Coriolis effect question for nerds

[godfrog 2]

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Just can't throw you off can I

As a pilot I've never made an adjustment while flying for the Coriolis effect, but then again I've never flown an F-14 at 1,000 knots from the equator to the north pole in still air.

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Maybe an SR 71

Experience is a difficult teacher, she gives you the test first and the lesson afterward

[Andy9o8 2]

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Ask Shah!

He'll chat up some engineering chicks and get you the answer ricky-tick!

Nah, he'd want to use them as the WDIs, but then he'd get bogged down on whether to use only skinny chicks.

[D22369 0]

OK, then we will just throw a student out first.Tongue***


figures someone would beat me to this answer, we are depraved people

Roy

They say I suffer from insanity.... But I actually enjoy it.

[kallend 1,920]

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Using the same numbers...2 jumpers do a high pull and one flys East and one flys West. How much further does the West flyer go?

In which reference frame are you measuring the distance? How fast are they falling? Are you accounting for air resistance?

...

The only sure way to survive a canopy collision is not to have one.

[JohnMitchell 16]

In long range gunnery I've read the Coriolis effect will throw a shell off by 100+ yds on 15-20 mile shots.

[quade 4]

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In long range gunnery I've read the Coriolis effect will throw a shell off by 100+ yds on 15-20 mile shots.

My gut tells me that's an exaggeration or an order of magnitude problem with misplacing a decimal. I can't for the life of me see how it would be that far off.

What's the flight time and latitude?

quade -
The World's Most Boring Skydiver

[csubl 0]

From the website

http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/movearth.html

"Another likely-sounding but ultimately fallacious Coriolis Force myth is set during the First World War at the Battle of the Falklands between British and German naval forces in the South Atlantic. The story goes that early in the battle, British shells kept falling consistently about 100 meters to the left of the German ships because the gunners neglected to take into account the opposite sign of the Coriolis effect in the Southern hemisphere (most of their experience was in the North), and so they were inadvertently correcting for the Coriolis effect in the wrong direction, resulting in twice the deflection that would occur if no correction had been made! While this story has made it into at least one physics textbook that I used back in the 1980s, and is now a common-place on the Internet, I've have never yet read an account of this story in historical accounts of the Battle of the Falklands, which was an important engagement and much discussed. A straightforward calculation, beyond the scope of this introductory course, readily shows that this is a bit of physics mythology."

[JohnMitchell 16]

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While this story has made it into at least one physics textbook that I used back in the 1980s,
................A straightforward calculation, beyond the scope of this introductory course, readily shows that this is a bit of physics mythology."

Well there you go. Thanks for setting me straight. I knew the effect could be calculated, but I sure didn't know how.

[Calvin19 0]

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From the website

http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/movearth.html

"Another likely-sounding but ultimately fallacious Coriolis Force myth is set during the First World War at the Battle of the Falklands between British and German naval forces in the South Atlantic. The story goes that early in the battle, British shells kept falling consistently about 100 meters to the left of the German ships because the gunners neglected to take into account the opposite sign of the Coriolis effect in the Southern hemisphere (most of their experience was in the North), and so they were inadvertently correcting for the Coriolis effect in the wrong direction, resulting in twice the deflection that would occur if no correction had been made! While this story has made it into at least one physics textbook that I used back in the 1980s, and is now a common-place on the Internet, I've have never yet read an account of this story in historical accounts of the Battle of the Falklands, which was an important engagement and much discussed. A straightforward calculation, beyond the scope of this introductory course, readily shows that this is a bit of physics mythology."

I'm no physicist, but I want someone who is to do the math. This seems plausible to my lesser brain.

the 8-13" guns of WWI used by British battleships had a muzzle velocity of ~800m/sec and a range of ~37km, that means ~40 seconds in the air.

Falklands are right around 45/50 degrees latitude (meaning a pretty high differential in rotation speed across a N/S line)

The difference in rotation speeds for that alleged story to be accurate or at least possible is less than 2m/s difference.

-SPACE-

[Calvin19 0]

From what I did in my head, the rotational speed difference between parallels 30km apart at around 45 degrees south is ~2.2m/sec.

So, A battleship 30km due north of it's target will be traveling ~2m/s faster to the east. With a 30second TOF this means the gunner will have to lead (is that the right word for a Coriolis correction?) by 80m.

Also, this is worst case scenario. Meaning I assumed the longest range weapons of that time at their longest TOF, as well as a N/S trajectory.

I think, and please correct me if i made a mistake, that the story of the big guns being 100m off due to a COS inversion is totally probable.

[csubl 0]

Just from some rough ballpark calculations, assuming earth is a perfect sphere of 25,000 miles circumference at the equator:

Circumference of 45 South Latitude is 17677.68 miles;

Rotation speed at that latitude is 1080.303 feet/second;

Moving 25 miles north (about 40 km), circumference is 17688.78 miles, and rotation is 1080.981 feet/second;

Assuming 1 minute flight time for a shell (due to speed of shell decreasing from 800 m/sec and increased distance due to trajectory), I get 60*.678 or about 40 feet offset due to Coriolis.

[csubl 0]

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Just from some rough ballpark calculations, assuming earth is a perfect sphere of 25,000 miles circumference at the equator:

Circumference of 45 South Latitude is 17677.68 miles;

Rotation speed at that latitude is 1080.303 feet/second;

Moving 25 miles north (about 40 km), circumference is 17688.78 miles, and rotation is 1080.981 feet/second;

Assuming 1 minute flight time for a shell (due to speed of shell decreasing from 800 m/sec and increased distance due to trajectory), I get 60*.678 or about 40 feet offset due to Coriolis.

After reading Calvin19's post, I see that he's right. I found a mistake in my calculations. Rotation at 25 miles north of 45 S. Lat should be 1087.069 feet/sec, so a one-minute flight time would result in a deflection of 60*6.763 or about 400 feet.

[Calvin19 0]

Don't assume anything I wrote was correct. I'm on middle school math here. They tried to teach me math in college but I passed anyway.

BUT, ballistics of that scale are really cool. I mean, holy crap. 18 year old kids in a 13 inch gun turret figuring out that the a target 20 miles away is being moved out of the way by the EARTH ROTATING?

It's even cool when one would realize that THEY are the ones moving, just as much as the target.

Space is pretty cool.

[/mindgasm]