Why are these caterpillars climbing over each other? The surprising science behind the swarm.

Imagine you’re deep in the Amazon rainforest, and you come across this.. thing. It’s a group of caterpillars, moving in a formation known as a rolling swarm.

rolling swarm caterpillars

If you’re anything like me, your first reaction might be to KILL IT WITH FIRE. Once this irrational fear subsides, your second reaction might be to understand what these caterpillars are up to. Why are they moving in this strange way? (they do it on flat ground as well, not just when going over a bump.)

You might guess that it has something to do with safety in numbers. While this might be part of the story, it turns out that there’s another really ingenious reason why these caterpillars climb over each other.

So here’s the scene. Destin, of the incredible YouTube video series Smarter Every Day, and Phil Torres, who’s a conservation biologist and intrepid rainforest explorer, come across this large, writhing ball of caterpillars in the Amazon rainforest. And seemingly immediately, Destin has an idea – what if the reason that the caterpillars are crawling over each other is to get a speed boost? So he goes home, and designs a wonderfully elegant experiment, using Lego, to prove his point. I just love how this simple Lego powered explanation gets right to the heart of this strange phenomenon.

It’s a simple, but totally mind-blowing idea. Anyone who’s been on one of those endless moving walkways at airports knows that if you walk on a moving belt, you’ll get to the end faster. And so these caterpillars have essentially built a caterpillar-powered conveyor belt. Unlike a typical conveyor belt, this one never runs out, because the caterpillars keep disassembling and re-assembling it.

The really surprising thing is that this entire rolling swarm of caterpillars moves faster than any single caterpillar can, as Destin taught us with his Lego race.

Isn’t that weird? I found it a little hard to swallow. I mean, sure, the caterpillars on the top are getting a speed boost. But the ones at the bottom are still trudging along at their regular speed. So why does the entire group get a speed boost?

Here’s the reason. Every caterpillar spends some time on each ‘floor’. At the ground floor, a caterpillar moves at normal speed. The next floor up, it’s moving at 2X speed, because the floor is moving forward and so is the caterpillar. The next layer up, it’s moving at 3X speed, because the floor is moving at 2X speed, and so on. Every single caterpillar has spent some time moving slowly in the first floor, and some time moving faster in the higher floors. On average, its speed is somewhere in between – faster than a lone caterpillar, but slower than the caterpillars on the top.

Just how much faster? Well, if you like math problems, this is a fun one. I’ll let you work out the details in the comments, if you’re so inclined. Spoilers ahead.

When there are two layers of caterpillars, it turns out that each caterpillar spends half its time on top layer (2X speed) and half its time on the bottom layer (1X speed). This means that the average speed is 1.5 X, or 1.5 times the speed of a lone caterpillar.

How does this compare with Destin’s Lego experiment? All we have to do is count Lego studs.

legorace

The two-layered Lego caterpillar is faster than the lone blue Lego caterpillar, by a factor of 28/19 = 1.47. That matches our prediction of 1.5.

So far so good. But here’s the real question. Does this simple Lego model accurately model the real behavior of these caterpillars? Well, the model makes a clear, testable prediction.

Model prediction (Destin’s idea): In a rolling swarm of caterpillars, every time you go a level higher, the caterpillars are moving faster (with respect to the ground). The second layer is twice as fast as the first layer, and the third layer is thrice as fast as the first layer.

OK, let’s test Destin’s idea. I took his video and used Tracker to track the speeds of a bunch of caterpillars in different layers. (Tracker is an easy to use video analysis and physics education tool created by Doug Brown.)

The tracking looks like this: (SCIENCE! CATERPILLARS! GRAPHS! It has it all.)

Now, we can plot the horizontal position of each caterpillar versus time. This will give us a straight line, and the slope of this line tells you the speed of each caterpillar. That’s the number we care about.

caterpillar speed
The position of a caterpillar versus time. The slope of this line (parameter A) is the speed of the caterpillar.

The speed is shown in some random units, but we don’t care about that, since we only want to compare speeds of different caterpillars (and ratios don’t depend on our choice of units).

I tracked two caterpillars on each layer. It was a little tricky to work out exactly which layer a caterpillar was on, and I had to make a few judgement calls. Here’s what I found.

Layer Speed of caterpillar 1  Speed of caterpillar 2
1 63 70
2 104 169
3 236 237

From these numbers, we can find out how much faster each layer is compared to the first layer. Does it match with the prediction?

Layer Speed (times faster than caterpillars in the first layer)
2 2.04 (1.6 to 2.5)
3 3.56

On average, the second layer of caterpillars is twice as fast as the first one, just as predicted. The third layer of caterpillars is 3.6 times as fast. That’s even faster than our prediction. I suspect the reason for this is we assumed that all the caterpillars were equally fast. But really, the ground floors caterpillars have to deal with the added weight of the floors above them, so they’re probably slower. Similarly, the caterpillars on the top can move faster, unhindered by any extra weight.

One last thing. How fast is this three-layered caterpillar train? A two layered caterpillar train is 1.5 times as fast as a lone caterpillar. I worked out the math, and a three-layered swarm of caterpillars is 15/8 (nearly 1.9) times as fast as a single caterpillar. A three-layered swarm of caterpillars should be twice as fast as a single caterpillar (see footnote). By working together, these caterpillars can move twice as fast as they would by themselves! This is probably why they crawl over each other in this rolling swarm.

So, your homework challenge, if you choose to accept it, is to run a Lego race with a three-layered caterpillar, like Destin did for two layers, and see if it matches the predicted speed boost (2X). If you’re really adventurous, stack some caterpillars on top of each other and measure their speed. Let me know what you find!

Very Geeky Footnote:

If you’re a math geek, one of the things you might think about is, hmm.. I wonder what’s the speed of N layers of caterpillars? Of course, this is a totally idealized problem, because with too many layers, you’ll ended up crushing the poor guys in the bottom of the pile! But if you’re a math geek, this minor practicality will probably not stop you from thinking about this problem.

I tried to derive an equation that tells you the speed of an N-layered caterpillar train (call this speed v_n). Here’s what I came up with (I call it the CATERPILLAR EQUATION).

For two layers, this gives a 1.5 fold speed boost, for three layers 1.875 fold, for four layers 2.2 fold, and so on. The speed multiplier (alpha_n) is always less than or equal to (n+1)/2.

Faraday Everyday pointed out the solution in the comments. The average speed of a caterpillar is just the average of its speed in all the different layers. For N layers, this V (N+1)/2, where V is the speed of a single caterpillar. The derivation is in this comment, with the added point that the sum of the first N integers = N(N+1)/2

Rosie Redfield adds that the Lego blocks don’t accurately model the caterpillars, because it ignores the forces of one layer on the other. Sigh. Any physics-oriented folk willing to take a stab at a full mathematical solution, taking this into account?

 

  • Faraday Everyday

    Awesome post, as always :)!

    One thing I’m not sure about, though, is the geeky footnote part. I have a much simpler (although probably wrong) equation for it.

    Vn=V0*a

    a=(1/N)*∑(from i=1 to N) of i

    What I did is just take the average speed of a caterpillar; I added up all its speeds, and divided by number of layers.

    How did you derive the real caterpillar formula?

    • http://www.empiricalzeal.com Aatish

      I’m not sure my solution is totally right, but I think it’s close. I didn’t have time to write it out carefully, but I can quickly describe it here (sorry it it’s overly cryptic).

      The issue, I think, is that the caterpillars spend different amounts of time in each layer in the general case. So the average is weighted by how much time a caterpillar spends in each layer. To get how much time each caterpillar spends in a layer, I reasoned as follows: In the case of three layers, the caterpillar in the top layer has speed = 3v. And it covers a distance L + 3/2 v t in a time t. The L is the length of the caterpillar layer beneath it. 3/2 v is the average speed of the two layers beneath it, and in time t, these two layers have moved a distance = 3/2 v t. So the caterpillar on top needs to cover the sum of these two distances = L + 3/2 v t. The equation is then 3 v = (L + 3/2 v t)/t , which you can solve for t, the time spent in the third layer. And so on for each new layer.

      • Faraday Everyday

        Oh, right. I forgot to account for different time spent in different layers! Thanks, I get it now :D

        • Mark

          The time spent on a layer is proportional only to the layer size. It doesn’t matter how fast the caterpillar is moving relative to the ground. What matters is how fast it gets to the end of its layer. That depends only on layer length and the caterpillar’s non-swarming walking speed. The formula provided by Faraday Everyday is the correct one (in the ideal case).

          • http://www.empiricalzeal.com Aatish

            Hmm, in hindsight I see that you’re right. I’ll make an edit to the footnote accordingly. So basically you get v = v_0 * (N+1)/2. Thanks for pointing this out, you two.

          • Faraday Everyday

            Oh, you’re right! Thanks, I should have thought it through more thoroughly…

  • Rosie Redfield

    One complication I think you’ve missed: The caterpillars on the bottom aren’t just pressed down by the caterpillars above them. They also experience a backwards force, since the caterpillars on the next layer are pushing back against them to get their forwards movement. The key question is how fast could the caterpillars go if they had no caterpillars walking forwards on top of them. I don’t think the Lego model accurately replicates this issue.

  • http://www.thinkingofthings.com/ ToT

    So, if cars could be built into becoming more caterpillaresque, a traffic jam could make them go faster instead of slower…?

  • Faraday Everyday

    I did some calculations and arrived at sort of a strange conclusion. I’ll upload the pictures now.

  • Faraday Everyday

    Number 1

  • Faraday Everyday

    Number two

  • Faraday Everyday

    Number three

  • Faraday Everyday

    Number four

  • Faraday Everyday

    Number five, with the weird conclusion.

