Exterior ballistics

Coriolis on a rifle bullet

The horizontal part depends on your latitude and the bullet's time of flight — not on which way you are pointed. The vertical part depends on which way you are pointed as well. At 500 yards, for a .308 match load, the whole thing is under an inch. It is still worth carrying, because it is a bias and it never cancels.

The short answer

Horizontal. Right in the northern hemisphere, left in the southern, on every shot, whatever direction you fire. Zero at the equator, maximum at the poles. Applied Ballistics publishes 2.5–3.0 inches to the right at 1000 yards near 45 °N for a small-arms trajectory.

Vertical. Shoot east and you hit high, west and you hit low, due north or south and there is no vertical component at all. Maximum at the equator, zero at the poles — the opposite of the horizontal. For a shot due east or west at 45 ° latitude the two components are the same size, which is geometry rather than coincidence, as the derivation shows.

Source: Bryan Litz, Gyroscopic (spin) Drift and Coriolis Effect, Applied Ballistics, 2021.

Where it comes from

The bullet flies straight as far as the universe is concerned. You, the rifle and the target are riding a rotating sphere, so in your frame the bullet appears to curve. Writing Newton's second law in a rotating frame produces the extra term:

a = −2 Ω × v Ω — the Earth's rotation vector, magnitude 7.292115 × 10⁻⁵ rad/s
v — the bullet's velocity in the Earth-fixed frame · × — vector cross product

That rate is a defining parameter of WGS 84, the datum GPS positions are given in (NGA.STND.0036, formerly NIMA TR8350.2). It is one turn relative to the stars: 2π ÷ 86,164.1 s returns the same number.

Split Ω into local east, north and up components at latitude φ. It has no east component anywhere; it points north with magnitude Ω cos φ and up with magnitude Ω sin φ. Those two pieces produce the two effects, and they behave nothing alike.

The horizontal component

Only the vertical part of Ω acts on horizontal motion. Crossing it into the bullet's horizontal velocity rotates that velocity 90 degrees:

a_horizontal = 2 Ω V sin φ , directed 90° to the right of the line of flight φ positive north, negative south — the sign flips in the southern hemisphere.

Because the cross product simply rotates whatever horizontal velocity it is handed, the firing azimuth does not appear. Fire north, south, east or west from the same spot and the bullet goes the same distance right. Shooters expect this one to vanish when they turn the rifle. It does not.

An acceleration that is always square to the flight path and proportional to the speed does one simple thing: it swings the direction of flight to the right at a steady 2Ω sin φ radians per second. Drag acts along the flight path, so it changes the speed and never the direction. After time t the path has turned through 2Ω t sin φ, and the deflection is that angle summed over the distance flown, writing t(x) for the time at which the bullet reaches distance x:

deflection = 2 Ω sin φ ∫ t(x) dx → ≈ Ω R T sin φ for constant speed R — range · T — time of flight

The right-hand form is the one usually quoted, Litz's book included. It is convenient and it reads high — 6 % at 500 yards for the load below, more as the bullet slows — for the reason given below. One term is dropped here: Ω cos φ also acts on the bullet's vertical velocity to give a small east-west push, but for flat fire the downward speed is of order 20 ft/s against thousands forward, so at mid latitudes that term is about 1 % of the one kept, or less.

The vertical component

The north-pointing part of Ω acts on the bullet's eastward velocity to push it up or down. With azimuth A measured clockwise from true north, the eastward velocity is V sin A, so:

a_up = 2 Ω V cos φ sin A

This is the same physics that makes an eastbound aircraft weigh fractionally less than a westbound one — the Eötvös effect. It is a vertical acceleration like gravity, so a convenient way to carry it is as a correction to gravity; McCoy derives exactly this factor for a flat-fire vacuum trajectory:

g_effective = g × [ 1 − (2 Ω V / g) cos φ sin A ] A = 90° due east → gravity effectively weaker → hits high
A = 270° due west → gravity effectively stronger → hits low
A = 0° or 180° → no vertical effect

Put numbers in it. At 2600 ft/s (792.5 m/s), 2ΩV/g = 0.0118. Firing due east on the equator, gravity is effectively 1.18 % weaker at that speed; at 45 ° latitude, 0.83 %. Drop scales with gravity, so at muzzle speed those would be the percentage errors in elevation if you ignored it.

Note what V is doing there. It is the bullet's current speed, not its muzzle velocity — and by 500 yards a .308 match load is 31 % slower than it left at. A solver that plugs muzzle velocity into this factor overstates the back half of the flight. The honest version integrates, and it gives the vertical deflection the same form as the horizontal: 2Ω cos φ sin A ∫ t(x) dx.

R. L. McCoy, Modern Exterior Ballistics, 2nd ed., section 8.8.

A worked example, from published data

Federal publishes the velocity of its Gold Medal 308 Win 175 gr Sierra MatchKing load (GM308M2): 2600 ft/s at the muzzle, then 2427, 2262, 2102, 1949 and 1803 ft/s at 100-yard steps to 500. Time of flight is the integral of dx/v over that table — by trapezoid on the six points, T = 0.694 s to 500 yards.

