A look at projectile motion

A mathematical look at projectile motion

David Coulson, 2014

A few weeks ago my students asked me to show them some mathematics about “things going splat” when they fall off a cliff and so on. So I showed them the five kinematic formulae, which pretty soon got to be called the “Splat formulae”.

And after some time we got round to looking at the classic cannon-shell problem.

The simplest (and least accurate) approach is to pretend that there is no air resistance and look at what gravity alone does to the cannon shell.

Gravity causes the projectile to travel in a great arch through the sky.The mathematical name for an arch is a ‘parabola’.

The mathematics of parabolae is understood very well these days, so we can plot the journey of that cannon-shell very precisely, including how high it goes and where it lands.

The most important thing for you to know at this point, however, is that the curve is symmetric. That means that the journey upwards is identical to the journey downwards.

And that means that the highest point in the journey is exactly halfway between the two ends.

We can find the highest part of the journey by noting very simply that at the highest part of the journey, the cannon shell has stopped rising (!) and hasn’t started to fall yet.

Let’s put some numbers in here. Let’s say that the cannon is slanted at 45 degrees to the horizon, and that its muzzle velocity is 1 kilometer per second.

That puts the cannon into the class of, say, a gun on a modern battleship, firing projectiles that weigh maybe half a tonne. These projectiles can travel right over the horizon to targets we can only see on a map.

For a cannon aimed at 45 degrees, the upwards velocity is given as the sin(45) of the total velocity.

Since the sine of 45 degrees is 0.7071, we can say that the upwards velocity is 70 percent of the muzzle velocity.

That’s 70 percent of 1000 m/s = 700 m/s.

Now go back to the five Splat formulae and find one that will help you find the top of the parabola.


This one will do.


We can also get the time to the top of the parabola.

The total flight time is double this number.Because the journey down is exactly like the journey up.

The impact velocity is easy.It’s the same as the muzzle velocity, because everything here is symmetric.

The shell hits the ground at 45 degrees, at 1000 m/s, 140 seconds after leaving the cannon.

Yes, but where does it land?

We know the flight time is 140 seconds. We can compute the horizontal distance travelled from the horizontal velocity.

So now I know everything there is to know about this cannon shell, except where it happens to be at any point in its journey.

I can get that from another Splat formula.

This breaks up into two formulae:one for the height at any timeand one for the horizontal distance covered in that same time.

22

10 - 700 tth td 700

22

10 - 700 tth td 700

I actually used these formulae to generate the picture that’s been on the screen throughout this lesson.

Now this may look like the end of the problem, but actually it’s only the beginning....

These numbers will be hopelessly wrong because I have ignored the effect of the atmosphere.

Air resistance will slow the cannon shell down, so that it won’t go nearly as high,

nor will it travel nearly as far.

Now this may look like the end of the problem, but actually it’s only the beginning....

To overcome that, I’m going to need a hell of a lot more physics and a hell of a lot more paper.

We start with the formula for fluid resistance around any object.

221 AVCF Ddrag

I can turn this into an acceleration formula by dividing out the mass of the object.

Doing so means I can add air drag as a vector that directly opposes the velocity vector at any instant in time.

gravity

Velocity

Air drag

At any instant in time, the cannon shell’s future will depend on the size of these three arrows.

This is what I’m talking about...

At this point I have to put some realistic numbers in.

Cd is the coefficient of drag which comes from a book and depends on the shape of the cannon shell’s front end.

For a bullet-shaped shell, with a cone angle of 30 degrees from the centre line,the coefficient is 0.55.

The A value is the frontal area of the cannon shell. That means how it looks if photographed from the front.This is a circle, and for a cannon shell of diameter 30 cm, the area can be worked out using pi-r-squared.


The mass (m) of the object can vary quite a lot from shell to shell. For this example, I decided to use a weight of 1 tonne.

That’s not unusual, by the way. Cannon shells can go up to 1.7 tonnes, I have read.


The funny-shaped ‘p’ in the formula is the Greek symbol “rho” which has come to mean ‘density’, measured usually in kilograms per cubic metre.

In this case we’re talking about the density of air, which at ground level is 1.2 kg/m3.

I’ll have a LOT more to say about that in a wee while.


The V in the formula is the speed at which the fluid is flowing around the object or the speed at which the object is flowing through the fluid.

This time around I’ll start with a muzzle velocity of just 500 m/s because (for reasons I’ll share with you later) I don’t want the cannon shell to rise too high.


The Velocity term is the tricky one because it will change radically and continuously throughout the journey, once gravity and wind resistance get a hold of the cannon shell.

