Author Topic: Math Question  (Read 10587 times)

Offline tweeter55

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Re: Math Question
« Reply #40 on: July 18, 2017, 12:01:02 PM »
Well, not really a valid question.

If you 'hover over a fixed spot', you would have to put in whatever corrections were needed to maintain that position and so the spot could not move away..... so to speak.

If you simply moved vertically, without any side- motion compensation, you would drift away rapidly but not due to the Coriolis effect but rather because of wind. Think hot air balloon: as soon as they break the tether they drift off.

Maybe the easier way to think of it would be if one were returning to the surface from a great altitude, maybe from orbit for example. If all things were correct and a path plotted that would put the object in a specific point when it landed, and the it was 'let go' and not directed in any way and did indeed follow the path predicted..... it would miss because the target would be in a different place than it was when the device was sent (let go) in the first place.

Brian

When a rocket takes off and does not  go straight up,  would that be the same thing?
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Offline jimmymac

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Re: Math Question
« Reply #41 on: July 18, 2017, 01:31:53 PM »
You guys are ate the frick up,man... I need an aspirin.
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Offline B.D.F.

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Re: Math Question
« Reply #42 on: July 18, 2017, 02:41:29 PM »
No.

This is a simple concept but I am finding it difficult to put into words. I think the problem is that it is not intuitive; we cannot normally see the Coriolis effect and any attempt to simulate it using optics or mechanical means fails.

Let's try this for an example: Say there are two lighthouses, one mile apart. And to make it easier, they are on a line of longitude, meaning one is directly north of the other, no east / west difference at all. And one of them is located at the equator and the other one is one- mile away from the equator, on a perfect north / south line.

Now, you are at the top of the southern lighthouse and you train a telescope on the north lighthouse and measure the angle of the telescope (relative to the Earth's axis, or the North Pole) and find it to be zero (of course). If the telescope is not moved, it will point north and look at a specific spot on the other lighthouse forever.

Then we set up a rifle next to the telescope and also point it perfectly, absolutely north. It is sighted directly on the other lighthouse, in a very  specific point. We will ignore vertical drop in this example to make it simple but again, the rifle is pointed directly at the other lighthouse, as is the telescope next to the rifle. We put a mark on the north lighthouse exactly where the rifle is aimed.

In this example, there is no wind, there is no inaccuracy in the rifle and the projectile will follow a path absolutely straight left / right path, due north, on a line that intersects the north lighthouse where we have marked that lighthouse.

We fire the rifle and wait for the projectile to impact the other lighthouse, directly in- line with the mark placed there. There will be some drop of course but the projectile WILL hit the lighthouse some distance exactly below our previously marked point. It cannot do otherwise. There are no tricks or gimmicks or hidden wrinkles in this example.

But the bullet impacts the north lighthouse SOME DISTANCE to the SIDE of the mark by some inches or feet. We check the rifle, and it is still pointed directly at the mark on the other lighthouse so in our confusion, we fire it again. And again it misses our mark but it impacts in exactly the same point the previous projectile did.

Now, just to make absolutely sure there is nothing wrong here, we now string a cable between the two lighthouses, exactly on the line between the rifle and the point on the other lighthouse that the rifle is aimed toward. The cable is pulled tight and we can verify that the two points on the two lighthouses are indeed directly in- line as shown by the cable (we do not care about any vertical sag in the cable, only the left- right relationship) and perfetly aligned north / south as we knew they had to be.

So the cable is straight. The rifle sends the projectile directly toward a point on the other lighthouse and it travels straight. But the impact point is off to one side, every-time it is tested.

