Page 2 of 3 FirstFirst 123 LastLast
Results 11 to 20 of 23

Thread: Coriolis effect! And it's application to corrections/dial-in's

  1. #11
    Master Sergeant hiddenmongoose's Avatar
    Join Date
    Aug 2010
    Location
    Roscommon.
    Posts
    556
    Rep Power
    9

    Default

    Pointless worrying about it unless your shooting roughly 1400m or more.even then a 1 mph error in judgment of the wind call has a greater effect in most rounds on target vs coriolis.when you start shooting 1400m regularly and have your wind calls down to 1 Mph(that includes downrange and not at the muzzle where your standing with your kestral,and 20 odd feet vertical off the ground as that is where the bullet is living)then it is seriously worth worrying about.Wind calls and trigger control will wash out coriolis affects the majority of the time.I would worry bout the basics first!

  2. #12

    Default

    Very valid point but I'm simply interested in it and want to understand it better! There is absolutely no harm in knowing about it and understanding the effect!

  3. #13
    Master Sergeant belowaverageiq's Avatar
    Join Date
    May 2009
    Posts
    542
    Rep Power
    10

    Default

    Quote Originally Posted by hiddenmongoose View Post
    Pointless worrying about it unless your shooting roughly 1400m or more.even then a 1 mph error in judgment of the wind call has a greater effect in most rounds on target vs coriolis.when you start shooting 1400m regularly and have your wind calls down to 1 Mph(that includes downrange and not at the muzzle where your standing with your kestral,and 20 odd feet vertical off the ground as that is where the bullet is living)then it is seriously worth worrying about.Wind calls and trigger control will wash out coriolis affects the majority of the time.I would worry bout the basics first!
    100%
    Plus, simply zero your scope marginly left of aim (as in cutting just left of the line or 1/8th moa) point at 100yrds and this will take care of spin drift / coriolis.

  4. #14
    Corporal tadhgocuilleain's Avatar
    Join Date
    Dec 2012
    Location
    Galway
    Posts
    115
    Rep Power
    3

    Default

    Quote Originally Posted by zxthinger View Post
    Just focusing on the shift to the right for a moment!

    Is the shift to the right constant then for all azimuths? It seem that this is what Im picking up from your angular momentum description!
    Yes, by my understanding of it.

    To be honest, I have a .22LR so I don't need to worry about making corrections for anything except gravity and wind, so I only started to look into it when you asked the question. Like you, I always assumed coriolis only affected north or south travelling objects.

    I haven't done out the maths yet to prove it mathematically to myself that the drift to the right (or left in southern hemisphere) is constant for all azimuths, but from what I read, it seems to be.

    Imagine an ideal, theoretical world, where there's no such thing as wind or air resistance. You have an air balloon that always hovers 10m above the earth's surface. You push it towards the north. Because there is no air resistance or anything else, it will always travel at a constant speed. As it starts to move north, coriolis causes it to start drifting toward the right, as we know. As it starts to drift to the right, it is now travelling less to the north and a little bit to the east. So there is less drift due to it getting closer to the axis of rotation (as it moves north), but more drift due to its increase in angular velocity (as it starts to move towards the east). On its new, slightly easterly heading, it continues to drift to the right, so eventually, as it continues to drift to the right, it will be travelling due east. At this point, it is not moving north or south, so it is not getting any closer to the axis of rotation, but it is moving due east and as I explained in the earlier post, it will drift to the right due to its angular velocity being lower relative to earth's. As it drifts right, it now starts moving south. This continues until it has moved in a full circle, back to where it started. It a theoretical world with no drag or other external forces, it will continue to travel around in circles forever.

    No matter what point on that circle you started at, it will always travel in that circle. In other words, no matter what the azimuth, it will always drift to the right. If it wasn't a constant drift, it would spin inward or outward in a spiral.

    That all assumes a perfect, ideal world. In the real world, the earth isn't a perfect sphere, it's fatter at the equator; there are trees, wind, air resistance, elevation etc that all affect the angular momentum of the bullet. But ballistic calculators help take the majority of these into account. As mentioned, 1mph error in wind estimation will be far more significant than coriolis effect but, as you said, it is interesting to understand what's happening and why. I've already had hours of fun trying to figure it out since you posed the question.

    EDIT:
    This video has a good explanation in it, although it doesn't explain why the drift is independant of azimuth. I found the part where it shows the baseball moving in a circle the most interesting, because it's a visual demonstration of what happens:

    http://www.youtube.com/watch?v=aeY9tY9vKgs
    Last edited by tadhgocuilleain; 12-03-2013 at 00:15. Reason: Inserted link to video

  5. #15

    Default

    Quote Originally Posted by zxthinger View Post
    So I'm stumped by the fact that the horz correction(or at least the few that i looked at) takes no account of bearing


    This thread is a couple of weeks old now, but maybe it's still being watched by some.