Therefore I can’t use the same easy process as I used before to solve the equation. The best I can do is set up a computer model of the journey, using the spreadsheet on my laptop.

After a bit of experimentation, I have decided to calculate the velocity after each half-second of flight, and use that number to calculate the air resistance acting on the cannon shell at that instant.

I can step through the journey a half second at a time, recalculating the air resistance as I go until the cannon shell hits the ground. It will require tens of thousands of calculations but that is no problem for a spreadsheet.

Since I’m using a spreadsheet to do the grunt work, I may as well reveal to you that air density is going to change radically throughout the flight too. By the time the cannon shell gets to the top of its arch, air density will be perhaps only a tenth what it is a ground level, which means air resistance will be only about a tenth as great as it would be at ground level. I will have to include this variation in my spreadsheet calculations.

- And that, my friend, will be a very tricky business all by itself!

Air density depends on how much atmosphere is pressing down on it from above, but it also depends on how hot it is. Hot air expands and tends to fight back against the weight of air around it.

Sunlight passes down through the atmosphere and heats it somewhat, but heats the ground a lot more. Therefore most of the heat in the atmosphere radiates back up from the ground, making a fairly linear decrease in heat as you go up to about 17 km, when there is a substantial shift in the composition of the atmosphere.

That’s why I didn’t want to send the cannon shell up that high. The maths is (erm, comparatively) easy in the lower part of the atmosphere.

Here’s the page of notes from Wikipedia that I used to calculate air pressure at a given height.

...And here’s how to turn it into density (also from Wikipedia):

B = 6.50 oC / km

T0 = 288 Kelvin

All this boils down to this two-step formula:

Altitude (m)

Here’s what the first 19 seconds of my calculation looks like:

The shell was in the air for about 65 seconds this time. That means that the full calculation is about three times the length of what you see here.

30

45

60

75

15

Here are some charts showing the course of the projectile, depending on the angle of elevation from horizontal.

30

45

60

75

15

You can see that the best result comes from a 45-degree angle (not surprisingly.

A 75-degree angle is no better than a 15-degree angle.And a 60-degree angle is no better than a 30-degree angle.

Heights and distances on the chart are measured in km.

This is what we would have if there was no atmosphere.

Interestingly, the trajectories still look very much like parabolae.

You can see that there’s quite a difference.

This changes when air-resistance is stronger.

In this experiment, I’ve flattened the nose of the shell until it’s completely blunt.

Now things get really interesting when we trade off velocity against cannon-shell weight for a given load of energy from the cannon.

Assuming the cannon can deliver a fixed amount of kinetic energy, we could put a small shell into it and send it off very fast, or put a heavy shell into it, which will travel more slowly but which will be more resistant (through inertia) to air resistance.

1000

2000

If I double the shell weight, the shell only goes half as far. That’s like you throwing a piano over a fence instead of a chair.

500

2000

1000

I can make the shell go very much further if I lighten it to half its previous weight.

Ooh! But look what happens when I make the shell too light!

Even though it leaves the cannon very much faster, air resistance pushes it back and causes it to fall short.

250

2000

1000

Clearly there is an optimal weight for a cannon shell that depends on the energy it can deliver.

125

2000

1000

250

2 x diameter4 x diameter 1 x diameter

A similar thing happens when we vary the diameter of the shell. The pointyness of the shell didn’t make much difference, but the fatness of it makes a huge difference.

Clearly cannon shells (and rockets?) work a lot better when they are shaped like pencils.

Are these results accurate?

They are as accurate as I have been able to make them, but even so they are more indicative than factual. By that I mean that they illustrate the factors that have a big effect on projectile motion but the numbers themselves are probably not so reliable.

I have not considered wind speed, nor have I considered the seasonal and daily weather variations that can affect air density.

Then there’s the matter of the Earth’s rotation. For a truly big weapon firing a cannon shell over dozens of kilometres, we need to consider what happens to the cannon shell while the Earth is rotating underneath it.

On that scale, it becomes clear that the trajectory is not a simple arch but an ellipse, like a cometry orbit that passes through the Earth’s surface at a couple of places.

If fired directly North or South, a cannon shell appears to move East or West slightly, depending on how its actual speed compares to the rotational speed of the ground underneath it.

In theory at least, it should be possible to fire a cannon shell very far out into space so that by the time it returns to the ground, the Earth has rotated enough so that your ship is there to be obliterated by it.

Now there’s a thought...

[END]

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A look at projectile motion