User:Stamcose/sandbox

Interference by user "FyzixFighter"

My original formula was

${\boldsymbol {a}}_{C}=-2\,{\boldsymbol {\Omega \times v}}+\omega ^{2}{\boldsymbol {r}}_{ort}\ +{\boldsymbol {f}}$

Because user "FyzixFighter" is "not a big fan of the force per unit mass formulation" because he thinks that "it tends to "muddy the distinction between force and acceleration" he changed it to

${\boldsymbol {F}}-m{\frac {\operatorname {d} {\boldsymbol {\Omega }}}{\operatorname {d} t}}\times {\boldsymbol {r}}-2m{\boldsymbol {\Omega }}\times {\boldsymbol {v'}}-m{\boldsymbol {\Omega }}\times ({\boldsymbol {\Omega }}\times {\boldsymbol {r}})$ $=m{\boldsymbol {a'}}$

1 - I disagree, there is no risk for any "muddy"

2 - If one really does not like " force per unit mass" simply write

${\boldsymbol {a}}_{C}=-2\,{\boldsymbol {\Omega \times v}}+\omega ^{2}{\boldsymbol {r}}_{ort}\ +{\frac {\boldsymbol {F}}{m}}$

or if preferred

${\boldsymbol {a'}}=-2\,{\boldsymbol {\Omega \times v'}}+\omega ^{2}{\boldsymbol {r}}_{ort}\ +{\frac {\boldsymbol {F}}{m}}$

3 - He even screved it up, what he meant was

${\boldsymbol {F}}-2m{\boldsymbol {\Omega }}\times {\boldsymbol {v'}}-m{\boldsymbol {\Omega }}\times ({\boldsymbol {\Omega }}\times {\boldsymbol {r}})$ $=m{\boldsymbol {a'}}$

but it is certainly better for the reader to see my

$\omega ^{2}{\boldsymbol {r}}_{ort}$

rather then the more mysterious "double cross product" what is the same, only more complicated!

${\boldsymbol {\Omega }}\times ({\boldsymbol {\Omega }}\times {\boldsymbol {r}})$ 

of user "FyzixFighter"

I would therefore propose to use (as a compromise!)

${\boldsymbol {a'}}=-2\,{\boldsymbol {\Omega \times v'}}+\omega ^{2}{\boldsymbol {r}}_{ort}\ +{\frac {\boldsymbol {F}}{m}}$

To multiply the Coriolis term and the Centrifugal term with the mass to be able call them "forces" is uttermost nonsense!

Am I right? If I do not receive any protests I will change it to this (correct and useful!) format

Formula

See also: Fictitious force

The Newtonian equation of motion for a mass point in an inertial system is

${\boldsymbol {a}}={\boldsymbol {f}}$

where ${\boldsymbol {a}}$ is the acceleration of the mass point relative to the inertial system and ${\boldsymbol {f}}$ is the force (per unit mass) acting on the mass point. Transforming this equation to a rotating system it takes the form

${\boldsymbol {a}}_{C}=-2\,{\boldsymbol {\Omega \times v}}+\omega ^{2}{\boldsymbol {r}}_{ort}\ +{\boldsymbol {f}}$

where

${\boldsymbol {a}}_{C}$ is the acceleration relative to the rotating reference system

${\boldsymbol {\Omega }}$ is the rotation vector of the rotating system relative to inertial space

${\boldsymbol {v}}$ is the velocity relative to the rotating reference system

$\omega$ is the (scalar) rotation rate of the rotating system relative to inertial space

${\boldsymbol {r}}_{ort}$ is the projection of the position vector ${\boldsymbol {r}}$ on the plane orthogonal to ${\boldsymbol {\Omega }}$

${\boldsymbol {f}}$ is the force (per unit mass)

and "$\times$" signifies vector cross product

See fictitious force or for a derivation.

The term

$\omega ^{2}{\boldsymbol {r}}_{ort}$

which (only) depends on the position in the rotating system is known as the "centrifugal force" and the term

$-2\,{\boldsymbol {\Omega \times v}}$

which (only) depends on the velocity relative to the rotating system is known as the "coriolis force"

The Coriolis effect is the behavior added by the Coriolis acceleration. The formula implies that the Coriolis acceleration is perpendicular both to the direction of the velocity of the moving mass and to the frame's rotation axis. So in particular:

• if the velocity is parallel to the rotation axis, the Coriolis acceleration is zero. (For example, on Earth, this situation occurs for a body on the equator moving north or south relative to Earth's surface.)
• if the velocity is straight inward to the axis, the acceleration is in the direction of local rotation. (For example, on Earth, this situation occurs for a body on the equator falling downward, as in the Dechales illustration above, where the falling ball travels further to the east than does the tower.)
• if the velocity is straight outward from the axis, the acceleration is against the direction of local rotation. (In the tower example, a ball launched upward would move toward the west.)
• if the velocity is in the direction of local rotation, the acceleration is outward from the axis. (For example, on Earth, this situation occurs for a body on the equator moving east relative to Earth's surface. It would move upward as seen by an observer on the surface. This effect (see Eötvös effect below) was discussed by Galileo Galilei in 1632 and by Riccioli in 1651.)
• if the velocity is against the direction of local rotation, the acceleration is inward to the axis. (On Earth, this situation occurs for a body on the equator moving west, which would deflect downward as seen by an observer.)

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The concept "Coriolis force" is specially suitable for the description of motion of atmosphere (i.e. winds) over the surface of the Earth. The Earth (like all rotating celestial bodies) has taken an oblate shape such that the gravitational force is slightly off-set towards the Earth axis as illustrated in the figure.

Balance between gravitational an centrifugal force on the Earth surface

For a mass point at rest on the Earth surface the horizontal component of the gravitation counteracts the "centrifugal force" preventing it to slide away towards the equator. This means that the vector term of the "equation of motion" above

$\omega ^{2}{\boldsymbol {r}}_{ort}\ +{\boldsymbol {f}}$

is directed straight down, orthogonal to the surface of the Earth. The force affecting the motion of air "sliding" over the Earth surface is therefore (only) the horizontal component of the Coriolis term

$-2\,{\boldsymbol {\Omega \times v}}$

This component is orthogonal to the velocity over the Earth surface and is given by the expression

$\omega \,v\ 2\,\sin \phi$

where

$\omega$ is the spin rate of the Earth

$\phi$ is the the latitude, positive in northern hemisphere and negative in the southern hemisphere

In the northern hemisphere where the sign is positive this force/acceleration is to the right of the direction of motion, in the southern hemisphere where the sign is negative this force/acceleration is to the left of the direction of motion

Intuitive explanation

If there is wind towards east the inertial velocity of the atmosphere (relative to the Earth center) is higher then the velocity of the Earth surface below. The horizontal component of the gravitational force is then not sufficient to prevent the atmosphere from being accelerated towards the equator. If there is wind towards west the the centrifugal effect is instead lower and the horizontal component of the gravitational force accelerates the atmosphere away from the equator. If there is wind in the direction "away from the equator" the atmosphere will reach areas where the eastward velocity of the ground is lower. Relative the ground the atmosphere therefore have an increasing eastward component. At the same time the centrifugal effect decreases the original velocity component "away from the equator". If there is wind in the direction "towards the equator" the atmosphere will reach areas where the eastward velocity of the ground is higher. Relative the ground the atmosphere therefore have an increasing westward component. At the same time the horizontal component of the gravitational force decreases the original velocity component towards the equator.

A detailed description of an "inertial circle" resulting from the "Coriolis effects" described above can be found in ref x