Talk:Precession

From Wikipedia, the free encyclopedia
Jump to: navigation, search
WikiProject Physics (Rated Start-class, High-importance)
WikiProject icon This article is within the scope of WikiProject Physics, a collaborative effort to improve the coverage of Physics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Start-Class article Start  This article has been rated as Start-Class on the project's quality scale.
 High  This article has been rated as High-importance on the project's importance scale.
 
WikiProject Astronomy (Rated C-class, High-importance)
WikiProject icon Precession is within the scope of WikiProject Astronomy, which collaborates on articles related to Astronomy on Wikipedia.
C-Class article C  This article has been rated as C-Class on the project's quality scale.
 High  This article has been rated as High-importance on the project's importance scale.
 

Discussion[edit]

The sentence in section "Precession of the equinoxes" reading

The brightest star to be North Star at any time in the foreseeable past or future is the brilliant Vega, which will be the pole star in 14000 AD.

fails to reflect the fact that "foreseeing" inherantly means future.

I think the concept being groped for here is that there are may be stars hidden from view and thus not subject to our extrapolations brighter than the pole stars we can extrapolate forward or back to; i don't know whether increasing errors with increasing time interval relative to the present become an issue. Setting forth actual numbers for the time range intended would be informative and more credible.

But i'm making a hack at a temporary improvement. --Jerzy(t) 04:28, 2004 Mar 27 (UTC)


Great article, though I have little idea about physics. I was coping until the example was given around torsional precession talking about gimbles and the like. Perhaps better labelling of the components of motion so that there is no ambiguity, and a sequence of diagrams illustrating the points may be helpful. thanks adam —Preceding unsigned comment added by 165.228.172.109 (talk) 00:52, 24 December 2009 (UTC)

Physics - first sentence[edit]

Precession is due to the fact that the resultant of the angular velocity of rotation and the angular velocity produced by the torque is an angular velocity about a line which makes an angle with the permanent rotation axis, and this angle lies in a plane at right angles to the plane of the couple producing the torque.

This sentence is quite confusing to me (and I believe many others). Could someone try to clarify it a bit? After reading it over several times and trying to think it through, I still don't believe that I have an adequate understanding to reword it myself.

I'd like to also make a nice diagram (like the one I made for Vortex ring), but first I've got to figure out the actual physical concept... ✈ James C. 05:48, 2004 Aug 22 (UTC)

Physics - first sentence revisited[edit]

Precession is an angular velocity produced by the resultant of the angular velocity of the rotating mass and the angular velocity generated by the torque. The axis of Precession lies at right angles to the permanent rotation axis (axis of the rotating mass) and the rotation axis resulting from the torque.

That is how I understand Precession. Is that correct?

The equation is also confusing me. Let me explain with an example; in a video at http://science.howstuffworks.com/gyroscope1.htm , I saw a bicycle wheel tied via one end of its axle to a string which was itself tied to the ceiling. Some person was holding the bicyle wheel such that the axle was horizontal (bicycle axle and string make a 90 degree angle), proceeds to spin the tire and let go.

Had the wheel not been spinning or spinning very slowly, intuition says the bicyle wheel would rotate such that the axle and string become parallel. However, because the bicyle wheel was rotating itself, instead of rotating such that axle and string align, the bicycle wheel instead begins to rotate about the string, axle remaining horizontal.

Back to the equation:

In which Is is the moment of inertia, Ts is the period of spin about the spin axis, and Q is the torque.

Ts is a measure of time defining a set rotation, say 360 degrees. Q is the Torque created by gravity pulling on the mass of the tire (approx mid axle) against the string (tied at one end of the axle).

Now, because of where Ts is located in the equation, keeping all other variables constant, increasing the duration of one 360 degree rotation decreases the duration of the precession spin. IE the precession spins faster about the string as the bicycle tire spins slower about its axle. That just doesn't make sense.

Where am I going wrong?

One way of interpreting that situation is to bear in mind that the faster the rate of rotation, the larger the angular momentum. The larger the angular momentum the smaller the response to a torque that is applied. If I hazard a guess I'd say that if the torque is increased proportional to the increase in angular momentum then the rate of precession remains the same. --Cleon Teunissen | Talk 22:50, 10 Mar 2005 (UTC)


Precession and Children of the Stones[edit]

The series Children of the Stones involves in part the prehistoric alignment of a stone circle with a black hole. The story is set in the present (well, 1977, though the series has not dated) - surely precession would have caused there to be a disparity of direction (for want of a better term)?

Thought experiment to help visualisation of the dynamics.[edit]

I am thinking about inserting the following section I wrote in the article, or replacing a section. Any comments are welcome. --Cleon Teunissen | Talk 22:38, 10 Mar 2005 (UTC)

Imagine a perfectly spherical spinning asteroid. A monorail track is build, in a perfect circle all around the asteroid. Initially the monorail track goes over the geographic north pole and south pole of the spinning asteroid, but it can slide, remaining a flat ring as the whole of the track slides. (The direction of the spin determines the geographic poles and the equator). Positioned on the monorail is a continuous train of wagons, all around the asteroid. Initially the monorail track and the train on it are co-rotating with the asteroid.
Next, imagine that on all wagons small rocket engines are started, exerting thrust. The combined action of those thrusters is a torque around an axis that passes through the equator. Then to visualize what will happen it helps to divide the picture into four quadrants. 1) a section of train that is moving from the north pole to the equator, 2) a section moving from the equator to the south pole, 3) a section moving from the south pole to the equator, 4) a section moving from the equator to the north pole.
Then in the northern hemisphere the coriolis effect will shift the track in one directon, and on the southern hemisphere in the other. The combined action of the coriolis effect in the northern and southern hemisphere is that there is a torque around an axis that passes through the equator, and that is perpendicular to the axis of the torque of the thrusters. The monorail track will slide over the asteroid, and eventually it will come to rest parallel to the equator.
In this visualisation only a part of the rotating system undergoes precession. Normally precession refers to a solid, spinning body on which a torque is applied. The 'coriolis effect' usually refers to the motion of a small mass relative to a much heavier rotating mass. This visualisation shows that precession and coriolis effect are related phenomena. Precession is manifestation of the coriolis effect in the case of a single solid body on which a torque is applied.

--Cleon Teunissen | Talk 22:38, 10 Mar 2005 (UTC)


Here [Presession of the Equinox] is a non-technical and easy to understand explanation of the phenomenon of precession. In the least, this explanation frames the issues and the apparent effects of precession. John Charles Webb July 21st 2005

It is not only non-technical, but also non-knowledgable. The precession of the Earth is independent of our method of time-keeping. What is happening is that the direction in which the Earth's axis of rotation is pointing is moving. The duration of one precession cycle is about 26.000 years, so in about 13.000 years the Earth's axis of rotation will point in a direction 46 degrees away from its current direction. (The tilt of the Earth's axis with respect to the plane of the Earth's orbit around the sun is about 23 degrees, the Earth's precession is (roughly) a motion with respect to the plane of orbit around the sun)

--Cleon Teunissen | Talk 07:15, 22 July 2005 (UTC)

Binary model of equinox precession[edit]

User:Ungtss had added material about a binary theory for explaining the precession of the equinoxes. I propose deleting that material.

The phenomenon is usually called 'precession of the equinoxes'. That is actually not the best name. A better name would be 'gyroscopic precession of the Earth'.

Gyroscopic precession of the Earth is not a hypothesis, it is a theorem of Newtonian dynamics. That is: if newtonian dynamics is correct then gyroscopic precession of the Earth must occur. Conversely, if gyroscopic precession of the Earth would be absent then that would constitute a glaring anomaly in Newtonian dynamics.

As it happens, the gyroscopic precession of the Earth was observed in astronomy long before it was theoretically predicted. But if gyroscopic precession of the Earth would have been unknown in Newton's time it would have been predicted anyway: the gyroscopic precession is an inescapable consequence of the entire dynamics involved.

