One library. Multiple articles. Completely insignificant.
The Physics of Spider-Man’s Webs
Still image from The Amazing Spider-Man 2. Image: Marvel
Perhaps the most distinguishing feature of Spider-Man is his ability to shoot webs. Now, let’s be clear. Spider-Man’s webs are a technology-based superpower. Forget what you saw in previous Spider-Man movies. His webs don’t just come out of special holes in his wrists. Those movies were wrong. No, Peter Parker developed these devices using his brain (or maybe he stole them).
The first thing to consider is the strength of these webs. There are several methods that could be used to estimate the web strength. Let me just consider a case from a previous movie that shows Spider-Man using his webs to catch a falling car. What kind of tension would the webs need so that they don’t break? Oh, just find the weight of a car? Nope. That’s not good enough. The webs not only support the car, but also slow the car down.
Let’s say that a falling car has a mass of 2,000 kg and for 1 second before being stopped. This means that I can use the momentum principle to find the momentum of the car in the downward direction.
Since the car starts from rest, the initial momentum is zero. Now, what about stopping the car? Once the web grabs onto the car, there will be two forces on the car: the downward gravitational force and the upward force from the web. Of course a web doesn’t instantaneously stop the car, it also takes some amount of time over which the web stretches. All materials stretch a little bit. For simplicity, I will assume a stopping time that is also 1 second long. The momentum principle looks the same as before except there are two forces on the car and the final momentum is zero.
This means that the web would have to have a tension of at least 39,200 Newtons.
Let’s use this value to make a comparison to other web-like options. The strength of a material can be describe by the ultimate tensile strength. This is the maximum tension per cross sectional area that the material can withstand before breaking and is measured in units of MPa (mega Pascals – or 106 Newtons/m2. In order to get a maximum tension, you need to know the cross sectional area of the wire since obviously thicker wires are stronger. Here comes the first wild estimate (ok, not the first). Let me approximate the web shot from Spider-Man as a cylindrical shape with a radius of 1 mm. If I replaced the web with real materials of the same size, this would be their maximum tension (based on the values from Wikipedia).
Steel cable: 6,503 Newtons
Nylon rope: 235 Newtons
Spider silk: 3,142 Newtons
Carbon nanotube rope: 1.98 x 105 Newtons
Based on these calculations, it looks like carbon nanotube rope is the only thing that would work. Well, the steel cable could work but it would have to be much thicker with a radius of 2.5 mm.
How Much Webbing Can Spider-Man Carry?
In the recent versions of Spider-Man, it seems that all the webbing “ammo” is contained in a small watch-sized wrist thing. In order to estimate the amount of webs, Spidey (he lets his close friends call him Spidey) can shoot, I need to first settle on the webs. I am going to go with carbon nanotube rope. According to Wikipedia, this could have a density of around 0.55 g/cm3 which I assume is the density for the nanotubes in the form of a cable.
How much webbing would Spider-Man need for just one shot? It seems like he primarily uses the webs for swinging. If I were Spider-Man (and I’m not saying either way), I would aim for a height of about 5 to 10 stories high. Let’s say this requires a web length of about 20 meters. Using my initial estimate of a 1 mm radius web, this would be a super skinny and long cylinder. The volume of this cylinder would be:
This would put the total web volume for one use at 6.28 x 10-5 m3. That might be a little difficult to visualize in terms of the size. How about a comparison to the volume of a standard pencil with a radius of 0.25 cm. If all of this webbing was put into a pencil, the pencil would be 3.2 m long. That’s a long pencil and remember, that’s for just one of his typical web shots.
Well, then how big of a container would he need to have a reasonable number of shots? Let’s say he wants 50 uses of the web for each hand. If I were Spider-Man, that’s what I would want. In that case, we can find the web volume estimation by a factor of 50. That gives a total volume (per hand) of 0.00314 m3.
What would this look like if it fit around a wrist? If I use my own wrist for a basis, then I find that it has a circumference of 16.5 cm. In my web container design, I will let the cartridge go back 10 cm along my arm. Now I can calculate the thickness of this container. Maybe a picture will help. Here is a look at my device looking down the arm.
Using the values from my estimates, I get a container radius of 9.6 cm or a height above the wrist of 7 cm. Here is what that would look like.
Where did I get such an awesome Spider-Man arm? This is my arm, I added the Spider-Man suit myself.
Yes. That looks a little awkward. But just imagine how large this thing would be the webs were something like nylon or steel cable instead of nanotube rope.
Web Speed and Range
I already said that it seems like these webs should be able to reach at least a 10 story building (about 30 meters). What kind of launch speed would a web need to get this high? Let’s just start with the assumption that that the front of the web is just a particle and that air resistance is negligible. Yes, that is obviously not realistic but I will proceed anyway. As a bonus, isn’t it great that I can say “not realistic” when talking about Spider-Man? This is what makes the Internet so great.
If a web is launched straight up, there will be only one force on it – the gravitational force. This constant force will make the vertical velocity decrease as it rises. At the highest point, the web velocity will be zero m/s (assuming it just barely makes it to the top). This will give an average vertical velocity of:
Since the web is slowing down with an acceleration of -g, I can find the total time to get to the top of the building using the definition of the acceleration.
Now I can use the average velocity and this time interval to get an expression for the change in vertical position.
And there is your expression for the launch speed of the web. Sure, you could have just used one of the kinematic equations but what fun would that be? Using a the value for the change in height of 30 meters, the web launch speed would be 24.2 m/s (54 mph). That doesn’t seem too bad, does it? But wait. What about air resistance.
I’ll admit that calculating the air resistance in this case can be quite tricky. I could use the typical model for air resistance that say the force from air is proportional to the square of the speed:
Here ρ is the density of air at about 1.2 kg/m3 and A is the cross sectional area of the web. The problem is with the value of C which is a coefficient that depends on the shape of the object. If a web is like a cylinder, a longer cylinder (as the web shoots out) has a different drag coefficient than a shorter web. This means that I will just have to guess at a value for C.
Here is the next problem. As the web rises, it goes slower. With a slower web there is also less air resistance. This means that there is a non-constant acceleration on this rising web. In cases like this, the only practical method for solving for the motion is to use a computer to create a numerical model. It’s not too difficult, but if you want the details check out this previous post.
For this simulation, I am going to assume carbon nanotube webs with a radius of 1 mm and a length of 2 meters in a cylindrical shape. The mass of this section of web can be found from the density of 0.55 g/cm3.
You can see from this plot that the web doesn’t quite go 30 meters high – but it’s pretty close. Ignoring air resistance isn’t such a bad assumption so that the web launch speed of 24 m/s seems legit.
What if Spidey wants to shoot his webs at a bad guy somewhere down the street? How far away horizontally could these webs go? I’ll spare you the math (but it’s here if you want it) and just give you the expression for the horizontal projectile motion distance when and object is fired on level ground at 45°.
Putting in an angle of 45°, Spider-Man gets a range of 58.8 meters. Oh, but maybe he can ramp up the launch speed up to 40 m/s for those special occasions. In that case, he would have a range of 163 meters.
And now for some preemptive comments and answers:
This is silly. Toby McGuire is the real Spider-Man, not this guy that looks like Anakin Skywalker. You might be correct.
I think you made a mistake. You assumed that the density of Spider-Man’s webs once it comes out of the shooter is the same as the density inside the shooter. Couldn’t it be packed in even tighter when inside? Yes, this is possible. However, it would be difficult to estimate the compression inside the shooter.
Why are you wasting your time on stupid posts like this? Don’t you have more important things to do as a physicist? Maybe you should be working on fusion or other clean power sources? You are probably correct, but I just can’t help myself.
I thought Spider-Man’s webs came out of his wrist and were just part of his super hero powers. No. You are very wrong. That was from the previous Spider-Man movies. I suspect they did that because they didn’t want to spend the time to show how Peter Parker developed the webs. If he did have webs as part of his super hero powers, the webs would probably come out of his butt and not his wrist. That would be weird.
What if Spider-Man’s webs are stored in another dimension and his web shooters just grab them and pull them into this dimension? Wouldn’t that explain how he can shoot so many webs? Yes. I think you are correct. This must be the real way that his webs work.
Did you just say “real way the webs work”? You are really disconnected from real life, aren’t you? Comic books aren’t real you massive dolt. If I realized I was disconnected from reality, would I be completely disconnected from reality? I think not. Spider-Man is real but Superman is not.