Trajectory

How the API flies a bullet: the drag it steps through, the air it flies in, how a drop becomes a path from the line of sight, and what it assumes.

The flight

A point-mass solution, the method public ballistic calculators use for flat fire: the bullet slows under drag read from the standard G1 or G7 table at its Mach number, scaled by the ballistic coefficient its load is published with, and falls under gravity for as long as it has been flying.

\frac{d v}{d x} = - \frac{ρ C_{d} (M) v}{2 B}, M = \frac{v}{a}

\frac{d t}{d x} = \frac{1}{v}, D_{bore} = \frac{1}{2} g t^{2}

v_{0} = v_{ref} {(\frac{L}{L_{ref}})}^{k}

where

$ρ, a, g$: air density 1.225 kg/m³, speed of sound 340.294 m/s and standard gravity 9.80665 m/s²: the ICAO standard atmosphere at sea level
$C_{d} (M)$: the drag coefficient of the standard G1 or G7 projectile at Mach M, from its published table
$B$: the load's ballistic coefficient against that table, in kg/m²
$v_{ref}, L_{ref}, k$: the load's published velocity, the barrel it was measured from, and how fast velocity changes with barrel length

Equations 1 and 2 are stepped one metre at a time with fourth-order Runge-Kutta, fine enough to follow the drag rise through the speed of sound, where a bullet loses speed fastest.

From the bore to the line of sight

Drop from the bore line grows from the muzzle, but no drop table is read that way: a gun is sighted so that the bullet, launched slightly upward, crosses the line of sight at the zero distance. Tipping the bore by the angle in Equation 4 does that, and Equation 5 is what every published drop table prints.

θ = \frac{D_{bore} (z) + h}{z}

y (x) = x θ - D_{bore} (x) - h

where

$z$: the zero distance, where the bullet crosses the line of sight
$h$: the sight height: the line of sight above the bore
$y$: the bullet's height relative to the line of sight, negative below

Three figures are assumptions, stated in every answer and open to a caller: the zero (25 m for a handgun, 100 m for a long gun), the sight height (20 mm for a handgun's sights, 38 mm for a long gun's optic), and the barrel exponent k where a load does not publish one. The air is the standard atmosphere at sea level, with no wind, spin drift or altitude.

Worked example: a Glock 17 with M882

M882 ball (8.04 g, G1 0.149, published at 350 m/s from a 102 mm barrel) from a Glock 17's 114 mm barrel, zeroed at 25 m with the sights 20 mm above the bore. Every figure below is the API's own solver.

$v_{0} = 350 \times {(\frac{114}{102})}^{0.22} = 359 m / s$

Distance	Velocity	Energy	Below the bore	From the line of sight
0 m	359 m/s	517 J	0.0 cm	-2.0 cm
25 m	332 m/s	444 J	2.6 cm	0.0 cm
50 m	312 m/s	390 J	11.1 cm	-3.9 cm
100 m	284 m/s	324 J	49.8 cm	-33.6 cm
200 m	246 m/s	243 J	239.2 cm	-204.6 cm

The bullet leaves just above the speed of sound, where drag peaks, so it sheds speed fastest in the first fifty metres. From the line of sight it sits at minus the sight height at the muzzle, crosses at the zero, and falls away after: the column a shooter reads.