# If a circling airplane (AC-130) fires a bullet how does the plane's trajectory alter the shell's flight path?

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!

• 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. – kevin Mar 27 '18 at 0:18
• I don’t believe they do compute a lead as there is no relative translation between the aircraft and the pylon. – Carlo Felicione Mar 27 '18 at 2:01
• 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. – Jim Garrison Mar 27 '18 at 4:42
• 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. – Trevor_G Mar 27 '18 at 16:45
• @Trevor_G No... there's no 'centrifugal force effect' at all, since that 'force' disappears the moment the shell is shot. – xxavier Jul 10 '19 at 15:36