Snipers and the Coriolis effect

Cecil quotes an author named Telander as asserting “…as a rifle is fired, its barrel heats up, the metal contracts and the bullets are propelled faster.” I believe heating metal causes it to expand. And heating a rifle barrel tends to make it sag, thus the bullet drops below the aim point.

In the following text, Cecil states “If you were firing due east, you’d have to aim six inches lower, since the earth is rotating towards you…” Here in New Mexico the earth rotates from west to east, thus the sun rises in the east. I’m not sure about what happens in Chicago.

Dear Cecil: In an original-message email dated Thursday, June 24, 2010 20:08 hours, the sender, one Maurice Cammack <phantom4597@gmail.com> [and/or] <mecammack@telepak.net>, attached a photocopy of an article “Scanned… from this month’s American Rifleman Magazine”
[page number missing], entitled “World-Record Shot” and claiming that “British sniper Corporal-of-Horse Craig HARRISON of the Household Cavalry had ‘dispatched’ two Taliban machine gunners from a distance of 2,707 yds. with an Accuracy International AWM, or military-designated L115A3, in .338Lapua Mag. in November 2009… in Helmand Province…”

:slight_smile:

Presume we’re discussing this column?

How does the earth’s rotation affect the path of a bullet?

I screwed this up. Cecil’s original text had this right, and I then proceeded to ball things up in a misguided attempt to clarify. I’ll fix.

It’s conceivable that the interior diameter of the barrel may decrease if it’s heated sustantially higher than the outside of the barrel. In that case, the metal on the ID expands, but the cooler metal farther outward constrains the expanding inner metal. I’m not quite sure if the numbers work out correctly for this to be feasible, but I offer it as a possibility.

Man, Cecil’s gonna be pissed at you.

Also, wouldn’t you have to aim left or right if you’re shooting in a north or south (or, maybe the other way) direction? How would a marksman compensate for a ssw shot? Seems pretty tricky.

Wouldn’t be the first time. Remember talking books for the deaf?

I’m not getting these same numbers. Assuming 38 degrees N latitude (approx. Sacramento), with a 1000-yard shot to the due North. Approximate the diameter of the Earth at 4000 miles. The Coriolis effect in this situation boils down to the fact that the shooter is going faster horizontally than his target, since he is farther from the Earth’s axis. Therefore the bullet will miss sideways by the difference in their speeds.

The shooter is moving sideways at 1210.28 ft/s, the target is moving sideways at 1210.24 ft/s. Assuming a 2000 ft/s bullet speed and a 1000 yard distance, the bullet will be in transit for 1.5 seconds.

During those 1.5 seconds, the target will have moved sideways w/ respect to the shooter by only 0.06 ft, or about 0.85 inches.

I also don’t buy the six inches up/down number for an East/West shot.

Possibly because you’ve confused the Earth’s diameter with its radius?

No, I just carelessly used the wrong word. I should have said “radius,” and my calculations were based on that.

Here are my calculations.

Using 1,000 yards as an example: the distance is 3,000 ft, the flight time for the bullet in question (from the Remington calculator) is 2.005 seconds at 25 ft elevation (altitude for Sacramento), so the average velocity is 3,000 ft/2.005 sec = 1,496.3 ft/sec.

Sacramento is at 38.556 degrees latitude. The horizontal Coriolis effect or “force” is given by the following equation:

F = 2 * Omega(angular velocity of the Earth) * sin(latitude)

= 2 * 0.00007292 rad/sec * sin(38.556) = 0.0000909 / sec

If this force acts on an item with velocity V, then the acceleration is:

A = FV, where V is the average speed over the time.

A = 0.0000909 / sec * 1,496.3 ft/sec = 0.136 ft/sec^2

Now we get the distance traveled over time by:

D = 0.5 * A * T^2

D = 0.5 * 0.136 ft/sec^2 * (2.005 sec)^2 = 0.273 ft = 3.28 inches


For the altitude difference, I used these calculations.

1 – 2*(rotation of the earth, rad/s * muzzle velocity / G)*cos(latitude)*sin(azimuth)), where azimuth is measured clockwise from due north.

