Precession of a Gyroscope

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 (downward 3) gravitational field, and where the gyro is supported at one end of the shaft (toward the right, but not shown in our drawing), 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 (vector), and then bend your fingers so that the fingers move toward the direction (vector) of the axis of the perturbing action, your THUMB will then be pointing in the correct direction (of the resulting vector) along the third axis.

This will seem simpler as we give this example. The gyro rotor is spinning such that its (another right hand rule: rotation, fingers curved following the spinning makes the right hand thumb then point in the correct direction [of the vector arrow]) 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 (first) right hand rule then has our thumb pointing upward, which tells us that something must happen regarding (3) where there is a vector 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 (body or frame) quickly accelerates to be precessing at a constant rate. It is usually rather slow. Down below, we will calculate the rate it will precess.

Is it obvious that if you would cause the rotor to spin the opposite direction, then the Precession will go the opposite way?

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 application of the right-hand-rule has our fingers now going from 1 to 3, which makes our thumb now go (into the paper/screen) along axis 2. When we calculate that, it turns out that the numbers are the SAME as in the first calculation. The result is that the new rearward Moment (vector) along axis 2 is EXACTLY the same size/strength/intensity as the weight-caused forward Moment (vector) we started with along axis 2. In real terms, this new moment (vector) exactly matches and cancels out the natural falling effect (vector). THIS is what causes the gyro to appear to defy gravity! We see any OTHER object FALL due to gravity, but this very unusual situation CREATES a force/torque/Moment (vector) which exactly balances out the effect of gravity! And so the gyro is able to "hang there", seemingly defying gravity! (This situation is called meta-stable. It can ONLY occur as long as the gyro rotor is spinning fast enough so that this cancelling Moment can be created. At some point, bearing friction and air friction will slow the rotor spin down to a speed where the cancelling effect will suddenly fail and the gyro then quickly drops off of its pedestal!)

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 (with the other half of its weight supported by your fingers), it actually quickly increases up to the entire weight of the gyro when we have the situation above. No weight "disappeared" or anything like that. No anti-gravity effect exists! The weight of the entire gyroscope just gets transferred to 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 motion occurs! But now you know even more than that, actually why it seems to defy 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 (or your fingers) 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 (a vector) H, we can see that it will be along axis 1 (to the left). We already determined that the gravity/falling effect (vector) is M, which is along axis 2. The actual analysis here involves Calculus, but we can discuss a simpler version of that here. We will now look at a brief interval of time, which we call dt. This drawing is showing that, during that brief interval, we have an "angular impulse" of M * dt (along axis 2). Because of the reasoning described above, this angular impulse IS the "perturbing action" which then applies to the initial rotational angular momentum H.

We can "translate" that vector to the left to put it in a location where it will be "added to the vector representing H". We see here that there is therefore 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. We therefore now have a NEW rotational angular momentum vector, shown as H'. 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.

You might think from this simplified discussion that the LENGTH of the H' vector is longer than the H vector was. That is NOT the case! When Calculus is used, the size of dt is made incredibly tiny, and the effect of the very tiny dH then ONLY changes the direction and does NOT affect the length of the H vector. The rotor therefore does NOT either speed up or slow down its spinning as a result. Therefore, the result of all this is that the H vector stays at its constant length, but simply ROTATES in space, twisting the axle of the rotor slowly, smoothly and continuously. THAT is Precession!

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 necessary to use Calculus 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. Like before, 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 can see that there are THREE separate terms (on the right) here. If we separate the second and third, we can continue.

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 third terms exactly cancel each other out. This exact match is actually why a horizontal gyroscope is able to hang there, seemingly disobeying gravity! The third term is actually creating its own UPWARD (torque/Moment) vector which exactly matches the first term's downward (torque/Moment) vector due to the pull of gravity on the gyroscope.

We then wind up with ONLY the middle term remaining in this differential equation.

This allows a great simplification of the original equation to: M₂ = I₁ * (W₃) * (w₁).

This is usually presented without the subscripts as: M = I * W * w.

(We have replaced the lower case omega with a capital omega to show the form that the popular simple equation for Precession is usually presented.)

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!

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 (sometimes!) different from simple scalar multiplication! If two VECTORS are multiplied together, the result is a Vector which is perpendicular to BOTH of them! In our very simple case, we wind up with Vectors along all three coordinate axes. That is actually called a CROSS PRODUCT, and the multiplication symbol is an x. Vectors can also be multiplied by SCALARS, which still results in a Vector, but now it is in the exact SAME direction as the original Vector was! This is called a DOT PRODUCT and the multiplication symbol is the more familiar .. Confused yet? The Euler Equations are therefore VECTOR equations, where the various parts are sometimes dot products and other times cross products, and remember, they are DIFFERENTIAL equations, so solving them for any situation more complicated than the very simple example we have been considering involves Vector Calculus.

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. Because of the peculiar characteristics of Cross Products, Vectors with components along all three axes show up all over the place in those equations! 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.

Note that each of the 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.

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. Notice that 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 (along 2) (of 0.01176) to be added to the original vector describing the angular momentum (0.06, and in a direction along the shaft, along 1). 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!

I can pretty much guarantee that you have NEVER even MET anyone who could actually explain why a slowing top or gyroscope INCREASES in precessing speed as the rotor slows down! But now YOU could explain this odd fact to strangers you meet on the street!

IF someone GUESSES at why it happens (like adults do when their kids as about this), their guess is usually that the spin energy of the gyro rotor is getting transferred to the precessional motion. Interesting guess, but totally wrong! The gyro rotor ONLY slows down due to air friction and the friction of its shaft bearings, and so the energy it loses in slowing down simply goes into heating up the air around it (a tiny bit) and the metal near its bearings! We will discuss (far) below just WHERE the energy for the precessional movement comes from, and THAT is actually also the source of the additional kinetic energy as the precession speeds up. Nice try, parents, but no cigar!

With the above approach, you should now 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 timed 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 discussed 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.

But we have already calculated what M and I * w are! Therefore, for our example, we have 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 fairly 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 (*sec)       M (*sec)
_________________ =  _____________________________________________  =  ________
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.

Precession of the Earth's Axis

Eventually, (after it was accepted around 1500 AD that the Earth rotates and moves around the Sun!) it was realized that the fact that the Earth's spin axis is significantly TILTED (which causes our Seasons), caused a situation where the Earth's motion represented an effect apparently resembling that of a child's top or gyroscope. The CAUSE of this effect is not quite the same as for a toy gyroscope on the Earth's surface. The premise is that the earth is a giant gyroscope which is affected by each of the Sun and Moon, and which therefore has a period for the precessional "wobble" of (currently) about 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. That is only a partially correct assumption! 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, and definitely somewhat misleading. Even though they LOOK to be the same, they are actually somewhat OPPOSITE!

A child's gyroscope is affected by the outside effect (gravity) where the gyroscope tries to fall over. The Earth is affected by the outside effect(s) (the separate gravitation of the Sun and Moon) where the Earth tries to stand up straighter!

If the Earth were perfectly spherical and uniform, there would be no precession at all! It only occurs BECAUSE the Earth has an equatorial bulge, that is, the Earth is deformed due to its rapid spinning.

It is pretty accurately known that the entire Earth has a moment of inertia of about 8.070 * 10³⁷ kg-m². But this number has no direct value in analyzing the precessional movement of the Earth! Because the Earth spins fairly fast on its axis, once every day, where locations on the Equator are rotating at over 1,000 miles per hour, the actual shape of the Earth is not precisely a sphere but it is slightly deformed. We sometimes say that the Earth is "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, for our purposes here, we can look at the Earth as being a composite of two different (spinning) objects. The first is a perfectly spherical planet, with a diameter of the Polar diameter (around 7899 miles [12,713 km]). This object, being spherical, has NO significance regarding any precession effects. The second component is 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" somewhat resembles a very distorted "donut" of material has its maximum thickness at the Equator, and tapers off in thickness both to the north and south. The significant fact here is that ONLY THE BELT represents an equivalent to our toy gyroscope. That belt of material is estimated to have an effective moment of inertia (I) 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! The Earth does NOT Precess at all on those days! Now consider June 21, the Summer Solstice, where the Sun is as far north of the Earth's Equator as it can ever 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 (on that day). The Sun's gravity would be pulling it, of course. The important part here is that part of that pull of the Sun on the Earth would be pulling THAT PART of the belt, not only TOWARD the Sun as we expect, but ALSO UPWARD. This would have the effect TOWARD flattening the tilt position of the belt out (or of causing the Earth's rotational axis to stand up more vertically). 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 (or of causing the Earth's rotational axis to tilt over farther). These two effects are opposite of each other, the first trying to level out 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. When they are combined, the TOTAL EFFECT is to act to try to flatten out the belt of material, or equivalently, to stand the Earth's spin axis up more vertically.

The effect is therefore actually a SECOND-ORDER effect, where the main (larger) effect tends to cancel out amd only a smaller effect remains to have any effect.

In case this was too confusing, the important part is that Sun's gravitational effect on the belt (on that day) would have the effect of creating a Moment (or torque) that would be trying to tilt the Earth's belt axis toward being more upright. The fact that the belt is rigidly attached to the rest of the Earth means this same effect occurs to the entire Earth. 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 therefore the same whether the Sun is looking at the top or the bottom of the equatorial belt.

Even though Summer and Winter have the Sun looking at opposite sides of our Equatorial belt, the effect turns out to be the same, where both have the effect of trying to cause the Earth's axis to become more vertical. So we have NO precession around March 21, a MAXIMUM around June 21, NONE again around September 21 and MAXIMUM again around December 21! It is NOT a constant effect like everyone has always told you!

OK. This is SOMEWHAT like a child's gyroscope, but with two major differences. (1) Where gravity on a toy 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, which is actually the important part!) 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 usually ASSUME that it is a CONSTANT process! They probably don't realize that they are actually talking about an AVERAGE rate, because it is constantly changing like crazy!

We have been discussing the effect of the Sun, but the Moon also has essentially identical effects. The Moon's motion is extremely complex and there are other (small) variations in the rate of precession due to the Moon. The Moon revolves around the Earth each month, and it is much closer to us than the Sun is. Even though the Moon is far less massive than the Sun, the gravitational effects regarding Precession are actually greater due to the Moon than due to the Sun (as calculated below). AND they change even faster, with the two maxima and two minima every 29.5 days!

The analysis of the Precession due to the Moon is more complicated, for several reasons. The Moon's orbit is tilted several degrees from the Ecliptic, its orbit is rather 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, one of the effects which are collectively 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!

Because the Moon has so many other effects involved, our calculations here will focus on the Sun and its precessional effects. The Moon is analyzed in the same way, but the mathematics is far more involved, and we feel it would be distracting here.

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. We have also discussed above that the estimated moment of inertia of the Earth's equatorial belt (I_b) is about 3.3 * 10³⁵. Any torque (or Moment) applied to the (belt or whole) Earth has to act on the whole solid Earth. 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 the TOTAL Moment of Inertia of the Earth times that rotation rate, or 5.866 * 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 possible to calculate the Moment/torque effect of the gravitational attraction of the Sun on the Equatorial BELT at any instant, since it is basically only dependent on the distance involved and the angle of the tilt at that instant. We know that it is ZERO around March 21 or September 21!

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 represents 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 as being 1.44 * 10²² nt-m.

We then have an (average) Moment (torque) applied of 1.44 * 10²² nt-m. This is the dH/dt of far above! H is the 5.866 * 10³³ nt-m-sec we just calculated. In the same was as we did with the toy gyroscope, we can now determine how great the effect is. The quotient of this is 2.45 * 10^-12 radians/second. This is the same as a Precession rate 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 (360 degrees divided by 15.94 arc-seconds or) 81,300 years, much more slowly than we know that it actually does!

By the way, we have simplified these calculations here to not include the effects of the varying distance of the Earth to the Sun due to its elliptical orbit, or that that ellipticity is even slowly changing!

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 (360 degrees divided by 34.45 arc-seconds or) 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.

. By the way, the fact that the Moon has these effects far more rapidly than the Sun does, every few days, and it is also of around twice the net effect, our associated presentation regarding the Violation of the Conservation of Angular Momentum includes the calculations which show that an average of around 63 million kiloWatts of power is being transferred into or out of the Earth's Lunar-driven Precessional motion! This 63,000 Megawatts is comparable to ALL the electrical power generated by ALL of the US Nuclear-powered electric powerplants. I do not see any possible way that we humans could ever accomplish where we might ever capture any of that power, but it is certainly an intriguing thought!

Scientists generally refer to a Luni-solar Precession, which is the total effect of these two, the 15.94 arc-seconds per year due to the Sun and the 34.45 arc-seconds per year due to the Moon, 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 any 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 a phenomenon called 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 are able to cause the same sort of precessional effect! And the other planets' gravity act to distort the orbit of the Earth, too, (Regression of the Nodes) slightly changing its shape and angle and orientation. Each planet causes its own effect, but the greatest is due to the massive 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 slightly reduced from the lunisolar precession and 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. (For clarification, all the way around means 360 degrees, which is 1,296,000 seconds of arc. Divide 1,296,000 / 50.29 to find out how many years that precession rate would take to go all the way around, which is about the 25,800 years.)

For math geeks, I would encourage trying to use the method presented above regarding the Sun, but for Jupiter, and calculate the precessional effect on Earth of Jupiter. You will find that it represents the majority of that entire Planetary Precession amount! (Rough values are: Jup -0.095; Sat -0.028; Ura -0.005; Nep -.004; Mar -.006; Ven +0.027; Mer +0.002; these values total to -0.109" of arc per year, moderately close to the measured total of -0.106" of arc per year. No other objects in the Solar System cause large enough precessional effects to be significant.)

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 generally lumped in the category called Nutation, and there are a variety of different other effects that have that same name. (Even the fact that snow and ice accumulates near the North Pole during winter, which transfers some mass up there, and then much of it melts the next summer and flows down into oceans, causes an effect regarding the Rotational Inertia of the Earth which can be measured, another process which is called Nutation.)

There are actually close to 200 different effects which are collectively called Nutation! Roughly half of those are due to the Sun or Moon and the other half are due to effects of the other planets and some of their moons. Only some of them due to the Sun and Moon are very large in effect. The four which have largest effects are (1) due to Regression of the Nodes of the Moon's orbit, 18.6 years; (2) due to the alteration of Solar Precession, 182.6 days or half a year; (3) due to the alteration of Lunar Precession, half a month; and (4) the revolution of the Moon's perigee, 9.3 years. The Sun's principal 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 the 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 poor circle, twice each year!

You probably already guessed that 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. Lunar Nutation has a maximum effect of around 17.2" of arc. Where Solar Nutation repeats about twice each year, Lunar Nutation repeats about twice every month! The combination of these two Nutation effects can therefore have a maximum of around 18.4" of arc of effect. This is usually described as a 9.2025" Nutation which can be either positive or negative from the calculated precession position, mostly for the simplicity of calculations. Of course, since the Moon has a rather elliptic orbit, and that orbit is constantly evolving, such as by Regression of the Nodes, the lunar effect actually varies to some extent. The full calculations are horrendously complex. (Which accounts for all the separate effects which are lumped into Nutation!) In fact, during the 1960s, when the US Government had committed to landing a man on the Moon, the available calculations were pretty terrible, and many scientists worried about whether the Moon would actually be at the location where the spacecraft would try to land!

The combination of these Nutation effects is that the Poles of the Earth Nutate around the "average" axis of rotation in a small wavy somewhat circular path. 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!

When explorers go to the North or South Pole, they probably do not get even very close to the ACTUAL axis of rotation ON THAT DAY, AT THAT MINUTE, and instead plant their flag at the point of the AVERAGE location of it, more than a city block away!

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 has lots more effects on it. It is not very constant! These effects are generally pretty small, and VERY slow, so there is no reason to get excited over them! But in the constant pursuit of science to achieve ever and ever more accurate data and info, these things are studied and researched. We are NOT going to discuss here the reasons for the many wiggles in this graph, except to say that the discussion above actually provides all the necessary info and even the equations. The scale on the left shows the SIZE of the variations away from an average value of around 50.27" of arc per year.

This is a far more accurate and more recent graph from Lasker (1986 and 1993) It has a smaller scale of only 100,000 years both in the past and in the future. It is drawn in the opposite direction of the graph immediately above it.

Such graphs are possible because of a mathematical procedure called the Fourier Transform, where extremely precise data can be analyzed for virtually invisible very long-term patterns. A different subject!

Deeper Math regarding When Precession Is First Beginning

A Clear Violation of the Conservation of Angular Momentum!

We need to now look at the third Euler Equation which is:
M₃ = I₃ * (dw₃/dt) + (I₂ - I₁) * w ₂ * w₁

This is the equation that describes the dynamics of the motion around axis 3, the precession. There is no external 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 small fraction of a second. (Once the Precession is up to its proper speed, even those two terms also become zero.)

If we (Vector) 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). Think about exactly what this means. The body of the gyroscope SINKS some distance (in the Earth's gravitational field) which therefore GIVES UP an amount of POTENTIAL ENERGY. THIS is our source for the energy that becomes the Kinetic energy of the Precession motion.

This result, of the mathematical Integration of the Euler Equation(s), clearly is extremely important to science, because of its many consequences, but as far as I know, it has never even been noticed before!

We therefore can calculate the TOTAL amount of Kinetic energy which is in the precession motion, and know that that amount is EXACTLY the same as the amount of Potential energy which was lost. Therefore, we can calculate exactly how far down the gyroscope body must drop to keep Conservation of Energy true!

If we consider the example toy gyro discussed above, we have the necessary numbers. We had calculated that I₃ = 4.8 * 10^-5 kg-m². We also calculated how fast the precession acts, w₃ = 0.1785 radians/second (in the form needed here). In rotary motion, the Kinetic Energy is given by 1/2 * I₃ * w ₃². For our example gyro, this results in 7.65 * 10^-7 nt-meter (or joule).

This amount of energy is equal to the Potential energy given by m * g * Dh. We learned above that m * g is 2.94 nt. Therefore we have Dh = 7.65 * 10^-7 nt-meter / 2.94 nt. This is therefore 0.00026 millimeter that the body of the gyroscope had to drop in order to release the necessary amount of Potential Energy.

This indicates that the gyroscope drops down a tiny fraction of a degree while the precession accelerates up to speed. (This represents around 0.00026 millimeter, a distance that would be nearly impossible to notice, nearly being on the atomic scale!) The precessional kinetic energy which appears in our toy gyroscope is about three-fourths of one one-millionth of a joule, which is necessarily the same as the amount of potential energy that was released as the gyroscope dropped that tiny fraction of a millimeter, in an example of Conservation of 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 IS 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!

We can even calculate how FAST the precession will get up to speed! (we are going to simplify here, in assuming that the angular acceleration is constant, where the reality is that it is not, but this calculation is simply to get a ball-park time interval. It is left to a reader who loves math to do the accurate calculation, as it is really not that much more difficult.)

We know that the Moment which applies to trying to make the gyro fall is 0.01176 nt-m (each second). So we have 7.65 * 10^-7 nt-meter / 1.176 * 10^-2 nt-meter/sec which would give less than a ten-thousandth of a second!

Of course, we could also calculate the time that gravity would take to have that weight fall the 0.00026 millimeter. d = 1/2 * g * t², so we have t² = 2.6 * 10^-7 meter * 2 / 9.8 meter / sec^-2. t² = 5.3 * 10^-8sec². This also results in around a ten-thousandth of a second!

A primary wrong assumption was made in the middle 1800s when several brilliant mathematicians assumed AM was always conserved, which therefore required that planets could perturb each others' orbits but NOT the actual orbital radius. That assumption was repeatedly built upon many times over the following 150 years, where quite a few errors in science were built up and which now still exist.

It turns out that there are now simple and obvious explanations for many things that no one has ever tried to explain! Like the amazing relationship between the orbits of the four giant moons of Jupiter, where their orbital periods are amazingly related. Also, Jupiter and Saturn have a Long Inequality which is the same. Similarly, Saturn's rings have gaps in very specific places. The asteroid belt around the Sun has Kirkwood Gaps that were first noticed 150 years ago. And on and on. Those phenomena could NOT be explained when the assumption of C of MA was rigidly believed, but they become relatively obvious CONSEQUENCES when this Violation of C of MA is applied!

It is now not only possible to tolerate such phenomena, but to actually CALCULATE them!

The same situation is necessarily true in Atomic Physics regarding the amazingly consistent orbits of electrons around atoms. During the 1930s, they were seen thus as it was ASSUMED that there were EXCLUDED orbits (Pauli and others)! In fact, that was then later built upon to create the entire field of Quantum Dynamics, upon which nearly all of modern Physics research is based! But when this Violation of C of MA is added to Nuclear Physics, electrons can now be seen to be able to PERTURB each other's orbits. It becomes clear that an electron could be simply tossed into any atom with any orbit whatever, and within a few million orbits (a tiny fraction of a billionth of a second), the electrons would perturb each other to APPEAR to us in the only way we ever see them! In principle, this might mean that the entire basis for the development of Quantum Dynamics might have been a wrong assumption!

There seem to be even more implications of the Violation of the Conservation of Angular Momentum which are beyond imagination!

You must certainly see that the fully accepted phenomenon of Planetary Precession IS an example of a Violation of the Conservation of Angular Momentum! NOT once it is in operation, but WHEN IT FIRST STARTED. Consider our tilted Earth, spinning away early in its existence. It would have to START precessing, right? But even more obviously, we know that Solar Precession is ZERO around March 21 of every year, but it BECOMES significant and a maximum around three months later, around June 21. We even SEE the herky-jerky motion of Precession which we call Nutation! We learned above that the ENERGY involved is explainable, and Conserved. But the simple fact that on March 21, the Earth has NO Solar Precessional motion, it has NO angular momentum in the Z-axis on that day. But just three months later, it has an immense amount of Z-axis angular momentum!

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. (We had calculated that exact amount up above.) 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 starmaps they use! 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 due 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, about seven days per year), 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.

If you are a really good math and Physics student, try to figure out how long Precession will take at that time (billions of years from now)! (Clue: it will be MUCH longer 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

Advanced Physics-related presentations in this Domain:

C Johnson, Physicist, Physics Degree from Univ of Chicago