Coriolis - apparent?

[VR-HDB]

Maybe someone can help me with this;
I (think I...) understand the idea that the CF is an apparent force that is a "product" of different frames of reference - the object is traveling in a straight line, but when viewed from the earth it appears to curve.


What I don't understand is:
If the CF is only an apparent force - how can it create the spinning motion of, for example Tropical Revolving Storms? They definitely seem to be rotating regardless of ones reference (earth or satellite).

[B Fraser]

The coriolis force is a difficult one to get your head around. It is a true vector force i.e. it has direction and magnitude, that is, it acts always at 90 degrees to your direction of travel and (correct me if I'm wrong) is proportional to your velocity.

Tropical storms, cyclones, anticyclones rotate due to the influence of the CF. The geostrophic wind (2000 feet say, outside the influence of surface friction) flows parallel to the isobars where the CF is balanced by the pressure gradient force. If it wasn't for the CF, there would not be much weather as any pressure imbalance would result in a bit of wind (sea breezes for example). Thanks to the good old CF, we have all sorts of interesting shennanigans to contend with.

There are some really good web pages to be found where the forces are explained alongside some animated graphics. Happy hunting.

[VR-HDB]

I can see the effects of the CF, but what is confusing me is that it is referred to as an "apparent force" .

English is not my first language, but doesn't this mean that it is not actually a real force - its just the way we see it from our reference point. Since we can clearly see the effects of the CF - isn't it a real force then?

[bookworm]

A very good question.

It's precisely becausethe results of the coriolis force are difficult to visualise in a non-rotating frame that it's useful to look at the situation in the frame rotating with the earth, even if you have to add apparent forces.

If you look at the school textbook problem of a stone twirling around on a string, it's easy to solve the problem in either a non-rotating frame, or a frame rotating with the stone, with centrifugal force added.

But when you start looking at objects moving in the rotating frame and at extra forces being applied that are dependent on displacements in the rotating frame too (e.g. the pressure gradients on the surface of the earth, which move with the surface of the earth as it rotates), it becomes almost impossible to model and solve the problem in a non-rotating reference frame. In a non-rotating reference frame, the motion of the air in a cyclone is not a simple rotation but very complex indeed.

[D120A]

Coriolis acceleration is the acceleration experienced by an object moving in a straight line on a rotating frame of reference. Let me use a simple example. An old-style gramophone record rotates (at 33 and a third rpm, if it is an old LP player). A fly, trying to walk in a straight line at a constant speed across the rotating record, would experience a Coriolis acceleration. The fly would have to dig his heels in to ensure that he developed the necessary force to be able to accelerate that way, where the force required equals his mass times the required acceleration. If he couldn’t, he wouldn’t be able to hold the straight line and he would wander off!

(To digress a moment, a similar situation exists when you consider centripetal acceleration and force; if you whirl a mass around your head on a piece of string, the tension in the string provides the ‘centripetal’ force that makes the mass seek (Latin, ‘peto’ – to seek) the centre of the circle. Cut the string and the mass would whirl away in a straight line like a slingshot because, then, no force is causing it to change its ‘state of uniform motion’ - which is to travel at constant speed in a straight line - Newton’s Law. Be careful not to use the term ‘centrifugal’ force in such analysis, because it doesn’t exist. ‘Fugo’ is Latin for ‘to flee’. There may be a centrifugal reaction of the string on your fingers as you whirl the mass round, but what makes the mass accelerate towards the centre is centripetal force. End of rant!)

Back to Coriolis. The value of the Coriolis acceleration is 2 times the angular velocity of the rotating frame of reference (in radians per second) times the speed of the object travelling across it. The direction of it is a vector product of the two, and that is fiendishly difficult to explain without waving hands in the air because it has both magnitude and direction, but I will have a go. Magnitude is easy, it is the product of 2 and the two values. For direction, however, we have to consider the vector directions of the two values and how they multiply – which will give us a vector in a third direction. Consider the turntable. If you look down upon it, suppose it rotates clockwise. Curl the fingers of your right hand in the direction of rotation; where is your thumb? Pointing downwards. That is the direction of the angular velocity vector of the record turntable.

Now the fly. If it is struggling to walk from the rim of the record to the centre, that direction is that of its velocity vector on the rotating frame of reference - the record. The vector product of two vectors, one down the spindle into the record player and the other at right angles to it, is again given by the right-hand-rule. Use the fingers to align the rotation vector with the velocity vector and where is the thumb? Out to the right as you look down on the turntable and that’s the direction of the Coriolis acceleration.

So, I reckon an aircraft tracking over the North Pole in a straight line at 500 mph (733 feet/sec), where the earth is rotating at 2-pi (6.283) radians per day with a vector vertically upwards, experiences a Coriolis acceleration of 0.1066 feet per sec per sec, out towards the port wing tip. Check my calculation, and remember a day has 24x60x60 seconds! So there, it will be ‘left wing down a bit’ (I think) to counter it, and that is the maximum value of Coriolis you will experience over the earth at 500 mph – it is only 0.33% of a ‘g’. But, in contrast, an aircraft travelling northwards at the Equator experiences zero Coriolis, because the angular velocity and aircraft velocity vectors are in the same direction and so their vector product is zero. The brave souls who first developed INS (the late Charles Stark Draper and his team, at MIT) had to work out the Coriolis equations to allow for all latitudes, speeds and three-dimensional flight paths. 0.33% of a ‘g’ is a big number in an inertial navigation system.