Firing due west, we get:

= 1 – 2*( (0.00007292 rad/s * 2820 ft/s / 32.2 ft/s^2) * cos(38.555556)sin(270) )
= 1 – 2
( 0.006386 * 0.782 * -1)
= 1 – (- 0.009988) = 1.009988

The drop for this round at the 25 ft altitude is normally 2.35 inches at 100 yards, and 559.4 inches at 1,000 yards. With the multiplier the new drops are 2.37 inches at 100 yards and 565 inches at 1,000 yards. So the shooter must raise her aim 0.02 inches at 100 yards, and 5.6 inches at 1,000 yards. For a due east firing, the only thing which is different in the above equation is the azimuth, which becomes 90. sin(90) = 1, so in effect we can quickly calculate that firing due east results in a multiplier of:

= 1 – (0.009988) = 0.990012

Thus the drop is now 2.35 * 0.990012 = 2.33 inches at 100 yards, and 559.4 * 0.990012 = 553.8 inches at 1,000 yards, for a lowered aim (you lower your aim if the bullet isn’t going to drop as much) of 0.02 and 5.6 inches respectively.

Edited to add: here is the link to the Remington calculator. I used the downloadable version of the software, which has more options, rather than the web-based one.

http://www.remington.com/pages/news-and-resources/ballistics.aspx

To refresh the people’s memory, from Who is this man called Cecil Adams?:

Una, Curt,

Slow down a little bit. Remington’s calculator is nice, but if it’s your intent to provide an accurate comparison, you need to state/provide the specific ammunition type used in your calculations. Even within the same caliber family, there can be as much as a 500fps difference in muzzle velocity with the bullet weight varying by as much as 25%. This will affect your calculations by a non-trivial amount. For most long distance applications, the muzzle velocity will be closer to 3000fps, not 2000fps. The calculations can get complicated, depending on environmental conditions, which is why modern ballistics calculators are so handy.

/signal-11 - embarrassingly poor shooter for an (former) soldier of his qualifications.

What about relativity? A shooter on earth’s surface, his gun, and the bullet are all moving the same relative to the earth. Presumably his target is as well, so the effect of the rotation of the earth would cancel itself out, no?

As far as heating of the barrel goes, both the inner and outer diameter of the barrel would increase with heating expansion, not decrease. This is how you can heat up a ring of metal and attach it to a rod with a slightly smaller diameter. When the ring cools it uniformly contracts, getting pinched onto the rod.

If there’s a non-negligible latitude difference, the rotational velocity of the target and firer will differ; hence “Coriolis”. And for longitude, there’s a slight up/down vector to consider. Whether the signal to noise ratio is high enough to count is another matter.

Yes, but what happens when the interior of the barrel heats, and before conduction has time to even it out? That’s what was being addressed.

My assumption was a Winchester .308 cartridge, metal case (UMC) slug.

1.) Just to check ; I assume Cecil is not talking about Coriolos affecting a shot directly east or west ; there wouldn’t be any Coriolois affect then right?

2.) I believe I saw an episode on History Channel which included the Canadian previous record-holder. I don’t recall them mentioning Coriolis affect. I think it may be the episode on YouTube at

http://www.youtube.com/watch?v=BbOyaPC5wBQ

but I haven’t (re)viewed all the way through to confirm.

OK, I see where I went wrong in my calculations above. When I was figuring the rotational speed of the target, I used 1000 feet instead of 1000 yards.

Carrying out with a little more precision, the shooter will be moving sideways at 1210.30 ft/s, the target is moving sideways at 1210.16 ft/s; a difference of 0.14 ft/s, over a 1.5 second bullet travel time corresponds to 0.2 feet or about 2.4 inches.

That’s just basically modeling the setup as a shooter going at one speed, aiming at a target going a slightly different speed, and figuring how much he’ll miss by.

Again, I’m basing that on a 2000 ft/s bullet speed. The only reason I picked that is because in my studies of the Kennedy assassination, Oswald’s rifle/rounds were around 2000 ft/s.