About Tides at Moodies Group Time (about 3220 Ma ago) and some toy
models
(This second posting is due the new google program
spoils my texts !!!)
(Author: Hannu K. J. Poropudas, Date: 8.1.2005.
This second iterate text of the real situation is modified from:
Hipkin, R. G., 1975.
Tides and The Rotation of the Earth.
In:
Rosenberg, G. D., and Runcorn, S. K., 1975.
Growth Rhytms and The History of The Earth's Rotation.
Printed in Great Britain by
William Clowes and Sons Ltd., London, Colchester and Beccles.
Copyright 1975 by John Wiley & Sons, Ltd.
559 pages, 319-336.)
Assumption of tidal parameters at Moodies Group time (3220 Ma)
(Moodies Group, Barberton Greenstone Belt, South Africa)
Suppose (without knowledge of accuracy) that day length is 10.5834192
present hours (h), hour length is 0.4409758 present hours and minute
length is 7.349596667 * 10^-3 present minutes at about 3220 Ma ago.
This would mean that angular velocity (w) of the Earth would be
34.10860333 degrees per hour (present value of w is 15.04106864).
Angular velocity of the Moon (n_m) would be 0.8296456598 degrees per
hour (43 mean solar days per synodic month which is about 41 mean solar
days per sideric month at Moodies time which is about 18.0800078 mean
solar days per sideric month at present time. Earth-Moon distance is
then about r = (f*M*T^2 / (4*Pi^2))^(1/3) = 291047583.4 m = 45.68297173
Earth's radii, present value of n_m is 0.54901653) and angular
velocity of the Earth (n_e) would be 0.00464184 degrees per hour
(present value due I don't know its ancient value). I don't also
know ancient values of two less important present periods of about 9
years and 18.6 years due to the eccentricity of the Moon's orbit and
to the inclination of this orbital plane to the plane of the Earth's
orbit about the Sun. ( 1 degree = 1.745329252 * 10^-2 radians). Suppose
also that seas are few kilometers shallower than present ones and
suppose we have no good knowledge of distribution of seeds of ancient
continents at Moodies time.
(?)-mark points should be tried to figure out with aid
of computer programs of Earth tides)
Types of partial tides
M_2. The largest (?) partial tide, by a factor of two (?), is the
semi-diurnal tide raised by the Moon, called M_2. It has a period of
10.82 h and a speed of
2*w - 2* n_m = 66.55791534 degrees per hour, which is twice the
rotation rate of the Earth with respect to the Earth-Moon line.
K_1. The second largest (?) partial tide is the diurnal one, K_1,
raised by the combined action of the Sun and the Moon. Its period,
10.55 hours, is at most exactly equal to the rotation period of the
Earth, not with respect to the Sun or Moon, but with respect to the
stars, consequently it is sometimes called the sideral diurnal tide.
Its speed is w = 34.10860333 degrees per hour.
S_2. Next (?) comes the semi-diurnal solar tide, S_2. Its period is
5.28 hours, corresponding to the rotation period of the Earth with
respect to the Earth-Sun line, which is the basis of "timekeeping".
Its speed is 2*w - 2*n_e = 68.20792298 degrees per hour.
Q_1, P_1. The fourth (?) and fifth (?) largest partial tides are the
lunar and solar diurnal tides, Q_1 and P_1 with periods of 11.09 hours
and 10.56 hours and speeds of w - 2*n_m = 32.44931201 degrees per
hour and w - 2*n_e = 34.09931965 degrees per hour. These figures
demonstrate the important points that there are no diurnal partial
tides with exactly twice of the period of M_2 or S_2 (?).
The speed of Q_1 is not w - n_m but w - 2*n_m so that its phase
gradually gets
more and more behind of M_2. For the same reason, the phase of P_1
gradually
falls behind of the S_2 component and therefore behind local solar
time.
N_2, M_f. At about 20 % (?) of the size of the M_2 tide comes a
semi-diurnal component, N_2, which depends upon the lunar eccentricity,
and one of the long-period tide, M_f, whose period is about one
fortnight. Their speeds are approximately (2*w - 3* n_m) =
65.72826968 degrees per hour and 2*n_m = 1.65929132 degrees per hour
respectively.
Real Earth responds to the tide-generating forces, of which the
equilibrium tide is a model, in an extremely irregular way. There are
two aspects to this irregularity: firstly, the size of the response in
a particular ocean basin depends very critically upon frequency, that
is the speed of the tide, so that some partial tides are suppressed and
others amplified; secondly, the tidal bulge is best modelled by a
sphere with warts rather than a simple ellipsoid.
"North Atlantic" toy model
The amplitudes of the diurnal tides are consistently less than those of
the equilibrium tide by a factor of about two (?), while the
semi-diurnal components are strongly amplified, preferentially at
slower end (?). The N_2 component, with speed of 65.72826968 degrees
per hour is amplified about six times (?) compared with the equilibrium
tide and nearly three times (?) compared with the only slightly faster
S_2 component whose speed is 68.20792298 degrees per hour. This
illustrates the tendency of each ocean basin to have certain preferred
frequencies of oscillation in the vicnicity which the tidal amplitude
is greatly enlarged. Characteristic feature of the "North Atlantic"
toy model is the suppression of diurnal tides. On the "Atlantic"
coasts of "Europe and North America" most of the tidal variation is
described by the semi-diurnal tides M_2 and S_2; these alternatively
reinforce each other and cancel each other with a period corresponding
to their difference in speed:
(2*w - 2* n_e) - (2*w - 2*n_m) = 2*(n_m - n_e) = 1.65000764
degrees per hour.