The difference between the impact points and the line- of- sight of the rifle is due to the Coriolis effect. It appears that both lighthouses are fixed and not moving but that is incorrect they are indeed moving, because they are rotating around the axis of the Earth at ~1,000 MPH. The key to this is the fact that both lighthouses (as well as the telescope, the rifle, you and everything else in sight) is fixed to the Earth and being pushed sideways at 1,000 MPH. Because EVERYTHING around you is also moving at this same speed, the movement is not apparent but it is this sideways motion, or rotation, that causes the sun to appear and disappear each day. But in our example, the instant the projectile leaves the barrel, it is NO LONGER being pushed sideways but is now free to travel in a truly straight path, which it does. And when it arrives one mile north, the northern lighthouse has MOVED SIDEWAYS since the projectile was fired and so it misses the point of aim by a substantial amount.

The amount of side movement is dependent on the time it takes the projectile to travel from one lighthouse to the other.

The Coriolis effect is only applicable when an object is sent on a path that will lead toward an object on the Earth's surface but not attached to the Earth's surface. It is a pretty simple math problem to figure out the deceleration rate of the projectile (it starts off moving 1,000 MPH sideways but begins to slow when it leaves the barrel), and the calculate the flight time to yield a specific distance.

And again, per the Paris gun, given the projectile's flight time of three minutes, the 'target' rotated away from the point of aim by almost exactly one kilometer.

Now that is the simplest example. The next thing is to think that this effect reduces as the angle changes toward one more in- line with the equator until it is no longer applicable but because a projectile will always travel in a arc, and the projectile must be launched at some upward angle to hit a distant target, the Coriolis effect still 'bites' the shooter, but instead of a left / right error, we now get a closer / farther error where the projectile lands in front of or behind the point where the arc (called a ballistic path in our case and not a true arc but close enough for this example) would have it land. Again, this if fairly simple to calculate. Of course in the real world, it is almost certain a firearm, an artillery piece, a rocket, a golf ball sent a long way or any other example would ever be on a perfect north / south or east / west line so both parts of the Coriolis effect must be calculated and then added together to get the left / right and too short / too far corrections in the launch azimuth and attitude (the angle of the gun, in rotation around in a circle and the angle the barrel is elevated to before firing).

That is the simplest example I can think of.

'Everything should be made as simple as possible, but not simpler.'
-an axiom attributed to A. Einstein   

What he meant by this is to reduce everything down to the simplest possible explanation or example but not so much that the example or explanation becomes wrong. In my example above, the temptation is to simply say that one seemingly fixed object moves in relationship to another seemingly fixed object but that is too simple and incorrect. So in my view, the simplest explanation is that both objects attached to the Earth are being moved sideways all of the time, and the projectile is NOT being moved sideways for the duration of its flight. That is as simple as I believe it can be made and still be correct and not some bad or wrong example.

The Coriolis effect is also why water swirls in a bowl as it drains out the hole in the bottom (round sink, toilets and so forth), and while the reason is absolutely identical to the above example, the explanation is more complex. So that can be an optional, no added cost, homework assignment.     ;D

The rocket traveling straight up is not sufficient to show the Coriolis effect but if the rocket did have sufficient power or fuel to leave the Earth's gravitation and fell straight back down (straight up, stopped, then straight back down) it would NOT hit the point from which it was launched. That would be an example of the Coriolis effect.

Next weeks' class will be on the Doppler shift (or Doppler effect.... to maintain a steady course curriculum) and how it is used to tell if a distant star is moving toward us, away from us or neither. And if it is moving toward or away from us, exactly how fast. That is a much longer, more detailed and more complex thing than the Coriolis effect is though.

And still on- topic: math answers.  :rotflmao:

Brian

When a rocket takes off and does not  go straight up,  would that be the same thing?
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Offline VirginiaJim

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Re: Math Question
« Reply #43 on: July 18, 2017, 03:14:52 PM »
You guys are ate the frick up,man... I need an aspirin.

Screw that, I'm going to start drinking.  My brane is hurting.
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Offline maxtog

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Re: Math Question
« Reply #44 on: July 18, 2017, 03:51:57 PM »
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Offline jimmymac

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Re: Math Question
« Reply #45 on: July 19, 2017, 07:51:03 AM »
The grass isn't always greener.

