Philosophy 167: Class 7 - Part 7 - Galileo's Parabola: Sublimity, and the Striking Result on Maximum Ranges.
Smith, George E. (George Edwin), 1938-
2014-10-14
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All right, we've got to this point and now the next slide the reason I broke here is I have to show you what the scaling parameter is for a parabola. Scaling parameter for a circle is of course it's radius. We know the radius, we know which circle. The scaling parameter for a parabola is this distance p, where the parabola is defined as points equidistant from a point, the focus in a straight line, the directrix.
And if you look, the distance from the directrix to the focus is 2p. And what's called the latus rectum is the line perpendicular to the ordinate y. The line that runs through the focus, that's true for all conic sections, it's the correlate of the diameter of a circle, the latus rectum of an ellipse is the lateral distance of a line through either focus.
Not the long axis or the shorter distance etc, well, here you'll notice it's 2p so the line I've drawn in there, it has a vertical pp and horizontal 2p, that's a 45 degree angle. Okay? And that's the mark, at the point where the latus rectum intersects a parabola, it's slope is always 45 degrees.
Okay, and now the parabola in our notation is simply, I’ll do it the way you’re more accustomed to it, y equals x squared over four p. That’s the algebraic equation for a parabola, that’s universal. And p therefore is the scaling parameter. You know p, you know which parabola you're dealing with.
Fair enough? Well, p turns out to be given a name by Galileo, he calls it the sublimity. It is the height from which you fall in order to acquire the initial horizontal velocity. Okay, and that's gonna scale the parabola in a way that is absolutely genius. So that's the same parabola I just gave you, but I've now renamed what p is using his word, sublimity, the height from which you acquire the initial horizontal velocity.
So if you look, I start the top line with exactly the same equations as I had before, but now v sub 0 has been replaced by square root of 2gp. And look what that gives you, you collapse the two quantities we have to know to say which parabola.
The initial speed in g, we collapse it into a single parameter, the sublimity. Does everybody see that? So, sure enough, y equals x squared over four p, on the page a couple of moments ago, I gave you the second half theory, g over 2v0 squared x squared. So indeed, using the height of fall as the measure for velocity, you don't need to know the horizontal velocity or the acceleration of gravity, they're both absorbed in that height.
And absorbed in such a way that the height takes care of everything. So if we continue that derivation now, you'll see that the tangent at any point, the y, the x is x over 2p. You see that the velocity is the same as before, but now it's 2gp 2gy which of course, they don't know g, so they can't really measure the velocity except in terms of the height.
But they have it, as you see at the bottom, they have the velocity at impact compared to the initial velocity, 1 over the square root of hp. But the really important relation is this one, a distance along the semi-parabola, half of it is equal to the mean proportion, that is the square root of the product of the height times the sublimity.
In other words, I know two of those three, I have the third. But look I can easily measure the height and I can easily measure the range if there's no air resistance, therefore I can get the sublimity. I can always infer what the speed, what the height had to be to get the initial speed, independently of knowing the actual initial speed or the acceleration of gravity.
So this is just eminently clever, to call it brilliant is an understatement because he totally bypasses the problem of having to know the values of speed and acceleration of gravity by going to this proxy for speed. And the parabola just falls totally out of it as cleanly as it possibly can, fair enough?
So for example, if I have a cannon and I keep using the same cannonball in it with the same amount of powder, I should be able to figure out what the equivalent muscle velocity is, by figuring out what sublimity goes with the range at the height I achieve.
You'll see that later when I do it with angles, but I repeat, calling it clever is an understatement, it's absolutely beautiful. Everybody follow that, because it's just stunning once you have it. So, this is what he says in proposition five, problem two, in the axis of a given parabola extended upward to find a high point from which a falling body describes the same parabola when deflected horizontally at its vertex, that of course is still.
So what's the problem here, the problem is define the sublimity. Okay? And he gives you a construction. I'm not gonna bother to do the construction because I've already gone through it and described what to do. But it comes out that we're gonna have a mean proportional in there that gives us the height.
And then he says something very striking in the corollary. From this it follows that one half the base or amplitude of a semi-parabola, which is one-quarter the amplitude of the whole parabola, is a mean proportional between its altitude in the sublimity from which a falling body would describe it.
Now, that's exactly the expression I gave to you here, algebraically, exactly the expression he's just described. And he's got to it by doing the construction that gives you the height off a known quantities of the parabola. Cuz if you know the parabola you can work backwards to what the sublimity had to be.
So he gives you that construction, and from that as a corollary he gives you that relationship. Okay, so it's the same thing, I just described algebraically. I'm now explaining it, y is 1 over 4px squared and a over 2 is the square root of hp, that's what he's saying.
I gave you the construction that gives you the answer, I did not prove it's correct. That's what you do with problems. You first describe the construction that gives you the answer, then you prove the answer is correct. I skipped the proof that the answer is correct, partly to save space on a page and partly because the proof is less interesting to us than the result itself.
Now, to just two propositions after that, he comes upon a fairly spectacular result and the results in the corollary are merely start with the proposition. In projectiles by which semi-parabola of the same amplitude are described. Less impetus is required for the describing of one whose amplitude is double its altitude, than any other.
Okay, it's a theorem, he gives you, I put it in here, the construction that he's using, he's comparing, of course, several different parabolas. He's actually comparing two parabolas, and then the proof is based on those triangles, but it's the corollary I want to get to. From this, it is clear that in reverse direction, through the semi parabola DB, The projectile from point D, now pause for a moment.
What's it mean reverse direction? The semi-parabola starts on height and lands at d. The reverse direction, it's starting at d and going back through the semi-parabola. Okay? So the projectile from point d requires less impetus than through any other having greater or smaller elevation than semi-parabola bd, which elevation is according to the tangent ad and contains one half a right angle with the horizontal.
Hence it follows, that if projections are made with the same impetus from point d but according to different elevations, the maximum projection or amplitude of semi-parabola, or a whole parabola, will be that corresponding to the elevation of half a right angle. The others made according to a larger, or smaller angle will be shorter.
I've brought this book for your entertainment. It's a beautiful book. The mathematics of projectiles in sport, okay? And there's a whole section on the Galileo result here. The reason this is so striking is he's getting a result out of the theory that was known to everybody in artillery, maximum range with a 45 degree initial line.
So he's just coming out of nowhere with a known result. Now, that known result was with air resistance. So again, going back to Isaac's question, we have a funny mixture here of real world effects with air resistance, and the theory that's giving you a result without it. Now, I gave you the explanation at the bottom, it's the tangent of the angle is a over 2p.
And that's really all you need to know, where is that tangent maxima. For given, what do I want to say here now? I want the p to be the highest. No, I want a to be the highest, and p to be fixed. So for a to be greatest, the tangent of the angle needs to be 45 degrees.
What's the tangent of 45 degrees, one? What's the tangent as you go greater after that? It's greater than one, right? What's gonna happen as I go, well, that's interesting I think my explanation is lousy at the moment. I'll have to think through what I mean by that. But that is the correct, as you've seen from two slides ago it's the tangent of the angle is a over 2p.
And what we want is the angle of an impact to be 45 degrees. So forget the fact that on my feet I'm not explaining it. Okay, the important thing here is we're now turning to questions of evidence. And we've got something striking. I'll come right back for you.
We're predicting something that was known from it was known Tartaglia a century earlier and we're getting it out of nothing it would seem that 45 degrees is right, and 45 degrees here is very much tied to this parabola. It's a feature of the parabola that it be 45 degrees.
