We will consider a toy gyroscope, as in this schematic drawing, where we will consider three perpendicular axes, 1, 2, and 3. For simplicity, it is spinning with its spin axis (1) horizontal, in a vertical (3) gravitational field, and where the gyro is supported at one end of the shaft, such that its gravitational weight attempts to rotate it (downward) (in a Moment or Couple) around the other horizontal (2) axis.
It is necessary to think of this in two separate stages, to really understand what happens. And also, it is necessary to know the "right hand rule" in Physics, where if a motion along one axis is perturbed (forced) along another axis, there is a resulting force/torque/Moment around the third (perpendicular) axis, as follows: If you use your RIGHT HAND and first point your (four) fingers in the direction of the (arrow of the) first motion, and then move your hand so that the fingers move toward the direction of the axis of the perturbing action, your THUMB will then be pointing in the correct direction along the third axis.
This will seem simpler as we give this example. The gyro rotor is spinning such that its (right hand rule rotation, fingers curved following the spinning and thumb then pointing in the correct direction) motion is defined as shown, going OUTWARD along axis 1. The weight of the gyro is trying to make it fall, which is a rotational action around axis 2, again right-hand-rule giving an outward direction (toward us).
Then we have the outward (1) being perturbed by (2), and again the right hand rule then has our thumb pointing upward, which tells us that something must happen regarding (3) that is upward. Your upward thumb actually then tells you which way the "precession" will occur! Since there is really nothing that keeps that torque/Moment from having effect, the gyro quickly accelerates to be precessing at a constant rate. Down below, we will calculate the rate it will precess.
THAT is the first part!
NOW, we have a precessional action (upward, 3) which now has to act on the spinning rotor (axis 1). Another right-hand-rule has our fingers going from 1 to 3, which makes our thumb now go (into the paper) along axis 2. The numbers involved turn out to be the same as in the first calculation. The result is that the new rearward Moment along axis 2 is EXACTLY the same size/strength/intensity as the weight-caused forward Moment we started with along axis 2. In real terms, this new moment exactly matches and cancels out the natural falling effect that we see in all other objects! And so the gyro is able to "hang there" seemingly defying gravity!
Just in case you might think something magical is happening, it isn't! If you measure the weight on the single support location, which would be half the total gyro weight when it is not rotating, it actually increases up to the entire weight of the gyro when we have the situation above. No weight "disappeared" or anything like that. Weight just gets transferred where the single support point is then supporting the entire weight.
Even advanced Physics textbooks seem to not mention this second process, where the weight of the gyro is therefore supported without falling. They seem to be satisfied at showing why the precession occurs! But now you know even more than that, actually why it seems to ignore gravity. And you know that it really does not! The weight force downward on the single point of support is actually that of the entire weight of the gyro, while if the gyro effect was not occurring, only half the weight would be supported there and a second support post would need to hold the outer end up, to keep it from falling.
Now we will explore some more mathematical treatment of this situation.
Maybe we can first do a partially-math approach!
If we call the rotational angular momentum of the spinning rotor H, we can see that it will be along axis 1. We already determined that the gravity/falling effect is M, which is along axis 2. We are now looking at a brief moment in time, which we call dt. This drawing is showing that, during that brief moment, we have an "angular impulse" of M * dt (along axis 2). We see here that there is a "change in the rotor's angular momentum" (called dH) that is exactly the same size and direction (this has to do with vectors!) as M * dt. This shows us two things. First, we can see that H has to change in a direction toward axis 2 (in this example) so we confirm that we know which direction that precession will occur. Second, we know the value of M, so we can calculate the rate of change of dH with time (or dH/dt), which is another way of calculating the rate of the precessional motion we are about the calculate below.
Our discussion below, and these comments above, are all about the most simple possible example of gyroscopic action. When a gyroscope is at an angle, things are a lot more complicated, and it turns out better to calculate dH/dt for more general situations. However, for us here, we intend to keep this as simple as possible, so we are going to only discuss that simplest possible case, where the gyro spin axis is exactly horizontal.
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, some of these differential terms are zero and can be ignored. 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.
The first Euler equation then applies to the axis around which the rotor is spinning. Since we are not trying to accelerate the speed of rotation and there is no friction slowing it down, this equation (only for this extremely simple situation) is 0 = 0 + 0. The third Euler equation applies to the vertical axis around which the gyroscope is precessing. Since we are taking the case where the precession is already up to its correct speed, and we have no external forces acting, this equation too is 0 = 0 + 0 (again, only because we have taken such a simple situation.)