Check it before using it. Crosswind deflection obeys the lag-time rule — wind speed times the difference between the real time of flight and the time a vacuum trajectory would take, D = W (T − R/V₀) — which goes back to Didion in 1859 and is in McCoy's Modern Exterior Ballistics. Federal publishes its own 10 mph crosswind figures for this load, so the rule can be scored against them:

Range	lag rule, using T from the velocity table / Federal published
100 yd	0.72 in / 0.6 in
200 yd	2.96 in / 2.9 in
300 yd	6.89 in / 6.9 in
400 yd	12.68 in / 12.5 in
500 yd	20.56 in / 20.3 in

Within 1.5 % from 300 yards out, and at 200 yards the gap is 0.06 inch, barely more than Federal's rounding to a tenth. The worst row is 100 yards, 0.12 inch high, and rounding covers less than half of that. A full point-mass integration that reproduces Federal's velocities to within 2 ft/s also gives 0.71 inch there, so the gap sits in Federal's wind column, not in the time of flight. The time of flight is good.

The same trapezoid gives the time at each 100-yard mark, and summing those times over distance gives ∫ t dx = 490 ft·s. At 45 °N:

Horizontal	2 × 7.292115×10⁻⁵ × sin 45° × 490 = 0.0505 ft = 0.61 inch right
Vertical, due east	the same integral times cos 45° = 0.61 inch high
Constant-speed shortcut	Ω R T sin φ = 0.64 inch — 6 % high
Full point-mass integration	Federal's published G1 0.505 in Army Standard Metro air — the combination that reproduces Federal's table — with `−2Ω×v` in the equations of motion: 0.60 inch

The shortcut is high because it assumes the bullet covers ground at a constant R/T. A real bullet is fastest at the start, so it reaches every intermediate distance sooner than that straight line says. The flight path has turned less by the time it gets there, and ∫ t dx comes out 6 % smaller than RT/2. McCoy's point-mass tables for the 7.62 mm M80 ball bullet at 45 °N give the same size — 0.6 inch at 500 yards, 2.8 at 1000 — and he notes that the average-speed shortcut over-predicts.

Scaling to 1000 yards needs a time of flight Federal does not publish. Carrying Federal's published G1 0.505 past the end of its table, in Army Standard Metro air — the combination that reproduces the published table to within 2 ft/s and 0.2 inch — gives 1.71 s and 2.8 inches of horizontal Coriolis at 45 °N. That is an extrapolation of Federal's model, not Federal's data; a G7 model of the same bullet gives 1.75 s and the same 2.8 inches. It lands inside Litz's published 2.5–3.0 inches by an independent route.

One result worth knowing

For a shot fired due east or due west, the horizontal deflection carries sin φ and the vertical carries cos φ, over the same integral. The size of the combined deflection is therefore proportional to √(sin²φ + cos²φ), which is 1 at every latitude on Earth.

The total Coriolis deflection on an east or west shot is the same everywhere. Only the split between elevation and windage moves: all elevation at the equator, all windage at the pole, half and half at 45 °. For the load above that total is 0.86 inch at 500 yards, in Ecuador or in Alaska.

When it stops being negligible

0.61 inch at 500 yards is 0.12 MOA, or 0.03 mrad — a minute of angle subtends 1.047 inches per 100 yards, a milliradian 3.6 inches. Federal's own table says a 10 mph full-value crosswind moves that bullet 20.3 inches at the same range, so the entire horizontal Coriolis correction is worth about a third of a mile per hour of wind call. At 1000 yards it is about 2.8 inches, roughly a quarter of a MOA, against about 100 inches of wind for the same 10 mph on the same extrapolated model — still only 0.3 mph.

So the case for correcting it is not size. It is that it is a bias with a known sign. A wind error is as likely one way as the other and averages out over a string; Coriolis puts every shot on the same side, exactly like spin drift, and the two stack. In the northern hemisphere with a right-twist barrel, Litz gives about 9 inches of spin drift plus 2.5 inches of Coriolis at 1000 yards — 11.5 inches right in a dead calm.

Two practical points that cost more than the physics does:

The azimuth is a true bearing. A magnetic compass reading must be corrected for declination first, or A is wrong by that angle.
The latitude is the firing position's latitude, and it is the only thing about the place that enters the horizontal term. The azimuth, the target and the rifle do not; the range and time of flight do, through the integral.

Work it out

The deflection needs ∫ t dx, and that needs the bullet's time at every distance. Two published numbers pin it down well enough: muzzle velocity, and either the velocity at the target or the time of flight. The defaults are Federal's GM308M2 at 500 yards, fired due east at 45 °N.

Latitude of the firing point degrees; north +, south − Firing azimuth, true degrees clockwise from north Range yards Muzzle velocity ft/s Downrange figure you have Velocity at the target ft/s

How it works: horizontal = 2Ω sin φ ∫ t dx, vertical = 2Ω cos φ sin A ∫ t dx, with Ω = 7.292115×10⁻⁵ rad/s. To get t(x) the calculator takes drag as proportional to speed squared, so speed falls exponentially with distance, and fits that to your two numbers. Level flat fire, no wind; the small push from the bullet's own vertical speed is left out. Against a full point-mass integration, over G1 and G7 bullets from 1800 to 4200 ft/s, while the bullet is still supersonic at the target: within 4 % from the target velocity for muzzle velocities up to 3000 ft/s and within 8 % above that, and within 3 % given the time of flight. It reads low once the bullet is well subsonic. It reproduces McCoy's published vacuum table to every printed digit, and his M80 point-mass figure at 500 yards. Use a true azimuth, not a magnetic one.

In the solver

Range Walkout Coming soon is our long-range ballistic solver for Android. You mark the firing point and the targets on a range map and it derives range, incline, bearing and latitude from those points — exactly the inputs the equations above need — then adds Coriolis and spin drift into the windage and elevation it gives you, instead of leaving them to be added by hand. Muzzle velocity can be back-computed from a known-range shot. Details on the products page →

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