A SIMPLE MODEL FOR THE 18.6 YEAR ATMOSPHERIC TIDAL OSILLATION

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 
tides.

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 = P_O exp [−gMdz/RT_O]_____________(4)

where P_O (in Pascals) and T_O (in Kelvin) are the pressure 

and temperature at the Earth’s surface and dz = (h – h_O) 

where h is the height in metres above some reference height h_O.

Now Newton’s Second Law tells us that:

________P = M_AT g / A_____________________(5)

where M_AT 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:

_____M_AT = [P_G A /g] x {1 − exp [−gMh/RT_G]}__(6)

where [P_G A /g] is the total mass of the atmosphere above 

a given area A of the Earth’s surface, P_G and T_G are the

 pressure and temperature at ground level  

(i.e. T_G = 288.15 K (15 C^O) and P_G = 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 = 4πR^R,

 R = 6.371 x 10⁶ m, we get:

Area of the Earth’s Surface (1014 m2)	Height (m)	Pressure (hPa)	TO (K)	Mass of atmosphere  above height h  (1018 Kg)	Mass Faction
5.1006	0	1013.25	288.15	5.270	1.000
5.1006	3000	709.98	268.66	1.577	0.299
5.1006	3000	701.20	268.66	1.623	0.308

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 T_O, 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 40^O ± 10^O South of the Equator.

__________Area = 2 π R h_______________(7)

where h = height of the zone = R (sin L_U – sin L_L) and L_U 

and L_L are the latitudes of the bottom and top arcs of the zone.

_________Area = 2 π R² (sin 50^O – sin 30^O)__(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)	Pressure (hPa)	TO (K)	Mass of atmosphere  above height h  (1017 Kg)	Mass Faction
6.7850	0	1013.25	288.15	7.010	1.000
6.7850	3000	709.98	268.66	2.098	0.299
6.7850	3000	701.20	268.66	2.159	0.308

B. Change in Angular Momentum of the Earth

CASE A: The Summer Southern Sub-Tropical 

High-Pressure Ridge shifts south by 1^O.

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 = 1^O = 0.017453293 radians and L = 40^O

ΔX = 71.48 km

(N.B. In winter time the corresponding numbers would be:

for ΔL = 1^O = 0.017453293 radians and L = 30^O                 

ΔX = 55.60 km)

Now the Total Angular Momentum of the Earth (L) is 

given by:

___________L = I ω_______________(11)

where I = M_AT X², ω = V_T/R is the angular frequency 

of rotation of the Earth and V_T  is the tangential velocity 

of rotation of the Earth’s surface (in m s^-1)  then:

__________L = M_AT X² ω__________(12)

____________= M_AT X V_{T___________}(13)

The movement of the Summer Southern Sub-Tropical 

High-Pressure Ridge 1^O 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): 

_____M_AT V_T  ΔR_AT = - M_E R_E ΔV_{T___}(14)

or_ΔV_T (Earth) 

___= - (M_AT/M_E) (ΔR_AT /R_E) V_T(Atmosphere)__(15)

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

where the nominal length-of-day (LOD) is 86400 seconds. 

So for L = 40^O V_T = 354.91773191286 m s^-1. 

Then if M_AT = 2.098 x 10¹⁷ kg, M_E = 5.9722 x 10²⁴ kg, 

R_E = 6.371 x 10⁶ m, ΔR_AT = 7.148 x 10⁴ m:

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

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

_____LOD^/ = 2 π R_E cos L /V_T ^/ (Earth)_____(17)

where the new lower velocity V_T ^/ (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 1^O at the 
same time.

At this stage we have not taken into account a comparable

shift (i.e. ~ 1^O) towards the equator by the northern 

sub-tropical high pressure ridge, which is likely to 

normally be located at its winter position 30^O North of 

the Equator. The movement of the Winter Northern 

Sub-Tropical High-Pressure Ridge 1^O 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 30^O North is:

___________V_T = 401.24013019223 m s^-1

So for L = 30^O V_T = 401.2401301922 m s^-1, 

if M_AT = 2.098 x 10¹⁷ kg, M_E = 5.9722 x 10²⁴ kg, 

R_E = 6.371 x 10⁶ m, ΔR_AT = 5.560 x 10⁴ m, then 

the Earth’s rotation speed slows down by:

__________ΔV_T (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:

_____ΔLOD_TOT = (-0.0341 + 0.0264) ms

______________= -0.0077 ms

______________= -7.7 μsec

Note that ΔLOD_TOT 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 INFLUENCE THROUGH LOD 

VARIATIONS ON THE TEMPORAL 

DISTRIBUTION OF EARTHQUAKE 

OCCURRENCES

P. VARGA1, D. GAMBIS2, Ch. BIZOUARD2, Z. BUS1, 

M. KISZELY1

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

Ssa_____________182.62_______0.1400__3.453

Msm_____________31.81_______0.0304__0.750

Mm______________27.55_______0.1589__3.920

Msf______________14.76_______0.0264__0.651

Mf ______________13.66_______0.3008__7.4200

Mstm______________9.56_______0.0109__0.2689

Mtm_______________9.13_______0.0576__1.4209

Msqm______________7.10_______0.0092__0.2269

Mqm_______________6.86_______0.0076__0.1875

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 ~ 1^O every

9.3/18.6 years, which is what we observed.