Then the second Euler Equation is the one we need to consider and it is:
M2 = I2 * (dw
2/dt) + (I1 - I3) *
w3 *
w1 (we are using the
letter d for the differential symbol.)
In our very simplified case, the term (dw2/dt) is exactly equal to the component w3 * w1 along the (2) axis. For the simple case of a body (rotor) of symmetry, I2 is equal to I3, and so the first and last terms cancel each other out. This exact match is actually why a horizontal gyroscope is able to hang there, seemingly disobeying gravity! It is actually creating its own UPWARD (torque) vector which exactly matches the downward (torque) due to the pull of gravity on the gyroscope.
This allows a great simplification of the original equation to: M2 = I1 * (w1) * (W3).
This is therefore the simplified theoretical equation that describes the precession of a horizontal gyroscope. M is the 'moment' meaning the sideways (along axis 2) moment (or torque) that is trying to change the direction of the horizontal (originally along axis 1) angular momentum vector of the spinning rotor (in other words, the fact that gravity is attempting ot make it fall, and therefore hinge around axis 2). I is something called the 'moment of inertia', a characteristic of the construction of the spinning rotor, w is the rate of spinning of the rotor, and W 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 * a, "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!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 * r2.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 (for a horizontal gyroscope) is therefore a Vector product of I and a. The rules of Vector multiplication are different from simple scalar multiplication, and the result is that, in order to actually calculate vector multiplication values (for any situation other than a horizontal gyroscope axis), three separate equations must be considered, along the three axes of the coordinate system. This is also 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. For most situations, the math can get extremely complex!
In our simplified 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 precession, again, essentially due to friction or external forces. That left us with just the second Euler equation to evaluate. (But again, ONLY because of all of our simplifications!)
If the gyroscope was tilted at any angle other than horizontal, then there would include terms in all three coordinate directions, for more complicated reasons. We chose a horizontal axis gyroscope in order to keep the math manageable here. Everything discussed here is also true for an angled gyroscope, but the math is much more complicated to understand and to solve.
The shape of the rotating part of a gyroscope is designed essentially as a ring 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.
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?) (It is actually a moment or torque around a horizontal axis, as is the gravitational effect here.) 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 effect mentioned is the one that causes precession, where the gyroscope slowly revolves around its support pedestal.
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/sec2 [the acceleration due to gravity] or 0.294 kg-m/sec2 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 0.01176 newton-meter. [This is M in the equation above]
We're getting there! Imagine that downward torque applied for a one-second interval. That would mean that we were applying a (torque) moment of 0.01176 newton-meter for one second. Writing this a different way, it is 0.01176 kg-m2/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 0.01176) 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 (like the hypotenuse) 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 0.01176 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 11.1 degrees horizontally during that second. This particular gyroscope would therefore take about 32.5 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 rotor from 12,000 rpm to half that (6,000 rpm), there is an interesting result. The sideways vector is still 0.01176, 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 21.4 degrees, so the gyroscope will now take only 16.8 seconds to complete an orbit, instead of the earlier 32.5 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. You could even work backwards and use the precession rate to determine the rate the gyro rotor must be spinning! If it is placed at any angle, all of the theory described is still true, but the mathematics gets more complicated. The simplifications 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 * w * W, where M is the perturbing torque moment, I is the moment of inertia of the spinning part, w is the rate of spin of the spinning part, and W is the rate of precession. For our example, 0.01176 nt-m = 4.8 * 10-5 kg-m2 * 1256/sec * W. This solves to 0.1951 radians/second, which gives a period for the precession of 32.2 seconds, in good agreement with our earlier rough calculation. This value is technically the correct one!
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 the rotor's 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. 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 * w
In our example, this (initially) gives 0.01176/0.06 = 0.01176/0.06
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.
Eventually, (after it was accepted around 1500 AD 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! It is certainly another application of Euler's equations, and very closely related, but different. 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 somewhat different sources.
It is pretty well known that the entire Earth has a moment of inertia of about 8.070 * 1037 kg-m2. But this number has no immediate value in analyzing the precessional movement of the Earth! If the Earth was perfectly spherical and uniform, there would be no precession!
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, with a maximum thickness of around 13 miles (21.4 km), around its middle that accounts for the Equatorial diameter of 7926 miles (12,756 km). This "belt" of material has its maximum thickness at the Equator, and ONLY THE BELT represents an equivalent to our toy gyroscope. That belt of material has an effective moment of inertia of about 3.3 * 1035 kg-m2, about 1/250 that of the whole Earth.
