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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
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    $\begingroup$ @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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    $\begingroup$ @SebastianLenartowicz, all the “ouch” is still focused in at most a few meters radius and the correction is by tens of meters at least, so it definitely does matter. Of course the gunner does not have to think about it, because the sight does it for them. $\endgroup$ – Jan Hudec Feb 9 at 12:00
  • $\begingroup$ Wouldn't the rotation of the aircraft put a torque on the bullet that might deflect it slightly in the direction of the rotation of the aircraft? $\endgroup$ – nick012000 Feb 9 at 20:09
  • $\begingroup$ No, The bullet is spinning and this creates a stability and resistance to any applied torque. But more relevant, there are two ways in which motion occurs, or in which motion can be changed. The velocity of an object can be changed by a linear force, and the rotation rate of an object can be changed by a torque. Torque on an object does not change it's velocity. It only changes it's rate of rotation. And so even if the bullet was a non-spinning sphere, and a torque did change it's rotation rate, it's velocity, (and trajectory), would not change and it would still impact at the same point. $\endgroup$ – Charles Bretana Feb 10 at 14:07
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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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The heart of the question is whether or not it matters that the gunship airplane is travelling in a circular path. No it does not. All that matters is the aircraft's instantaneous linear velocity at the instant the bullet leaves the muzzle of the gun.

Some of the answers addressed the effect of the spinning barrel assembly of a Gatling-gun type weapon. This doesn't really appear to be what the question was about, but the same principal applies: the instantaneous linear velocity of the barrel at the instant the bullet leaves the muzzle will affect the bullet's trajectory. To know what direction the barrel is travelling at the instant the bullet leaves the gun, we have to know how long it takes the bullet or shell to pass through the barrel, and how quickly the barrel assembly is spinning. If rate of spin of the barrel assembly is not constant, the effect of the barrel's motion on the bullet's trajectory will not be constant.

This Wikipedia page mentions a firing rate of 1,800 rounds per minute for one of the Gatling-gun weapons used on the AC-130U. Since there are 5 barrels, the rate of spin of the barrel assembly must one fifth of this figure, or 360 revolutions per minute. Assuming that each barrel assembly rotates around a circular path with a diameter of 8 inches, i.e. a radius of 4 inches (a wild guess), the instantaneous linear speed of each barrel due to this rotation is about 9043 inches per minute or 8.6 mph. This appears to be a rather trivial number compared to the flight speed of the aircraft. (To correct this estimate for a different known radius of spin, divide by 4 and multiply by actual radius of spin in inches.)

If we assume that the bullet is fired at either the top or the bottom of barrel's circular path, and we assume that the barrel only travels a short distance before the bullet exits the barrel, then the barrel would be travelling roughly horizontally when the bullet leaves the barrel. This would mean that the instantaneous linear velocity of the barrel due to rotation of the barrel assembly should be either added to, or subtracted from, the instantaneous linear velocity of the aircraft, depending on whether the bullet is exiting the barrel at the "top" or the "bottom" of the spinning barrel assembly, and which direction the barrel assembly is spinning, and on which side of the aircraft the gun is mounted (the left side in the case of the AC-130U.)

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  • $\begingroup$ 2000 SPM divided by 5 barrels would give the barrel assembly rotating at 400 RPM, not 2,000. Otherwise, I agree with the central point of the answer. $\endgroup$ – Ralph J Feb 8 at 19:44
  • $\begingroup$ @RalphJ -- good point, will edit $\endgroup$ – quiet flyer Feb 8 at 20:13
  • $\begingroup$ @RalphJ --also my intention was to use 6 inches as the diameter of rotation, not the radius. A 6 inch radius might be too much and might give too high a linear velocity. Have reworked using a 8 inch diameter / 4 inch radius. $\endgroup$ – quiet flyer Feb 8 at 20:27
  • $\begingroup$ That 8" number is the right ballpark. I'd have guessed a little wider than that, but it's been a while since I've been close enough to a gunship. For purposes of discussion, that's plenty accurate. $\endgroup$ – Ralph J Feb 9 at 5:10
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This is more of a physics question than an aviation one. As other answers say, the effect on the trajectory of the bullet is determined by just the instantaneous velocity. However, if the purpose of taking into account the effect is to calculate lead time, then circular travel will be different from linear. If you decide you need to fire 1 second earlier, then with circular motion the velocity of the aircraft when you otherwise would have fired is different from the velocity 1 second earlier. So with a tight circle, you can't just start at the time you otherwise would fire, take the velocity at that point, and calculate how much earlier you need to fire based on that. If you want an exact solution, you'll have to set up a formula for the bullet's trajectory and solve numerically.

Also, when you watch the bullet, it will appear to curve from your frame of reference (it's really you who is turning, but it looks like the bullet's direction is changing), which can make it difficult to learn to fire from a circling vehicle.

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  • $\begingroup$ Gunships adjust fire by seeing the impacts, not by watching the bullet trajectory - which can't be observed anyway, since they don't use tracer rounds. Also, everything is about computing an impact for pressing the trigger "right now" rather than at some defined time in the future. $\endgroup$ – Ralph J Feb 9 at 5:06
  • $\begingroup$ @RalphJ OP expressed it in terms of firing early. $\endgroup$ – Acccumulation Feb 9 at 5:12
  • $\begingroup$ Good points here in this answer. Took almost 3 years for anyone to get to this! This is highly relevant to points raised in the body of the original question, if not in the actual title. $\endgroup$ – quiet flyer Feb 9 at 14:24

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