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Coriolis Effect on Projectiles

  1. Jun 21, 2007 #1
    I've been searching all over the web and on this forum for an answer, and I haven't found it (or I may have found it and not understood...). If this is a re-hash, I'm sorry....

    I'm working on some ballistics calulations for long range rifle shots. I've got pretty much everything worked out except for the Coriolis Effect. It may have such a small impact it's not worth dealing with, but I need to know. I've never done rotational dynamics like this, so I'm a little out of my element.

    I know that the main equation is Fc = -2m(WxV) where F is the end force, m is the mass of the item in movement, W is the angular velocity of the frame of reference, and V is the velocity of the item in movement. That's pretty much all I understand....

    Here's my example: I'm sitting at a laditude of 36.728 North and I'm firing a 750 grain 0.510 caliber bullet at 4000 fps straight North. It goes a distance of 1760 yards (one mile) in 2.082 seconds. It's traveling 1607 FPS when it get's there. How many inches has it been pushed off course due to the Coriolis effect (pushed to the right I believe)?

    I've read that in the northern hemisphere the pull is always to the right. Is this correct, even if I'm firing Northwest for example? I need to know if I have to adjust for direction of fire as well as longitude....


  2. jcsd
  3. Jun 21, 2007 #2
    I understand the concepts, I'm just having trouble putting math to it....

    I know that the velocity of the latitude I mentioned before (36.728 North) is 622.934 miles per hour (25000 mile circumference of earth * sin (latitude) / 24 hours). 622.934 MPH is 913.6365 fps. 750 grains is 0.1071429 lbs.

    Using the equation I mentioned before (Fc = -2m(WxV)), does this mean the force is:

    -2 * 0.1071429 * (913.6365 * 4000) = -783,117.3 lb-ft^2/sec^2

    This can't be right.....

  4. Jun 21, 2007 #3


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    Homework Helper

    By WxV they mean the cross product, which is not the same as the ordinary product of the lengths unless the vectors are perpendicular, which they aren't here.
  5. Jun 21, 2007 #4
    Ok. That explains why I got such a big number.... Thanks.

  6. Jun 21, 2007 #5
    I've looked at this, and it still doesn't make sense. If someone can help, I'd appreciate it....
  7. Jun 21, 2007 #6


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    You're gonna have to clarify what you mean by "it doesn't make sense." If you ultimately want to find the total displacement, you need to integrate the acceleration = force/mass twice.
  8. Jun 21, 2007 #7
    I'm at the same point I was when I put up the first post. The vectors are perpendicular. The bullet is going North, and the rotation is "East". Using the right hand rule, this puts the new vector straight up, which makes no sense....

    Using the data in my example, can someone show me how the formulas go together? I've found a lot of information on the web about what the effect is, but I've found no practical applications. If I see the math, I can probably make it work, but I haven't yet.

    Last edited: Jun 21, 2007
  9. Jun 22, 2007 #8
    In [w, v], w is the angular velocity of the Earth which is directed from South to North, along the rotation axis, using the right hand rule. You're confused with the direction of the rotation, that's why you ended up with the upward vector :-)
  10. Jun 22, 2007 #9
    OK, now I'm really confused. If the vecor is from South to North, then it's parallel with my velocity vector. The Sin of 0 is 0 so there'd be no acceleration due to this effect? I know that's not true....

    To me, the angular velocity vector should be "up". This would put the new vector to the right of my velocity vector, which would make sense. The only problem with this is that the angle between them will always be 90, so the vector is the same regardless of direction.

    This would also mean that my previous math is correct because you just multiply by the Sin of 90, which is 1. Something doesn't jive....

  11. Jun 22, 2007 #10
    I still don't know the direction issue, but I think I may have figured out part of the math issues. I was using the linear velocity of earth instead of the angular velocity. The angular velocity of earth is 4.17 x 10^-3 deg/sec. Now the acceleration comes to:

    ac = -2 * 4.17 x 10^-3 * 4000 * sin(deg) = -33.36 * Sin(deg) deg-ft/sec^2

    It still doesn't look quite like acceleration because degrees are still in there, but it's closer. I can believe 33. If this is right, it's about the same as the pull of gravity....

    I just had a thought on this. I know the effect is dependent on the latitude you're at. The angular velocity will be the same regardless of latitude won't it?

    Last edited: Jun 22, 2007
  12. Jun 22, 2007 #11


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    No, the earth's angular momentum vector is not parallel with the bullet's velocity vector, except at the equator. The earth's angular momentum vector is along the earth's axis. To get the direction of the bullet's velocity vector with respect to the earth's axis, consider that the plane of the horizon at the gun's location is at an angle with respect to the earth's axis, depending on the latitude.
  13. Jun 22, 2007 #12
    OK. At the equator the effect is 0, but it gets bigger as you near the poles.... That makes sense.