This all makes excellent sense. The next part gets a little more complicated. Imagine JUST the Sun for a moment (and forget the effects of the Moon). On March 21 or September 21, on an Equinox, the Sun is directly above the Equator. At this time, there is NO precession effect! Now consider June 21, the Summer Solstice, where the Sun is as far north of the Equator as it can get. Try to ONLY think about that belt now, from the Sun's point of view. The belt would look tilted downward, 23.45° toward the Sun. (Someone on the Sun would sort of think they were looking at the 'top' of the oval-looking belt.) Now just think about the nearest part of the belt to the Sun. The Sun's gravity would be pulling it, of course. The important part here is that part of that pull would be pulling THAT PART of the belt (upward) TOWARD flattening the tilt position of the belt out. Now, if we consider the farthest part of the belt from the Sun, the effect would be to also pull it upward, but that would act to make the belt tilt even more. These two effects are opposite of each other, the first trying to flatten the belt and the second trying to tilt it more. It might seem that these two effects would exactly cancel and nothing would happen. Almost! The nearer part of the belt is nearer the Sun, and the effect of gravity depends on distance (the inverse-square law). The result of this is that the effect on the front portion is stronger than the similar effect on the rear section. So the "flattening" effect is stronger/greater than the "increased tilting" effect. In case this was too confusing, the important part is that the belt would have the effect of feeling a moment (or torque) that would be trying to flatten it out, trying to tilt the Earth's axis toward being more upright. The technical name for the tilt of the Earth's axis is the Obliquity of the Ecliptic.
If you followed this, you might want to think through what happens at December 21, the Winter Solstice. You should find that, again, the net effect is to try to cause the Earth's axis to become more vertical. This (differential) effect of the Sun is the same whether the Sun is looking at the top or the bottom of the equatorial belt.
OK. This is somewhat like a child's gyroscope, but with two major differences. (1) Where gravity on a gyroscope is trying to INCREASE the tilt angle of the axis with vertical (make it fall over), the situation with the Sun and the belt of the Earth is trying to do the opposite, to stand the Earth up straighter! (But both are applying a Moment (torque) to try to tilt the axis!) And (2), the precession effect of the Sun is NOT CONSTANT! Every year, it completely stops at March 21 and September 21 (when the Sun is directly over the Equator and the belt) and becomes maximum on June 21 and December 21. When people speak of the precession of the Earth, they are actually talking about an AVERAGE rate, because it is constantly changing like crazy! There are other variations in the rate of precession due to the Moon. The Moon's orbit is tilted several degrees from the Ecliptic, its orbit is elliptic (the Moon coming nearer and going farther from us every single month, and worse!) and the plane of that orbit constantly Regresses in an action similar to precession, in a period of around 19 years. This variation on the orientation of the Moon's orbit sometimes enhances the Sun's Precessional effect and sometimes fights it, so there is a prominent variation of precession in around a 19-year period, an effect called nutation.
I hope you can see why I am a little bothered by using the same word, precession, to describe a gyroscope's (constant, fall-over) motion and the Earth's (herky-jerky, stand-up) motion. But, yes, they certainly have many similarities, especially the end effect of the wobble!
The magnitude of the Sun-sourced force vector (and Moment) must be calculated in the same manner as for the gyroscope. The Earth's total moment of inertia (I) is 8.07 * 1037. Any torque (or Moment) applied to the (whole) Earth has to act on that. The Earth's daily rotation rate w is 7.292 * 10-5 radians per second, so the magnitude of the Earth's angular momentum vector (H) is the product of these, or 5.9 * 1033 kg-m2/sec or nt-m-sec. The mass and distance of the Sun are known, and so is the 23.45° angle of the tilt of the Earth's axis. It is therefore possible to calculate the Moment/torque effect of the Sun on the Equatorial BELT at any instant. It is pretty easy to calculate the instantaneous Torque on June 21 for the tiny part of the equatorial belt directly under the Sun and that part exactly opposite. It is about 5.73 * 1022 nt-m. One must mathematically Integrate the Moment effect over the entire belt and over an entire year (fairly easy Calculus) to get the net annual average torque effect due to the Sun 1.44 * 1022 nt-m.
In one (average) second, we then have an angular impulse of 1.44 * 1022 nt-m-sec. This is the dH/dt of far above! H is the 5.866 * 1033 nt-m-sec. The quotient of this 2.45 * 10-12 radians/second. This is the same as a precession of 15.94" arc-seconds per year. Note that we have ONLY considered the Sun, and resulted in a precession effect on the Earth, completely due to the Earth's non-spherical shape (due to the Earth spinning on its axis) and its effect of creating a belt around the Equator. The Moon does NOT even need to exist for this Precession to occur! However, this precession due to only the Sun (the 15.94 arc-seconds per year) would mean that the Earth would precess entirely around in 81,300 years, much more slowly than we know that it actually does!
It turns out that if we ONLY consider the Moon, using entirely different numbers but essentially the same logic, we get another precessional effect on the Earth. The Moon's orbit is tilted from the Earth's equator (and also has another 5° of orbital inclination from the Ecliptic) and so all the same stuff applies. The Moon is much less massive than the Sun, but it is also much closer. The calculations (like above) give us an average monthly average torque as 3.1 * 1022 nt-m, slightly more than double the effect of the Sun. With all the exact same calculations as above, we get a result that if only the Moon were causing precession, the Earth would precess entirely around in 37,600 years, or alternately, we can say that the Moon causes an annual precession of 34.45 arc-seconds.
The Moon's orbit is not very circular (it has eccentricity, and that eccentricity has constant changes of its own!) and so the calculations depend on where the Moon is in its orbit, so the math is much more complicated, but otherwise the same.
Scientists generally refer to a Luni-solar Precession, which is the total effect of these two, the 15.94 arc-seconds per year and the 34.45 arc-seconds per year, or a net effect of 50.39 arc-seconds per year.
All this has been presented to remind you that the ACTUAL precessing effect at an instant is dependent on a huge number of variables, and that the normal reference to "precession" is really just a long-term average of a lot of complex effects!
The Earth's precessional movement is actually an effect almost identical to the Regression of the Nodes of the orbit of the Moon. If you are familiar with that phenomenon, hopefully you see the many similarities. If not, never mind! (Or consult a good textbook on Regression of the Nodes.)
We just noted that when the (annual) precession of the Earth due to the Sun and that due to the Moon are combined, the calculated effect is a (lunisolar) precession of around 50.39" of arc per year. (This would indicate that precession would take around 25,700 years.) However, because the Earth's axis is tilted, even the other planets cause the same sort of precessional effect! And the other planets' gravity act to distort the orbit of the Earth, too, slightly changing its shape and angle and orientation. Each planet causes its own effect, but the greatest is due to Jupiter. Since we "pass" Jupiter in our smaller orbit, its precessional effect is actually in the reverse direction of that due to the Sun or Moon. When all the planets are combined, their combined precessional effect (planetary precession) on the Earth is (currently) around -0.106" of arc per year. SOOO! The TOTAL of all precessional effects on the Earth is (currently) around 50.29" per year (and slowly but temporarily increasing, due to a variety of causes). This results in a full precession cycle as being (currently) very close to 25,800 years.
Again, for simplicity, consider just the Sun's effect for a moment. We noted above that there is no precession effect at all on March 21 or September 21, and a maximum precession effect on June 21 and December 21. This causes the herky-jerky advancing of the precession effect. If we chart the extended position of the axis of the Earth on the sky, yes, it advances around 50.29" of arc each year. But that movement seems to speed up and slow down! Such variations of the Earth's precession are effects called nutation. The Sun's effect is called solar nutation and is a maximum of around 1.2" of longitude (different from what a constant precession would have caused). It turns out that these on-off precession effects, which are caused by the Sun trying to flatten out the Earth's equatorial plane and stand up our axis of rotation, remember, also have a brief effect at accomplishing that. So the tilt of the Earth's axis is also slightly (temporarily) affected by solar nutation. The effect of these two things is that if we carefully tracked the exact axis of the Earth's rotation, it has a really tiny (around 1.2" of arc) effect. This causes the Earth's axis to shift by around 120 feet, in essentially a circle, twice each year! When explorers go to the North or South Pole, they probably do not get to the actual axis of rotation ON THAT DAY, and instead plant their flag at the point of the AVERAGE location of it!
There is also lunar nutation, due to the "inequality" of the variations in the precession due to the Moon. In the same way that lunar precession has a greater effect than solar precession, so does lunar nutation. It's maximum effect is around 17.2" of arc. The combination of the two nutation effects can therefore have around 18.4" of arc of effect. This is usually described as a 9.2025" nutation that is either positive or negative from the calculated precession position, mostly for the simplicity of calculations.
The combination of these two effects is that the Poles of the Earth nutate around the "average" axis of rotation in a small wavy circle. Interestingly, the ACTUAL axis of rotation on any day is NEVER exactly at the place that we call the North Pole, but somewhere on that wavy circle around 900 feet away from it! For maps, we must use a specific location, and so the average location of the axis is identified and used. This brings up an interesting point! Since, on any given day, the ACTUAL "North Pole" (the actual axis of the Earth's rotation on that day) is around 900 feet away, and moving nearly a foot an hour, the reality might easily be that NO ONE has yet actually been to the "real" North or South Pole!