Offline Classvino

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Re: Math Question
« Reply #46 on: July 19, 2017, 10:20:27 AM »
This is a simple concept but I am finding it difficult to put into words. I think the problem is that it is not intuitive; we cannot normally see the Coriolis effect and any attempt to simulate it using optics or mechanical means fails. ...
Brian

A hypothetical question  ( since you seem to be waxing professorial...)

Since the distance from the equator also affects the speed that everything is moving sideways - imagine a bicycle wheel - an item 1 inch from the centre point and an item 15 inches away from the center move at exactly the same RPM, but the item 1 inch from the centre travels 6.28 inches in one rotation, while the object 15 inches from the centre travels 47.12 inches in one rotation...

So - using your lighthouse example - the distance that the bullet landed from the aim point would be different when one was shooting north, rather than south, given identical conditions - since the south lighthouse would have been travelling faster than the northern one - since the southern lighthouse's 'latitude' would be further away from the earth's axis (assuming northern hemisphere) - right?  (may have to extend the distance to be more easily measured)

Jamie
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Offline tweeter55

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Re: Math Question
« Reply #47 on: July 19, 2017, 11:40:58 AM »
A hypothetical question  ( since you seem to be waxing professorial...)

Since the distance from the equator also affects the speed that everything is moving sideways - imagine a bicycle wheel - an item 1 inch from the centre point and an item 15 inches away from the center move at exactly the same RPM, but the item 1 inch from the centre travels 6.28 inches in one rotation, while the object 15 inches from the centre travels 47.12 inches in one rotation...

So - using your lighthouse example - the distance that the bullet landed from the aim point would be different when one was shooting north, rather than south, given identical conditions - since the south lighthouse would have been travelling faster than the northern one - since the southern lighthouse's 'latitude' would be further away from the earth's axis (assuming northern hemisphere) - right?  (may have to extend the distance to be more easily measured)

Jamie

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Offline maxtog

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Re: Math Question
« Reply #48 on: July 19, 2017, 03:43:05 PM »
That was awesome! LOL!

Hadn't you seen that before?  Was all over the place forever.  Besides being extremely silly/crazy, the autotune music video of the news report was just incredibly well done.  Kinda catchy too.

If you like that one, there is the other one from the same news report:  https://www.youtube.com/watch?v=hMtZfW2z9dw

"only" 136 MILLION views :)

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Offline B.D.F.

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Re: Math Question
« Reply #49 on: July 19, 2017, 03:54:17 PM »
Yeah, I understand it appears that way (professorial) and it may be exactly what it is; I find this 'junk' fascinating and tend to enjoy both learning it as well as trying to find examples that are (believe it or not) as simple and direct as possible.

And yep, what you say is exactly what would happen because one lighthouse is moving faster than the other. To alter that example so that it would not matter which way the projectile traveled, the lighthouses would have to be the same distance away from and on either side of the equator; one 1/2 mile north of the equator and the other 1/2 mile south, to maintain the same spacing between them and yet have the same rotational speed. I did not bother with that complication because I really was shooting (pun marginally intended, I cannot find an place to put an Easy Boys! anywhere in this thread..... yet) for the simplest example I could think of.

Brian

A hypothetical question  ( since you seem to be waxing professorial...)

Since the distance from the equator also affects the speed that everything is moving sideways - imagine a bicycle wheel - an item 1 inch from the centre point and an item 15 inches away from the center move at exactly the same RPM, but the item 1 inch from the centre travels 6.28 inches in one rotation, while the object 15 inches from the centre travels 47.12 inches in one rotation...

So - using your lighthouse example - the distance that the bullet landed from the aim point would be different when one was shooting north, rather than south, given identical conditions - since the south lighthouse would have been travelling faster than the northern one - since the southern lighthouse's 'latitude' would be further away from the earth's axis (assuming northern hemisphere) - right?  (may have to extend the distance to be more easily measured)

Jamie
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Offline B.D.F.