    As noted by several people, the rotation-of-Earth effect will always be swamped by other effects, so there is no practical value in trying to take it in into account.

    The question posted is: how to understand that the effect is (to a good approximation) the same in all directions?

    I will of course use the simplification of disregarding air friction.




    The first thing to recognize is that during its flight the mechanics of the bullet trajectory is orbit mechanics. During its flight the bullet moves like a miniature satellite, in orbit around the Earth's center of mass. It's just that the orbit is very, very close to the Earth's surface.

    Every satellite orbit is planar. If the satellite is high enough and fast enough to go all the way around the orbit loops back onto itself, and that orbit is in a single plane. The groundtrack of that orbit is a circle that loops around the Earth. The center of the Earth is in the plane of that circle. In other words, the groundtrack of a satellite is a Great Circle.

    Once a bullet is in flight it is no longer affected by the Earth's rotation. Once in flight the bullet is on a particular orbit trajectory, and the plane of that orbit has a fixed orientation.
    The total starting velocity of that orbit trajectory is the sum of two components:
    - The velocity of co-rotating with the Earth before the bullets is fired
    - The muzzle velocity.

    The clearest case is at the equator. A point on the equator moves east at about 465 meters per second. So if you were to take a gun that fires its bullets at about that velocity, then firing due east the two components add up, but firing due west the two components drop away against each other. (Modern rifles have a far faster muzzle velocity than that of course, but still shooting with the Earth's rotation or shooting against it does make a difference.)

    Finally: the line of sight as a shooter takes aim is also a great circle.


    Once a bullet is in flight the Earth rotates underneath that orbit, and the original line of sight of taking aim rotates with the Earth underneath that orbit.



    There is a Java applet on my website that you can use to explore these things. You can set up any latitude, any velocity, any bearing, any elevation, and any rotation rate for the 'planet'. The applet computes and shows the trajectory. You can set a very low value for gravity, so that the projectile will move slowly.

    The applet shows two perspectives: a stationary point of view, so that you can see that the orbit lies in a plane, and co-rotating perspective, to see the deviation due to the rotation effect:
    The applet is called Ballistics and orbits

    The Java applet is available as applet-in-a-webpage, but it's also available for download as standalone applet. (The webpage-applet requires the Java browser plug-in, the standalone applet requires Java, but not a browser to run.)




    tadhgocuilleain mentioned another case, an air balloon.

    I need to point out that the mechanics of the motion is different in that case.
    The weight of an air balloon is carried. A balloon is buoyant; its weight is carried by the air it is immersed in. In turn the weight of the Earth's atmosphere rests on the Earth.

    Because of that the motion of an air balloon with a velocity with respect to the Earth is not like satellite orbit mechanics.
    (Motion of buoyant objects (including the atmosphere itself) is also discussed on my website. It's the subject of the article Inertial oscillations )
    Last edited by Cleonis; 27-03-2013 at 22:47. Reason: fixed an ypto

  6. #16
    Corporal tadhgocuilleain's Avatar
    Join Date
    Dec 2012
    Location
    Galway
    Posts
    115
    Rep Power
    3

    Default

    Quote Originally Posted by Cleonis View Post
    The first thing to recognize is that during its flight the mechanics of the bullet trajectory is orbit mechanics.
    Orbital mechanics only deals with the effect of gravity on the trajectory of the projectile, does it not?

    In your above post, you do not refer to Coriolis at all, that I can see.

    Quote Originally Posted by Cleonis View Post
    Once a bullet is in flight it is no longer affected by the Earth's rotation
    This statement is misleading. It is true that the surface of the earth no longer has an effect on the bullet as it travels, but if we disregard air friction, the projectile will continue to have the same rotational velocity as the point on earth from which it was fired (until Coriolis changes it, of course).

    Quote Originally Posted by Cleonis View Post
    The clearest case is at the equator. A point on the equator moves east at about 465 meters per second. So if you were to take a gun that fires its bullets at about that velocity, then firing due east the two components add up, but firing due west the two components drop away against each other. (Modern rifles have a far faster muzzle velocity than that of course, but still shooting with the Earth's rotation or shooting against it does make a difference.)
    This is Eötvös effect and only applies to shooting due east or due west.

    Quote Originally Posted by Cleonis View Post
    Finally: the line of sight as a shooter takes aim is also a great circle.
    Can you please explain that?
    As I see it, line of sight is a straight line from the shooter to the target. The corrections a shooter dials in cause the bullet to travel in a curve (caused by various forces such as gravity, gyroscopic drift etc), which intersects the line of sight at the target (hopefully ). This curve usually travels to the left and high (in the northern hemisphere), before curving back to the target. This would mean the bullet's path is actually a small circle, as the plane does not pass through the centre of the earth.

