Question on Applied Ballistics App and Coriolis

[Timber338]

I am wondering if the Applied Ballistics App uses shooting angle (as in shooting uphill/downhill) along with Azimuth/Latitude to calculate Coriolis drift?

I have checked every app I own as well as a few online, and none of them use shooting angle as a variable in the Coriolis solver.

there are a few real world shots locations where I hunt with steep shooting angles where this could be a factor, and I would like an app that gets me as accurate of a shooting solution as possible. Can anybody who owns the app verify this one way or another? Or does anybody know of any ballistics software that uses shooting angle in the Coriolis solver? Thanks!

 

[Pdvdh]

I am wondering if the Applied Ballistics App uses shooting angle (as in shooting uphill/downhill) along with Azimuth/Latitude to calculate Coriolis drift?

Or does anybody know of any ballistics software that uses shooting angle in the Coriolis solver? Thanks!

Perhaps the inclination/declination angle of shot as entered to calculate drops is automatically incorporated into the coriolis solution when coriolis is activated. Do you know that it isn't?

Wouldn't make much sense to have to input the angle of shot a 2nd time, solely to calculate coriolis.

 

[Timber338]

Perhaps the inclination/declination angle of shot as entered to calculate drops is automatically incorporated into the coriolis solution when coriolis is activated. Do you know that it isn't?

Wouldn't make much sense to have to input the angle of shot a 2nd time, solely to calculate coriolis.

No it certainly would not be entered a second time. I have verified all of the apps that I have tried treat the Coriolis drift as though the shot is perfectly flat (no uphill/downhill angle), regardless of shooting angle. Based on math/physics, not accounting for shooting angle in Coriolis is incorrect.

Specifically, the Coriolis function should just grab the shooting angle and use it in the calculation of Coriolis drift.

 

[Pdvdh]

I use ColdBore 1.0, by Patagonia Ballistics. If my coriolis solutions are correct, this is something you don't need to worry about.

Two horizontal coriolis solutions for the same load with a 215gr Hybrid from my 30/375 S.I., a MV=3060fps. Azimuth = 0 degrees. N. Latitude = 60 degrees.

1st solution at 40-degree inclination. 2nd solution at 0-degree inclination.

Here's the coriolis x-axis (horizontal) drift solution for both.
Inclination -- 40*- - - - 0*
1000yds __ 2.79" __ 2.77"
2000yds _ 13.44" _ 13.34"
3000yds _ 32.58" _ 32.19"
4000yds _ 61.44" _ 60.36"

____________________________________

Two vertical coriolis solutions for the same load with a 215gr Hybrid from my 30/375 S.I., with a MV=3060fps. Azimuth = 90 degrees. N. Latitude = 60 degrees.

1st solution at 40-degree inclination. 2nd solution at 0-degree inclination.

Here's the coriolis y-axis (vertical) drift solution for both.
Inclination -- 40*- - - - 0*
1000yds __ 1.61" __ 1.60"
2000yds __ 7.76" __ 7.70"
3000yds _ 18.81" _ 18.58"
4000yds _ 35.47" _ 34.85"

 

[Timber338]

I briefly read over your numbers and I believe they are proving that your software also does not use shooting angle in Coriolis calculations.

Your example with 60 deg azimuth shooting due north at 40 deg inclination should have much less Coriolis drift than the 0 deg inclination shot.

Rerun your numbers with a 60 deg inclination at 60 deg azimuth and the Coriolis drift should be zero as long as you are shooting somewhere north.

If you are shooting south at an uphill inclination of 60 deg at 60 deg azimuth, Coriolis drift would be twice as much compared to a flat shot at 0 inclination.

 

[Timber338]

I should also note that shooting inclination should have no impact on the elevation component (east/west shooting direction) of the shooting solution. I.e., shooting east/west only needs time of flight like the Coriolis solver is currently working. Inclination/declination would only impact the horizontal drift with the the north/south component of the shot.

 

[Pdvdh]

I briefly read over your numbers and I believe they are proving that your software also does not use shooting angle in Coriolis calculations.

Your example with 60 deg azimuth shooting due north at 40 deg inclination should have much less Coriolis drift than the 0 deg inclination shot.
Both runs were at 60* north latitude. My 1st comparison was shooting due north (0* azimuth), and my 2nd comparison was shooting due east (90* azimuth). Under those criteria, I ran one at 40* inclination of bore and then one at 0* inclination (with the bore horizontal).

Rerun your numbers with a 60 deg inclination at 60 deg azimuth and the Coriolis drift should be zero as long as you are shooting somewhere north.
Not following. Provide the latitude, azimuth, and angles of inclination you would like me to compare?

If you are shooting south (south is 180* azimuth - not 60* azimuth) at an uphill inclination of 60 deg at 60 deg azimuth, Coriolis drift would be twice as much compared to a flat shot at 0 inclination.

To help ensure we're communicating; 0* direction of fire (azimuth) is due north, 90* azimuth is due east, 180* azimuth is due south, and 270* azimuth is due west.

Inclination is angle (in degrees) the bore is pointed above a horizontal plane.

So which latitude, azimuth, and angles of inclination would you like me to compare?

 

[Timber338]

Crap, I said (typed) azimuth when I was thinking latitude. Glad you clarified.

Ok. So let's just stay in the same latitude of 60 degrees. I'm assuming this is about where you are up in Alaska? That's really up there!

lets keep shooting azimuth due north at 0*.

The only two variables you'll want to change in the two shooting scenarios is inclination... let's say 0* and 60*.

Shooting north at 60* inclination should have zero windage drift due to Coriolis. Every ballistics solver I have used gets Coriolis wrong for a shot with any inclination/declination other than 0*. Or, the Coriolis solver is only correct when you are in fact shooting at 0* inclination, or due east/west at any shooting inclination.

From your initial numbers, I think your solver is going to show that 60* and 0* inclination have about the same wind drift due to Coriolis, and this is incorrect.

 

[Pdvdh]

Here's my understanding of the vertical and horizontal components of Coriolis "drift" as they affect bullet flight:
There is no vertical coriolis component affecting the vertical point of bullet impact when the shooter's direction of fire (azimuth) is 0* due north or 180* due south. There is a vertical component when shooting any direction other than due north or due south, no matter the shooter's location on the earth (expressed in degrees latitude).

There is always a horizontal component of coriolis drift when shooting any direction of fire (azimuth), unless the shooter is at the earth's equator - which is 0* latitude. This horizontal component is rightward in the northern hemisphere, and leftward in the southern hemisphere. The coriolis-caused horizontal component increases in magnitude as latitude increases from 0* (equator) to 90* (north or south latitude). The farther the shooter is from the equator, either north or south, the larger the horizontal component of coriolis.

Below are three horizontal coriolis solutions for a 215gr Hybrid from my 30/375 S.I. at MV=3060fps. Azimuth = 0*. N. Latitude = 60*. There is no vertical component of coriolis since the shooting direction is due north (0* azimuth).

1st solution @ 0* inclination. 2nd @ 30* inclination. 3rd @ 60* inclination:

Coriolis x-axis (horizontal) drift solutions for the three differing angles of inclination.
Inclination -- 0* ---- 30* ---- 60*
1000yds __ 2.77" __ 2.78" __ 2.79" horizontal component of coriolis
Velocity _ 1860.3 _ 1841.9 _ 1828.2 fps
Flight Time 1.257 __ 1.286 __ 1.307 seconds

2000yds _ 13.34" _ 13.42" __ 13.47"
Velocity _ 1031.2 _ 1006.3 __ 990.2
Flight Time 3.477 __ 3.658 __ 3.805

3000yds _ 32.19" _ 32.49" __ 13.73"
Velocity __ 754.2 __ 718.9 __ 691.0
Flight Time 6.903 __ 7.541 __ 8.092

4000yds _ 60.36" _ 61.19" __ 61.85"
Velocity __ 555.4 __ 506.9 __ 468.6
TOF ____ 11.557__ 13.244 _ 14.823

It looks like the horizontal component of coriolis increases due to the increasing time of bullet flight caused by the increasing angle of inclination. I'm not sure why increasing angles of fire increase bullet time of flight. These are ColdBore 1.0 ballistic software solutions.

Why does the bullet speed drop, and the time of flight increase, with increasing shot angles? Distance of bullet flight is equal at 1000, 2000, 3000, and 4000yds. This is a puzzler... If anything, I'd expect bullet speed to increase slightly due to lesser air density at higher altitude. I'm going to run the same scenario, except with downward angles of fire (declination). Then see if the increasing downward angles of fire cause bullet speed to increase, and time of flight to decrease when shooting downward with gravity, rather than upward against gravity. lightbulb

EDIT: Yes indeed. When I change the angles of fire to -30* and -60*, velocity increases with increasing declination angle, and time of flight decreases with increasing downward angles of fire. Firing at an upward angle (against the ever present force of gravity) slows bullets slightly compared to firing at a downward angle (in the direction of the force of gravity). This in spite of the lesser air density above, and the greater air density below. This is the "engineer's explanation".

 

[DocUSMCRetired]

Let me see if I can help here:

• Coriolis - Has two completely separate components Vertical and Horizontal.
• Vertical Coriolis - Dependent on both Latitude and Azimuth(DOF - Direction of Fire).
• Horizontal Coriolis - Dependent only on Latitude, Azimuth(DOF) doesn't affect it.
• Aerodynamic Jump - Has a vertical and horizontal component. To remove this, set wind to 0. It is included in the firing solution (Missing on AB Mobile Android, but will be in it in the future). All other systems have it.
• Inclination - Calculated separately, our system takes in to account the slowing of the bullet over distance, or drag, to correctly calculate this.

Our systems incorporates all of this independently. That includes Analytics, Tactical, AB Mobile, Kestrels, Raptar-S, Rem 100/200, LRFs, Computational Weapons Optic(DARPA CWO) etc.

 

[Timber338]

Doc, just read your response as I was finishing up the below post. Not sure I follow that azimuth does not effect horizontal coriolis. a shot due east/west would have zero horizontal Coriolis drift... but I'll read your post in more detail...