[whazup]

It is common to see the Coriolis force described as "making it look like a force is acting upon the object, but actually there is no real force acting on the object". This prompts the question, "what is a real force"? From the viewpoint of general relativity, all coordinate systems are equivalent in describing physical processes, but in changing from one system to another things that look like forces will arise. For example, at the surface of Earth it is possible to (locally) remove the gravitational force by changing to a coordinate system accelerating towards the centre of Earth. But no-one would call gravity "fictitious".

Good scientific explanation: http://www.physics.ohio-state.edu/~dvandom/Edu/newcor.html

[Flight Safety]

I'll give this a try.

The Coriolis effect is referred to as an "apparent force" because it actually isn't real, but appears to be real to an observer in an accelerating reference frame. It's actually a correction factor that has to be applied to a force equation, in order to work out the real forces acting on an object, where the observations of the behavoir of that object, are distorted by the accelerating reference frame.

Here's an example to illustrate.

Assume a merry-go-round, with a cup to receive a golf ball in the middle. Also, assume a grid of lines on the merry-go-round surface, layed out as a series of radius lines extending from the center outward to the edge, and a series of concentric circles with the cup as the center of the circles. Assume the grid lines will be used to measure the forces acting on a golf ball.

Also assume that a level platform is located right next to the edge of the merry-go-round, with a surface height equal to the height of the merry-go-round's surface. Assume also that there's a chair on the edge of the merry-go-round facing the cup, to allow an observer to watch the actions of a golf ball.

Now we putt a golf ball from the platform to the cup, with an observer sitting in the chair, while the merry-go-round is stationary. In this case the observer can easily see that the only force acting on the golf ball, was the linear force applied to the ball by the putter. No other force was apparent to the observer, as the golf ball moved in a straight line from the platform to the cup. The observer watched the motion of the golf ball, and used his grid measurement system that was NOT in accelerating motion, to observe the motion of the golf ball.

Now lets assume that the merry-go-round is put in motion at a constant rate of rotation. Rotation is an accelerating reference frame because all rotating objects are undergoing directional acceleration, even when the rate of rotation remains constant.

Now we again putt the golf ball from the platform to the cup, with an observer in the chair, and we'll assume that the surface of the merry-go-round is basically frictionless. Because the observer of the event is now in an accelerating frame of reference, the golf ball will appear to him to have mutliple forces acting on it. An "apparent force" will appear to the observer to cause the golf ball to move in an inwardly directed spiral towards the cup. The observer's accelerating reference frame has now distorted his observation, because both he and his grid measurement system are part of the accelerating reference frame.

So now the observer sees a "coriolis force" acting on the golf ball. The force is not real, but it's apparent to him. So from his vantage point, he has to apply the "coriolis force" as a correction factor to a force equation, in order to figure out what really happened to the golf ball, from his acceleration distorted viewpoint.

I hope this makes sense.

[chrisN]

Alternatively, and to get back to the question of cyclones etc., consider the model where the golfer is on the edge of the merry-go-round and the observer is on the fixed chair/platform.

With the merry-go-round stationary, the golfer hits the ball; it goes straight along one of the radial lines (and over a fixed line on the ground) into the cup. Suppose the observer, and the golfer at that point, are at the south edge of the merry-go-round.

With the merry-go-round going anti clockwise (a model of the earth's west to east rotation viewed from above the North Pole), if the golfer hits the ball from the south edge and it goes into the cup, it still travels in a straight line as viewed from an external frame of reference (the observer sees it go over a fixed line on the ground). To do this, the golfer would have to incorporate into his swing a backward component - to him, the ball would appear not follow a radial line on the moving merry-go-round, but the spiral path mentioned above.

Alternatively, if the golfer just hit the ball along the northwards radial line, it would miss the cup and go past it on the right. The northward hit of the golfer, combined with the eastward motion of the edge of the merry-go-round, would take it on a route seen by the observer to go through a vertical plane angled north-eastwards, not directly north. Similarly, a parcel of air at the equator in a high pressure region, trying to fill a depression to its north, is impelled by the pressure gradient to go north plus a component to the east provided by the earth's rotation, so it misses the centre of the depression and passes it to the right.

Meanwhile, one can follow a similar analogy for a golfer at the centre hitting a ball to the edge, and a parcel of air at the North Pole trying to fill a depression to its south. It will miss and pass by on the right side as viewed by the golfer/from the North Pole - i.e. to its west.

So air trying to reach a depression in the northern hemisphere goes up the eastern side and down the western side, imparting anticlockwise rotation to a cyclone/low pressure. Air going towards it from other points of the compass is similarly deflected to varying degrees, which happen to work to give an entirely circular motion. Viewed from outside the earth (the observer outside the merry-go-round), there is no real acceleration in any direction except the initial impetus. But viewed by somebody rotating with the merry-go-round/earth, the ball/air seems to acquire an acceleration making go in a spiral - the coriolis acceleration. That is why it is an apparent, not real, acceleration.

Hope that helps.

Chris N.

[stalklow]

http://www.ukdivers.net/meteorology/coriolis.htm

This link may help