1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Coriolis effect on eastward\westward movement on earth

  1. Jan 12, 2014 #1
    Given an example of a rifle bullet traveling east or west, why does the bullet deviates HORIZONTALLY? (right on the northern hemisphere, left on the southern hemisphere)

    As rotation speed in the same latitude coordinates stays roughly the same, I wouldn't expect the bullet, or any other object for that matter, to deviate horizontally, as it does when traveling north or south. Could someone explain this?

    Thanks :)
  2. jcsd
  3. Jan 12, 2014 #2


    Staff: Mentor

    the deviation is due to latitude lines not being great circle lines aka the straight lines on a spherical surface. The bullet travel in a plane perpendicular to the ground affected only by gravity and air resistance.

    You see this on a globe if you plot a straight it will deviate away from the latitude line unless its equatorial line. Similarly a bullet shot east or west on the equator wont deviate because it too is a great circle line.

  4. Jan 13, 2014 #3

    Oh, I understand. However, how come the deviation caused by those latitude lines curving is exactly the same as caused by the velocity differences? Is it coincidence or is it the nature of all spheres?
  5. Jan 13, 2014 #4


    User Avatar
    Science Advisor

    The deviation from a "straight line" path due to the fact that lines of latitude are curved is not the same thing as the Coriolis deflection.

    In the Northern hemisphere, the Coriolis deflection is always clockwise. In the Northern hemisphere, the deviation of a great circle path from a line of latitude is either zero, clockwise or counter-clockwise depending on whether you are facing north, east or west respectively.

    For me, the easiest way to intuit the action of the Coriolis effect in the middle latitudes is to first envision the Coriolis effect at the North pole (full effect) and at the equator (no effect) and to see that in the middle latitudes the effect should be somewhere in between.

    The conundrum in the OP was:

    The rotation speed stays the same if you are looking east or west. But the direction changes. If you look toward the east, the land in the distance is moving down and to your left. If you look toward the north, the land in the distance is moving to your left. If you look toward the west, the land in the distance is moving up and to your left. If you look toward the south, the land in the distance is moving to your left. Ignoring the up and down components because you are only asking about horizontal deviation, the effect is always to move the land leftward so that a bullet travelling in a straight line path will appear to be deflected rightward.
  6. Jan 13, 2014 #5
    Thanks! Got it. However, looking at some sniper data, it seems bullets will deviate at a given point exactly the same amount/distance, when either shooting towards the east/west or north/south.
    My followup question is - why? As you said, the latitude curving is not the same as the coriolis effect, so how come they effect the bullet exactly the same?
  7. Jan 13, 2014 #6


    User Avatar
    Science Advisor

    As I said, they don't have the same effect. Latitude curving will cause an apparent southward deviation. Coriolis will cause an apparent leftward deviation.

    Perhaps your question is why the magnitude of the Coriolis effect is the same as latitude curving. A moment's thought leads me to conclude that it is not the same. The Coriolis effect scales with the rotation rate of the planet. Latitude curving scales with the size of the planet. There would be no reason to expect the two to be identical.

    jedishrfu's response was simply incorrect.
  8. Jan 13, 2014 #7
    Yes, I'm sorry for not being clear. My question is about the effects magnitude.

    Looking at sniping sheets, the horizontal deviation is calculated as follows:
    HDeviation = (Ω * Rangeft2 * sin(latitude)) / VelocityAvg

    Direction of shooting is not taken into account. Is it merely a coincidence that on earth, the difference between the effects magnitude can be negated and considered roughly the same?
  9. Jan 13, 2014 #8


    User Avatar
    Science Advisor

    I think that you are really asking about the magnitude of the Coriolis deflection when shooting east/west versus north/south. This is reaching the boundaries of what I feel competent to explain clearly, but I'll give it a shot.

    When shooting north/south, the resulting Coriolis deflection is purely in the east/west direction. The relevant component of the muzzle velocity is the component that goes inward, toward the Earth's axis of rotation. The component of the muzzle velocity that goes up and north along the axis of the earth is irrelevant. So there's the sin(latitude) term for a north/south shot.

    When shooting east/west the muzzle velocity is at right angles to the axis of rotation so the Coriolis acceleration is maximized. But the resulting deflection of the shot has both a north/south and an up/down component. You are concentrating on windage and ignoring elevation, so the up/down component may be ignored. So there's the sin(latitude) term for an east/west shot.

    In between, you get a combination of both and still get a net sin(latitude) term.
  10. Jan 14, 2014 #9
    I still don't quite get it. Perhaps I don't quite understand your explanation. I mean, both forces' magnitude are dependent on latitude, and since the sin(latitude).

    However, I would expect the other values in the equation to be different and calculated separately for each force - Ω, for example, which value is 0.0000729, as far as I know, is related to the Earth rotation speed. How does it relate to its spherical curve? I guess it is possible the equation was simplified for sniping purposes, but it still means the forces magnitudes are very similar...
  11. Jan 14, 2014 #10


    Staff: Mentor

  12. Jan 14, 2014 #11

    Let's say we have a rifle with a mile long barrel. We are trying to shoot a hole on a piece of tape covering the other end of the barrel.

    It's impossible to miss, so we know that coriolis force is countered by a force exerted by the barrel.

    Regardless of the direction of the barrell, the barrel turns 180 degrees in 12 hours, which turning rate dictates the force that the barrel exerts on the bullet, which force is opposite to the coriolis force.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook