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Coriolis Effect


Bodhi

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Posted
  • Location: Moray UK
  • Location: Moray UK

OK I am very happy with a projectile being sent North or South veering to its right in the N Hemisphere (and vice versa down under). It is clear that rotation speeds at each latitude are different and therefore the East West vector of speed of the projectile while 0 to the firer changes as it moves North or South.

However, that doesn't describe why it is said to be a force proportional to wind speed. The effect I have described is absolute for any speed as the Northerly projectile being fired has no East/West component except that given by the coriolis effect, which also has no North/South vector.

On an East/West projectile therefore there is no coriolis effect at all by this description.

Secondly the effect, usually subsumed into the coriolis one, of an East/West projectile veering to the LEFT doesn't make sense to me. The explanation given that the lines of latitude are not straight is not very good. I can cut a completely flat plane through a globe along a line of latitude. The ring is formed is perfectly straight as it goes round the curved plane ie the line of latitude. A projectile or wind on this line could quite happily follow this path without veering off.

Any help on where I am mistaken, with explanation, would be very useful. Thanks alot to the brave soul(s) who will answer this one :)

All the best

Rich

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Posted
  • Location: Crowborough, East Sussex 180mASL
  • Location: Crowborough, East Sussex 180mASL

It's because the Coriolis force is an apparent force and not a real force. That is to say the Coriolis force does not arise from any physical interaction but rather from the acceleration of the reference frame (the earths rotation) itself.

Think of what happens in the Foucault pendulum example where a body of large mass and hence inertia is set in motion. Newton says the body will keep swinging back and forth in a straight line as there are no other forces other than gravity (for all intents and purposes) acting on it. But the earth is rotating underneath it. The pendulum stays constant but because the earth rotates so the pendulum rotates slowly in relation to the ground underneath and appears to be acted upon by an external force. IT IS NOT.

The reference frame for the pendulum (i.e. the earth has moved) not the back and forth alignment of the pendulum itself. But in order to describe the effect mathematically, the rotation is attributed to an imaginary or apparent force that would have the same effect.

Because of the earths rotation and hence the differing velocity as one moves toward the poles, the air mass will rotate. Within that mass there will be a real air pressure gradient which is a very real force.

It's now easy to see that the pressure gradient (real) force is added to the corriolis apparent force, the apparent vector sum will always thus be proportional to the wind speed.

Try this explanation with diagrams if it helps: http://www.aos.wisc....os101/wk11.html

ffO.

Edited by full_frontal_occlusion
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Posted
  • Location: Camborne
  • Location: Camborne

Well ffo has already answered but I'll just carry on anyway as I've done it now. A brief PDF taken from Atmosphere, Weather and Climate (seventh edition) by Roger G. Barry and Richard J. Chorley.

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Posted
  • Location: Moray UK
  • Location: Moray UK

Hi folks

Thanks for the replies. I titled the post the Coriolis Effect because I am very much aware of the fact it isn't a real force. This however does not answer my question at all I am afraid. I also describe the changing velocities towrds the poles but again this does not describe why a moving body should have a change of motion directly proportional to its velocity.

The pdf is better but the presuppositions in it of why the maths used is as it is is what I would like a fuller explanation of please?

Cheers

All the best

Rich

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Posted
  • Location: Moray UK
  • Location: Moray UK

Afternote: I had seen the webpage before in my searching for an answer. It just says "The magnitude of the coriolis force is proportional to the speed of the wind." and that is the justification for using it further on. I am not disputing that it may be so I am just asking HOW exactly my intial explanation is wrong.

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Posted
  • Location: Moray UK
  • Location: Moray UK

OK I'll start with what I know:

1) The only forces actually involved with an object on the Earth are the gravitational attraction between its mass and the Earth and the physical barrier of the surface boundaries of each pushing against each other. In other words the object is pulled by a force m1g (where m1 is its mass and g is the acceleration due to gravity- 9.81 ms-2 approx)

2) An object at the equator has all of this force directed towards the rotational axis at 90 degrees, whereas at any other altitiude the force directed at the centre is not directed directly at the axis of rotation. The difference in degrees is equal to the latitude itself.

3) Point 2 suggest that a centripetal force can be calculated at that latitude by using- m1g cosA -(where A is degree of latitude). This centripetal force is real but merely part of the vector breakdown of m1g. It is also directed inwards of course. It must by nature get smaller to towards the poles and be at amaximum at the equator. Yet for the coriolis effect the acceleration must increase a latitudinal velocity and so the a= v2/r where a is reducing and r is reducing as we go further North we must still get an increase in v.

