*I shot an arrow into the air,
It fell to earth, I knew not where;
For, so swiftly it flew, the sight
Could not follow it in its flight.*

- Henry Wadsworth Longfellow

This is what happens when you don't pay attention in freshman physics. You don't need to see where the arrow goes to find out where it lands, you just need the initial speed and angle. Two simple accelerated motion equations later, you have the range. The method isn't perfect since it fails to account for air resistance and other factors, but for finding an arrow it's a lot better than shrugging and giving up once you lose sight of it.

In the early days of physics the question of refining the predictions of projectile motion was not merely one of academic or poetic interest. If you're commander Wellington of the noble British army wishing to lob cannon shells at the nefarious French (or vice versa), these other more delicate physics corrections to the projectile equations can mean the difference between life and death - or at least which side experiences which.

One of these corrections is due to the fact that the earth is not standing still. When a projectile leaves the ground, the earth keeps on rotating right under it. While this is a tiny effect for (say) a line drive in a baseball stadium, it can be substantial for artillery firing over longer ranges. The effect is called the Coriolis force, with the usual caveat that it's not a force as such but rather a "fictitious" force due to the fact that the earth is an accelerated coordinate system.

The acceleration of the projectile from the path it would have taken on a non-rotating earth is to first order:

Where v is the velocity in the inertial frame. For simplicity, let's assume we're just throwing the projectile straight up into the air. On a stationary earth, it would just land exactly where it started. On the actual moving earth, it'll have that additional acceleration deflecting it to the side as a function of its velocity. Plug in the velocity in the above equation:

Now omega is along the earth's axis, and z is the upward direction at the point of the throw. Thus the cross product will be in the y direction, defining y as east:

Integrate that twice to get the position:

Though there's not much to be gained by going through the algebra directly, it's not so difficult to find t and the initial velocity in terms of h. Doing this, we get a total deflection of:

in the westerly direction. Theta is the angle between omega and your local z - in other words, your colatitude. Plugging in a value of h = 100 meters, the earth's rotational angular frequency, and my colatitude of roughly 60 degrees, we get a total deflection of 7.6 centimeters. Very small considering the total distance of travel, but not microscopic either. It would be especially difficult to test due to the difficulty of throwing a projectile exactly straight upward.

The experiment has been done with the much more simple method of dropping a projectile off of a tall building, therefore automatically causing it to fall exactly straight. Bizarrely, if you work out the calculation the total deflection is exactly 1/8th of what we get for the up-and-down case - in the opposite direction (east). Nobody said being a modern major general would be easy.

(This problem more-or-less happens to be Fetter and Walecka 2.5. Great book, btw.)

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More than 20 years ago, my English teacher once recited this poem - it made a deep impression on me, but I never could find out what is was, and by whom - thanks!

The first couplet is quite famous, but this is the first time I have seen the second couplet. The lines would have made sense at the time, since your average Lake District poet would not have had the equipment to measure the arrow's velocity. Back then, field use of long range artillery tended to be a bit empirical: you didn't always know the exit velocity a standard shot of powder would give your projectile (among other things, weather could be a factor), so you saw where the shot landed and adjusted accordingly.

Heinlein wrote a famous parody of the opening couplet: "I shot an error into the air. It's still going--everywhere."

"This is what happens when you don't pay attention in freshman physics."

It's also what happens when you live on a planet with an atmosphere.

Even if the air is completely still, I imagine that this is going a bit beyond freshman physics. (Remember, unlike a cannonball, the force of air resistance on an arrow will depend not just on the velocity of the arrow but also its orientation. That's the point of fletching, after all.)

Oh, sorry. I see you mentioned air resistance. Still, the effect of air resistance is big enough that shrugging and giving up might be just as effective as looking for the arrow near the spot where that arrow would land in a vacuum.

I had a professor (complex analysis) who told a story of his college days, when in a physics class the professor would drop a baseball off the top of the tallest building on campus (15 or so stories, if I remember correctly) and award points to any of his students who managed to catch it.

According to the story, no student ever did, because

- the professor rarely managed to drop it straight, so the ball traveled a good distance from the building

- having been dropped with no spin, the ball tended to behave like a knuckle ball (flutter)

- none of the students were dumb enough to try and catch a baseball that had been dropped from that height

Must be fun to be a professor of physics.

Even when you understand the Coriolis effect, it is still somewhat difficult to understand the rotation of cyclonic storms. Winds traveling north, away from the equator, along the eastern side of the cyclone, should turn to the right, but the winds that turn right create equal but opposite forces that push the northern part of cyclone to the left, resulting in counterclockwise rotation. That's my approximate understanding, but I'm not sure.

Isn't that the same rhyming scheme as:

Yesterday upon the stair,I saw a man who wasn't there.

He wasn't there again today.

I wish that man would go away!

When

Mythbustersfired .30-06 rounds vertical to see where they landed and how hard, an estimated 10,000 foot drop at ~140 mph terminal velocity could not have helped subsequent recovery re inertial referece frames - at least 80 meters deflection right there.Don't say that he's hypocritical,

Say rather that he's apolitical.

