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Spherical Coordinates
This is the coordinate system that is most natural to use - for obvious reasons (e.g. NWP etc.).
λ = longitude (λ increases toward east)
φ = latitude (φ increases toward north)
z = radial coordinate, local vertical (z increases in local "up" dirn)
are unit vectors pointing east, north and up, respectively.
u, v, w are spherical velocity components (latitudinal, longitudinal, and vertical)
[Note: sometimes use same symbols u, v, w for Cartesian components. Should be clear from con-
text whether u, v, w are spherical or Cartesian.]
Let's relate u, v, w to rate of change of a parcel's spherical coords.
Intuitively, u should depend on rate of change of longitude λ. So consider a parcel that travels
eastward a distance δx in a small time δt:
From the figure we have:
We know that
where we assume that a~r.
i
j
k
v ui
vj
wk
δ x = r cosφδλ
φ
φ
r
rcosφ
δλ
rcos
φ
rcos
φ
δx
x
(out of the page)
ui
Similarly, for the N-S component in the spherical system, we’d intuit that it depends on the lati-
tude φ,
From our figure we have,.
and for the vertical velocity
In order to use the conventional ‘dx’, ‘dy’, and ‘dz’ we must use the following transformations:
With these definitions, our velocities are
Don’t be fooled into thinking that this is a Cartesian system however (even though it is made to
look like one with these transformations). The system remains spherical as the unit vectors, which
do not remain constant, are a function of spherical positions. Thus
u lim
δ t → 0
δ x
δ t
≡ ----- lim
δ t → 0
r φ
δλ
δ t
cos ------ φ
D λ
Dt
= ≈acos -------
r δφ =δ y
v lim
δ t → 0
δ y
δ t
≡ lim
δ t → 0
r
δφ
δ t
r
D φ
Dt
a
D φ
Dt
w
Dr
Dt
D
Dt
( a + z )
Dz
Dt
Dx = acos φ D λ , Dy = rD φ , Dz=Dr
u
Dx
Dt
, v
Dy
Dt
, w
Dz
Dt
Du
Dt
D
Dt
ui
vj
wk
( + + ) i
Du
Dt
j
Dv
Dt
k
Dw
Dt
u
Di
Dt
v
Dj
Dt
w
Dk
Dt
δφ
δy
vj
φ
r
r
From our figure then,
But we want
Of course we want , thus
OK, now for the change in the unit vector
Draw our pictures & play with our stickpins and Michaels $1.49 styrofoam globe-surrogates:
For this case we note that changes both latitudinally and longitudinally - so one more degree of
freedom than the component. How does this play out when describing?
So now we need spherical relationships for both , and ....
Off we go, as in Holton but with a little extra info...
δ i
δ i
φ j
δ i
φ k
= sin – cos
∂ i
⁄∂ x
∂ i
∂ x
lim
δ x → 0
δλ φ j
δλ φ k
sin – cos
δ x
lim
δλ → 0
δλ φ j
δλ φ k
sin – cos
acos φδλ
a
φ j
k
a
= = = tan –
Di
Dt
u
∂ i
∂ x
Di
Dt
------ u
∂ i
∂ x
u
a
--- φ j
k
= = ( tan – )
j
j
i
Dj
Dt
Dj
Dt
∂ j
∂ t
u
∂ j
∂ x
v
∂ j
∂ y
w
∂ j
∂ z
∂ j
⁄ ∂ x ∂ j
⁄∂ y
From the diagram we know that what direction does point?
Hint δθ is a good indicator from the top of the diagram.
Thus points due west (and is to ). What is the scalar magnitude or length of?
So what is δθ? From the big figure above,
This leads to the next question....|AB| =?
Looking at triangle ABC, we get
Thus,
δ j
j
( x +δ x ) j
= – ( ) x δ j
δ j
⊥ j
δ j
δ j
δθ j
AB δθ = δ x
AB
acosφ
sin φ
= = acotφ
δx
δλ
δ j
φ
acosφ
j
( ) x
j
( ) x
j
( x +δ x )
φ
90−φ
90−φ
φ
δθ
A
C
B
δθ
λ
λ +δλ
a
δθ
j
j
δx
grouping all of our terms,
Du
Dt
D
Dt
------ ui
vj
wk
( + + ) i
Du
Dt
------- j
Dv
Dt
------- k
Dw
Dt
-------- u
Di
Dt
------ v
Dj
Dt
------ w
Dk
Dt
i
Du
Dt
uv
a
wu
a
Dv
Dt
u
2
a
φ
wv
a
Dw
Dt
u
2
a
v
2
a