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Re: Math Question
« Reply #50 on: July 19, 2017, 04:03:10 PM »
That is how I feel one page into anything written by Shakespeare. Or poetry. And the great novels.

I like things with repeatable, physical 'rules' but seem immune to kultur.

With a couple of exceptions: Henry V was pretty good IMO and I think the only one of the very few plays of Shakespeare I have read (forced to do so) that I liked, probably because 1) I saw it as a study of human tenacity and I love tenacity, and 2) 'cause the English took France. What English speaker, anywhere in the world, does not like that?

 :rotflmao:

Brian

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Offline Rhino

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Re: Math Question
« Reply #51 on: July 20, 2017, 06:24:32 AM »
Had insomnia last night. Started thinking of the coriolis effect. I believe it is entirely due to the difference in angular movement between 2 points. Shooting directly east or west there is no difference. I do not believe it will make any difference which direction (east or west) as to weather or not the shot will be long or short. Same is true shooting north and south across the equator to a point equal distance from the equator. But lets consider the Paris gun shooting 100 miles to a target directly on the north pole. The target will be stationary. But the gun will be moving along a circle ~628 miles in circumference (100*2*pi) at the rate of 1 revolution per day. That is about 26 mph to the right (left if south pole) from the perspective of the gun looking at the target. If the projectile flight time is 3 minutes that would be ~1.3 miles. The gunner would have to aim at a point 1.3 miles to the left of the target.

Now we know the equation to compensate for Coriolis effect for the Paris gun. Paris is on the 48.8 parallel. cos(latitude)*(radius of earth)*2*pi per day is the angular speed of Paris. ( cos(48.8 )*4000*2*3.14 )/24 = About 690 mph. A point 100 miles south of Paris is about 1 degree south and moving at about 703 mph. A difference of 13 mph. 3 min flight time would be a difference of about .65 miles. Very close to what you said about being off by 1/2 mile. But I'm sure the gun was not in Saint-Benoît-sur-Loire. Apply these equations to any 2 points 100 miles apart on or near the equator and the difference is negligible.

That's my math and I'm sticking to it!

Offline gPink

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Re: Math Question
« Reply #52 on: July 20, 2017, 06:33:31 AM »
Are you talking actual north or magnetic north?

Offline Rhino

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Re: Math Question
« Reply #53 on: July 20, 2017, 06:39:02 AM »
Are you talking actual north or magnetic north?

For Coriolis effect the magnetic pole does not matter. True north is what matters.

Offline B.D.F.

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Re: Math Question
« Reply #54 on: July 20, 2017, 07:08:33 AM »
Sorry to hear about the insomnia.

The Coriolis effect is not due to the difference in rotational (or angular) velocities on different points on Earth, it is due entirely to the projectile slowing down only in relationship to the Earth's rotation, and the flight time to first point of impact (POI). Again, the only thing that can cause the projectile to strike anyplace other than where it was expected to strike is that the projectile itself is no longer connected to the two points, launch and impact. The cable example shows this I believe: if the cable is straight, and the projectile path is straight, what is happening that is causing the projectile to deviate from the cable's line.

If the projectile trajectory is on a line of latitude, with the equator being the line that would have the greatest effect, the Coriolis effect is far more noticable because it is no longer dependent only on the projectile's deceleration.

A quick search turned up this:https://loadoutroom.com/thearmsguide/external-ballistics-the-coriolis-effect-6-theory-section/

And from this site, I quote: "Despite being associated with Coriolis, the phenomenon that actually affects the vertical component of the trajectory is called Eötvös Effect. The rotation of the Earth generates a centrifugal force, the same that pushes you to the side when you make a sharp turn with your car. This force acts perpendicular to the Earth rotatory axis, adding or subtracting to the gravity force. When an object flies eastward, in the same direction of Earth’s rotation, centrifugal force acts opposite of gravity, pushing it away from the Earth’s surface. If the object flies westward, in the opposite direction of the Earth rotation, centrifugal force pushes the object toward the ground concurrently to gravity force. Thus, bullets fired to the east always fly a little higher, and, conversely, bullets fired to the west always travel somewhat low.

The amount of drop change is in function of:

Latitude – The linear velocity of a point on the Earth’s surface, and thus the amount of centrifugal force, is maximum at the equator and decreases going toward the poles, where it is null.

Shooting direction, or azimuth – The amount of drop change is highest when shooting east or west, and as the trajectory angles north or south,  the amount of drop change decreases, becoming null, as the angle points toward either pole. "

The key here is that this site has actually disassembled the two components of the Coriolis effect into the north / south component, properly called the Eötvös Effect, and the simpler, direct component of the Coriolis effect, which is what occurs on a line of latitude (east / west direction).

Again, there is no direction which this effect does not apply to: purely east / west, the point of impact is either moving toward or away from the point of launch.

A simple drawing showing the effect's results using a north / south trajectory: http://www.dictionary.com/browse/coriolis-effect

Brian

Had insomnia last night. Started thinking of the coriolis effect. I believe it is entirely due to the difference in angular movement between 2 points. Shooting directly east or west there is no difference. I do not believe it will make any difference which direction (east or west) as to weather or not the shot will be long or short. Same is true shooting north and south across the equator to a point equal distance from the equator. But lets consider the Paris gun shooting 100 miles to a target directly on the north pole. The target will be stationary. But the gun will be moving along a circle ~628 miles in circumference (100*2*pi) at the rate of 1 revolution per day. That is about 26 mph to the right (left if south pole) from the perspective of the gun looking at the target. If the projectile flight time is 3 minutes that would be ~1.3 miles. The gunner would have to aim at a point 1.3 miles to the left of the target.

Now we know the equation to compensate for Coriolis effect for the Paris gun. Paris is on the 48.8 parallel. cos(latitude)*(radius of earth)*2*pi per day is the angular speed of Paris. cos(48.8)*4000*2*3.14 = About 690 mph. A point 100 miles south of Paris is about 1 degree south and moving at about 703 mph. A difference of 13 mph. 3 min flight time would be a difference of about .65 miles. Very close to what you said about being off by 1/2 mile. But I'm sure the gun was not in Saint-Benoît-sur-Loire. Apply these equations to any 2 points 100 miles apart on or near the equator and the difference is negligible.

That's my math and I'm sticking to it!
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Offline B.D.F.

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Re: Math Question
« Reply #55 on: July 20, 2017, 07:10:54 AM »
True north, the actual lines of longitude that run from the north to the south poles.

Magnetic north is not a physical reality, merely a 'cheap parlor trick' we humans use with our crude efforts to navigate with a compass.

Brian

Are you talking actual north or magnetic north?
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Offline jimmymac

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Re: Math Question
« Reply #56 on: July 20, 2017, 07:24:51 AM »
Hadn't you seen that before?  Was all over the place forever.  Besides being extremely silly/crazy, the autotune music video of the news report was just incredibly well done.  Kinda catchy too.

If you like that one, there is the other one from the same news report:  https://www.youtube.com/watch?v=hMtZfW2z9dw

"only" 136 MILLION views :)
I have no idea how I missed it... I did catch the original interview though... Great stuff, Max.
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Offline Rhino

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Re: Math Question
« Reply #57 on: July 20, 2017, 11:22:14 AM »
Interesting. I will have to study the Eötvös Effect more. This is why NASA launches to the east the vast majority of the time. Extra 1000 mph is a lot of rocket fuel.