"Once the rockets are up, who cares where they come down?

That's not my department," says Wernher von Braun.

Tom Lehrer, "Wernher Von Braun"

This reminded me of an old article about the Paris Gun at http://blog.modernmechanix.com/2008/10/09/secrets-of-the-mystery-gun-th….

It states "The shell, traveling at an average speed of 30 miles a minuteâor sixty times as fast as the usual legal rate for automobiles â took three minutes to complete its aerial flight of 90 miles. It remained away from the earth so long, in fact, that the old world revolved on in space while the projectile was away, so the gunners had to aim a half mile east of the target in order that the target might be there when the shell arrived to hit it."

I won't vouch for the figures, but the article and the site it comes from are interesting.

@Robert: What's really happening is that the wind wants to blow toward the low pressure center (west, in your example), but the Coriolis force prevents it from blowing straight in. It still feels the force of the pressure differential, which eventually wins out over the Coriolis force, but it has to spiral in. Contrast with high pressure systems, where the corresponding force pushes the wind outward. Coriolis force kicks in and pulls the wind around to the right (in the Northern Hemisphere), giving you the observed clockwise rotation around high pressure.

Great post Matt!

I don't think the last result you mentioned is bizarre. Here's why:

Let's compare the two experiments - one throwing a ball straight up, the other dropping it from a tall building - from an inertial frame in which the Earth is spinning. In this frame, the guiding principle will be that the ball's velocity in the azimuthal direction (East-West) is constant. However, due to its changing height, its azimuthal angular velocity, which is v/r, changes. The azimuthal angular velocity is lower for a point on the Earth's surface than for a point at the top of the building.

When we throw the ball up, it comes down in the direction opposite the rotation of the Earth. That's because its original azimuthal velocity is that of the Earth - the low value. When it gets high up, the azimuthal velocity of a reference point on top the building is faster than it, so the ball falls behind.

On the other hand, for the dropped ball, it starts with high azimuthal velocity, and so gets ahead of a reference point on the surface of the Earth. So the dropped ball is deflected with the direction of Earth's rotation.

That shows why the two deflections are in opposite directions. The factor of 1/8 remains to be explained. One might expect a factor of 1/2, since the up-and-down trip is seemingly just two "down" trips back to back, with one in the reverse direction.

This isn't the case, because the up-and-down trip and the falling trip get their angular displacements from different parts of their journey.

When you throw the ball up and down, its angular velocity relative to the surface of the Earth is zero when it's at the surface of the Earth, and highest when it's gotten up to the top of the building. The ball also spends the most time near the top of the building. So the up-and-down trip has the ball spending lots of time in the part of the journey that has high deflection.

The dropped ball doesn't gain any deflection at all while it's at the top of the building - its initial angular velocity is matched to the Earth's there. It gains angular deflection when it's near the surface of the Earth. But the ball is moving quickly near the surface of the Earth, and so spends less time there, and so gets deflected less. So really, we do expect the dropped ball to have much less angular deflection than the up-and-down one.

We can even get the factor 1/8 without too much trouble. The average height of the ball, for either trip, is a distance 2/3 of the way up the building, which is fairly intuitive but can be proven with an integral.

The acceleration of the ball is proportional to the height above the ground for the ball thrown up, and to the height below the building top for the one dropped. So the ball thrown up has an average acceleration double that of the one dropped. It also has double the travel time. So its deflection is proportional to a*t^2. That gives three total factors of two, for an overall factor of 8 in the deflections.

On my website there is a Java applet called 'ballistics and orbits', for simulation of projectile trajectories. (Effects from air resistance are not included.)

http://www.cleonis.nl/physics/ejs/ballistics_and_orbits_simulation.php

You can adjust the following variables:

- Latitude where the projectile is fired from.

- Compass direction of firing (north, east, etc)

- Elevation of firing

- Nozzle velocity (down to zero nozzle velocity

- Altitude of release (so zero nozzle velocity with high altitude is drop from height.)

Purpose of the simulation: to check your intuitions.

(Actually, in artillary aerodynamic and wind effects are way larger than the rotation-of-Earth effect. In firing ballistic projectiles any rotation-of-Earth effect is completely swamped.)

Most important facts: (if rotation-of-Earth effect isn't swamped)

- In ballistics: on the northern hemisphere, if you fire due West, the projectile will land _to the south_ of the latitude that you fired from.

(If you fire due East the projectile will also land to the south of the starting latitude: both to East and to West it lands south.)

- In ballistics: if you use only the coriolis term your calculation result will be wrong.

(Simpler simulation, more focused on a particular aspect of the rotation of Earth effect:

http://www.cleonis.nl/ejs/great_circles_simulation.php)

Cleon Teunissen

Oops, the correct URL for the 'Great circles' simulation is:

http://www.cleonis.nl/physics/ejs/great_circles_simulation.php

Cleon Teunissen

We can even get the factor 1/8 without too much trouble. The average height of the ball, for either trip, is a distance 2/3 of the way up the building, which is fairly intuitive but can be proven with an integral.