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Precession of a Gyroscope

and of the Earth's Axis

A gyroscope seems to be able to defy gravity, all the while having its
axis slowly weave its way around a cone-shaped path. Everyone has seen
a child's spinning top or gyroscope do these mysterious things. It can
seem to hang in space, apparently defying the laws of science. It's
surprising how few people seem to actually understand why it is able
to behave in such a way. There are even some web sites, where
purported "experts" "explain" this action, but all of those web sites
I have seen make enormous errors in their explanations. This page is
an attempt to develop some clarity regarding gyroscopes and
precession.

We all know that Newton's Laws of motion apply to all objects. His
laws were refined by Euler into a set of three generalized
differential equations for when an object is moving in a
three-dimensional motion. The three equations are of the form:

Euler's Equations

If we consider a gyroscope that is not slowing down (no friction) and
there are no other external forces acting and the gyroscope is a
symmetric body, two of the three equations can be ignored, and we are
left with a single equation of the form above. Simplifying this even
further, if we consider a gyroscope that is exactly horizontal
(gravity being vertical) then our three axes are all perpendicular to
each other. We will call axis 1 as the (horizontal) axis of the
gyroscope rotor, axis 2 is the (horizontal) axis around which gravity
is trying to make it fall, and axis 3 is the vertical axis. In that
simplified case, the term (dw[2]/dt) is exactly equal to the component
omega[3] * omega[1] along the (2) axis, and they cancel each other
out. This allows simplification of the original equation to: M = I *
(omega[1]) * (OMEGA[3]).

This is therefore the simplified theoretical equation that describes
the precession of a horizontal gyroscope. M is the 'moment' meaning
the weight of the whole gyroscope trying to tilt it downward around
its point of support. I is something called the 'moment of inertia', a
characteristic of the construction of the spinning rotor, omega is the
rate of spinning of the rotor, and OMEGA is the rate of precession
that will occur.

One additional comment about the Euler equations: Everyone knows
that Newton established that Force equals Mass times acceleration,
the famous F = ma. In circular motion, in a polar coordinate
system, this is described as M = I * (alpha), "Moment"
(tangentially acting Force) equals "Rotational Inertia" (the
rotating equivalent of Mass) times "Angular acceleration". That
might suggest that the equation should actually remain very simple!

However, for rotating objects, and for general objects which are
rotating in a three-dimensional coordinate system, each of these
three terms must be considered as a Vector quantity, having both
magnitude (size) and direction. The simple single equation above is
therefore a Vector product of I and alpha. The rules of Vector
multiplication are different from simple scalar multiplication, and
the result is that, in order to actually calculate values, three
separate equations must be considered, along the three axes of the
coordinate system. This is why there are THREE Euler equations, to
represent the Vector component of the Moment along the three axes.
It is also why these components each include terms that depend on
factors which are along all three axes.

In our case, the first Euler equation is essentially regarding any
frictional slowing down of the gyroscope rotor, which we have
eliminated. The third is essentially regarding a slowing down of
the gyroscopic motion, again, essentially due to friction. That
left us with just the second Euler equation to evaluate.

A spinning gyroscope or top is designed with the greatest possible
moment of inertia. In the case of a gyroscope, as much of the mass is
arranged as far as possible from the axis it will spin on. The laws of
physics define moment of inertia for each tiny particle of that object
(or any object) as being the product of the mass of that particle
times the square of the distance between the particle and the axis of
rotation (radius), squared. Therefore, the moment of inertia I = m *
r^2.

The shape of the rotating part of a gyroscope is designed so that most
of the rotating mass (of metal) is very close to R, the maximum radius
of the gyroscope. Looking at the equation above, you can see that this
would give the maximum total moment of inertia I for the whole object.

When this structure is spun, it has angular momentum, also called
moment of momentum. Again, from the definitions of Physics, the
angular momentum is given by the product of the moment of inertia
times the rate of spin. This rate of spin is usually measured as a
certain number of radians per second of rotation. A full circle
contains 6.28 radians, so an object that is spinning at one revolution
per second is rotating at 6.28 radians per second.