So anyone who wants to support the binary theory has a huge task at hand. Supporting the binary theory means that the theoretical framework of Newtonian dynamics is not available.

Anyone who wants to support the binary theory must develop an alternative to newtonian dynamics that reproduces almost everything of newtonian dynamics, except the gyroscopic precession of the Earth. That is a very, very daunting task indeed. --Cleon Teunissen | Talk 21:26, 15 August 2005 (UTC)

while i understand you're rather convinced that the newtonian conclusion is correct, that does not alter the fact that other interpretations exist. npov suggests all significant views be presented. if indeed you think the newtonian view is correct and the binary view is incorrect, i suggest adding information supporting the newtonian view, rather than deleting the binary view entirely. we're building an encyclopedia here -- let's have an article that describes and discusses all the views, yes? Ungtss 15:04, 16 August 2005 (UTC)
I present the body of knowledge of newtonian dynamics as evidence. The body of knowledge of Newtonian dynamics allows calculation of trajectories of space missions to exceedingly high accuracy, allowing for example a comet to be hit in midflight.
Gyrosopic precession of planets is an inescapable consequence of the dynamics as a whole. Gyroscopic precession is not like, for example, rings around a planet. Saturn has prominent rings, the other giants have rings too, but much fainter; rings around a planet are "optional". Gyroscopic precession isn't "optional", it will happen anyway.
The measured amount of gyroscopic precession is accounted for. In the case of theorizing about how Saturn's rings formed there is room voor multiple theories, emphasizing this or that aspect. In the case of gyroscopic precession there is no such room.
Another example of "no room" is precession of the perihelion of the inner planets. In the past there have been proposals for an as yet undiscovered inner planet, that would possibly account for the anomalies, particularly in the precession of the perihelion of Mercury. Currently the precessions of the perihelia of the inner planets are accounted for, so there is no room for another inner planet.
To suppose another inner planet would bring with it the necessity to formulate a completely new theory of motion, that would have to reproduce everything newtonian dynamics predicts, except a couple of things. That is a very very unlikely scenario. Likewise in the case of gyroscopic precession. That is accounted for, and there is no room for "tweaking". To suppose another factor that would affect precession brings with it the necessity to formulate a completely new theory of motion.
So it's not a case here of preferring this or that interpretation: there is no room for any "tweaking", or emphasizing. You either accept that gyroscopic precession of the planets is accounted for, or you reject newtonian dynamics as a whole, those are the two options in this particular case. --Cleon Teunissen | Talk 16:56, 16 August 2005 (UTC)
how do newtonian dynamics account for the fact that the equinox precesses relative to objects outside the solar system, but not for objects inside the solar system? Ungtss 17:06, 16 August 2005 (UTC)
To begin with basics about logic, the fact that the Newton model predicts a precession doesn't affect at all the possibility that there is a second active cause. It's not Newton XOR Alternative idea, it's at least OR: there could be subtle effects not perfectly predicted by Newton. This is the way science progresses, by the way, by observing better the not predicted. Newton mechanic was superseded by Relativistic Mechanic, but seems perfectly fit for non relativistic cases. It's not rejecting Newton but adding to it for completion and detail predictability.
I agree with the plurality of voices as long as they seem based on scientific reasoning and confirmed data. This site seems to be explaining and claiming scientific validity and deserves at least an inquiry and a check, not an aprioristic debuke. I'm not an expert to decide it by myself but I can recognize when an idea is being given a well deserved opportunity - a fair trial - or when it is being lynched, and when it is without basis and deserves its removal. You need to read and demystify the arguments of this theory if they're clearly wrong and if you have the knowledge to precisely refute them, Cleon_Teunissen. Apparently, this is not about an inner planet but an outer stellar non-visible object and would not change anything about Newtonian mechanic inside the solar system. Pronoein (talk) 12:10, 28 March 2010 (UTC)

Gyroscopic precession relative to what?[edit]

how do newtonian dynamics account for the fact that the equinox precesses relative to objects outside the solar system, but not for objects inside the solar system? Ungtss 17:06, 16 August 2005 (UTC)

I'm not sure I understand your question correctly here. It seems to be a question relative to what the precession is defined.

Take each individual planetary orbit of the planets of our solar system. It is by good approximation an elliptical trajectory, with the Sun at a focus of that ellipse. If you draw a line through both of the foci of that ellipse then you find that that line always keeps pointing in the same direction. (That is: you do need to take perturbations by other planets of the solar system into account. These perturbations are minute, but not zero.)

You can call the line connecting the foci an axis of the elliptical orbit. For each planet that axis keeps pointing in the same direction with respect to the system of fixed stars.

Take the plane of the Earth's equator, and extend it into space. Take the plane of the Earth's orbit around the Sun and extend it into space. Take the line of intersection of those to planes. The Sun is located at the spring/autumn equinox, when the Sun is located on that line of intersection of the Earth's equatorial plane and the Earth's orbital plane.

Returning to the first question, my guestimate of what your question is:
The gyroscopic precession of the Earth is relative to the foci-connecting axis of each planet, for the foci-connecting axis is what is the constant factor for each planetary orbit.

So there is, to my knowledge, no conflict between measuring the gyroscopic precession relative to the fixed stars, or relative to the other planets of the solar system.

This may or may not address your question. You may have something else in mind. Your question is too terse to assess that. --Cleon Teunissen | Talk 18:59, 16 August 2005 (UTC)

thank you for taking time with me:). let me try to articulate my question a little better. the lunisolar theory holds that precession is caused by a "wobble" in the earth's rotational axis. it seems to me that if this were true, the wobble in the axis would cause everything in the sky to appear to move, including the sun. but our observations show that the location of the sun and the outside stars change relative to each other. the same thing with the planets. to my knowledge, while calculations of the location of the stars take precession into account, calculations of the location of the sun and planets do not. on the contrary, while the equinox appears to precess relative to the stars, it does not appear to precess relative to the sun or planets. this, within the binary model, implies that the entire solar system is moving in a curved path. here is a link -- i'm far from an expert and i would greatly appreciate your thoughts on this. [1] Ungtss 19:11, 16 August 2005 (UTC)

Gyroscopic precession of the Earth is unrelated to relative motion of Sun and Earth[edit]

I think it should be emphasized that 'precession of the Equinoxes' is a misnomer. It suggests a connection with motion of the Sun that isn't there. The Gyroscopic precession of the Earth is unrelated to relative motion of the Sun and Earth.

As seen from the perspective of the Earth the Sun takes a year to go around the Earth's orbital plane (the ecliptica).
Two times a year the Sun happens to be on the line of intersection of the Earth's equatorial plane and the ecliptica.

The gyroscopic precession is measured by counting how long it takes for the line of intersection of the Earth's equatorial plane and the ecliptica to complete a full 360 turn. That takes about 26.000 years. It's that line of intersection that counts. --Cleon Teunissen | Talk 19:35, 16 August 2005 (UTC)

How star positions have been charted throuhout history[edit]

Here is what I know.

Throughout the history of astonomy, astronomers have had to make decisions about what to use as "anchor" for their coordinate system.

One of the most obvious, and thus one of the oldest, is the line of intersection of the Earth's equator, and the plane of the ecliptica.

The plane of the ecliptica is what we today know as the plane of the orbit of the Earth around the Sun. When the Sun, Earth an Moon are aligned, then there is an eclipse. (When the Sun, Earth an Moon are aligned, then the Moon is in the plane of the ecliptica.

As far as the ancient astronomers could tell, that line of intersection did not budge. So they extended that line into space, and the point in the zodiac of one end they called zero degrees, and the other end of that line they called 180 degrees. Thus, one of the signs of the zodiac came to be called the first sign in the sequence of signs of the zodiac; the line of intersection had pointed it out.