  • Faraday Everyday

    To clarify why my conclusion is strange: For a three-layered swarm, to break even (go at the same speed as one caterpillar on its own):
    -the top layer would have to exert force F (same as a caterpillar on its own)
    -the middle layer would exert force 7F/4
    -the bottom layer would exert force 10F/4 (or 5F/2)

    (Not to mention the lower layers having to support extra weight).

    So it wouldn’t be worth it at all to move in swarms with layers (for speed purposes at least). Maybe after all, the only reason they move in swarms is that there’s strength in numbers! (Or the much more likely conclusion that I made mistakes in my analysis of the problem…). Can you tell me what’s wrong with what I did?

  • Faraday Everyday

    Oh, whoops. I just realized that in giving the velocity, I used the distance formula instead :S. It doesn’t change anyting except for the constant k, though, so the rest of the analysis isn’t affected by it.
    To clarify: the constant k should be t/m, not 1/2(t^2)/m
    (This is the constant used in pages 3, 4, and part of 5)

  • deturing

    I’m a bit confused about the force comment. A caterpillar that changes layers has to accelerate and decelerate, from and to the speed of the lower layer (or vice versa) and on average exerts zero force, no?

    • Faraday Everyday

      I’m not sure what you mean by the force comment, but for a caterpillar to move, it has to exert a force on the ground (or the layer of caterpillars below it) in the direction opposite to where it wants to walk. This is the way we walk too; with our feet, we push the floor back, which pushes us forward :)

      • deturing

        Hi. While I agree that friction is needed to walk, it’s not friction per se that does any pushing, since there is no relative motion between foot and road. The net force on your foot is zero if you add normal force, friction from road, and the force being relayed via the bones and muscles. You can walk because you have movable joints, and you can fall forward on the next foot. I don’t think its legit to translate forces around non-rigid bodies.

        There are some neat videos (http://www.youtube.com/watch?v=88PmGWk_RNQ) that demonstrate that a cart moving at constant velocity doesn’t exert any force on the surface. In any case, we exert a back force on the surface as we speed up and a forward force as we slow down. So, the forces exerted by the caterpillars should depend on how they walk/speed up/slow down. I imagine this would be fairly complex to model!!

  • BillCornelius

    cripes! just time a single caterpillar and time the swarm over the same distance, that’s gives difference in speed so it shows how efficient it is to move w/ the swarm. The hard part will be finding a use for that info.

    • Faraday Everyday

      While I agree that actually observing them would be the quickest/easiest (as well as most reliable) way to find out, I disagree with the last part of your statement.
      We do it because it’s fun :D! Not because there’s necessarily a practical use for it.

  • Fred

    It’s no free lunch. Energy expended = force * distance and the force at the caterpillar’s feet will be proportional to its weight; additionally there will be a backward force imparted by those moving above Those at the bottom will really be feeling the burn. A stationary caterpillar might be able to resist the latter force without expending much energy, with good timing and sticky feet, but efficiency overall probably won’t be as good as a lone caterpillar, especially as hill climbing is involved. It may well be faster though.

    • Joshua Murphy

      Maybe faster is all that matters to the caterpillars. One might compare the scenario to a weightlifter vs. a sprinter. While I don’t disagree with your statement, I think it’s important to consider that there is wasted strength if caterpillars move singularly, and when they work together, that strength is used more efficiently.

      I also think that it’s quite interesting that the caterpillars on the bottom who are “feeling the burn” will get time to rest when they are in higher layers. I kind of wonder if these factors mean that it is actually more efficient to travel as a group…your thoughts are more than welcome if you notice this at all.

  • Keith

    We flew over protection like it was not a big deal!! Camoflage is the most obivous benefit (where does one caterpillar begin and one end?!!); Next is probably the benefit of all your caterpillar buddies spotting a predator (hence all the stopping and starting propogating from different positions within the swarm). But there is another emergent property of the “living conveyor belt” that is quantifiable. A single caterpillar will spend more time (IN AGGREGATE) with living armor on top of them then moving across the top where they are more easily eaten!! So in addition to moving faster, they have the added benefit of a couple protective layers on top of them most of the time. Haven’t had time yet but… what is the probability that you are exposed on top or protected on bottom when a predator comes around (arbitrary time and idealized)? Any interested math nerds out there want to give it a try before I take a crack? Remember (from the caterpillars perspective) it isn’t efficient if it is eaten (so we need to think of efficiency over the long term)!!

  • Nikki

    Sorry this is a bit late but but I wonder if there really is an increase in speed overall as the caterpillars do spend quite a bit of time stopped “thinking”. Its quite hard to tell from the video what portion of time the group spends immobile. The phenomenon of increased speed in this conveyer belt mechanism is quite interesting but it seems more likely that the explanation for this would be more along the lines of safety in numbers rather than a significant increase in ground speed

    • Joshua Murphy

      Hehe, keeping in mind the “this is a bit late” sentiment. I think a good question for you to ask would also be if caterpillars on their own “stop to think.” If they do, I’m sure we could still make the argument that it takes them less time, but I definitely think that it’s an important variable to consider…

  • Jorge

    thats an orgy in nature life