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ON THE POSSIBILITY OF VERY RAPID SHIFTS OF THE POLES

Flavio Barbiero

(if you want a zipped .doc version, click here)

Summary : - Evidence exists that the poles have changed position
during the past ages. This possibility, however, so far has been
disregarded on the basis that such a phenomenon is thought to be
physically impossible. The following article shows the possibility of
very rapid shifts of the poles due to the impact of astronomical
objects as small as a half-kilometer diameter asteroid.

Introduction

In his book "The Path of the Pole" (Chilton Book, Philadelphia, 1970)
Charles Hapgood expresses the hypothesis that the poles have changed
their position three times during the recent past. From the Greenland
Sea, where it shifted about seventy thousand years ago, the north pole
moved to Hudson Bay fifty thousand years ago, and finally to its
presents position 11.600 years ago, at the end of Pleistocene. To
support his hypothesis, Hapgood presents an impressive quantity of
evidence which can be summarised as follows:
a) the presence of ice caps in North America and Northern Europe,
highly eccentrical compared to the present north pole.
b) The contemporaneous absence of ice caps from Siberia which was
actually populated to its northernmost regions by an impressive
zoological community.
c) The arctic Sea was warmer than it is today, and there were human
beings living in the New Siberia Islands.
d) Antarctica was partially free of ice.
e) The general climatic situation of the Earth was coherent with a
different position of the poles.

The hypothesis that the inclination of the terrestrial axis in
relation to the ecliptic and that the position of the poles might
change has been taken into consideration since last century. Some of
the greatest geologists of the time, including J.C.Maxwell and Sir
George Darwin (son of the famous Charles Darwin), considered this
problem and decided that the stabilising effect of the equatorial
bulge was so great that no conceivable force originating within the
Earth could make it shifting on its axis, except for the collision
with another planet. They therefore dismissed the idea of any shift of
the poles as impossible and, in fact, not worth discussing. Their
influence has been so highly felt that to this day no one has
seriously considered such an hypothesis.
Hapgood too accepts un-critically the assumption that only a
"planetary collision" is capable of displacing the axis of rotation.
Therefore he proposes a theory that explains the shift of the poles as
the result of the shift of the whole Earth's crust. Based on the
research of the Russian scientist V.V. Beloussov, he assumes that at a
depth of approximately hundred miles in the upper mantle there is a
layer of liquid rock which behaves as a bearing allowing the whole
crust to "shift" when subjected to a displacing force. In Hapgood's
opinion this force is provided by the centrifugal momentum of ice caps
eccentrical to the poles. In this way the Earth would keep its axis of
rotation unchanged, but the poles and the whole Earth's surface would
shift and change latitude.
The evidence proving that the poles where in different positions
during the Pleistocene era is quite impressive, and this explains why
Hapgood's theory was approved by scientists such as Einstein and K.F.
Mather. But it meets with so many difficulties that it appears highly
controversial. Above all, it is not compatible with other geological
theories which are widely accepted today, such as the drift of the
continents and related theories.
Furthermore the theory does not explain some of the most significant
peculiarities of Pleistocene's climate changes, first of all the speed
with which these changes appear to have taken place. According to
Hapgood it took the north pole at least two thousand years to move
from its previous position to the present. The evidence we have,
however, are in favour of a definitely much faster climatic change. It
was Hapgood himself who underlined the enormous amount of evidence
proving the high speed at which the shift of the poles appears to have
happened; speed which the mechanism he proposes is unable to explain.
The only way to completely and coherently explain what took place at
the end of Pleistocene appears to be that of admitting the possibility
of a shift of the poles of the same magnitude Hapgood hypothesizes,
but in a much shorter time: not more than a few days. This possibility
is openly refused, only because no convincing explanation has been
forwarded so far.
The following work analyses the behaviour of a gyroscope subjected to
a disturbing force, and shows that the torque generated by the impact
of a relatively small asteroid is capable of causing almost
instantaneous changes of the axis of rotation and therefore
instantaneous shifts of the poles in any direction and of any
amplitude.

