Smith, George E. (George Edwin), 1938-

2014-12-02

What he goes on to do in the last problems and I don't want to keep you too long, so I'm going to be very sketchy here. Now put yourself in a position where you can compare Galileo with curvature of the Earth. An inverse square. With Galileo, flat earth, uniform gravity.

He's now gonna look at the case of Galileo with air resistance. And I'm gonna go fast through these. I'm not gonna even try to explain it. The basic idea of these is to figure out how to represent what's going on. If resistance is proportional to velocity, then the increments decrements of velocity are proportional to velocity.

And when that's the case, the velocities have to go in a geometric progression, in arithmetic progression of time. That is, in equal increments of time, you get geometric increments in decrease in velocity. It's called logarithms to us now, but that's what they thought of it geometrically. And it turns out, to make a long story short, the areas under a rectangular hyperbola increase in arithmetic progression if the distances along the axis progress in geometric progression.

So rather than do this symbolically as I've done it down here. This is the solution you get form the differential equation using Euler's constant. Euler wasn't born yet, so obviously this is anachronistic. Newton sits it up as a problem involving rectangular hyperboles. And again, I'm gonna run over probably five minutes, but I haven't even then, I have to make a few points.

That's the problem of uniform motion in a straight line and a resisting media that's resistance proportional to velocity. Next, he does vertical motion, and there we've got two things. We've got gravity pulling it down If I shoot it up, gravity and resistance are both gonna be slowing it all the way up til it gets to a peak, and this is gonna turn around and fall.

Gravity's gonna be accelerating it. Resistance is going to be stopping it. They will reach an equilibrium point that's called terminal velocity. And he gives you the full solution representing it. I'm not gonna take the time to go through this. You can see terminal velocity is by this. But he sets the problem up entirely geometrically where all the relations between the various parameters can be seen geometrically.

That's one of the nice things about geometric solutions, you can simply look at the relations among the parameters. You're not just looking at symbols. You'll see that dramatically when we get to the Principia. I've given you, knowing I'd run out of time to do the math. I've given you from a footnote to the correspondence a very first rate by Turnval first rate description of the math in modern form on those two propositions.

The beautiful proposition then is projectile with resistance bearing as velocity and again, for interest in time you can read it on your own. But the actual curve is the warped parabola that goes up here at this angle with the initial velocity, that long hypotenuse reaches a peak, comes down at a sharper angle, just as they knew it did.

So, and that's the trajectory under a resisting force. And it's again a perfectly good solution. All you need to know is one thing to do it. You need to know the ratio of the initial resistance to gravity. That's what you need to set up the rectangular hyperbole. Okay?

Again I'm going fast but I want to get this done because there's something striking at this point. And then it gives us a way to measure that. One thing we don't know, the proportion of resistance to gravity. Here's the idea. You know the initial angle, you shoot something off with canon, let's say, using the same kind of canon ball and the same amount of powder.

And you measure two things, you measure the total distance and the angle on impact. The angle on impact is not the same as the angle that you take off with. From those two and again you can just work through this. You can get for the cannonball you're using now, you can get the ratio of resistance to gravity.

So we have a way of quantifying this. To extend it to other spheres, he says something really weird. The deceleration from resistance varies directly as the density of the fluid find. The more dense the fluid, the stronger the resistance. The surface area of the sphere. He's picturing the fluid clinging to the surface area.

The more surface area, the more it can cling. Resistance doesn't work that way. Okay? But that's the picture he has at this point. And finally, the velocity and inversely is the weight of the sphere. Now, what he really needs here is mass. He doesn't have the concept yet.

So he's simply saying, heavier the ball, the less the deceleration. Now, two points and then we'll get to the last slide, and I'll get you out of here only five minutes late. Point number one, Huygens had all these results in 1669, did not publish them because he did some experiments and decided resistance does not vary as velocity.

Newton doesn't know this at this point. However Newton has set up a way that we can find out whether resistance varies as velocity. Do the measurement of this that I just described but do it with different angles of firing. You should get a constant for the ratio of the resistance to gravity.

Shouldn't matter what angle you use. You wont. The angles go all over. I don't think Newton did those experiments. He decided on opriora grounds in the Principia resistance can't theory this way. So when we get to the Principia we will see him saying this is all very nice mathematics, this is not the real world.

Okay? And book two of the Principia tries to do for resistance forces what books one and three do for gravity. Okay? To measure them and give the laws governing them.