Say you have a normal child's gyroscope and you get it spinning very
rapidly and then place it horizontally where one end is supported by a
small tower that has a dent in its top, and the other end is just left
out in space with nothing supporting it. Logically, it seems like it
should immediately fall off its pedestal, but it doesn't. It does two
unexpected things. It remains with its shaft of rotation horizontal,
and that shaft slowly moves the gyroscope in a circle around its
supported end.

Several hundred years ago, a mathematician named Euler developed
Newton's Laws of motion into the more generalized equations shown
above. Among many other things, these equations describe the motion of
a gyroscope. The math gets somewhat complicated for most real
situations, (as was hinted at above) so we'll minimize use of it here.
Suffice it to say that his analysis shows an inter-connection between
all three axes in space for such a situation.

* The angular momentum of the fast spinning object is defined as
being along that axis of rotation (it is horizontal in our case).
Physicists use something called the "right-hand rule" to identify
which of the two possible directions it could be pointed. If you
use your right hand, and rotate your hand so that your fingers
advance in the direction of the rotation of the gyroscope, your
thumb will point in the direction of this angular momentum Vector.
In our case, if our gyroscope was rotating counter-clockwise as
seen from its freely hanging outer end, this right-hand rule
indicates that the angular momentum Vector is then pointing
outward along that axle shaft. Since our gyroscope is horizontal,
this Vector is horizontal.
* The gravitational force (weight) is attempting to move/tilt
(perturb) that axle is in a downward direction. This is seen as a
ROTATING force (Moment) that is acting to rotate the body of the
gyroscope (downward) around its point of support. This Moment is
described as being along an invisible line that would represent a
"hinge" at the point of support. In our case, that Moment is
therefore also horizontal, but at a right-angle to the angular
momentum Vector we just identified above. Again, the right-hand
rule is used to determine which of the two possible directions
this Vector could be pointed along that axis. Again, in having the
fingers follow the downward rotation of the motion that gravity is
attempting to cause, the right thumb points in the correct
direction.
* The remaining axis is at right angles to both of these horizontal
axes. Therefore, in our case, it must be vertical. This is the
axis that the precession motion must revolve around. That handy
right-hand rule can even be used to know which direction a
gyroscope will precess!

Note that each of these three Vectors is described as being in a
direction that seems to be different from the actual motion involved!
That is because the actual quantities being considered are rotational
Vectors and not linear ones, and these are each defined as being along
the axis of that possible movement.

Euler's equations show that for this horizontally spinning gyroscope,
supported at one end in the earth's gravity, TWO new unexpected forces
(technically, they are each actually Moments or Torques) appear. These
are because of that three-dimensional inter-relationship of the Vector
components. One of the new forces is upward, which exactly balances
the perturbing force of the pull of gravity downward. (Remember above,
when we mentioned that two terms in the Euler equation were exactly
equal and cancelled each other out?) Since it exactly effectively
cancels out the downward effect of gravity, the gyroscope does not
fall. The second new term in the Moment equation is along that third
axis (vertical).

This last force mentioned is the one that causes precession, where the
gyroscope slowly revolves around its support pedestal.
_________________________________________________________________

Let's put some numbers in this, to help understand the concepts. Say
the toy gyroscope has a moving rotor that has an effective radius of 4
cm (a diameter of around 3 inches) and that it weighs 30 grams (about
one ounce). Its shaft is 8 cm long, and we get it to spin at 12,000
rpm. Let's analyze. The moment of inertia is M * r^2 which is 0.030 kg
* 0.04 m * 0.04 m or 4.8 * 10^-5 kg-m^2. A rotation speed of 12,000
rpm is the same as 200 revolutions per second, which is 200 * 6.28 or
1256 radians per second. Multiplying 4.8 * 10^-5 by 1256, we get about
0.06 kg-m^2/sec as being the actual angular momentum of the rotating
part. [This is I * omega in the equation above]

Since our gyroscope is horizontal, the force of gravity is directed
straight downward and is equal to the weight of the (whole) gyroscope.
Gyroscopes are always designed with the outer frame as light as
possible, so we are going to fudge a little here (for simplicity of
concept) and say it has no weight. If we would have considered the
weight of that framework here, the concept is still exactly the same,
but the math gets a little more complicated, that's all. So, the
weight of our gyroscope is 0.030 kg * 9.8 m/sec^2 [the acceleration
due to gravity] or 0.294 kg-m/sec^2 or 0.294 newton.

That weight is going to try to tilt the spinning shaft downward,
because we have the one end supported on its pedestal. This means we
are (unintentionally) applying a downward Moment or torque on the
shaft, around the point at the top of the pedestal. Moment or Torque
is force times radius arm, so we have the torque as being 0.294 newton
times half of the shaft length or 0.04 meter for a torque of 0.01076
newton-meter. [This is M in the equation above]

We're getting there! Say that downward torque applied for a one-second
interval. That would mean that we were applying a (torque) moment of
0.01076 newton-meter for one second. Writing this a different way, it
is 0.01076 kg-m^2/sec. This is in exactly the same form that we
described our angular momentum of the rotating part of the gyroscope.

Euler's equations tell us that this (torque) Moment (horizontally,
along that "hinge line", due to the action of gravity) is precisely
identical to the CHANGE in the vector that describes the gyroscope's
angular momentum. It is in that direction, horizontally sideways to
the gyroscope's angular momentum spin vector. (Getting too
complicated? Sorry!) This makes the gyroscope's angular momentum get
shifted toward that direction, which is what precession is.

In our one second interval of applying that perturbing force of
gravity, we have caused a sideways vector (of 0.01076) to be added to
the original vector describing the angular momentum (0.06, and in a
direction along the shaft). These two vectors are at right angles to
each other, and we can quickly determine the resultant vector that
will now describe the rotational inertia. It will still be horizontal
and level, but its direction will have changed. The axis of rotation
will have changed, which acts to move the gyroscope around its support
point. In our case, the one-second change was 0.01076 sideways
addition to an original vector that had an amplitude of 0.06. Using
the tangent of that angle, we easily discover that the shaft had to
turn about 10.2 degrees horizontally during that second. This
particular gyroscope would therefore take about 35.4 seconds to
revolve a full circle of 360 degrees, so it would have fairly slow
precession.

For the sake of argument and clarity, we picked an interval of one
second. In reality, we would pick an infinitely small interval and
integrate the results over that second or whatever interval we might
consider. The important point here is that the new sideways vector
does NOT change the magnitude of the vector that describes the angular
momentum but just changes its direction. The gyroscope spin rate would
therefore not be affected by this effect.

Notice also that, after friction of the bearings has slowed the
gyroscope from 12,000 rpm to half that (6,000 rpm), there is an
interesting result. The sideways vector is still 0.01076, because it
is due to the torque moment due to the weight of the gyroscope and the
force of the Earth's gravity. But the vector that describes the
angular momentum, still in a direction along the shaft, is now only
half as large, at 0.03. Doing the vector addition now gives a change
of direction in that one second of around 19.7 degrees, so the
gyroscope will now take only 18.2 seconds to complete an orbit,
instead of the earlier 35.4 seconds.

This explains another characteristic that everyone has seen with
gyroscopes and tops. As bearing friction makes their rotation slow
down, they start "wobbling" faster and faster. Now you know why that
happens!

With the above approach, you should be able to calculate a prediction
for how fast any toy gyroscope will precess when placed horizontally
on its pedestal. If it is placed at any angle, all of the theory
described is still true, but the mathematics gets more complicated.
The simplification mentioned earlier to Euler's equation would not
apply, since the Vector product then involves a variety of angle
terms.