    Quote Originally Posted by Cleonis View Post
    Once a bullet is in flight the Earth rotates underneath that orbit, and the original line of sight of taking aim rotates with the Earth underneath that orbit.
    If I understand what you are trying to say correctly, that would mean that the POI of a bullet would fall to the left of the target when shooting north, and to the right when shooting south, which isn't the case. If I misunderstood your intention, can you please elaborate?

    Quote Originally Posted by Cleonis View Post
    tadhgocuilleain mentioned another case, an air balloon.

    I need to point out that the mechanics of the motion is different in that case.
    The weight of an air balloon is carried. A balloon is buoyant; its weight is carried by the air it is immersed in. In turn the weight of the Earth's atmosphere rests on the Earth.

    Because of that the motion of an air balloon with a velocity with respect to the Earth is not like satellite orbit mechanics.
    I did qualify the example by saying that we were ignoring the effects of air. I simply used the example of an air balloon as a way to visualize something that was not affected by earth's gravity. Of course the flight of an air balloon is a lot more complex than the simplified version I used.
    Last edited by tadhgocuilleain; 28-03-2013 at 08:34.

  7. #17
    First Lieutenant Spunk84's Avatar
    Join Date
    Feb 2011
    Location
    Galway beside John
    Posts
    1,767
    Rep Power
    16

    Default

    Ive just lost 5 mins of my life reading that, i will never regain those 5 minutes of my life

    im still confused to why you dont add eggs in that recipe!
    I don't suffer from insanity; I enjoy every minute of it!

    I love Halloween, the one time of year when everyone wears a mask, not just me. People think its fun to pretend you’re a monster. Me, I spend my life pretending I’m not.

    http://www.irishshooter.com because i cant get in trouble for it here

  8. #18

    Default

    Quote Originally Posted by tadhgocuilleain View Post
    In your above post, you do not refer to Coriolis at all, that I can see.
    The reason for that is that I put the focus on the motion of the bullet as seen from a stationary point of view. I compare it to looking from space to a satellite orbiting the Earth. The satellite orbits, the Earth rotates underneath that orbit.

    Conversely, when the focus is on the motion as seen from a co-rotating point of view the deviation can can be expressed (in the equation of motion) as arising from a Coriolis term.



    Quote Originally Posted by tadhgocuilleain View Post
    Quote Originally Posted by Cleonis View Post
    Once a bullet is in flight it is no longer affected by the Earth's rotation. Once in flight the bullet is on a particular orbit trajectory, and the plane of that orbit has a fixed orientation.
    This statement is misleading. It is true that the surface of the earth no longer has an effect on the bullet as it travels, but if we disregard air friction, the projectile will continue to have the same rotational velocity as the point on earth from which it was fired (until Coriolis changes it, of course).
    Let me repeat what I wrote in my first post:
    The total starting velocity of that orbit trajectory is the sum of two components:
    - The velocity of co-rotating with the Earth before the bullets is fired
    - The muzzle velocity.

    Example:
    A bullet is fired due north, from the equator.
    For simplicity let's say the muzzle velocity is 465 meters per second, the same as the velocity of the equator. Pythagoras' theorem gives the total resultant velocity; 657 meters per second, and the motion (as seen from a stationary point of view) will be at an angle of 45 degrees wrt the equator.

    Of course, as seen from a co-rotating point of view the bullet is seen to travel in the direction that it was fired.


    So of course there is a lasting effect of the rotational velocity of the Earth at the latitude from where the bullet is fired. It's just that once the bullet is on its way with the resultant velocity it is otherwise not affected by the Earth's rotation.



    About the line of sight.
    I need to assume here that the description in the opening section of the wikipedia Great Circle article is read and understood
    On a sphere the line of sight from one point to another is parallel to a great circle through those two points. The line of sight is a section of a great circle.


    Given your response, I think I need to re-confirm the direction of the effect.

    At this level of simplification we have that the deviation (as seen from a co-rotating point of view) is the same for all compass bearings. As we know, the effect is too small, and it will always be swamped, but if it would be a detectable effect then when a rifle is dialed in with the bullets fired south-to-north, that correction will work fine for all other directions.
    Last edited by Cleonis; 28-03-2013 at 17:17. Reason: More precise formulation

  9. #19
    Team Irishshooter dundee's Avatar
    Join Date
    Oct 2009
    Location
    kilkenny
    Posts
    106
    Rep Power
    6

    Default

    Talk about a load of s!!!!goggle must be working over time!!!

  10. #20

    Default

    Quote Originally Posted by dundee View Post
    Talk about a load of s!!!!goggle must be working over time!!!
    Just because you have no interest in it doesn't mean others don't I don't have any interest in it either so I just refrain from posting

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •