Question on Applied Ballistics App and Coriolis

[KYpatriot]

I was just about to weigh in with this exact "north pole" explanation, but I saw that you had already done so. Great post.

To add to this description just a little bit- I believe one reason for this effect, is that the shooter/bullet trajectory is perpendicular to the surface of the earth (parallel to the axis of gravity), regardless of geographical location, and the only place on earth that this perpendicular axis is completely parallel to the rotation axis of the earth, is shooting in an east/west direction, right at the equator. Here's a quick sketch that was drawn by my kid in grade 1 (cough), illustrating this point. At the equator when shooting E-W, there is no horizontal component of the earth's motion, relative to the trajectory of the bullet. When shooting the same direction further north, the axis' are no longer completely parallel, introducing a horizontal component to the earth's rotational motion, relative to the axis of the bullet trajectory (parallel to gravity's axis).

The circle is the earth, the black lines intersecting the earth's axis' represent the axis of the poles, and the equator. The red triangle and circle are the shooter's position, and the red arcs are the bullet's trajectory at two different geographical locations. I made the northern trajectory's E-W shooting direction exaggerated to more easily illustrate the point. The black line intersecting the northern shooter and the center of the earth is the axis of gravity.

I agree with your equator shot description, there is n horizontal coriolis along the equator (which is why hurricanes cannot form there) but on the northern one I dont believe there is a gravitational effect to horizontal coriolis, coriolis in the horizontal is purely due to the geometry of a rotating sphere. I think you are veering into the eotvos effect, which some people also call "vertical" Coriolis, which is gravitational, and a the force of gravity on a projectile chamges as result of going "with" or "against" the earths rotation. The extreme example is a projectile fired with "escape" velocity; gravity has been offset completely.

By the way, that top path in red is very close to an illustration of a "great circle" route, the true shortest distance between two points on a sphere. Aircraft, being steerable, can and should take advantage of that and fly a curved path to their "target" to save fuel and time. Their trajectories are measured in hundreds or thousands of miles instead of yards so it actually makes a difference. Dont ask me to do the math but a great circle route, bending the tractory toward a pole instead of a straight line, over a thousand yard line of sight would probably only save a few inches.

 

[Timber338]

You guys are right on, and I know this is going to make me sound like a complete nerd, but I really am having fun thinking about this. you guys nailed some very complex thinking with very simple pictures and diagrams. I love that about physics. And even though this thread points out that I'm not that smart, I never cared about that from the beginning. I just like figuring this stuff out, and it really is fun to discuss. I tried to keep my snarky comments to a minimum, but I have a very sarcastic/snarky personality, so I appreciate how you guys refrained from calling me stupid because I really do understand this now from the diagrams and explanations you've posted.

 

[KYpatriot]

Hey if it wasnt fun I wouldnt do it, and the older I get the more I realize I wish I was as smart As I thought I was...

Anyway, I havent seen a model or any discussion of the fact that all these concepts hang on a truly independent projectile, ie a vacuum. We know that the earth drags part of the atmosphere with it because of friction. Thank goodness it does, 1200 fps winds would not be fun.

I'm just wondering if that is accounted for in actual coriolis calculations in any solver out there, the fact that the bullet is flying through a giant airmass that may have the effect of partially carrying the bullet along the same path as the target, muting the effect.

Maybe no one has bothered with it because we havent had projectiles with long times of flight that are repeatedly precise enough for it to matter. With our best non guided long range artillery having CEPs in the 50meter range or more, with first round hits not that critical (thats what firward observers are for lol) I wonder if they just said "close enough", like horseshoes.

 

[Timber338]

You very well could be correct with your thoughts on how the moving air mass might mute the effect. I bet that Bryan Litz could weigh in on that, he really has another level of understanding when it comes to all of this and how to apply the equations. He might be able to shed some light on how and why the solvers are set up, and maybe the difference for small arms is just too small and good enough like you have mentioned.

I also wonder if maybe you were to write up equations to model the entire system with spinning air masses, if it would be too heavy of a code for an app and/or smartphone to accomodate.

Either way, I'm not smart enough to figure all of that out, nor am I a good enough shot to make a difference. Also interesting that with your own shooting you have been more accurate with coriolis turned off. I don't think I have enough data/experience to make that call either way. My train of thought is to turn on Coriolis because I kind of inherently think that even if the math isn't perfect, it is far less flawed than my shooting ability and I think will get me closer to start with. That could very well be wrong, but at least is a starting point for me.

 

[KYpatriot]

Sure, and keeping a logbook with all your data will be useful, as well as a very good "tall target" scope test to make sure the clicks are consistent and repeatable. I hold windage with the reticle so I dont care so much about windage clicks, but if the scope doesnt elevate internally "plumb" or if it is canted in the mount, or the rifle is canted, some of the elevation is in effect converted to windage and your shot will if course be off in the direction of cant, and a little low. In my case, either my body position produces a consistent error or I just cant call wind good enough to see the 3-4 inches the solver predicts. Probably the latter, or a little of both.

 

[Jordan Smith]

I agree with your equator shot description, there is n horizontal coriolis along the equator (which is why hurricanes cannot form there) but on the northern one I dont believe there is a gravitational effect to horizontal coriolis, coriolis in the horizontal is purely due to the geometry of a rotating sphere. I think you are veering into the eotvos effect, which some people also call "vertical" Coriolis, which is gravitational, and a the force of gravity on a projectile chamges as result of going "with" or "against" the earths rotation. The extreme example is a projectile fired with "escape" velocity; gravity has been offset completely.

By the way, that top path in red is very close to an illustration of a "great circle" route, the true shortest distance between two points on a sphere. Aircraft, being steerable, can and should take advantage of that and fly a curved path to their "target" to save fuel and time. Their trajectories are measured in hundreds or thousands of miles instead of yards so it actually makes a difference. Dont ask me to do the math but a great circle route, bending the tractory toward a pole instead of a straight line, over a thousand yard line of sight would probably only save a few inches.

Sorry, I might have given the wrong impression. I'm not implying that there is a gravitational horizontal component to Coriolis, simply that the trajectory of the bullet is in plane with the axis of gravity, regardless of geographic location, while the rotating sphere only follows that same plane at the equator.

 

[Pdvdh]

I think you are veering into the eotvos effect, which some people also call "vertical" Coriolis, which is gravitational, and the force of gravity on a projectile changes as result of going "with" or "against" the earths rotation. The extreme example is a projectile fired with "escape" velocity; gravity has been offset completely.

I've been lurking on this Forum since 2005. This is by far and away the most informative Thread covering coriolis. It's much easier to activate coriolis functions in the ballistic program than it is to understand it. I find the understanding helpful when it comes time to entering the required data in my program.

Smart money is betting heavily that none of us pursued a Bachelor of Arts degree.

Eotvos indeed. I recall eotvos while researching coriolis 7-8 years ago, and it was a minor factor in the vertical component of coriolis. I believe the speed of the bullet, direction of travel with or against earth's direction of rotation, centrifugal force due to bullet speed, and gravitational force were all involved. The centrifugal force had a slight effect, additive or subtractive, from gravitational force. The reference article provided the mathematical expression of eotvos, as well as an associated narrative discussion. I tried to find that reference material last night on my PCs, but no such luck. I've replaced my computers since then and couldn't locate anything on the newer PCs using the search functions.

My understanding of horizontal coriolis for east/west directions of fire has been restored, thanks to the helpful visuals and discussion that's been provided. Just when I thought this Thread was dying out, it came back to life very positively. Appreciate the collective efforts. Good job to all!

 

[KYpatriot]

We are geeks with rifles lol.

 

[Pdvdh]

I got my hands on a globe. Looked at it last evening and could "sorta" visualize the horizontal coriolis, but I still wasn't getting the full perspective.

I just finished looking it over again and this time, bingo! If you have access to a globe, hold a planar straight edge perpendicular to the the surface of the globe, with east to west alignment, along the latitude lines printed on the globe. Then shift the planar straight edge from the equator up to the pole along the latitude lines, and watch how the latitude lines bend off at an increasingly greater angle from the straight edge as the straight edge approaches the pole (90* Latitude).

What made the huge difference for me was the perspective of holding the straight edge perpendicular to the globe (parallel to gravitational force) and then looking right along the planar straight edge - directly down onto the surface of the globe. The light bulb got much brighter! lightbulb

This perspective completely obliterated my prior mental block. And from this perspective, horizontal coriolis really is easier to visualize than vertical coriolis. Pretty sure I'll never ever have a problem visualizing the cause of horizontal coriolis - again.

I imagine it would be plain as day from the international space station, if latitude lines were visible on the ground surface of earth, and the earth rotated along its N-S axis.

 

[Jordan Smith]

I got my hands on a globe. Looked at it last evening and could "sorta" visualize the horizontal coriolis, but I still wasn't getting the full perspective.

I just finished looking it over again and this time, bingo! If you have access to a globe, hold a planar straight edge perpendicular to the the surface of the globe, with east to west alignment, along the latitude lines printed on the globe. Then shift the planar straight edge from the equator up to the pole along the latitude lines, and watch how the latitude lines bend off at an increasingly greater angle from the straight edge as the straight edge approaches the pole (90* Latitude).

What made the huge difference for me was the perspective of holding the straight edge perpendicular to the globe (parallel to gravitational force) and then looking right along the planar straight edge - directly down onto the surface of the globe. The light bulb got much brighter! lightbulb

This perspective completely obliterated my prior mental block. And from this perspective, horizontal coriolis really is easier to visualize than vertical coriolis. Pretty sure I'll never ever have a problem visualizing the cause of horizontal coriolis - again.

I imagine it would be plain as day from the international space station, if latitude lines were visible on the ground surface of earth, and the earth rotated along its N-S axis.

You got it! Sometimes verbal explanations of physical properties or principles can be pretty tough to imagine. Seeing it with your own eyes can make things come together and connect all the dots a bit easier than trying to visualize words

 

[Timber338]

Absolutely. Understanding the basic principals are not simple, but when you put your mind to something it's really not the brain drain we often expect it to be. I think the only downside to this forum is that it is a forum. There is a somewhat slow turn around in discussion. But we still got there with pictures to help. If we were all sitting around a table, the conversation would have taken about an hour.

after I read KYpatriots explanation, the 2-D paper still made it tough to see horizontal shooting east, but it was enough to make me grab the globe. That was the final key for me too. Now it's easy to visualize exactly what he was describing.

I think there's enough information in this thread for others to read it through and also understand Coriolis.

 

[Pdvdh]

Watching the globe rotate on the north/south poles axis, the reason the vertical component of coriolis drift is largest at the equator, and null and void at the poles, also become obvious - in a manner not likely to be forgotten anytime soon.

The equator is the farthest radius from the north/south axis. When the earth spins on that axis, the surface of the earth is rotating at the highest rate of speed at the equator - where the earth's surface is the farthest from the axis of rotation. Which means your target/game animal travels a greater distance between the time the bullet leaves the muzzle, and the bullet strikes the target/game animal.

It'll now be much easier for me to remember the primary affecting aspects of both both components of coriolis. If I find myself fuzzy over horizontal and vertical coriolis in the future, I'll just remember the rotating globe.

The eotvos component of vertical coriolis is pretty minor - not worth researching, unless you're a real nerd/geek. For all you nerds/geeks, it should be easy to follow this explanation from Wikipedia!

"The Eötvös effect is the change in perceived gravitational force caused by the change in centrifugal acceleration resulting from eastbound or westbound velocity. When moving eastbound, the object's angular velocity is increased (in addition to the earth's rotation), and thus the centrifugal force also increases, causing a perceived reduction in gravitational force.

In the early 1900s (decade), a German team from the Institute of Geodesy in Potsdam carried out gravity measurements on moving ships in the Atlantic, Indian and Pacific Oceans. While studying their results the Hungarian nobleman and physicist Baron Roland von Eötvös (see Loránd Eötvös) (1848–1919) noticed that the readings were lower when the boat moved eastwards, higher when it moved westward. He identified this as primarily a consequence of the rotation of the earth. In 1908 new measurements were made in the Black Sea on two ships, one moving eastward and one westward. The results substantiated Eötvös' claim. Since then geodesists use the following formula to correct for velocity relative to the Earth during a measurement run.

a_r = 2 \Omega u \cos \phi + \frac{u^2 + v^2}{R}.

