video back to normal speed and measuring the actual time it took to spin. Doing various trials got me an average angular velocity of 65.3 rad/sec.

Now, it’s another plug and chug situation. We take the formula from before (\(\frac{1}{2}Iw^2\)) and just plug in the values. This give us a rotational kinetic energy of 1376.40 J. Kinetic energy is additive, so we add the translational and rotational kinetic energies to get a total of 2338.30 J. That’s a lot of kinetic energy dispersed over a small area. But, to clarify this value, let’s take a look at the force that is applied to our unlucky villain.

To do this, we need to take a look at momentum. Because the time that the force is applied is constant, we can use the formula F = (\(\frac{mv}{t}\)). This formula comes from Newton’s first law (F = ma) as a is equal to v/t. To find the time Captain America’s shield is in contact with the person, I again took various shots to get an average time of 0.015 seconds. So, the average force that our man would be experiencing is a tremendous 6,818.13 N.

Now, to find the force on the guy at a specific point, we would use Ft = Impulse = Δp = FΔt. This might seem like we’re repeating the same steps as before but this time we are looking at the velocities directly before and after the collision. The average I got for an velocity right before impact was about 26.4 m/s and the shield’s velocity right after the collision was 5 .5 m/s using the Open Source Physics software once again. Here’s what’s happening:

So, Δp = p\(_{f}\) – p\(_{i}\) = mv\(_{f}\) – mv\(_{i}\) and if the left positive direction is positive, we can keep both velocities to be positive (both velocities are heading to the left). Now, mv\(_{f}\) – mv\(_{i}\)= (5.44 kg)(5.5 m/s) – (5.44 kg)(26.4 m/s) = -113.696 kg*m/s. This is the impulse of the man on the shield so the impulse of the shield on the man is 113.696 kg*m/s.

Impulse equals the force time the change of time so now force is easy to find. The average time of contact for the shield hitting the person is 0.021 seconds. So now we put that into the equation to get 113.969 kg*m/s = F\(_{N}\)(0.015 sec) to get a force of 7,597.93 N. That number makes total sense as it’s a 12 lb disk going 42 mph (5.44 kg disk going 18.77 m/s). To put that number in perspective, normal human being has a 25% chance of breaking a bone at 4,000 N of force.^{[note 1]} The shield, which hits someone with 7,597.93 N, has a 47.48% chance of breaking bones which is basically taking a 50/50 chance at not getting hurt. But, one thing that we fail to take into account by these numbers is the shield is effectively hitting you at a single point, which in turn means a lot of force will hit you within a small area. This means the shield is definitely breaking bones and if it hits anywhere on your body, you’re not going to be too happy about it and, well, you’ll probably be dead.

In conclusion, if you were hit by Captain America’s shield, you definitely would not make it out alive, so it’s better to not to take on the Star-Spangled man with a plan.