Rotational components in a disturbed gyroscope

The rotational components in a disturbed gyroscope are connected to
each other by the following equation, due to Laplace, which expresses
the principle of conservation of energy:

1) Jo 2 = Jo 2 + Jp p2 = Ji i 2

where:
= speed of rotation of the undisturbed gyroscope
= speed of rotation of the gyroscope around its main axis
p = speed of precession
i = speed of instantaneous rotation
Jo = main momentum of inertia
Jp = momentum of inertia related to the precession axis
Ji = momentum of inertia related to the axis of instantaneous rotation

The value of the torque developed by a disturbing force Fp, applied to
the main axis of the gyroscope with an angle b, is evidently given by:

2) Cp = R Fp sen

where R is the arm of the force, that is the distance of his point of
application from the centre of the gyroscope.

Instant by instant the gyroscope precedes around an equatorial axis,
but the resulting motion of he main axis describes a cone, with the
axis parallel to the force, an opening angle of 2 and its vertex at
the centre of the gyroscope. The main axis, therefore, appears to
rotate with angular speed pa around an axis parallel to the disturbing
force (see fig. 1). The value of pa is given by the following
equation:

3)

fig.1

Equations 1), 2) and 3) allow us to study exhaustively the behaviour
of a disturbed gyroscope, by means of an essentially graphic method.

Given a gyroscope let's draw, on the basis of its inertia ellipse,
another ellipse whose semi-axis are respectively:

Every radius of the ellipse, r( ) , where: = 0 - 2 , would obviously
have the value:
where J is the momentum of inertia of an axis forming an angle with
the main axis.
If we put 2 = 1 , for equation 1) every radius r( ) is proportional to
the speed of rotation that the gyroscope has to have around axis to
keep its initial energy unchanged.
The end of the arrows representing and i, therefore, always fall on
the ellipse, while all the other rotational components have to be
found inside the ellipse. This ellipse allows us to analyse
exhaustively the behaviour of all the rotational components of the
gyroscope, bound as they are by equation 1) (see fig.2).

fig.2

The meaning of the rotational components shown in fig. 2 is easily
understood. A gyroscope subjected to a disturbing torque reacts
generating an exactly equal and opposed torque. This is achieved by
means of a precession movement, p, around an equatorial axis, which
makes the gyroscope rotate "unbalanced", that is rotate instant by
instant around an axis, which forms with the main axis an angle
proportional to the disturbing torque. The instantaneous rotation, i,
is given by the sum of the rotation around the main axis, , plus the
rotation of precession, p. In every moment we have:
i2 = 2 + p2.
When a gyroscope is subjected to a disturbing force Fp, of increasing
value, p grows and as a consequence i moves towards pa.
When Fp reaches a certain value Fpa (see calculations further on), we
will have:

i = pa

At that precise moment the axis of instantaneous rotation coincides
with the axis of apparent precession, and becomes fixed with respect
to both, the space and the gyroscope. This is a very special condition
in which the system composed by the gyroscope and the disturbing
torque behaves like a non-disturbed gyroscope, with only a single
rotational component, '. This axis, therefore, becomes the new axis of
rotation of the system.
If at this point force Fp diminishes again, the system behaves like a
gyroscope to which is applied a torque of value:

C'p = Cpa - Cp

Therefore the new axis of rotation begins to precede around the main
axis, moving on the surface of a cone as shown in fig. 3.

fig. 3

As a consequence i' moves back towards the main axis, following the
same path it has run along in the previous journey. Value and
direction of the gyroscope's rotational components in this case are
represented in fig.4

fig.4

Due to the principle of conservation of energy we will evidently have:

Jpa '2 = Jo '2 + J'p' p'2 = Ji i'2 = Jo 2

For each value of the disturbing force, Fp, the speed of the
instantaneous rotation is exactly the same both ways, there and back,
that is i' = i. The other rotational components, instead, change
considerably and 'p has direction opposite to that of p . This is
justified by the fact that while Fp is growing, the main axis rotates
around axis pa. In the "return journey " the contrary happens: it is
the axis of ' (now fixed in respect to the body of the gyroscope) that
rotates around the main axis.

The most important fact is that along the ' axis we have a rotational
component which is fixed in respect to the gyroscope. This means that
the gyroscope keeps "memory" of the position of the new axis of
rotation. That rotational component, therefore the "memory", is
cancelled only if and when Fp is completely zeroed. If Fp should not
be zeroed , the gyroscope would keep this rotational component, and
therefore the "memory", indefinitely.

Behaviour of a semifluid gyroscope like the Earth

The behaviour of the Earth when subjected to a disturbing torque is
exactly the same as that of a gyroscope, with a fundamental difference
due to the fact that the planet is not a homogenous and rigid solid,
made up, as it is, of liquid parts inside and outside an intermediate
plastic shell. Every rotational component of the planet exercises on
its parts a centrifugal force, which causes deformations and/or
displacements of them.

If we force a gyroscope to rotate around an axis different from the
main, it develops a reaction torque constant in time . The Earth too,
forced to rotate around an axis different from the main, would at
first develop a reaction torque. The same centrifugal force
responsible for this torque, however, would act on solid and liquid
masses causing deformations and /or displacements tending to restore
the equatorial bulge around the new axis of rotation. As a consequence
the reaction torque would decrease, until completely spent after a
while.

We do not know forces capable of making the Earth rotate around an
axis different from the main, for a time long enough to complete such
a process. But we do know that the planet is periodically hit by large
celestial bodies at high speed, which develop an impulsive torque,
that can have a very high peak value, as high as the highest reaction
torque possibly developed by Earth (see following paragraph and
relative calculus).

Graphics of fig.2 and fig.4, can help us to understand what happens in
this case.

As soon as the torque developed by the impact starts growing, the i
moves in the direction of pa, parallel to the direction of impact. If
the impact develops a torque of sufficient value, i will coincides
with pa. On that instant the axis of pa becomes axis of permanent
rotation. As soon as the torque value decreases, the axis of i returns
quickly towards the main axis, but following a different path as shown
in fig. 4. As soon as the shock ceases, an instant later, the Earth
should again return to rotate around its natural axis, without any
further repercussion. But it is not necessarily so.

To cancel the "memory" of the new axis of rotation, and have the
gyroscope rotating again around the main axis, it is necessary that
the torque be completely spent. Unfortunately, there are good
probabilities that this may not happen. We know that the Earth is
permanently subjected to a torque generated by the gravitational
forces of the sun and the moon on the equatorial bulge. This torque is
millions of times smaller than the one developed by the impact, but
its role is fundamental.

If at that moment it has a different direction than the one developed
by the impact itself, as soon as the shock is exhausted, the Earth
instantly recovers its previous axis of rotation and all ends there.
The only consequences would be the destruction resulting from the
impact.

If, however, the torque due to the Sun-Moon attraction has the same
direction of the torque caused by the celestial body, it is added to
this, and contributes in its small way to the instantaneous change of
the position of the poles. A few instants later the shock exhausts
itself while the Sun-Moon gravitational attraction continues, and
however small, it nonetheless develops a torque higher than zero.
Therefore the "memory" of the axis around which the Earth has rotated
during the impact, even for a very short moment, cannot be cancelled.

In this case the Earth actually behaves like a gyroscope whose main
axis coincides with the one adopted during the impact, subjected to a
disturbing torque equal but opposite to the torque generated by the
impact. The overall motion is apparently exactly the same, but in
reality there are fundamental differences, as illustrated in fig.5 .

fig.5

Graphics n. 5.a and 5.b represent the situation of Earth's rotational
components immediately before (5.a) and after (5.b) the impact, in the
case in which the Sun-Moon disturbing force has the same direction of
the force developed by the impact. (To make it easier to represent
them, the precession rotational components in the illustration are
greatly exaggerated; in reality they are more than one million times
smaller than the main rotation. The rationale however does not
change).

Apparently the situation has not changed, because i is exactly equal
to 'i, and ' has the same magnitude as the previous precession speed
pa. There is however a crucial difference: at this point ' is the only
rotational component "fixed" with respect to the Earth's body. Thus,
the axis of ' has become axis of permanent rotation. The rotation
around it is extremely small (as much as millions times smaller than
the main rotation), but it develops in any case a centrifugal force
strong enough to form an equatorial bulge (of a few meters) around the
new axis of rotation.

If the Earth was a solid gyroscope, this situation would last
indefinitely unchanged. The planet, however, is covered by a thin
layer of water, which reacts immediately to any change of motion.

Sea water begins to move towards the new equator, and as this happens,
the value of ' increases again, therefore increasing the force which
makes the water move towards the new equator. This process gradually
accelerates, until the centrifugal force developed by ' grows strong
enough to induce deformations of the Earth's mantle.

From here on the equatorial bulge is quickly reformed around the new
axis of rotation and Earth will soon be stable again, with a different
axis of rotation and different poles.

This mechanism shows that the Earth's poles, contrary to what has
always been postulated, can "jump" almost instantaneously from a
location to another thousands kilometres away, due to the combined
effects of forces at first sight irrelevant, such as the impact of a
medium size asteroid and the Sun-Moon gravitational attraction on the
equatorial bulge.

Let's now evaluate the probability of such an event. According to
Hapgood, in the short lapse of time represented by the last 100.000
years, no less than three "jumps" happened. This means that the
probability of such an event should be extremely high.

In order to be able to estimate it, we must first determine the value
of the reaction torque developed by Earth. This value will allow us to
determine the dimensions and speed a celestial body must possess in
order to be able to overcome such a torque.

Value of the reaction torque developed by Earth

The value of the reaction torque developed by a gyroscope, when
rotating around an axis different from the main, can be calculated
(see fig, 6) reckoning the torque developed by the element of mass,
dm, rotating around the axis of i:
Ci = Fi b

fig. 6

where:

Fi = dm i2 r i = dm i2 ro cos is the centrifugal force;

b = ro sen is the arm of the torque.

We have therefore:
Ci = dm ro2 i2 sen cos = dJo i2 sen cos = ½ dJo i2 sen2

where dJo = dm ro2 is the momentum of inertia of mass dm with respect
to the main axis.

For a ellipsoid of revolution we will have therefore (see fig. 7):

4) C = (Jo - Jp) i2 sen cos = ½ Jr i2 sen2

where Jr = (Jo - Jp) is the momentum of inertia of the bulge.

fig. 7

Equation 4) shows that a gyroscope may develop a reaction torque only
if Jo Jp. In the case of it being perfectly spherical, it would rotate
indifferently around whatever axis and it wouldn't have any stability.
This is due to the fact that in a rotating homogenous sphere, all
centrifugal forces balance each other and there is no reaction torque,
no matter what the axis of rotation is. Only the equatorial bulge can
develop a reaction torque

Value of the stabilising torque developed by the equatorial bulge

From equation 4) we see that the maximum reaction torque possibly
developed by a gyroscope is reached when = 45o:

Cm = ½ Jr i2

For Earth the value of i is almost equal to that of the main rotation,
so we can assume that:

i2 (2 / 8,64)2 10-10 = 5,28 . 10-9 sec.-2

The calculation of Jr can be made by using the value of the
centrifugal force, Fo, developed by the equatorial bulge due to
Earth's rotation, as calculated by Gallen and Deininger for Hapgood
(see insert at the end):

Fo = 4,1192. 1019 kg.

For an approximate calculation we can put:

Jr Mr Ro2

Fo Mr i2 Ro = Jr i2 / Ro

where Mr is the mass of the bulge and Ro the radius of the Earth.

We have then:

Jr Fo Ro / i2 5 1034 kgmt2

And finally, thanks to equation 4) we have:

4') C = ½ Jr i2 sen2 = 1,38 1026 sen2 kgmt

For = 45o we have :

C 1,38 1026 kgmt

which is the maximum reaction torque possibly developed by Earth.

Calculation of the size an asteroid should have to cause the shifting of the
poles

According to equation 4) to displace the axis of rotation of for
instance 20o, an asteroid hitting the Earth must develop an impulsive
torque of the following value:

C20° = 8,87 . 10 25 Kgmt.

It is therefore easy to calculate the size and speed that such an
asteroid must have.

The impulsive force Fi developed on impact with Earth by the asteroid
is given by:

Fi = Ma.a

where:

a = dv/dt is the acceleration the asteroid undergoes on impact

Ma is the mass of the asteroid

To calculate the acceleration, a, we can assume the asteroid has, on
impact, a speed:

v = 5 . 10 4 mt/sec.

To calculate dt we have to rely on an estimate. In a very conservative
way, considering the depth of known craters, we can presume that the
depth of the crater caused by that impact to be 500 m, which means
that the speed of the asteroid decreases from 5.104 to 0 mt/sec, in a
space of 500 meters. The time in which this happens is approximately
one hundredth of a second, that is:

dt = 0,01 sec.

The average acceleration of the asteroid will therefore be:

am = dv/dt = 5.104 / 0,01 = 5. 106 m/sec2

The acceleration peak is certainly much higher. In a conservative
calculation we can assume it to be double the average value. We will
have then:

a = 5.10 4 / 0.005 = 10 7 mt/sec2

And therefore:

Fi = Ma . 10 7 kg

The torque developed by this force will obviously be:

Ci = Fi . Ri

where

Ri is the arm of the torque.

The value of Ri can be between 0 and Ro 6,4 106 mt, that is the radius
of the Earth. For statistical reasons we can put:

Ri = ½ Ro = 3,2 106 mt

The mass of the asteroid will therefore be:

= 2,77 1012 kg

If the density of the asteroid is of 3 Kg/dm3, we will have a volume
of:

Va = 0,92 km3

that is then a lithic asteroid of approximately a 1000 metres
diameter. This calculation is very conservative. We can realistically
suppose that an object of half that size is enough to develop a torque
of sufficient value for a huge shift of the poles.

Probability of a shifting of the Poles due to an asteroid impact

Following are the essential conditions necessary to cause a shift of
the poles by an asteroid falling on Earth:
1. The torque developed by the asteroid on collision must be equal to
the maximum Earth reaction torque, even if for only one instant.
This means that the asteroid must not only have size and speed of
sufficient magnitude, but the arm of the torque, as well, has to
be sufficiently long.
2. The force of the Sun-Moon gravitational attraction on the
equatorial bulge must have the same direction of the force
developed by the impact.

This second condition has no more than 50% probability of being
verified. Therefore, the probability that an impact results in a
shifting of the poles is smaller than 50%, regardless of the size of
the asteroid. The size of the object has also little influence on the
probability that it generates a torque of adequate value, the main
influence residing with the length of the arm of the torque. If the
impact is directed exactly towards the centre of the Earth, there is
no torque at all, regardless of the size and/or the speed of the
asteroid. On the other hand, even a very small object can develop a
very high torque if it hits the Earth at an almost tangent angle to
the surface. Of much importance to the peak value of the torque
developed, should be the density of the object and the nature of the
surface it falls on.

These variables make it impossible to calculate with precision the
probability that an impact might develop a torque sufficient to cause
a shifting of the poles. However, we can reasonably expect this
probability to be in the range of 20 to 30% for objects of a diameter
of 500 m or more; under this size the probability should fall sharply.

The Apollo objects

We can expect the chances of a celestial body as large as 500 mt
colliding with Earth to be rather high, in the range of several
collisions each 100.000 years period. The majority of these collisions
is caused by a class of celestial bodies named by astronomers "Apollo
objects", that is a class of asteroids whose perihelion lies inside
the orbit of the Earth.

The first of these objects was discovered by Reinmuth on 1932 and
named Apollo, which gave the name to the class. At present,
approximately one hundred Apollos of a diameter of at least one
kilometre are known. The largest discovered so far, Hephaistos, has a
diameter of ten kilometres. The total number of Apollo objects with a
diameter of one kilometre or more is estimated to be between 1.000 and
2.000.

As the perihelion of the Apollos lies inside the orbit of the Earth,
it follows that periodically they have the chance to collide with it.
The probability of such an event is estimated at 5. 10-9 per year per
single Apollo. Therefore we have a probability of at least 4
collisions each million year period with objects as large as one
kilometre or more. As the size of the objects becomes smaller, this
probability grows exponentially to become of one impact every few
centuries for objects of 100 to 200 metres diameter.

The calculated probability of 4 collisions every one million year
period is coherent with the statistic of one-kilometre-wide asteroids
fallen on Earth during the last 600 millions years (see: G.W.
Wetherill, "The Apollo Objects", Scientific American May 79; and Tom
Gehrels, "Collision with comets and asteroids", Scientific American,
March 96) . If the Earth didn't have oceans and atmosphere its surface
would be marked with craters like the Moon and Mercury . On our
planet, instead, erosion and sedimentary processes delete very quickly
the marks of collisions by meteorites. Only where recent ice sheets
have scraped the surface, thus uncovering the traces of ancient
collisions as in Canada, it is possible to count the craters
accurately. Based on this count G.W. Wetherill has estimated that in
the last 600 million years the planet has been hit at least by 1500
objects with a diameter larger than one kilometre.

Major events during a shifting of the poles

The effects of a collision with an Apollo-like object are devastating.
Gehrels estimates that a one-kilometre-wide object, colliding with the
Earth at a speed of 20 kilometres per second, would liberate an energy
equivalent to ten billions of Hiroshima-type nuclear bombs.

These effects, however, although disastrous, are not even comparable
with the destruction brought about by an instantaneous shifting of the
poles. Following is a short analysis of what could happen.

Suppose the Earth has been hit by an asteroid and that the conditions
for a shift of the poles have been met. On the basis of the
adjustments necessary to re-shape the equatorial bulge around the new
axis of rotation, and the consequent re-establishment of the isostatic
equilibrium of the crust, we can predict what the phenomena would be.

Some areas of the Earth's crust would be driven to move upward, others
downward. The up and down movements necessary to re-shape the bulge,
would be different from site to site of the Earth . For a shift of 20o
the movements would be of 3 or 4 kilometres at the most. Very small,
compared with the diameter of the Earth, but nonetheless of great
consequence on the surface. We know that the mechanisms which maintain
the isostatic equilibrium of the crust are very effective; there is no
doubt that after a while the equilibrium would be re-established
around the new axis of rotation, with poles and equator in different
positions.

It is important to evaluate how long it would take for this to happen.
We know that the layers of the crust, when subjected to a force over a
certain limit, break suddenly, causing an earthquake. In the situation
we have hypothesised, at the beginning only sea water would be
displaced, with a very slow and gradual increase of the speed of
rotation around the new axis. When the rotational speed reaches a
certain critical value, sudden adjustments of the crust would begin to
happen and from that moment the process would be sharply accelerated
and the re-shaping of the bulge would be completed in a very short
time.

How short? days or hours? Impossible to say. A simulation with a
mathematical model should give reliable results. The process of
re-shaping the equatorial bulge should follow a course of exponential
type: after the initial sharp peak, it should decrease very quickly.
Adjustment phenomena, however, are expected to continue for a long
time, as the isostatic equilibrium is re-established more and more
accurately.

Obviously, readjustments of that size of the equatorial bulge cannot
happen without causing extensive fractures of the crust, which would
provoke earthquakes of such a tremendous magnitude as to dwarf the
most devastating known today. A sudden incredibly strong burst of
volcanic activity in all areas subjected to strain would also be
inevitable.