Here,

a_r is the relative acceleration
\Omega is the rotation rate of the Earth
u is the velocity in latitudinal direction (east-west)
\phi is the latitude where the measurements are taken.
v is the velocity in longitudinal direction (north-south)
R is the radius of the Earth

The first term in the formula, 2Ωu cos(φ), corresponds to the Eötvös effect. The second term is a refinement that under normal circumstances is much smaller than the Eötvös effect"

 

[Timber338]

That's interesting stuff. Up until now, I had never even heard of the Eotvos effect. It does make sense though and the principals are very much in line with how Coriolis effects our bullet. Also think it's great that through all of the discussion of Coriolis, physics, etc, and our reference frames are still "target/game animal". We all definitely are still thinking of the same end result, even if the effects are very hard to see in the real world.

The other thing that is still bit mind blowing to me, is that we are all moving around in a circle at about the speed of sound, of course depending on your latitude. Up there in Alaska you get a bit of a break being so close to Santa. But when you're near the equator, you better be careful when you stick your head out the car window, especially if you're driving west.

 

[JasonOlech]

I tried the keep the post as simple as possible. I highly recommend reading Applied Ballistics for Long Range Shooting 3rd Edition. In Particular Pages 113-118. To keep it simple here:

• Coriolis - Horizontal and Vertical are seperate from each other.
• Horizontal - Only dependent on your Latitude. Has maximum deflection at the Poles. - Horizontal Deflection = Ω* X * sin(Lat) * tof
  Ω = 0.00007292 rad/s (rotation rate of earth)
  X = Range
  Lat = Latitude
  tof = Time of Flight
• Vertical Coriolis - Dependent on Latitude and Azimuth (DOF). Has maximum deflection at the equator and for Azimuths that are parallel to the equator. f = 1 - ((2 * Ω * MV)/g)cos(Lat)*sin(Az)
  f= gravity correction factor
  g=(acceleration of gravity 32.2 ft/s^2)
  Az = Azimuth
• Aerodynamic Jump - Has both Horizontal and Vertical deflection, so you need to set wind to 0 when trying to isolate Coriolis from it.

Remember These play a very small part in the firing solution, SD (Spin Drift) is generally much more pronounced.

Doc,

I am trying to understand how to calculate the Eotvos (aka vertical coriolis) on my bullet and I am having a very hard time getting this particular formula to make sense.

For the equasion:

F = 1 - ((2 * Ω *MV/g)*Cos(latitude)*Sin(Azimuth))/g)cos(Lat)*sin(Az)

Could you perhaps provide an example with numbers so that I could plug and chug to see what I am going wrong?

My latitude is 42.185710*N
For argument sake let's say my azimuth us 75*NE from due North

Let's say my muzzle Velocity is 2900fps

F = 1 - ((2 * 0.0000729*2900/32.2)*Cos(42.185710)*Sin(75))

F= 1-((0.42282/32.2)*(0.7409721054)*(0.9659258263)

F= 1-((0.0131310559)*(0.7409721054)*(0.9659258263))

F = 1 - 0.0093982131

F=0.9906017869

1) is this correct?

2) is there any other math applied to this correction value before it is dialed into the scope?

3) Does MV = muzzle Velocity or is M = Mass and V= Velocity as it would be written in physics?

4) What are the specific units of measure for the velocity variable? or for mass*Velocity?

5) (BIG Question) How exactly does one determine vertical coriolis at 1000m, versus 1500m, versus at 2000m?

This equasion does not seem to include any mathematical descriptors for variation at different ranges.

And this is where I am lost.

Every time I try to apply this formula I seem to get 0.999 something for every number I plug in.

It's driving me to madness......

Please help! I would be very grateful if you can point me in the right direction.

My email is [email protected]

Many thanks