This is a revision of an old post. I’m going to recycle posts sometimes. I can polish them then, and we read too much new stuff anyway. Better to reread my good old substacks than to read an inferior new one. Also, I’m going to publish Part 2 soon.
The 7th and 8th graders made a trebuchet, a kind of catapult.1 It isn’t as big as in Figure 1; theirs is about eight feet tall. They have about 100 pounds of weights, with a mousetrap-style trigger. They watched VirtualTrebuchet.com, which talks about estimating range. This was happening just as they’re learning to plot and solve equations in my math class, so when Principal DeWayne Pinkney2 told me about it, I improvised a new lecture and 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. We’ll call this the Triangle Model, since it creates a triangle when we look at it in 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 the angle of going straight up.3 It also assume that the trebuchet shoots the missile diagonally at a speed of 1.4 feet per second.4 For the first second of travel, the hypotenuse in Figure 2, the long side of the triangle, is about √ 2 feet, so the Pythagorean Theorem tells us the short sides are each about one foot long.5 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 for t = 0 is (d, h) = (0, 5).
At time 1 second, distance = 1′ and height = 6′, so the point for t = 1 is (d, h) = (1, 6).
At time 2 seconds, distance = 2′ and height = 7′, so the point 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 for t = 600 would be (d, h) = (600, 605), if the path didn’t go off the page to where we couldn’t see it any more.
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. 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 , it is made of plastic, not metal, and it has no insides— it just “models” what a plane looks like. A tiny model trebuchet might be made of plastic too, and it might or might not throw anything. Even if it didn’t, it would still be a good model of what a trebuchet looks like. 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 how it throws a missile.
The Triangle Model is unsatisfactory, though. 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. If you go further in time, though, 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.6
That something is gravity. Our first model ignored gravity, which pulls things towards the center of the Earth. We need to add gravity to the model so it looks something like Figure 3.
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 for t = 0 is (d, h) = (0, 5).
At time 1 second, distance = 1′ and height = 5′ + 1′ - .5′ = 5.5′, so the point for t = 1 is (d, h) = (1, 5.5).
At time 2 seconds, distance = 2′ and height = 5′ + 2′ - 2′= 5′, so the point 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 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 for t = 5 is (d, h) = (5, -2.5).
That last point shows a problem with the model, doesn’t it? How can the height be -2.5′, which is less than zero? 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.7 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 is good for showing how the missile starts to fall, instead of hitting the moon, but it has a new problem— it predicts that the missile will dig a deep hole into the Earth’s crust. We need to fix the model so the missile stops when it hits the ground. That’s actually not hard. We just need to say that if the Gravity Model’s equations say that the height is negative, change it so that once it hits the ground, height and distance don’t change any more. This, the Stopping Model is:
This Stopping Model is better, but it still 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 Stopping Model is a pretty good model if we’re just having fun with the trebuchet rather than trying to knock something down.
The series of models— the Triangle, Gravity, and Stopping models illustrate a routine trade-off with models. As you make a model more realistic and so it incorporates more features of the thing it’s modelling, the model gets more complicated. You have to decide whether the extra complexity is worth the extra realism. “A model should be as simple as possible, but no simpler” as the saying goes.
Plus, I’d like to get to Tuesday’s homework. That homework, inquiring readers was this:
Write down more realistic values for the variables. You have no idea about g, the gravity strength, might be, so use g = 32.8 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:
Pronounceable as either “trebuhshet” or “trebooshay”, English or French.
Oddly enough, there are two prominent citizens of Bloomington with very similar names, both in leadership positions in education, both wearing glasses, andboth carrying more more hair on their chins than on their heads, though one earns almost exactly 10 times the salary of the other. Dwayne Pinkney is Executive Vice President for Finance and Administration at Indiana University. DeWayne Pinkney is an elder of Trinity Reformed Church and Principal of Cedars Christian School. I think they might even both be from North Carolina. Is the Clone Army about to emerge?
There was no school Monday because of Eclipse Day. H in Bloomington, as in Cleveland and Buffalo, we were 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.