Trebuchet Math and Modelling Global Warming
Part I: The Diagonal, Gravity, and Improved Gravity Models
The 7th and 8th graders are making a trebuchet, a kind of catapult. They’ve made good progress. It isn’t as big as in Figure 1; theirs is about eight feet tall. The latest report is that they’ve got about 100 pounds of weightlifting weights with a mousetrap-style trigger, and liftoff has been achieved, so now it’s just a matter of improvement. They’ve been watching VirtualTrebuchet.com, a nifty site for estimating range. And this is happening just as they’re learning to plot and solve equations, so when Principal DeWayne Pinkney1 told me about, I improvised a new lecture and homework. Now I’ll write it up and give it the kids for their Tuesday homework.
Let’s think about how a trebuchet missile goes into the air. It shoots out at a certain speed. There are several ways you could think about how far it goes. You might be interested in:
1. How far it goes horizontally.
2. How far it goes vertically.
3. How far it goes on the arc of its pathway as it curves up and then down.
Thinking about the curve of its flight is a little bit hard, so let’s start with a simpler problem: how the flight starts out as a straight diagonal line up and across. It will go like Figure 2:
Figure 2 imagines that we put the trebuchet on a platform 5 feet high and that we aim at a 45 degree angle, which is half of going straight up.2 It also assume that the trebuchet shoots the missile diagonally at a speed of 1.4 feet per second.3 For the first second of travel, the hypotenuse in Figure 1 is about √ 2 feet, so the Pythagorean Theorem tells us the short sides are each about one foot long.4 Thus, the diagonal speed is 1.4 feet per second, the horizontal speed is 1 foot per second, and the vertical speed is 1 foot per second.
At time 0 seconds, distance = 0′ and height = 5′, so the point in Figure 2 for t = 0 is (d, h) = (0, 5).
At time 1 second, distance = 1′ and height = 6′, so the point in Figure 2 for t = 1 is (d, h) = (1, 6).
At time 2 seconds, distance = 2′ and height = 7′, so the point in Figure 2 for t = 2 is (d, h) = (2, 7).
You get the idea. At time 600 seconds, which is 10 minutes, distance = 600′ and height = 605′, so the point in Figure 1 for t = 600 would be (d, h) = (600, 605), except the path goes off the page and we can’t see it.
The path can be described by two equations as:
distance = speed · time
height = starting height + speed · time,
which is, if we shorten the variable names,
The diagram, numbers, and equation, are a mathematical model of a trebuchet, what we’ll call “The Triangle Model”. The kids are building a real trebuchet, albeit a small one.The kind of model we usually think of is like a model airplane, a small physical copy. A model is not an exact copy, though. It is made for a purpose, and to be simpler than the original thing. A model airplane is tiny and it is made of plastic, not metal, and it has no insides— it just “models” what a plane looks like. A model trebuchet would be tiny might be made of plastic too, not wood and metal, and it might or might not throw anything. The variables, diagram, and equations above are a model of the trebuchet too. The mathematical model tells you almost nothing about what a trebuchet looks like, but it tells you about how it throws a missile.
But the Triangle Model is unsatisfactory. It says that if you throw a missile at 45 degrees, it will go in a straight line upwards and diagonally. That’s good enough for a model of the first few feet of the missile’s path. But the model also says that the missile keeps on going in a straight line and eventually would end up in outer space, perhaps hitting the Moon if you aimed it right. It is a good model for the first few feet, a bad model for the entire path of the missile.
We know from experience that if you throw a ball up at 45 degrees, it first goes in a diagonal, then starts flattening out, then starts falling faster and faster, and then hits the ground, unless something is in the way. Why is that? Why does it flatten out and start falling, instead of keeping going? The real world has something missing from the Triangle Model.5
That something is gravity. Our first model ignores gravity, which pulls things towards the center of the Earth. We need to add gravity to the model so it looks something like Figure 2.
We’ll keep the variable values the same but use a new equation that will get us the right shape of path. The distance equation won’t change, because gravity only pulls the missile down, not back. The height equation, though, has to be modified because gravity does pull the missile down.
distance = speed · time
height = starting height + speed · time - gravity · time · time.
The two equations can be written using short variable names as:
The Gravity Model, as we will call it, has a new variable, g for gravity. We need to assign a value to g. Let’s use g = 0.5.
At time 0 seconds, distance = 0′ and height = 5′, so the point in Figure 3 for t = 0 is (d, h) = (0, 5).
At time 1 second, distance = 1′ and height = 5′ + 1′ - .5′ = 5.5′, so the point in Figure 3 for t = 1 is (d, h) = (1, 5.5).