Actually, for this rather simple configuration, as mentioned earlier,
Euler's equations simplify into: M = I * omega * OMEGA, where M is the
perturbing torque moment, I is the moment of inertia of the spinning
part, omega is the rate of spin of the spinning part, and OMEGA is the
rate of precession. For our example, 0.01076 nt-m = 4.8 * 10^-5 kg-m^2
* 1256/sec * OMEGA. This solves to 0.1785 radians/second, which gives
a period for the precession of 35.2 seconds, in good agreement with
our earlier rough calculation. This value is technically the correct
one!
_________________________________________________________________

I propose a slightly modified description of gyroscope motion. As in
all popular discussions of gyroscopes, the discussion above has
effectively ignored the first of the two newly created forces, the one
that counter-balances the perturbative force of gravity and keeps the
horizontal gyroscope from falling. The following modification includes
consideration of this aspect.

In our discussion, we only briefly mentioned that two out of the three
terms in the Euler equation cancelled each other out, and so we spent
our efforts on looking at the effects of the remaining term, the one
that causes the precession effect. The first of these ignored terms is
the Moment due to the normal effect of gravity, and the other is the
new term that effectively cancels it out, which keeps the gyroscope
from falling.

I suggest that a simple way of visualizing this exists. If we choose
to visualize Forces instead of Moments (easier to do for most
people!), then, looking inward at the outer end of the gyroscope
shaft, we would see a force vector that includes both upward and
forward components. (I am referring to forces rather than moments or
torques, which is actually a more proper way of describing these
effects. I do so here because I think the conceptualizing of the force
vectors is easier than dealing with the torques which are described as
being along directions that are less intuitive for initial
understanding. After the concept is grasped, changing to a moment or
torque analysis is straightforward.)

This composite force vector is perpendicular to the axis of rotation,
but has components along the two other axes we have been discussing,
the vertical and the forward. The vertical component is always exactly
identical to the downward force of gravity due to the weight of the
gyroscope in the Earth's gravity, which keeps it from falling. (This
is the two terms that cancel each other out in the equation) The
forward component is the precession-causing component we have been
discussing.

The following relationship is proposed as holding meta-stably true:

forward component    gravitationally caused downward moment       M
_________________ =  _____________________________________  =  ________
upward component         angular momentum of gyroscope         I * omega

In our example, this (initially) gives 0.01076/0.06 = 0.01076/0.06

This, therefore, does not change anything in the discussion above. But
it allows the addition of another insight into gyroscopes. As a
gyroscope slows down as a result of friction, its angular momentum
continually lessens, while the gravitationally caused downward moment
stays constant. The upward component must similarly remain constant,
which keeps the gyroscope from falling. As the right hand term above
therefore becomes larger due to frictional slowing, so must the left
term. This then has the effect already noted of increasing the forward
component as the gyroscope slows, which increases the precession rate.
(The cross product of the relationship above remains constant: as the
angular momentum lessens, the forward component must equally
increase). But this new perspective DOES add one new insight. As the
angular momentum of the gyroscope becomes less, at some point, it
becomes so low as to require an extremely high precession rate to
maintain this relationship. Therefore, it becomes too low to maintain
the relationship just presented. At that point, the gyroscope is no
longer capable of counter-acting the force of gravity and the
gyroscope suddenly falls. This relationship allows an analysis to
determine when that sudden breakdown will occur. This fact indicates
that the above relationship is not actually an equality but a
meta-stable equality that is only true when the gyroscope's angular
momentum is greater than a certain amount.

Experimental study might be appropriate to determine the point at
which the meta-stability breaks down. For example, it might come at a
point when the (horizontal) precessional component of the force vector
is equal to the (necessarily constant) vertical component. Whatever
circumstance would be found regarding the breakdown of the
meta-stability, seems likely to apply generally to all (horizontal)
gyroscopes.

It seems likely that a thorough study of the Euler equations and their
solution probably contains the mathematical proof of that situation.
_________________________________________________________________

Precession of the Earth's Axis

When ancient people looked at the stars at night, they saw a different
"Pole Star" than the Polaris that we have today as our Pole Star. Even
the ancient Greeks (specifically Hipparchus) had noticed this effect.
Of course, at that time they didn't realize that the Earth moved or
that it went around the Sun, so they didn't have any analysis or
explanation for what they were recording.

Eventually, (after it was accepted around 1500 that the Earth rotates
and moves around the Sun!) it was realized that the Earth's motion in
this way represented an effect apparently identical to that of a
child's top or gyroscope. The premise is that the earth is a giant
gyroscope that has a period for the precessional orbit of 25,800
years.

Since the motion of the Earth seems so similar to that of a gyroscope,
it seems that it has been assumed that the same mechanism is at work.
I am tempted to think otherwise! I think that using the same term,
precession, to describe the action of a gyroscope and the Earth's
motion, may be technically incorrect. Even though they LOOK to be the
same, I suggest that they have rather different sources.

First, the commonly accepted explanation. It is pretty well known that
the entire Earth has a moment of inertia of about 8.070 * 10^37
kg-m^2. But this number has no value in analyzing the precessional
movement of the Earth!

Because the Earth spins fairly fast on its axis, the actual shape of
the Earth is not precisely a sphere but it is slightly "flattened" at
the Poles. There are a lot of complex ways of describing and
explaining this flattening (officially called oblateness) but we will
just accept it as a fact here. Effectively, what this amounts to is a
planet that is perfectly spherical, with a diameter of the Polar
diameter (around 7899 miles [12,713 km]) with an extra 'belt' of
material around its middle that accounts for the Equatorial diameter
of 7926 miles (12,756 km). This "belt" of material, which has a
maximum thickness of 13 miles (21.4 km) at the Equator, represents an
equivalent to our toy gyroscope. That belt of material has an
effective moment of inertia of about 3.3 * 10^35 kg-m^2, about 1/250
that of the whole Earth.

This all makes excellent sense. It is at this point that I tend to
disagree with the normally accepted explanation regarding the cause
for the Earth's precessional movement. IF the Earth is considered as a
normal gyroscope, in order to explain the specific precessional
movement we measure, there would have to be an effective net Force
vector (essentially the equivalent to the Earth's gravity for a
gyroscope) exactly normal to the plane of the Ecliptic, downward
(South-ward). As a matter of fact, the gyroscopic action of the Earth
would also have to "counteract" this effect (like the gyroscope not
falling) and so there's even another problem with the "usual"
explanation!

The magnitude of this force vector must be calculated in the same
manner as for the gyroscope. The belt's effective moment of inertia
(I) is 3.3 * 10^35. The Earth's daily rotation rate (omega) is 7.292 *
10^-5 radians per second, so the magnitude of the rotational inertia
vector is the product of these, or 2.3 * 10^31 kg-m^2/sec. A
precessional period of 25800 years is equal to about 8.2 * 10^11
seconds (providing us with OMEGA). This implies that an Ecliptically
South-ward acting torque moment (M) that must be acting on the
Earth-gyroscope has a magnitude of about 2.9 * 10^19 newton-meters.

The difficult part to me is that this necessary torque moment must be
directed normal to the plane of the Ecliptic, and I do not see any
obvious mechanism to account for this magnitude torque moment to be
directed in that direction. The magnitude of this value seems
reasonable. The Sun's total gravitational attraction for the whole
Earth is around 3.6 * 10^22 newtons, and the Moon's total
gravitational attraction of the Earth is about 1.9 * 10^20 newtons (a
ratio of around 200:1). The attraction for the "belt" ring will be
substantially less, of course.

I am more tempted to think of the Earth's precessional movement to be
caused by an entirely different mechanism, but having an identical
result to that of gyroscope-like precession. It is still totally in
agreement with the Euler equations. I think it is far more accurate to
think of the Earth's "precession" to actually be an effect almost
identical to the Regression of the Nodes of the orbit of the Moon.

The 23.45 degree tilt of the earth's axis causes the Sun and the Moon
(separately) to sometimes be above the plane of the Earth's Equator
and sometimes below it. In the traditional thinking, this allows the
Sun or Moon to gravitationally pull differentially upward or downward
on that belt of material that is the Earth's effective gyroscope. The
thinking is that the component of that force that is normal to the
Ecliptic represents that 2.9 * 10^19 newton-meter torque moment. This
logic seems to have weaknesses though, because that normal component
would seem to symmetrically alternate upward and downward through the
month and year.