But I still maintain that the Coriolis Effect is near zero at the equator and more pronounced at the poles due the the difference in the rate of angular movement. I did find a website that agrees with me:

https://www.quora.com/Why-is-the-Coriolis-force-zero-at-the-equator-and-maximum-at-the-poles

The Coriolis force acts in a direction perpendicular to the rotation axis and to the velocity of the body in the rotating frame and is proportional to the object's speed in the rotating frame.
The Coriolis effect is caused by the rotation of the Earth and the inertia of the mass experiencing the effect. Because the Earth completes only one rotation per day, the Coriolis force is quite small, and its effects generally become noticeable only for motions occurring over large distances and long periods of time, such as large-scale movement of air in the atmosphere or water in the ocean. Such motions are constrained by the surface of the earth, so only the horizontal component of the Coriolis force is generally important.
 The horizontal deflection effect is greater near the poles and smallest at the equator, since the rate of change in the diameter of the circles of latitude when travelling north or south, increases the closer the object is to the poles.

Hey! My insomnia fueled brain farts can't be wrong!


Sorry to hear about the insomnia.

The Coriolis effect is not due to the difference in rotational (or angular) velocities on different points on Earth, it is due entirely to the projectile slowing down only in relationship to the Earth's rotation, and the flight time to first point of impact (POI). Again, the only thing that can cause the projectile to strike anyplace other than where it was expected to strike is that the projectile itself is no longer connected to the two points, launch and impact. The cable example shows this I believe: if the cable is straight, and the projectile path is straight, what is happening that is causing the projectile to deviate from the cable's line.

If the projectile trajectory is on a line of latitude, with the equator being the line that would have the greatest effect, the Coriolis effect is far more noticable because it is no longer dependent only on the projectile's deceleration.

A quick search turned up this:https://loadoutroom.com/thearmsguide/external-ballistics-the-coriolis-effect-6-theory-section/

And from this site, I quote: "Despite being associated with Coriolis, the phenomenon that actually affects the vertical component of the trajectory is called Eötvös Effect. The rotation of the Earth generates a centrifugal force, the same that pushes you to the side when you make a sharp turn with your car. This force acts perpendicular to the Earth rotatory axis, adding or subtracting to the gravity force. When an object flies eastward, in the same direction of Earth’s rotation, centrifugal force acts opposite of gravity, pushing it away from the Earth’s surface. If the object flies westward, in the opposite direction of the Earth rotation, centrifugal force pushes the object toward the ground concurrently to gravity force. Thus, bullets fired to the east always fly a little higher, and, conversely, bullets fired to the west always travel somewhat low.

The amount of drop change is in function of:

Latitude – The linear velocity of a point on the Earth’s surface, and thus the amount of centrifugal force, is maximum at the equator and decreases going toward the poles, where it is null.

Shooting direction, or azimuth – The amount of drop change is highest when shooting east or west, and as the trajectory angles north or south,  the amount of drop change decreases, becoming null, as the angle points toward either pole. "

The key here is that this site has actually disassembled the two components of the Coriolis effect into the north / south component, properly called the Eötvös Effect, and the simpler, direct component of the Coriolis effect, which is what occurs on a line of latitude (east / west direction).

Again, there is no direction which this effect does not apply to: purely east / west, the point of impact is either moving toward or away from the point of launch.

A simple drawing showing the effect's results using a north / south trajectory: http://www.dictionary.com/browse/coriolis-effect

Brian

Offline Tree

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Re: Math Question
« Reply #58 on: July 20, 2017, 11:34:30 AM »
Right. And maybe even more importantly, he may not have been where he thought he was when he filled up his fuel tank in the first place.

The plot thickens....

Brian

I totally confused Coriolis with Vortex.  Would it have made a difference if I were travelling in a north/south direction?  On the other hand, it doesn't matter because it has nothing to do with coconuts.

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Re: Math Question
« Reply #59 on: July 20, 2017, 12:11:07 PM »
Ok, you said coconuts...

http://youtu.be/Z47NljuPBFE
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