    So, if I fire at a direction other than north or south, will I have multiple angle vectors to take into account? If I fire East for example, I'd be at 90 deg to the axis of the earth. But firing NE, I'd have components of both.... If I do have multiple vectors, how do I figure the effects?


  14. Jun 22, 2007 #13
    That is 7.27 x 10^-5 sec^-1 (1 deg ~ 0.0175 rad).

    Simply split velocity into "north" (angle to Earth's axis depending on latitude) and "east" (always 90 deg to axis) components, and compute and work with Coriollis forces for both. You can do this because vector product is distributive, i.e. [tex](\vec{v}_{north}} + \vec{v}_{east}) \times \omega = \vec{v}_{north}} \times \omega + \vec{v}_{east}} \times \omega[/tex].

    The "north" force component will push sideways (to east if you fire north, to west if you fire south), and the "east" component will push upwards (that's why they like to blast off eastbound near equator).

    Chusslove Illich (Часлав Илић)
    Last edited: Jun 22, 2007
  15. Jun 24, 2007 #14
    OK, I put the pen down and got the result. You have to suppose that Coriolis effect is small. It is because the angular velocity of the Earth is small, 1 turn a day. Maybe the below picture helps you.
  16. Jun 25, 2007 #15
    Thanks Chusslove. From that, I'm guessing that based on my values above, I'd get:

    A_North = -2 * V_North * V_Earth * sin(lat)
    A_North = -2 * 4000 fps * 7.27 x 10-5 rad/sec * sin(36.728)
    A_North = 0.3478 rad-ft/sec^2 "Up"

    A_East = -2 * V_North * V_Earth * sin(direction of fire)
    A_East = -2 * 4000 fps * 7.27 x 10-5 rad/sec * sin(0)
    A_East = 0

    If I shifted my firing direction "East", I'd get:

    A_North = -2 * V_North * V_Earth * sin(lat)
    A_North = -2 * 4000 fps * 7.27 x 10-5 rad/sec * sin(36.728)
    A_North = 0.3478 rad-ft/sec^2 "Up"

    A_East = -2 * V_North * V_Earth * sin(direction of fire)
    A_East = -2 * 4000 fps * 7.27 x 10-5 rad/sec * sin(90)
    A_East = 0.5816 rad-ft/sec^2 "South"

    I'm not real sure on the A_North when firing other than North. Do I just do a sin of the firing angle to get the effect, or will it always be the same like I put here? What do I need to do to loose the radians, or do I just ignore them?

    Weimin, I think you forgot the picture....


  17. Jun 26, 2007 #16
    Hmm, for some reason, the attached did not show up. I do it again.

    I suppose that the up-north speed is constant since the Coriolis effect is small. So the force towards the East is also constant. The trajectory of the bullet can be treated as a thing being thrown horizontally with a small gravity.

    But look at the values you gave,I wonder why the bullet is slow down too much after 2 sec.

    Attached Files:

    • pf1.GIF
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  18. Dec 27, 2010 #17
    It just looks like it would be about 11 inches. I don't think your math has anything to do with it. I again think your premise is all wrong. Look at the 2 min youtube video at Google search, "Coriolis effect".
  19. Dec 31, 2010 #18
    Legoman - These are some of the items that 'I think' about your question. 1. It is really impossible to come up with an answer except an estimate. For constant velocity things a reasonable result can be calculated, where a fired projectile is constantly changing velocity, so some empirical data are usually required to start with, like using a 'Boulenge' chronograph and even then I don't think that a constant velocity will give the same results as with fired projectiles average velocity or time of flight. 2. The weight or caliber of the projectile has nothing to do with it, unless you are trying to incorporate the exterior ballistics into the problem. It looked like to me you may have confused 'F' with the projectile force. Notice in your formula 'F' had a sub 'c'. This is Coriolis Force. 3. Basically it will make no difference in which direction you will be firing-the amount of deflection will be about the same. Notice I said 'basically'. This is because theoretically the defections will be the same for only 90 degrees and 270 degrees. This is because only at these two angles will the latitude remain somewhat constant. Can you see where the latitude will constantly be changing at any other angles. But I don't think you want to get in to this correction (it would be so small), so for all practical purposes lets just assume the same latitude...and here again I think you were somehow thinking that the deflection depended on the firing compass angle-not true, the basic deflection will be the same for any firing angle (compass angle), okay. 4. Years ago I made some charts for an example projectile firing. I would put the image in a post if I figured out how. 5. Here is an example of the estimate for your one mile range with a .308 Winchester, Muzzle velocity of 2860 fps-10.9 inches at 45 degrees latitude, and of course the amount of the defection will be the same in either the north or south hemisphere, just the direction will be opposite. I have the charts and deflection amounts but I don't have the actual trig work that I did at the time. I am trying to come up with a formula where we can just stick in figures and crank out an answer-I will post it if and when...
    Last edited: Dec 31, 2010
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