Conceptualizing the situation in a way similar to the Regression of
the Nodes of the Moon's orbit seems to make far more sense. (See any
good text on that Regression for further information on the concept.)
The idea would be that each part of that belt of extra material the
Earth has around its middle would act in that way. As each portion
approached a (node) crossing point of the plane of the Ecliptic or the
Moon's orbit, as the Earth rotated, it would generally arrive at that
point at an angle, from either above or below. As with the Regression
of the nodes for the Moon, the angle of that approach would be very
slightly increased just before the node, which would make the exact
point of crossing just slightly before the point where it normally
would have been. After the crossing, the angle effect is reversed such
that the original angle of approach is restored. The only lasting
result of this action is that the (node) crossing point is very
slightly earlier in each revolution or rotation. The location of the
node moves slowly backwards, or regresses.

In the case of the Moon's orbit, this phenomenon is well documented.
Regarding its application to the rotation of the Earth as an
explanation for Precession, a complication is that the Earth is a
solid body. This effect of the regression process cannot actually
cause that part of the Earth to briefly move in a different direction,
but rather it applies a perturbative force that must act on the entire
Earth.

This manner of explanation would imply that Regression rather than
Precession would be a more correct term for the phenomenon. The effect
is the same, however. In 25,800 years, the nodes referred to here
would regress entirely around the Earth. The effect is to have the
Earth "wobble" in its precession-like way.
_________________________________________________________________

Comments are welcome, at the e-mail address below.
_________________________________________________________________

Assorted comments have been received since this page was first placed
on the Internet in 1998. I have been somewhat amazed to find that
there are many people who claim that there is no such motion of the
Earth, that it is too slow of a process to have ever been confirmed!
Wow! I have been sometimes tempted to ask if those people also believe
in a "flat Earth"!

Every amateur and professional astronomer is VERY aware of the effects
of the precession of the Earth! It is VERY annoying! Accurate star
maps are printed for a precise date, like 1950.0 or 2000.0, and the
exact place in the sky that is above 90°North Latitude (our North
Pole) is precisely marked. (All other stars are marked and shown
relative to that location.) Unfortunately, due to the precession of
the Earth, the point in the sky (stars) that is directly above the
North Pole seems to MOVE! Every year, it proceeds by a little over 50"
of arc. Since even amateur astronomers often look at Nebula and other
objects that are only around ONE second of arc in diameter, if it has
moved by 50 times that far in a year since a map was printed, that's a
problem!

As a result of this, all astronomers regularly have to consult
"Precession tables" which give the amount of correction that is needed
for any date different from the Epoch of the maps! Except for a single
day, all star maps are always wrong! And Precession it the greatest
cause of that!

We might as well discuss another subject! The period that is described
regarding the Earth's Precession was not just "made up"! And you can
prove it yourself! Say you become an amateur astronomer, and you VERY
accurately locate that point in the North sky where all the stars seem
to circle around it every day. If you just put a good camera on a
telescope, and point it near where you think that point is, and make a
time exposure for a whole night, all the stars will record as arcs of
circles. The point you want is EXACTLY in the center of those circles!
(Easy, huh?) Now, being REALLY patient, you wait exactly ONE YEAR and
do it again! It will be in a noticeably different spot in the sky!

Countless researchers have done this (but a lot more precisely!) and
they have always gotten the same distance between the two points,
around 50.2" of arc. Well, you're virtually done! In order for that
apparent movement to eventually get back to where it started, it has
to proceed through 360° (of a great circle, like Longitude on Earth).
At 50.2" each year, a simple calculation (360 * 60 * 60 / 50.2) shows
that it will take a little over 25,800 years! THAT'S why astronomers
and astrophysicists know how long the Earth's Precession takes!
(Simpler than you had thought!)
_________________________________________________________________

This presentation was first placed on the Internet in April 1998.
_________________________________________________________________

Link to the Index of these Public Service Pages

( http://mb-soft.com/public/othersci.html )
_________________________________________________________________

E-mail to: Public1 at mb-soft.com

C Johnson, Physicist, Univ of Chicago