____________

So I think that most of us do not take the time to really understand Coriolis since in the real world it has a small effect on shooting solution (this includes myself). Over the past several months I have been able to dial in my 338 RUM shooting the 300 Elite Hunter so well that I am seeing these small differences at the distances I am shooting. So I took the time to fully understand what was going on. The more I read shooting articles the more confused I became, it just didn't add up with the way other shooters were explaining Coriolis, at least not in my own mind.

Once I understood what was going on, it was like a lightbulb went off, and that is when I started looking at ballistics solvers because I really do think they are missing the ball on horizontal Coriolis on anything but flat shots.

Gaspard-Gustave Coriolis was the guy who first documented the effect in 1835 and is why we call this phenomenon the Coriolis Force. But most articles discuss how this force effects weather patterns. So I finally just started drawing diagrams on my white board to figure it out for shooting. And it is so simple.

In my mind I have broken it down to the very basics and have been explaining it with the following analogy to other people, and it has been working well to get the point across. Let's continue to use the due-north shot for simplicity:

When you shoot a bullet, the bullet is also traveling eastbound with the surface of the earth. It's like driving down the freeway eastbound at 60 mph and throwing a rock sideways out the left window. The rock travels out the window and flies across the other lanes of the freeway but it is also moving in the direction of the car at 60mph. If you throw the rock at a guy driving next to you, the rock hits the guy if he is also traveling at 60 mph. But if the guy next to you is driving slower than you, the rock is going to fly out in front of him … you miss the guy to the right (shooting north). Now if another guy is driving on your right side, and he is driving faster than you are, the rock you throw at him is going to fly behind him… you also miss to the right (shooting south).

The only way you hit any guy driving near you is if he is driving the same speed as you… or if you adjust your point of aim appropriately. Now pay attention to the first part… you do NOT adjust your point of aim if your target is traveling at the same speed as you.

Now back to the spinning earth and shooting. When you shoot north, you are shooting from a position of the earth that has a particular velocity, and the target you are shooting to is closer to the spin axis of the earth, therefore it must be moving slower than you are. You let loose the bullet, and it remains to travel eastward at the velocity of the shooter, but the target is moving slower. So the bullet impacts right.

Flip around and shoot south… you are shooting to a target which is sitting on the earth that is further away from the spin axis of the earth, so the target is moving faster in the eastward direction than the bullet from where it was fired… so you also impact to the right.

These are just the basic physical principals of the horizontal component of the Coriolis Effect on a bullet.

Now… all of this is implying a perfectly flat shot with zero inclination/declination. I happen to hunt in a location where a very steep long range shot is possible. That location is about 40 degrees Latitude. Let's say I take that 1200 yard shot that is aiming to the top of a huge rim, and my inclination is also 40 degrees. With that singular shot scenario, I am shooting to a target that is at the same exact distance away from the spin axis of the earth as I am shooting from. Therefore they both are moving at the same exact speed, and I do NOT adjust my point of aim. The horizontal Coriolis effect on a bullet is zero when both the shooter and the target are moving at the same speed.

Does this make sense? I've been wrong before, and I could be wrong here, so I'm all ears if you or anybody else can prove otherwise. But I think I'm right and I also think that ballistics solvers are leaving out a key variable when it comes to the horizontal component of the Coriolis calculation. I agree that this does not have an impact on the vertical component of Coriolis… (and think about that ... there is no horizontal coriolis in a east/west shot because of the very fact that the targets have the same initial velocity!)

The scenario where your shooting inclination matches your shooting latitude is the easiest to see if any particular ballistics solver gets this right. I currently have not found a solver that does this, but I also do not own Applied Ballistics… because the focus and strength of this app is to provide the most accurate shooting solution, I would think it would be my best bet. If it does not currently do this, it could be added fairly easily with an update to the software. I could even help write the script to include inclination/declination correctly into the app.

 

[DocUSMCRetired]

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.

 

[Pdvdh]

Doc, just read your response as I was finishing up the below post. Not sure I follow that azimuth does not effect horizontal coriolis. A shot due east/west would have zero horizontal Coriolis drift... but I'll read your post in more detail...

Azimuth doesn't affect horizontal coriolis. The only input that affects the horizontal component of coriolis is latitude, or more specifically, the distance the shooter is from the equator. The horizontal component of coriolis is only zero when the shooter is located at the earth's equator (0* latitude). The magnitude of the horizontal component of coriolis increases with increasing distance from the equator. The magnitude of the horizontal component of coriolis is maximum at the north and south poles (90* latitude, north or south).

 

[Timber338]

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.

Got it. I think the Horizontal deflection equation is incorrect.

I dont' fully understand how you are integrating the equations into the code, but i think it should be:

• Horizontal - Only dependent on your Latitude. Has maximum deflection at the Poles. - Horizontal Deflection = Ω* X * sin(Lat-inc) * tof
  Ω = 0.00007292 rad/s (rotation rate of earth)
  X = Range
  Lat = Latitude
  tof = Time of Flight
  inc = shot inclination