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Background: An AC-130 is a military aircraft that has a cannon that fires out the side of its fuselage.

If an AC-130 (or similar aircraft) is traveling in straight, level flight at a constant rate of speed calculating gun lead is straightforward. For example if the plane is traveling forward at 100 meters per second, the target is 1000 meters to the side and its cannon shell travels at 1000 meters per second simply fire 1 second or 100 meters early to score a hit.

However what if the AC-130 is traveling in a circular orbit at 100 meters per second around its target instead of beside the target? I'm not sure how to calculate gun lead in that circumstance.

Hopefully the image I attached below will help this make sense. Thanks so much in advance for any input!

enter image description here

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    $\begingroup$ This is a purely math-based question, but the solution is straight forward. Given any two extremely close time instance, the plane can be considered travelling along a straight line in that timeframe (for more details, see circular motion topics in Physics). Therefore by simply taking a snapshot of the position and orientation of the plane at the instant the bullet is fired, the solution is the same as it is in the straight line case. $\endgroup$ – kevin Mar 27 '18 at 0:18
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    $\begingroup$ I don’t believe they do compute a lead as there is no relative translation between the aircraft and the pylon. $\endgroup$ – Carlo Felicione Mar 27 '18 at 2:01
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    $\begingroup$ The scenario is almost the same as in your first diagram. In the millisecond or two that the shell takes to travel the length of the barrel the plane turns only a very slight amount. Once the shell leaves the barrel it’s all linear physics and the continuing turn of the plane has no effect on the shell. There might be a few centimeters difference at the target. $\endgroup$ – Jim Garrison Mar 27 '18 at 4:42
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    $\begingroup$ As others have commented the instantaneous motion of the aircraft at time of firing is the same as linear motion. One thing that does make a slight difference however is the outward G-force acting on the shell as the aircraft turns. That would act to reduce the muzzel velocity very slightly which will change the landing point a smidge but likely not enough to bother about. $\endgroup$ – Trevor_G Mar 27 '18 at 16:45
  • $\begingroup$ @Trevor_G No... there's no 'centrifugal force effect' at all, since that 'force' disappears the moment the shell is shot. $\endgroup$ – xxavier Jul 10 '19 at 15:36
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The bullet leaves the barrel with the combined velocity of barrel muzzle velocity plus the instantaneous aircraft true air speed velocity (actually, the instantaneous true air speed velocity of the tip of the muzzle). That puts the velocity vector of the bullet ahead of the line of the barrel (ahead of the 90 degree wing line). The fact that the barrel is rotating in inertial space is irrelevant. Whether the aircraft is straight and level, turning towards the tank, or away from it, if, at the moment the bullet leaves the muzzle, it is traveling in inertial space towards the tank, (and sufficiently above it for gravity drop) it will hit it.

And since the bullet's velocity leaving the barrel is a simple vector sum of the muzzle velocity, and the aircraft's instantaneous velocity at that instant. The velocity vector of the aircraft before, (or after), the bullet leaves the barrel is of no relevance.

So if the aircraft is moving due North at 100m/s and the muzzle velocity is 1000m/s due East relative to the aircraft, the bullet is actually traveling SQR(100^2 + 1000^2) = 1004.98 ~1005m/s in a direction slightly North of due east by an amount equal to the arctangent of 100/1000 or ArcTan(0.1) = 6.345 degrees So the bullet is traveling 1004 m/s on a heading of 90-6.34 degrees = 83.66 degrees true. So, to hit the tank, the pilot has to fly so that the gun barrel is pointed behind the tank by 6.34 degrees (and above it by a sufficient amount to account for gravity drop)

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  • $\begingroup$ Though, to a certain degree, I wouldn't expect this to matter much in practice - the AC-130's main armament puts enough "ouch" downrange per round (or, for the GAU, "ouch" per second) that small drifts in accuracy, especially at the low altitudes it flies at, likely wouldn't have much of an effect. $\endgroup$ – Sebastian Lenartowicz Sep 8 '20 at 8:23
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When a projectile is fired from a gun in a fixed wing aircraft or helicopter in a direction perpendicular to the direction of flight, the airflow across the muzzle has an effect on the trajectory which is known as "bullet jump", the deflection can be upwards or downwards depending on the direction of the rifling of the gun (normally right hand or clockwise) and the direction of the airflow. In the case of the AC130, in which the weapons are mounted on the port side the deflection would be in a downward direction. This occurs because the combination of the rotation of the projectile and the airflow causes a difference in air pressure between the upper and lower sides of the projectile. Technically this is known as the Magnus Effect.

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In the case of a gatling type weapon the detailed design of the gun and mounting and the rate of fire would all have to be taken into consideration as they would determine the direction and magnitude of the velocity vector imparted to the projectile by the rotation of the barrels while it was being loaded and fired, as well as the internal ballistics of the ammunition and the individual barrels. AC130s were formerly armed with 7.62mm, 20mm and 25mm gatling guns, but now as I understand it these have all been replaced with 40mm single barreled guns.

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