Because the Earth's orbit is elliptic and (the elliptic orbit) is also very slowly revolving around the Sun, and other similar effects, such as variations in the Obliquity of the Ecliptic, the Precession motion of the Earth is not very constant!
We need to now look at the third Euler Equation which is:
M3 = I3 *
(dw3/dt)
+ (I2 - I1) * w
2 * w1 (we are again using
the letter d for the differential symbol.)
This is the equation that describes the dynamics of the motion around axis 3, the precession. There is no externam Moment applied, so M3 = 0. The other two terms must therefore always add to zero. The first term involves the angular acceleration of the precession (dw3/dt) which is what we need to determine. The second term includes three terms that cannot change and one which could (w 2). Both of these terms become non-zero for a brief period, in the case of a toy gyroscope, usually a fraction of a second.
If we Integrate both terms over the time interval of the precession acceleration, we wind up with terms which include w3 (the actual final precessional rate) and q2 (a change of angle of the tilt of the gyro axis). If we consider the example gyro discussed above, we have the necessary numbers. We calculated that I1 = 4.8 * 10-5 kg-m2. w1 = 1256 radians/second. w3 = 0.1785 radians/second. Because there is necessarily a frame surrounding and supporting the gyro rotor, we are going to say that I3 = 6.0 * 10-5 kg-m2 (with I2 being the same). We then have 6.0 * 10-5 kg-m2 * 0.1785 radians/second = (6.0 * 10-5 kg-m2 - 4.8 * 10-5 kg-m2) * 1256 radians/second * q2. This gives q2 as around 0.0007 radians or around 0.04 degrees.
This indicates that the gyroscope drops down a tiny fraction of a degree while the precession accelerates up to speed. (This represents around 0.03 millimeter, a distance that would be hard to notice!) The precessional kinetic energy which appears in our toy gyroscope is about one one-millionth of a newton, roughly the same as the amount of potential energy that was released as the gyroscope dropped that tiny fraction of an inch, which Conserves Energy.
So even though the precessional motion appears to begin without any source of energy, it actually has a source in the potential energy in the vertical axis (in the gravitational field).
However, Conservation of Angular Momentum appears to be violated, where it is always true otherwise. As the precessional motion begins, angular momentum "appears" where it had not existed before. This is unique in the field of Physics!
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. That is actually a lot, roughly the entire diameter of the Moon in 40 years! 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! If you set a clock-drive of a telescope to a specific position in the sky, you will probably not see what you are looking for! Until you correct for the precessional effect!
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! It might seem that star maps are star maps, they never change! Pretty much, that is true, except for the effect of precession. If someone now tried to use a 1950.0 star map to find some nebula, they would probably never find it, as it would be more than a full Moon diameter away from where that map showed it to be!
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 small 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 distance between the two points to be 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 around 25,800 years! THAT'S why astronomers and astrophysicists know how long the Earth's Precession takes! (Simpler than you had thought!)
It strikes me as mind-boggling that any person with a small telescope and a standard camera (and a year) can accurately determine an effect that takes over 25,000 years to occur. We humans have only known how to write for less than 5,000 years, and 'little you' can definitively determine a process that has not even close to completed itself in all the time of human civilization. See? You're smarter than you realized!
By the way, 25,800 years per wobble seems pretty slow, right? But the Earth is so old that the Earth has wobbled like that around 200,000 times already! Another interesting thing is that where toy gyroscopes precess faster and faster as the rotor slows down, the opposite effect will happen to the Earth. As friction of the oceans to the ocean tides (caused by the gravitation of the Moon and Sun) gradually slow the Earth's rotation down (around 22 seconds longer days a million years from now) the reversed action of the Earth precession effects (to try to stand the Earth straighter up) will slow down the precession. At a REALLY distant time in the future, the Earth's rotation will have slowed down to about 1/50 of what it is now (a day then will take 50 times as long as now), and the Moon's rotation and revolution will be the same 50 days, the Earth and Moon will rotate and revolve together, always with the same side facing the other. At that time, the Earth's "axis of rotation" will be in the plane of the Moon's orbit. So then, no further precession could occur due to the Moon, and the Sun's precession effect will be different (and much smaller because the Earth will have a far smaller equatorial bulge) than now.
Link to a thorough mathematical discussion of tidal issues, is on Tides
Link to a thorough mathematical discussion of assorted Moon issues, is on Origin of the Moon
( http://mb-soft.com/public/othersci.html )
C Johnson, Physicist, Physics Degree from Univ of Chicago