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When shooting north or south in the northern hemisphere, your shot will always land to the right, due to the Coriolis effect.

Bizarre.

Jordan covered it but a good read nonetheless...

http://www.appliedballisticsllc.com/Articles/ABDOC108_GyroscopicAndCoriolis.pdf

When shooting north, isn't the target moving right and POI will be left of POA? Either N or S hemisphere, same/same. (ignoring spin drift)

Another way of saying it is, when shooting North or South POI will be WEST of POA, either N or S hemisphere (again ignoring spindrift which is a greater effect)

A cheesy but good visual....

Starts at about 1:10.

https://www.youtube.com/watch?v=i2mec3vgeaI

No. Think of it this way- a man standing near the North pole does a complete rotation in 24 hours, and a man on the equator does a full rotation in 24 hours. Who is moving at a higher speed? The man at the equator must be moving at a much higher speed in order to travel much further in 24 hours than the man at the north pole.

Okay, so if we’re standing on the equator, and shooting a target near the North pole (new world record?), The bullet starts out going eastward faster than the target is. So in the time it takes the bullet to arrive, it travels east more than the target does, and hits right of POA.

On the other hand, if we’re standing near the North pole and shoot at a target at the equator, the target is moving east faster than the bullet, so while the bullet is in flight the target is moving to the “left” faster than the bullet is, and the bullet hits right of POA.

When shooting south, the target outruns the bullet to the left, and when shooting north, the bullet outruns the target to the right.

The opposite is true in the southern hemisphere.

^^^^^^

Good explanation!

I read the Litz thing above and I am not on the equator nor pole nor shooting quite that far..... Consider the following.

So, dude standing at 45 degrees N. Fires due north 1 mile. Ignore spin drift. Surely the target is moving to the east, just like it was when firing east? Hit is to the west, left of POA. Dude turns around and fires due South, target moves east, hit is west, right of POA.


So, downunder Dude standing at 45 degrees South. Fires due North 1 mile (not toward the south pole, lol). Ignore spin drift. Surely that target is moving to the east, just like it was when firing east? Hit is again west, left of POA. He does a 180, fires due south, target moves east as it always does, hit is west, right of POA.


In the N hemisphere, couple spindrift with shooting north or south, since its a larger effect than Coriolis, yes hit is alway right....

In the S hemisphere, same diff, hits are right.

So, how fugged is this line of thought?

Ignore spin drift.

The guy shooting to the north has a bullet that is moving to the east faster than the target is, so his bullet hits to the right of POA.

The guy in the southern hemisphere does hit left of POA in every case. If you add in spin drift to the right, that will negate the effect of Coriolis at some distance, which pushes the bullet to the left regardless of which direction you're pointing (except for due east or west).

Read my previous post again. It explains Coriolis without taking spin drift into account.

So, that's what I was tripping on....
Thought we established the bullet is going "straight" (as it is not in contact with the Earth) whereas the target moves with the Earth.... But we did not, though that 'xplains why Coriolis is not centrifugal....

I broke down and watched the corny storm vid and grasp the concept of conservation of initial "lateral" motion.

Thanks for the continued help Jordan, 32 20fan and anyone else with a clue, including the King, LOL.

Some really interesting info in this thread. Thanks for sharing!

Pasta Mann,
We've exchanged on 264 WM password RL33. I have a Sense to and just couldn't get 140he to work with 1:9 barrel. Finally switched to 130gr ELD and found a great load, 72.0gr RL33, producing 0.25 MOA at 100yd., 1 MOA @ 650yds.
I'm teaming and turning necks and bushing size to dust I/D to ream then clean up turn and bushing size. Total removal 0.0008" I/ D, 0.0008-0.0010" O/ D to remove brass creep each firing. Wondering if you've ever tried this as neck tension is super excited uniform.

You seem to have worked your 264 quite a bit, wondered how you prep your brass.

Mile

Pasta Mann,
We've exchanged on 264 WM password RL33. I have a Sense to and just couldn't get 140he to work with 1:9 barrel. Finally switched to 130gr ELD and found a great load, 72.0gr RL33, producing 0.25 MOA at 100yd., 1 MOA @ 650yds.
I'm teaming and turning necks and bushing size to dust I/D to ream then clean up turn and bushing size. Total removal 0.0008" I/ D, 0.0008-0.0010" O/ D to remove brass creep each firing. Wondering if you've ever tried this as neck tension is super excited uniform.

You seem to have worked your 264 quite a bit, wondered how you prep your brass.

Mile

Darned spell checker. Remington Sendero, reaming & turning....
Mike

Good thread

LOL! Shoot across a time zone and the bullet gets there an hour before you shot it!

Tag

Just saw that this thread was bumped.

I should clarify something that I said 2 years ago: Coriolis causes the bullet to strike to the right in the northern hemisphere and to the left in the southern hemisphere, regardless of which direction you're shooting. The only place where there is no lateral drift caused by the CE is if you are standing on the equator shooting due E or W, in which case there is only a vertical shift in POI (due to the Eotvos effect).

The max CE for a 308 winn at 1000 M is 1/4" not enough to matter

jwp,

No, the windage deflection due to the CE depends on your latitude, and can be 5-8" at 1000 m for .308 Win factory loads.