At time 2 seconds, distance = 2′ and height = 5′ + 2′ - 2′= 5′, so the point in Figure 3 for t = 2 is (d, h) = (2, 5).
At time 3 seconds, distance = 3′ and height = 5′ + 3′ - 4.5′= 3.5′, so the point in Figure 3 for t = 3 is (d, h) = (3, 3.5).
At time 4 seconds, distance = 4′ and height = 5′ + 4′ - 8′= 1′, so the point in Figure 3 for t = 4 is (d, h) = (4, 1).
At time 5 seconds, distance = 5′ and height = 5′ + 5′ - 12.5′ = -2.5′, so the point in Figure 3 for t = 5 is (d, h) = (5, -2.5).
Oops. How can the height be -2.5′? How can it be less than zero? Well, the missile could dig a hole in the ground, maybe. But if we keep going to t = 10, the height would be 5′ + 10′ - 50′ = -35′, and it doesn’t seem like a trebuchet could dig that far.6 Plus, going negative doesn’t fit our diagram, which has the curve going all the way to the wall and then only falling to the ground, not making a crater.
So the Gravity Model has a problem too. We need to change things so the missile stops when it hits the ground. Change it so that the Improved Gravity Model is:
This Improved Gravity Model finally works, except it doesn’t incorporate the possibility of the castle wall. I’ll leave that off for now, perhaps creating a Wall Model in a future Substack, since the Improved Gravity Model is a pretty good model if we’re just having fun with the trebuchet rather than trying to knock something down. Plus, I’d like to get to Tuesday’s homework. That homework, dear 7th graders and inquiring readers, is the following:
Write down more realistic values for the variables. You have no idea about g, the gravity strength, might be, so use g = 32.7 You need to specify values for s_up, s_across, and h_0.
What is the diagonal speed, the speed along its path as it leave the trebuchet? You can compute it using the values you specified in question 1.
Draw a graph on graph paper of the path of the missile according to the Diagonal Model.
Draw graph on graph paper of the path of the missile. It may be tricky getting the scale of units right— whether to have one graph-paper square be 1 foot, or 1/10 foot, or 5 feet, or 1/100 feet. You should try to get the peak height of the missile about in the middle of the graph. So you might have to do this twice.
Estimate the value of t_max and d(t_max), the length of time the missile travels. Plug them into the equation for t_max to see how close you’ve come.
At time (1/2)t_max, how high is the missile, and how far has it gone?
Optional: At the time when the height come back down to its original level of h_0, what is the distance travelled?
Optional: How can you use python to figure out t_max and d(t_max)
Footnotes:
Oddly enough, there are two prominent citizens of Bloomington with very similar names, both in leadership positions in education, both wearing glasses, both carrying more more hair on their chins than on their heads, though I also note that one earns almost exactly 10 times the salary of the other. Dwayne Pinkney is Executive Vice President for Finance and Administration at Indiana University. Elder DeWayne Pinkney is the Principal of Cedars Christian School. I think they might even both be from North Carolina. Is the Clone Army about to emerge?
There’s no school Monday because it’s Eclipse Day, and here in Bloomington, as in Cleveland and Buffalo, we’re located just right to get maximum totality.
What happen if we aim it straight up, a 90-degree angle?
Why do I choose the speed to be the square root of two per second? –-Because it actually makes the arithmetic easier, even though it is an irrational number. If you think for a minute, though, you will realize that to make the arithmetic simpler, I’ve given up realism. One foot per second would be a pretty sad trebuchet. It might not be able to get through a curtain, much less a curtain wall.
The Theorem says a2 + b2 = c2. Note that 1.4·1.4 = 1.96. More precisely, √ 2 ≈ 1.41421356237, and 1.414213562372 = 1.99999999999.
Actually, that’s not quite true. The Triangle Model is fine, so long as you locate your trebuchet on an asteroid, and the right asteroid. The asteroid Nemesis is only 50 miles across, and has very little gravity. Ceres, the biggest asteroid, has only 3% as much gravity as on the surface of the Earth, but it’s 583 miles across, so it might be that the trebuchet missile would still land on the surface, or at least go into orbit rather than achieving escape velocity. I haven’t done the calculations.
Unless the missile was filled with high explosive, in which case a crater 35 feet deep is going to destroy the trebuchet too, since we’d expect the crater to be at least 35 feet across too, probably more, and the trebuchet is only 10 feet away.
A useful rule of thumb I figured out a while back is that 1.5 feet/second--close to sqrt(2) fps--is pretty close to 1 mile/hour. Comes in handy.