4) I'll try to get some accurate figures to input in this- as it might be becoming clearer just through this process of writing it out.

Please feel free to chip in anyone :)

All the best

Rich

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Posted
  • Location: Camborne
  • Location: Camborne

I'll be honest and admit I'm not quite sure what you getting at I thought the Coiriolis had been explained. I think you are wrong about the forces involved because in atmospheric motion they are the pressure gradient force, gravitational force, and frictional force. Perhaps this might be better considered as whole. I'll attach a brief PDF courtesy K. Mohanakumar, Stratosphere Troposphere Interactions. It does contain some basic maths so any questions concerning the equations please address to higher authorities in the forum.smile.png

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Posted
  • Location: Crowborough, East Sussex 180mASL
  • Location: Crowborough, East Sussex 180mASL

Just noticed this still ongoing.

Let's try a different tack? Please forgive me if I take this back to basics (I know you already understand, but it helps the thought process - and useful for others who do not)

1) Start with the givens:

  • Our numerical weather problem, we need to determine the forces acting on a finite element of the atmosphere at any given point in time within a 3-dimensional co-ordinate system. (u,v,w).
  • By determining the forces acting on the finite element of the atmosphere, we should be able to predict its future vector - i.e. velocity magnitude and direction and hence it's futire position.
  • Our 3-dimensional co-oridnate system (frame of reference) is non-inertial. i.e. it is itself one that is undergoing acceleration due to the earths rotation.

2) Newtons laws of motion are defined with respect to a fixed reference frame. i.e. one that is not moving.

3) Newton describes acceleration as the rate of change of momentum or direction of an object w.r.t. that fixed reference frame. In other words, measurements made w.r.t. the earth.

Big problem:

  • Unfortunately, the earth is not a fixed reference frame as it is rotating. To successfully solve our weather problem, we must therefore include a description which defines that reference frame acceleration (elminate the error) or.......
  • We must measure the effect of the earths rotation on that element of the atmosphere and describe the error arising as an apparent (imaginary) force exactly compensating for that error.
  • That compensation is described as the Coriolis force.

6) The velocity vector component arising from the Coriolis force at any temporal point within the 3-dimensional co-ordinate system must now be defined:

7) So, referring to that finite element of the atmosphere:

  • The inertia of any element is directly related to its velocity. i.e. a larger force is required to change its velocity or direction of travel as its velcity increases.
  • If we measure its velocity vector relative to our co-ordinate system to be zero, then by association, the Coriolis effect at that point must also be zero. i.e. there is no relative motion.
  • As the elements velocity vector increases, then its inertia also increases and the Coriolis compensation also needs to increase accordingly to null the earths rotation effect.
  • The Coriolis compensation increases with increasing latitude. i.e. zero at the equator and maximum at the poles.

In other words, the Coriolis compensation force that must be applied to correctly define the earths rotation and all other real forces acting on that element is directly proportional to that finite elements velocity vector which is itself defined by:

a.) The pressure gradient (independent of the earths rotation)
b.) Frictional force (independenat of the earths rotation)
c.) Gravity. (independent of the earths rotation)
d.) Latitude

NB The Coriolis effect cannot generate a wind, it can only change its direction.

By extrapolation, the Coriolis force is therefore always proportional to wind speed and with a latitude component defined by:

F
coriolis
= 146x10
-6
v.sin(latitude)
Where v is the wind velocity (m/s) and the constant 146x10
-6
as defined by the rotation rate of the earth.

As such, the rotation of the earth is only in one direction and hence the Coriolis force always acts perpendicular to the direction of motion of the earth. The force acts to the right when looking north in the northern hemisphere which describes the motion of the atmsophere as veering to the left (equal and opposite according to Newton).

I hope this aids your understanding. I don't believe I have made any mistakes. Try not to get bogged down by the full mathematical description, they are not intuitive in this process.

Try to envisage what's happening physically first and then relate that to the full mathematical description.

Best wishes.

ffO.
Edited by full_frontal_occlusion
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Posted
  • Location: Crowborough, East Sussex 180mASL
  • Location: Crowborough, East Sussex 180mASL

Having read through my own answer, which helped me to crystalise why Bodhi is having difficulty understanding why the Coriolis force is always proportional to wind speed, I can simplify the whole process by stating:

The Coriolis force is neither real nor can it give rise to a wind.