When it became firmly established that the line of intersection of the two planes rotates with respect to the fixed stars it was abandoned as an "anchor point" for coordinate system.

I'm not an astronomer, I assume that averages of positions of a large number of stars is used as anchor now. Whatever moves the least

As star postions were charted more and more precise, astronomers had to take the motion of the Earth around the Sun into account, so as a new zero point they adopted the center of mass of the solar system as zero point of the standardised astronomical coordinate system.

When an astronomer wants to detect whether a distant Sun has a (heavy) planet orbiting it, then he must correct his measurements for the following things: he must subtract the motion of the Earth-Moon-system with respect to the center of mass of the solar system; the motion of the Earth with respect to the center of mass of the Earth-Moon system; the motion of the observatory with respect to the Earth's axis (and maybe there are one or two more, I forget.)

In performing calculations on either star positions or planetary positions an Earth-independent coordinate system is used. --Cleon Teunissen | Talk 20:22, 16 August 2005 (UTC)

thanks for all the very interesting information. i don't dispute any of it, and i don't think binary theorists do either. the argument is simply physical: the lunisolar theory holds that the sun does not curve in inertial space, while the binary theory holds that it does. the strongest evidence for this is the fact that, if precession were caused by a "wobble" in Earth's orbit relative to inertial space, we would expect ALL objects in the sky to appear to move in the sky, because it is WE that are wobbling. instead, the stars OUTSIDE the solar system appear to move relative to objects inside the solar system, but objects inside the system do not appear to move relative to each other. here's another article, about venus. [2] Ungtss 20:55, 16 August 2005 (UTC)
You lost me here, I think.
I expect the axis of ellipticity of each planet to keep pointing at the same fixed stars. Suppose our Sun and another non-luminous object, about as heavy as the Sun, are orbiting their common center of mass, then I would just as well expect the axis of ellipticity of each planet to keep pointing at the same fixed stars.
As I understand astrodynamics, the presence of another non-luminous object, with our Sun and that other object orbiting their common center of mass, would not show up in the way that you describe.
If our Sun would have a companion, it would be possible to detect that, but it would not manifest itself in the way that you describe. --Cleon Teunissen | Talk 21:48, 16 August 2005 (UTC)
What I mean is: it is unclear to me what is is that you have in mind when you refer to 'precession of the Earth'. What I have in mind is that the Earth's axis, tilted about 23 degrees with respect to the ecliptica, traces out a cone, and this tracing out of a cone has a period of about 26.000 years. --Cleon Teunissen | Talk 21:56, 16 August 2005 (UTC)
i'm sorry i'm not being very clear:). what i mean by precession is that the sun appears, every year, to rise in on the vernal equinox at a slightly different position relative to the background stars, and that the direction of the pole relative to the background stars changes at a comparable rate. those are the observables.
the question is whether those observables are caused by a gyroscopic motion in the earth, whereby the earth changes its angle relative to inertial space, or whether those observables are caused by a curved motion by the sun in inertial space. both would cause the same observables -- the sun appearing to move relative to the background stars and the pole appearing to move. but there is one key difference. the lunisolar model holds that the sidereal year is the true year, while the binary model holds that the tropical year is the true year. and the way to test which is the true year is to ask whether the location of objects within the solar system are predicted by the tropical year or the sidereal year. it appears to me that we determine the location of objects within the solar system, including the moon, venus, and other planets, by using the tropical year, rather than the sidereal year. this implies that the tropical year is the true year, and that the sidereal year is mere artifact, because precession is not caused by a motion in the Earth's axis, but by a motion in the sun. at least that's what my studies up to this point have led to me believe:). have i made myself any clearer? Ungtss 16:40, 17 August 2005 (UTC)

Solar day and sidereal day[edit]

Recapitulating:
The solar day is the amount of time between two consecutive sightings of the Sun by an observatorium, at exactly the same angle. For example the Sun passing straight overhead over the Greenwich meridian, and the next pasage the next day.

The siderial day is the amount of time it takes for the Earth to complete a full rotation, a sidereal day is 23 hours 56 minututes, 4 second.


In for example meteorological models the rotation rate of the Earth is taken into account. A meteorologist told me he had at one time, out of curiosity, tried what would happen if he inserted the wrong rotation rate into the model. Sure enough, when he used the solar day, the quality of the projections was less good than with the correct rotation rate.

Something similar is the case with the duration of the year. The sidereal year is the true year. The tropical year is an artifact of circumstances, and in that sense it is not particularly interesting from a physics point of view. --Cleon Teunissen | Talk 22:42, 16 August 2005 (UTC)

you've said that the "sidereal day/year" are the "real" day and year, and the tropical day/year are merely artifacts of circumstances. the heart of the binary argument is that the tropical day/year are the real day/year, and the sidereal day/year are the artifacts of circumstances. i understand that the mainstream view holds to the former, but i don't see how they can prove that, without disregarding a substantial amount of evidence regarding the fact that all objects within the solar system are predicted using the tropical year, while only objects outside the solar system are predicted by using the sideral year. that fact appears to me to falsify your conclusions about which is the "true year." Ungtss 16:26, 17 August 2005 (UTC)

In general: Newton's laws of motion hold good if the sidereal day and year etc. are used. The solar day is no use in gravitational computations.

The solar day is the time between two sightings of the sun straight overhead the same meridian. The tropical year is the time between two sightings of the Sun on the line of intersection of the plane of the Earth's equator and the plane of the ecliptica. The tropical year retains the line of intersection as a zero point of coordinate system, while astronomy has abandoned its use as zero point of coordinate system.

Tycho Brahe proposed a solar system model with the Earth at the center, and the orbits of each of the planets a combination of two circular motions: one circular motion with a period of a year and one circular motion with the period of that particular planet. In retrospect we see that that common period of a year, shared by all planets according to the model, is an artifact of using the Earth as zero point of the coordinate system.

If you use the tropical year, you can no longer use gravitational computations for the planetary orbits, you have to use tables, then. If you use the tropical year, and you use tables that give the positions of the planets with respect to the tropical year, then sure enough the precession rate keeps popping up wherever you look. --Cleon Teunissen | Talk 07:05, 17 August 2005 (UTC)

as i understand it, the locations of objects within the solar system, including solar eclipses, are always calculated with respect to the tropical year. is this not true? Ungtss 16:26, 17 August 2005 (UTC)

Neptune's moon Nereid[edit]

Quote from site about planets and their moons

Of all moons in the solar system, Nereid has the most elliptical orbit around its planet. When Nereid is at its closest to Neptune, its distance from the planet is only about 1.35 million km (about 812,000 mi). At the farthest point in its orbit around Neptune, Nereid is about 9.624 million km (about 5.77 million mi) from the planet. Nereid’s orbit is tilted 28° relative to Neptune’s equator.
[end quote]

I expect the axis of the elliptical orbit of Nereid to keep pointing to the same fixed stars (With any deviation accounted for in terms of expected perturbation)

The fact that Neptune is orbiting the Sun does not in itself influence the direction of the axis of the elliptical orbit of Nereid.

Likewise any gyroscopic precession of any of the solar system's moons. Those gyroscopic precessions are independent of their parent planet's orbit around the Sun. --Cleon Teunissen | Talk 08:38, 17 August 2005 (UTC)

precisely. the binary argument is that the moons and planets do not tilt relative to the objects around which they are revolving. the argument is that they do not change their tilt to any substantial degree at all. instead, the argument is that because of the revolution of neptune, the location of neptune in Nereid's sky on its "vernal equinox" with neptune will appear to move slightly relative to the background stars on every "year" of nereid's orbit, because of neptune's revolution around the sun. they're saying there is no gyroscopic motion at all, but that the observed precession from the earth's surface is not a result of motion in the axis of the earth, but a result of curved motion by the sun. is this any clearer? Ungtss 16:48, 17 August 2005 (UTC)

The discussion has become somewhat fragmented, partly because of me starting new entries.

