Tuesday, June 26, 2012

Singular Spectral Analysis of the Summer (DJF) Median Maximum Temperature for Adelaide between 1888 and 2011

Figure 1 (below) shows the maximum daily temperature 
for the South Australian capital city of Adelaide for the
summer of 2009-10. The data comes from the High-
Quality Australian Daily Temperature Data Set

Figure 1:

Summers in Adelaide are characterized by frequent
hot-spells with maximum temperatures exceeding 35
degrees celsius. These hot-spells generally last for a
day or 
two before temperatures are moderated by 
flows of colder air from the south. 

The highly variable nature of maximum summer 
temperatures in Adelaide means that the best way
to characterize the average summer [December/
January/February (DJF)] maximum temperature 
is to use the median rather than the mean. 
 Figure 2 (below) shows the summer-time (DJF)
median maximum temperature for Adelaide between
1888 and 2011.  

Figure 2

Singular Spectral Analysis (SSA)

Preliminary Results:
If you use SSA to investigate the de-trended maximum 
temperature time series (see figure 2), you find that there 
are spectral peaks at all sub-multiples of the 22.3 year
Hale (H) cycle from H/2 to H/10. The most prominent
sub-harmonics are those at H/3, H/6, H/9 and H/10.

This result strongly suggests that the long-term median 
summer time maximum temperatures in Adelaide are
primarily being driven by factors that are associated 
with the 22.3 year Solar Hale Cycle. 

The presence of sub-harmonics in the temperature record
is indicative of the fact that the ~ 22 year forcing term
must have a broadened temporal structure that is 
triangular like in appearance.     

Figure 3

This result is broad general agreement with the results of
Thresher (2002) who finds that the variability in the strength
of the zonal west winds (along their northern margins) broadly
correlate with the 22 year sunspot cycle [see abstract below]. 

Hence, the most likely causation sequence is:

22 Hale cycle ---> strength of zonal west winds --->
the median summer maximum temperatures in Adelaide

with the strength of the zonal west winds depending directly
on the strength of the wind vorticity around low and high 
pressure cells in the Southern Hemisphere.


Int. J. Climatol. 22: 901–915 (2002)

Royal Meteorological Society.



Atmospheric circulation in the southern mid-latitudes is
dominated by strong circum-Antarctic zonal west winds
(ZWW) over the latitude range of 35 to 60 °S. These 
winds exhibit coherent seasonal and interannual variability, 
which has been related both to Antarctic (e.g. polar ice) 
and low-latitude climate (e.g. El Ni˜no–southern oscillation) 
parameters. Historical and recent studies suggest that, 
at its northern margins, variability in the ZWW also has 
a marked quasi-decadal component. Analysis of sea-level 
pressure and rainfall data for the Australian region, South 
Africa and South America confirms frequent indications 
of quasi-decadal variability in parameters associated with 
the ZWW, which appears to be in phase around the 
hemisphere. This variation broadly correlates with the 
sunspot cycle, and specifically appears to reflect 
sunspot-correlated, seasonally modulated shifts in the 
latitude range each year of the sub-tropical ridge over 
eastern Australia. Sunspot-correlated variability in the 
southern mid-latitudes is likely to have substantial effects 
on temperate climate and ecology and is consistent with 
recent models of solar effects on upper atmospheric climate, 
though the mechanisms that link these to winds and rainfall 
at sea level remain obscure. 

Wednesday, June 20, 2012


Hypothesis: The -12.57 μsec change in the length-of-day 
(LOD) associated with the 18.6 year Draconic lunar tides
is a direct result of a systematic one degree shift to the 
South/North by the Southern/Northern Summer Sub-Tropical 
High Pressure Ridge that is produced by lunar atmospheric 

A. The Mass of Summer Sub-Tropical 
_______High-Pressure Ridge

The ideal gas equation states that:

__________P = ρRT/M___________(1)

where P is the atmospheric pressure, ρ is the atmospheric 
density, R is the Universal Gas Constant (= 8.31432 
Nm/Mole), T is the temperature and M is the molar 
mass of air (= 2.89644 x 10-2).

If we assume that in the vertical direction (z) the 
atmosphere is hydrostatic equilibrium then:

_________dP = −ρ g dz___________(2)

where dP is the change in pressure with change in 
height dz and g (=9.80 m/s2) is the acceleration due 
to gravity (N.B. g = 9.80665 m/sec2 at 0 m, 
g = 9.797 m/sec2 at 3000 m, and g =  9.776 m/sec2 
at 10000 m).

Hence, from equations (1) and (2) we find that:

________dP/P = −ρ g dz M / ρRT = −gMdz / RT___(3)

which gives the barometric formula as its solution:

________P = PO exp [−gMdz/RTO]_____________(4)

where PO (in Pascals) and TO (in Kelvin) are the pressure 
and temperature at the Earth’s surface and dz = (h – hO
where h is the height in metres above some reference height hO.

Now Newton’s Second Law tells us that:

________P = MAT g / A_____________________(5)

where MAT is the mass of atmosphere above a given area 
A of the Earth’s surface, above a given height h. 

Combining equations (4) and (5) we get:

_____MAT = [PG A /g] x {1 − exp [−gMh/RTG]}__(6)

where [PG A /g] is the total mass of the atmosphere above 
a given area A of the Earth’s surface, PG and TG are the
 pressure and temperature at ground level  
(i.e. TG = 288.15 K (15 CO) and PG = 101,325 Pascals), 
respectively, and h is the height above ground level in metres.

If we take A to be the surface area of the whole Earth = 4πRR,
 R = 6.371 x 106 m, we get:

Area of the Earth’s Surface
(1014 m2)
Height (m)
Mass of atmosphere  above height h 
(1018 Kg)
Mass Faction

where the data in the second last row uses the pressure 
value derived from the barometric formula while the data
 in the last row uses the value for the US Standard Atmosphere. 
All of the values for TO, are those for the US standard atmosphere.
     The actual NCAR value for the total mass of the atmosphere 
that is derived from sophisticated models is 5.1480 x 1018 kg for 
wet air (+/- 1.2 to 1.5 x 1015 kg due to water vapour) and 
5.1352 x 1018 kg for dry air. Our simple model gives a value 
of 5.270 x 1018 kg  most of the difference explained by the 
reduction of volume due to mountain ranges.
     The surface area of the Earth’s southern summer sub-tropical
 high pressure ridges can be obtained by assuming that bottom 
surface area this body of air is a zone of a sphere with the radius 
of the Earth, centred at 40O ± 10O South of the Equator.

__________Area = 2 π R h_______________(7)

where h = height of the zone = R (sin LU – sin LL) and LU 
and LL are the latitudes of the bottom and top arcs of the zone.

_________Area = 2 π R2 (sin 50O – sin 30O)__(8)

which gives and area of 6.7850 x 1013 m2 (approximately 
13.3024 % of the area of the whole Earth’s surface).

Area Earth’s Sub-Tropical high pressure ridge.
(1013 m2)
Height (m)
Mass of atmosphere  above height h 
(1017 Kg)
Mass Faction

B. Change in Angular Momentum of the Earth

CASE A: The Summer Southern Sub-Tropical 
High-Pressure Ridge shifts south by 1O.

Let X = the radius of circle of latitude of the Earth at latitude L.

____________X = R cos L___________(9)

___________ΔX = -R sin L ΔL________(10)

for ΔL = 1O = 0.017453293 radians and L = 40O
ΔX = 71.48 km

(N.B. In winter time the corresponding numbers would be:
for ΔL = 1O = 0.017453293 radians and L = 30O                 
ΔX = 55.60 km)

Now the Total Angular Momentum of the Earth (L) is 
given by:

___________L = I ω_______________(11)

where I = MAT X2, ω = VT/R is the angular frequency 
of rotation of the Earth and VT  is the tangential velocity 
of rotation of the Earth’s surface (in m s-1)  then:

__________L = MAT X2 ω__________(12)

____________= MAT X VT___________(13)

The movement of the Summer Southern Sub-Tropical 
High-Pressure Ridge 1O south means that there is 
decrease in the total angular momentum of the Earth’s 
atmosphere since part its mass (i.e. the sub-tropical ridge) 
is now rotating about the Earth’s axis of rotation at a 
radius that is 71.48 km smaller. This decrease in 
atmospheric angular momentum is compensated 
for by an increase in the angular momentum of the 
solid-Earth so that the total change in the Earth’s 
overall angular momentum (ΔL) remains unchanged 
(i.e. ΔL = 0): 
_____MAT VT  ΔRAT = - ME RE ΔVT___(14)
or_ΔVT (Earth) 
___= - (MAT/ME) (ΔRAT /RE) VT(Atmosphere)__(15)

Now_______VT = (2 π R cos L)/LOD______(16)

where the nominal length-of-day (LOD) is 86400 seconds. 
So for L = 40O VT = 354.91773191286 m s-1
Then if MAT = 2.098 x 1017 kg, ME = 5.9722 x 1024 kg, 
RE = 6.371 x 106 m, ΔRAT = 7.148 x 104 m:

____ΔVT (Earth) = 1.398865 x 10-7 m s-1

and since the new length-of-day (LOD /) is given by:

_____LOD/ = 2 π RE cos L /VT / (Earth)_____(17)

where the new lower velocity VT / (Earth) is equal to 
354.91773191286 m s-1 + 1.398865 x 10-7 m s-1
then the change in LOD (ΔLOD) is:

____ΔLOD = LOD / - 86400 = - 0.0341 ms____(18)

CASE B: The Winter Northern Sub-Tropical 
High-Pressure Ridge shifts south by 1O at the 
same time.

At this stage we have not taken into account a comparable
shift (i.e. ~ 1O) towards the equator by the northern 
sub-tropical high pressure ridge, which is likely to 
normally be located at its winter position 30O North of 
the Equator. The movement of the Winter Northern 
Sub-Tropical High-Pressure Ridge 1O south means 
that there is increase in the total angular momentum 
of the Earth’s atmosphere since part its mass (i.e. 
the sub-tropical ridge) is now rotating about the Earth’s 
axis of rotation at a radius that is 55.60 km larger. 
This increase in atmospheric angular momentum is 
compensated for by an decrease in the angular 
momentum of the solid-Earth so that the total 
change in the Earth’s overall angular momentum 
(ΔL) remains unchanged (i.e. ΔL = 0):  

From equation 16 the nominal speed of rotation 
of the Earth at 30O North is:

___________VT = 401.24013019223 m s-1

So for L = 30O VT = 401.2401301922 m s-1
if MAT = 2.098 x 1017 kg, ME = 5.9722 x 1024 kg, 
RE = 6.371 x 106 m, ΔRAT = 5.560 x 104 m, then 
the Earth’s rotation speed slows down by:
__________ΔVT (Earth) = 1.2301064334 x 10-7 m s-1

Therefore from equation 17:

_________LOD / = 86400.000026488012 seconds
and so____ΔLOD = LOD - 86400 = +0.0264 ms

Hence, the combined movements of both the Northern 
and Southern Tropical High-Pressure ridges will produce 
a change in the Earth’s rotation rate at the 9.3/18.6
year time scale of:

_____ΔLODTOT = (-0.0341 + 0.0264) ms
______________= -0.0077 ms
______________= -7.7 μsec

Note that ΔLODTOT would be zero if both the northern 
and southern sub-tropical ridges where at the same 
absolute latitudes (i.e. in Spring and Autumn/Fall). However, 
this is not the case during the  Northern and Southern 
Summers and Winters, so it will probably be non-zero 
during these seasons. This compares with the -12.57 μsec 
change in the length-of-day that is associated with the 
effect of 18.6 years lunar tides upon the earth’s rotation.  
Remember that this is only a back-of-the-envelope 
calculations with many assumptions thrown in, so it is 
amazing the it come out to within a factor of two.



Tidal wave___Period (days)__Δ_LOD___Erot
 ______________________(10-4 sec )_(1020 J)
18.6 year____6798.37______-0.1257__3.100
Sa _____________365.26_______0.0222__0.547
Mf ______________13.66_______0.3008__7.4200

So the 18.6 year period changes to the Earth’s LOD 
could produce movements of the sub-tropical high 
pressure ridges in both hemispheres by ~ 1O every
9.3/18.6 years, which is what we observed.                                             

Tuesday, June 19, 2012

LOD (AAM) and Land Surface Temperatures in California

The following discussion is based upon a WUWT post
that highlighted a paper called:

Persistence in California Weather Patterns
by Jim Goodridge - State Climatologist (Retired)
which can obtained from:

In this paper, Goodridge notes that the accumulated departure 
from average of 47 Californian de-trended temperature records
(figure 1) shows a remarkable similarity with the accumulated 
departure from the average Pacific Decadal Oscillation (PDO)
(figure 2).

Figure 1:

Figure 2:

Goodridge notes that link between the PDO and Californian
land surface temperatures is provided by the prevailing coastal
wind directions on the Californian coast that are associated 
with the rising and declining phases in the accumulated departure 
from the average PDO. 

"The accumulated departure from average plot of the PDO 
index has a peak in about 1944 as well as during the current 
period of heating starting in 1975. The rising limb of this 
PDO index represents a warming period of SST on the 
West Coast coast, where as a declining trend represents 
a time of cooling. In general, cooling on the US Pacific 
Coast occurs when coastal North winds induce offshore 
coriolis force that result in off shore winds that lower tides 
that induce cold-water upwelling. This reflects the delicate 
hydrostatic balance of the oceans. Warming SST is associated 
with South winds that induce an onshore coriolis force with 
higher tides, which suppress cold-water upwelling."

In addition, Goodridge notes that the variations in the Earth's 
Length-of-Day (LOD), a quantity that is inversely related
to the Earth rotation rate, appears to closely match the 
observed changes in the accumulated departure from the 
average PDO (compare figure3 with figures 1 & 2):

Figure 3:

The remarkable thing is that the changes in the LOD appear to
preceed those observed in the accumulated departure from the 
average PDO [and the accumulated departure from average of 
47 Californian de-trended temperature records] by about seven
to 10 years! This means that the latter are causally linked to the

Goodridge provides a possible explanation for this causal link: 

"The PDO index is a measure of the East West sea surface 
temperature difference in the North Pacific Ocean. The 
AAM index [closley linked to the LOD] is a measure of the 
ratio of East West vs. North-South winds on the planet 
that affect the earth rotation rate.