# How should a closed-ended terrestrial trajectory be corrected for the Coriolis effect?

I have tried verifying the numerical integration of the Coriolis effect for 1000 to 2000-yard rifle fire by switching ON/OFF the Coriolis correction of a good ballistic simulator program (PRODAS). The program integrates an instantaneously evaluated Coriolis acceleration along with the aerodynamic and gravitational accelerations. My calculations yield about 10 percent larger Coriolis effects than the delta after the same number of timeslice intervals are computed in two otherwise identical simulation runs. I theorize that the integrated Coriolis effect should be independent of the path between the two end-points of any segment of a terrestrial trajectory. If so, the velocity used in the instantaneous Coriolis acceleration can be replaced with the displacement vector of the projectile divided by the time-of-flight over the flight segment. The displacement is just the vector difference between the projectile positions at the ends of the flight segment. And the cumulative Coriolis effect over a segment becomes just the displacement vector crossed with the earth rotation rate vector multiplied by the time-of-flight. I speculate that the directly integrated "instantaneous Coriolis force" produces some along-track component that should not be allowed. An additional "normalizing constraint" is needed. The constraint should be perpendicularity to the displacement vector as above. I can e-mail a 3200-word tech note about this question as an attachment in Word or PDF file format to anyone who would like it.

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A projectile is just in a low orbit, and the earth turns under it. Like if shot toward the north pole from a high latitude, the earth's surface is moving slower as you get toward the pole, so the projectile will seem to be displaced toward the right. Aside from windage, I don't see the big deal. – Mike Dunlavey Nov 23 '11 at 1:35
@MikeD The vector cross-product also produces a vertical Coriolis effect that is max upward when firing eastward. It really is no big deal unless you need to disable an enemy battleship 25 miles away. We might have been calculating the Coriolis effects incorrectly since the dawn of digital computing.--Jim Boatright – James Boatright Nov 23 '11 at 4:22
Isn't that accounted for by simply modeling the motion of the earth? My understanding of coriolis acceleration is it only applies to bodies if their motion is considered to be constrained, like dropping an elevator. Like with weather patterns, you could say a northward-traveling air mass (in the northern hemisphere) experiences an eastward acceleration (trade wind), or you could just say the earth underneath it is moving eastward more slowly. Keep it simple, no? – Mike Dunlavey Nov 23 '11 at 13:48
@MikeD No the object's motion neeed not be constrained. The Coriolis effect is due entirely to the motion of the observer in his rotating frame of reference. The only reason the object needs to be moving is so that the Coriolis effect is not identically zero. The Coriolis effect in terrestrial ballistics is very real, but there is no such thing as a "Coriolis force." I believe that there is a good academic paper here for anyone who wants to work on it.--Jim – James Boatright Nov 23 '11 at 18:54