This corresponds to two sets of higher and lower tides, spring and
neap tides, in the period of revolution of the Moon with respect to the
Earth-Sun line, in astronomical terms, this period is the synodic
month.
Because the response of the real ocean may lag or lead the equilibrium
tide and the extent to which it does so varies markedly from place to
place, the time when M_2 and S_2 reinforce each other, the spring tide,
does not usually coincide exactly with full or new Moon. It can be
displayed by as much as (?) days (present value is 8 days).
"Pasific Ocean", "South China Sea" toy model
A tidal cycle to the synodic month is not a necessary feature: in
extensive areas of the "Pasific Ocean", the "Atlantic"
situation is reversed and the diurnal tides are amplified at the
expence of the semi-diurnal ones. Where this relative amplification
exceeds a factor of three (?) or four (?), for example in the "South
China Sea", the total tide is predominantly diurnal and it is the
interference of K_1 and Q_1 partial tides which govern the
"observed" elevation. The difference in their speed is
w - (w - 2* n_m) = 2*n_m = 1.65929132 degrees per hour. Spring and
neap tides now occur twice every siderial month, resulting (x+2 ?)
fortnightly patterns per year instead of (x ?) (present case x = 24.74
so here is two fortnights difference per year).
"Oregon coast" toy model
For a lesser degree of amplification, the "observed" tide will be a
rather more compicated "mixed semi-diurnal-diurnal" tide in which
the main pattern may be controlled by M_2 and K_1. In this case spring
and neap tides again occur twice every sideral month.
***
Peaks from fig 3A (ref.1) could be perhaps better understood with
above first iterate of mine of the real situation. Below the peaks
(foreset numbers) which I consider could be important:
1
(9-10)
14
(20-21)
25 or (25-27)
(29-30)
(35-42) (min even 34, max even 43 or 44)
----------------------------------------
(47-49)
(51-53)
(62-64)
(67-69)
(71-73)
(77-80)
(82-86)
(88-93)
----------------------------------------
95
(98-102)
(104-106)
(108-118)
---------(end of fig 3A)----------------
Second possible interpretation (if semi-diurnal tides
case would be correct so one day foreset numbers corresponds
twice "one sand layer and one very thin mud layer"
1............. 0.5 (?)
(9-10)........ (4.5-5) (0.5 week sign ?)
14............ 7 (1. week sign ?)
(20-21)....... (10-10.5) (1. week sign ?)
25 or (25-27). 12.5 or (12.5-13.5) (1.5 week sign ?)
(29-30)....... (14.5-15) (1.5 week sign ?)
(35-42) (min even 34, max even 43 or 44)..... (17.5-21) (2. week sign
?)
----------------------------------------
(47-49)....... (23.5-24.5) (2.5 week sign ?)
(51-53)....... (25.5-26.5) (2.5 week sign ?)
(62-64)....... (31-32) (3. week sign ?)
(67-69)....... (33.5-34.5) (3.5 week sign ?)
(71-73)....... (35.5-36.5) (3.5 week sign ?)
(77-80)....... (38.5-40) (4. week sign ?)
(82-86)....... (41-43) (4. week sign ?)
(88-93)....... (44-46.5) (4. week sign ?)
----------------------------------------
95
(98-102)
(104-106)
(108-118)
---------(end of fig 3A)----------------
So in this second possibility case I would also get
about same numer of lunar days per synodic month that
I got in my first possibility interpretation, namely
I would have now
38.5 - 46.5 lunar days per synodic month at Moodies Group
time (about 3220 Ma ago).
First interpretation of mine was in ref. 2 and 3.
References:
1.
Eriksson Kenneth A., Simpson Edward L. 2000.
Quantifying the oldest tidal record: The 3.2 Ga Moodies
Group, Barberton Greenstone Belt, South Africa.
Geology, vol. 28, no. 9, September 2000, p.831-834, 5 figures.
2.
Poropudas, H. K. J., 2004
Ancient time data from The Moodies Group (3220 Ma),
Barberton Greenstone Belt, South Africa
<1104317390.609149.239200@z14g2000cwz.googlegroups.com>
Date: 29 Dec 2004 02:49:50 -0800
(Summary of all 6 articles of mine between dates:
10.12.2004-23.12.2004)
3.
Poropudas, H. K. J., 2005.
Re: Ancient time data from The Moodies Group (3220 Ma),
Barberton Greenstone Belt, South Africa
<1105196622.880537.292490@c13g2000cwb.googlegroups.com>
Date: 8 Jan 2005 07:03:42 -0800
Comments please !!!
Best Regards,
Hannu Poropudas
Vesaisentie 9E,
90900 Kiiminki
Finland
.
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