Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
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
add comment

1 Answer

The cumulative Coriolis effect must be independent of the path between the two end-points of the flight segment to which it applies. The Coriolis effect is entirely due to the motion of the observer and not to that of the object. Direct integration of the instantaneous Coriolis acceleration over a curved trajectory introduces error in the resulting Coriolis effect to the same extent that it introduces path-dependence. That commonly used approach only works for straight-line trajectories. My path-independent approach to calculating the Coriolis effect (described in the question), a direct integration of Coriolis acceleration approach, and, indeed, a "conservation of angular momentum" approach to calculating the eastward deflection of an object that is carefully released to fall within a vacuum tower 50 meters in height located at sea-level on the mean equator all produce the same answer: 11.643 mm east of plumb. I used 6,378,137 m as the earth-centric radius to the base of the tower, 9.8066 m/s/s as the effective acceleration of gravity (including the "centrifugal acceleration" effect), and 7.2921 times ten to the minus five rad/s as the rotation rate of the earth. All calculations are carried to five significant figures. It might help in seeing all of this to consider the instantaneous Coriolis acceleration (a function of the instantaneous apparent velocity of the body) as the limiting case of a time-averaged Coriolis acceleration (a function of the object's time-averaged apparent velocity) over shorter time and distance intervals. The time-averaged apparent velocity of the object is just the apparent spatial displacement vector of the object divided by its temporal displacement between the two state vectors bounding the flight segment being considered. My tech note on this has grown to 5500 words and is available upon request. Addendum: I just worked out the example "tower drop" problem much more carefully. The eastward deflection is proportional to the 3/2 power of the height of the drop or to the cube of the fall time. I used a hybrid height-times-fall-time, but it is trickier. Both a "conservation of angular momentum" approach and a "direct integration of Coriolis acceleration" approach yield 2/3 of my previous result, or 7.7620 mm. The difference would vary with other types of problems. This counter-example blows up my theorem of "independence of path" for the Coriolis effect. The integration of the Coriolis acceleration is still the only way to go in the general case. The tech note mentioned has been withdrawn. "Nevermind..." Jim

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.