The beginning of adjustments of the crust would start not only
earthquakes and volcanic activity, but even a dreadful hurricane all
over the continents, with violent winds and torrential rains. A
gigantic water avalanche would devastate the valleys and plains of the
world. On the whole the oceans' water and the atmosphere follow the
rotational movement of the Earth, but they are not tied to it. If the
Earth should suddenly change the direction of its rotation, they
would, at first, thanks to their inertia , keep up their previous
motion; only after a while the attrition with the Earth's surface
would force them to follow the new movement.

The continents would be swept by hurricane force winds, reaching
speeds of hundreds of kilometres per hour. The water of the oceans
would play a much greater destructive role.

If we abruptly change the movement of a bowl full of water, we see the
level of the water rising on one side and getting lower on the
opposite. Something like this would happen on Earth. We must expect
wide fluctuation of sea levels in many parts of the world. Presumably
an enormous tide, dozens, or even hundreds of meters high, would
slowly move around the globe.

The same reasoning applies where the core of the Earth is concerned.
It being liquid it would at first continue on its motion, naturally
undergoing a strong attrition in the core-mantle boundary region.
Obviously, after a certain time the core would adopt the same
rotational motion of the mantle, but the attrition would create a
turbulence which might have important effects. According to the last
theories, the liquid core is the site of electrical currents, that are
responsible for the magnetic field. This turbulence could influence
the behaviour of the magnetic field, and it could break the balance,
thus provoking perturbation on the magnetic field that might lead even
to an inversion of the magnetic poles.

An important element in order to evaluate the climatic conditions
following a shift of the poles, is the inclination that the new axis
of rotation will assume with respect to the ecliptic. This has a
tremendous effect on the climate. According to the mechanism we have
spelt out so far the axis of rotation that the Earth would acquire at
the moment of impact should be parallel to the direction of the hit.
It is impossible to predict which would be the actual direction of the
new axis once stabilised. It is certain that it would not be the same
as the previous one, except for a fortuitous chance. Therefore,
following a shift of the poles the course of the seasons would very
likely be different .

For example, in the hypothesis that the axis of rotation is almost
vertical with respect to the ecliptic, there would be an enormous
growth of ice at high latitudes and altitudes, with subsequent
lowering of sea level. On the other hand the climate would be much
more stable then it is today, with very limited (or non-existent)
seasonal climatic differences and an uninterrupted vegetation's
growth. This would bring about the disruption of today's climatic
barriers, with subsequent spreading of tropical species towards
northern regions and viceversa. There would also be the maximum
possible development of ecological communities.

This appears to be exactly the existing situation in the Pleistocene
era, when formidable ice caps covered North America and North Europe,
and there were enormous glaciers on all mountains. In the immediate
surroundings of these masses of ice one of the most impressive
zoological communities of all times thrived. Millions (more than 40
millions, according to F.C. Hibben) of mammoths roamed Siberia and
Alaska, large animals the size of which can be found today only in
tropical regions, or in those areas where the supply of fodder is
guaranteed all the year round.

It's against common sense that precisely during the ice age, one of
the largest zoological communities since the dinosaurs existed in
those very areas which are today reputed, due to their extreme
climatic conditions, as the most hostile on Earth. With the mammoths
there were dozens other animal species, the majority of which are
extinct today. Of these species we have a great number of skeletons,
several complete animals that have been perfectly preserved in the
permafrost, and many wonderful paintings in Palaeolithic caves. The
oldest amongst them is the "Chauvet" cave, in France, which is
believed to have been painted 30.000 years ago, precisely in the
middle of the ice age. They are pictures of breathtaking beauty. The
unknown artists, with a few strokes, have represented to perfection
animals which at the time were living in the plains of central Europe
(and at the same time in Siberia and Alaska) . But the beauty of the
paintings makes the zoologist wonder in more than one way: how could
such a varied assembly of animals coexist? To what a bizarre
ecological environment could such a motley fauna belong? We find the
rein-deer next to rhinoceros, the mammoth, with its woolly mantle,
near the hippopotamus, bears with lions, the leopard and Brezalwski
horses. There were also giant beavers and sloths, big horn deers,
camels, sabre teeth tigers, buffaloes, aurochs bulls and many more.

It's an incredible mixture which leaves us puzzled and astonished.
Arctic and tropical fauna together, on the same plain, in perfect
balance with the environment! Such an extraordinarily varied and
numerous animal community disagrees with whatever current opinion we
might have on climatic conditions during the ice age. And definitely
such a community couldn't exist anywhere on Earth today.

This community suddenly disappeared at the end of the Pleistocene era,
when, according to today's theories, the climatic conditions were
supposed to have become milder.

A shift of the poles occurred 11.600 years ago, with all its related
destructive phenomena, could explain completely and coherently the
climatic situation before that date, and the situation brought on
after that date.

Flavio BARBIERO

Rue E. Claus, 13

1050 Brussels

Gallen's calculation of the stabilising centrifugal effect of the
equatorial bulge of the Earth

Let the equations of the sphere and the ellipsoid of revolution be:
1) x2 + y2 + z2 = b2
2)
where the axis of y is the axis of revolution. Take as the element of
mass, dM, the ring generated by revolving the rectangle dxdy about the
axis of y. We have:

3) dM = 2 x dxdy
where is the density. For each particle of the ring the centrifugal
acceleration is the same, being equal to 2x, where is the constant
angular velocity in radiants per second. The element of centrifugal
force, dF, exerted by the ring is then:

4) dF = 2x dM = 2 2x2 dxdy
Integrating equation (4) with respect to x and y, there results:

5)

In equation (5) F is expressed in dynes when is given in grams per
cubic centimeter, and a and b in centimeters. The quantity may be
replaced by 2n, where n is revolutions per second. The Earth makes one
complete revolution in 86,164.09 mean solar seconds.

Mrs. Deininger's computation based on Gallen's calculus

Computation of centrifugal force produced by rotation of the bulge,
A. Essential data:
1. The attached formula should apply to the bulge taken as 13.3443
miles at the equator, not the bulge as it would be if there were
no flattening at the poles.
2. In making the calculation, Hapgood asked Mrs Harriet Deininger, of
the Smith College faculty, to subtract three miles from the depth
of the bulge, because he was concerned with a purely mechanical
action of stabilisation, in which water could not have effect. (He
later recognised that he subtracted about three miles too much,
because he had disregarded isostasy, which in this case makes it
probable that the rock under the oceans has a density higher than
the density of the rock of the continents; so he should have
subtracted the weight rather than the volume of the water. This
however is a minor correction)
3. Mrs. Deininger actually took the depth of the bulge as nine miles,
without the water.

B. Calculation:
1)

where s = density in gm/cc

a = radius of Earth at bulge in cm

b = radius of Earth at poles in cm

w = 2 - n r = rps

2) F = 4sn2. b(a3 - b3)

where = 3,1415

s = 2,7 gm/cm3

n = 1/86.164

b = 6,402 . 108 cm

a = 6,4165 . 108 cm (using nine miles or 1,450,000 cm as depth of
bulge)

3) F = 4,0368 . 1025 dine = 4,1192. 1019 kg. 
_________________________________________________________________

* Charles Hapgood, "The Path of the Pole", Chilton Book Co,
Philadefphia, 1970
* R. F. Deimel, "Mechanics of the Gyroscope. The dinamics of
Rotation", Dover Edition, 1950
* G.W. Wetherill, " The Apollo objects"'Scientific American, May 79
* Tom Geherels, "Collision with Comets and Asteroids"' Scientific
American, March 96
* E. Spedicato, "Apollo objects, Atlantis and the deluge: a
catastrophical scenario for the end of the last glaciation", Quad.
90/22, 1990, Bergamo University, Italy

Work performed in the framework of a 60% MURST research.