Any mass of object, no matter how big or small or how fast it is moving, must by definition give rise to a proportional Coriolis component since the earth will move irrespective of those variables.

That is to say a mass of 1gm needs a different compensation factor than one of say 100,000 tonnes. Both will appear to move in exactly the same way when the earth rotates underneath it. It matters not a jot how big or fast it is travelling.

However the Coriolis force must therefore always be proportional to exactly compensate that mass.

If we then consider that the inertia of an already moving object (in our case the wind speed of any given mass of air) is subject to exactly the same argument, it should now be crystal clear that the Coriolis force is indeed proportional to the wind speed.

I really do hope this will help you and perhaps others who have struggled to grasp the concept.

ffO.

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Posted
  • Location: Moray UK
  • Location: Moray UK

Firstly thank you FFO for continuing with this thread.

I still don't think we have a definite answer here unfortunately. Why?

I am with you for your initial points until "Big Problem"

The compensation for an object travelling north from the equator to an observer on the ground would seem to just be based on its EW velocity picked up at the equator from where it started. This compensation is a direct one that relates not to the objects NS velocity but only to its EW component. This compensation is fixed irrespective of the NS therefore.

The coriolis force is non-intuitive it seems and probably does require the maths to explain it I am thinking....

The pdf is good but again assumes the coriolis effect. Is there no simple step by step explanation of the coriolis term on the net, as I can't find anything that doesn't give me a headache!

Cheers

Rich

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Posted
  • Location: Crowborough, East Sussex 180mASL
  • Location: Crowborough, East Sussex 180mASL

Firstly thank you FFO for continuing with this thread.

The compensation for an object travelling north from the equator to an observer on the ground would seem to just be based on its EW velocity picked up at the equator from where it started.

True. In addition, the Coriolis component is proprtional to the latitude of the object.

This compensation is a direct one that relates not to the objects NS velocity but only to its EW component. This compensation is fixed irrespective of the NS therefore.

Not true. As the latitude of the object increases away from the equator then to conserve the angular momentum (omega) of that object, the spin rate must also increase.

You have missed out the conservation of momentum laws:

The angular momentum is the product of mass x spin_velocity x radius.

If the radius of spin decreases then it's rate of spin increases. This is the crucial factor I think you are missing.

Think of an ice skater and the conservation of momentum - the skater spins with arms outstretched (large radius), then pulls their arms inwards (decreases radius) and the spin rate increases substantially.

Going back to our problem: as the object moves north it decrease its radius of spin closer to the spin pole. it's spin rate must also change to conserve momentum. hence the Coriolis force must also change accordingly.

That also means the rate of change of Coriolis force is dependent on the rate of change of distance from the equator (velocity).

ffO.

Edited by full_frontal_occlusion
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Posted
  • Location: Moray UK
  • Location: Moray UK

Hi FFO

Thanks for replying. Yes, I agree that angular momentum wasn't taken into consideration. If we look at the formula for this we have:

Angular Momentum = Iw where I is inertia and w is the angular velocity.

This also equals mvr where m is the mass, v is the instantenous velocity (or general speed if you like) and r is the radius.

Given that m will be the same for the air parcel at the equator and all the way up through the SN trajectory we can see that to conserve angular momentum we will need to increase v as r decreases.

If we take r at the equator to be 6372km and, by my calc, the earths surface speed at that point to be 463ms-1 we get the following speed increases at the following latitudes:

20N 492ms-1

45N 655ms-1

62N 986ms-1

85N 5316ms-1

Two points:

The spin speed increase is still only proportional to the EW beginning velocity and so still unaffected in the NS direction (ie the angular momentum is all at 90degrees to the NS intial velocity).

The velocity at 85N would require a centripetal acceleration of 51ms-2 to hold it in place. Even if gravity were fully acting in the direction of centripetal acceleration (which it is not by a huge margin) it could only provide just under 10ms-2 of that 51. As it happens it has such a weak centripetal component at 85N that it is hugely under 1ms-2. Ie the air parcel would fly off into space.........

For the coriolis effect the angular momentum conservation isn't going to solve it.

I did these calculations before looking at the (kindly provided) web address and nothing I have said in this post contradicts the sites very good explanation.

So we are still needing a reason for the NS velocity to affect the EW component and increase curvature when we fire a shot (or move an air parcel) due North....

Back to you

Cheers and all the best

Rich

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