I am confident in recognizing the sidereal day as the true day, and the solar day as an artifact. The recognition of the sidereal day as the true day is corroborated by for example measuring the rotation of the Earth with respect to space itself. For example, the university of Canterbury New Zealand has an observatory deep inside a cave. Insulated as much as possible from outside influences there is a very sensitive ring laser interferometer. A ring laser interferometer is a special application of what is called Sagnac interferormetry That ring interferometer in that cave gives the same value for a full rotation as astronomical measurement: 23 hours 56 minutes and 4.09etc. seconds

Al forms of measurement that measure the rotation of Earth with respect to space, (this is often called: measuring the Earth's rotation with respect to inertial space) mutually agree, and find the period of 23 hours 56 minutes 4etc. seconds as rotation period.


I cannot from your words infer what exactly you have in mind when you refer to 'performing calculations'. Possibly you refer to gravitational calculations, with orbital mechanics, but maybe you are referring to adding and subtracting calender dates and calender periods to calculate when something will occur again.


Contemplating the adoption of the solar year as the true year instead of the sidereal year has vast consequences. The recognition of the sidereal day and sidereal year as the true year is in many ways interconnected/corroborated in other measurements.

You write:

the binary argument is that the moons and planets do not tilt relative to the objects around which they are revolving. Ungtss 16:48, 17 August 2005 (UTC)

I am unable to reconstruct what you have in mind. For example, in the case of the Earth the Earth's axis is not perpendicular to the Earths orbital plane of its orbit around the Sun, there is an angle of 23 degrees between the Earth's axis and the line that is perpendicular to the Earth's orbital plane. It is unclear to me what your mental picture of the solar system looks like. It doesn't seem to add up. --Cleon Teunissen | Talk 05:46, 19 August 2005 (UTC)

So far I have automatically assumed that the binary companion of the Sun (if it is there) is assumed to be very far away. So far I have assumed that the binary theory uses a distance to the companion that is many times the distance of Pluto to the Sun. Let's talk numbers. What is the distance between the Sun and the companion to the Sun? --Cleon Teunissen | Talk 05:59, 19 August 2005 (UTC)


thanks for your patience again:). i don't know that i can address your concerns any better than this website can: http://www.binaryresearchinstitute.org/introduction/curvature.shtml

with regard to the interferometer, to my understanding, those findings are consistent with the binary theory model. according to binary theory, the earth makes a full rotation in inertial space, and relative to the outside stars, in a sidereal day, just like the lasers show. nevertheless, because the sun is moving, the earth makes a full rotation relative to the sun in a tropical day.

in the same way, because the sun is moving, the earth makes a full revolution around the sun in a tropical year, but makes a full revolution with respect to inertial space and the outside stars in a sidereal year.

imagine the sun moving, and the earth revolving around the sun. every noon relative to the SUN, the sun will be in a slightly different place. this is why the day relative to inertial space (as described by the interferometer) is different than the day relative to the sun. because the sidereal day and sidereal year reflect the motion of the earth relative to inertial space, while the tropical day and year reflect the motion of the earth relative to the sun.

with regard to the location of the binary companion, the binary research institute has some estimates for where we should look for it. another organization thinks that it is the star Sirius, and that is why Sirius was reported red in ancient texts but is blue today, because it was moving away from us in ancient times, but is currently moving toward us. but the bottom line is, neither claims to have a lock on the binary companion. Ungtss 14:17, 19 August 2005 (UTC)

Celestial dynamics[edit]

The orbit of the Moon around the common center of mass of the Earth-Moon system has a quite rare feature.

There is a 1:1 resonance of the sidereal orbit of the Moon around the Earth and the sidereal rotation of the Moon around its own axis. Generally the ratio of satellite rotation rate and orbiting rate is not a ratio of two whole numbers.

One of the consequences is that the sidereal precession rate of the rotation of the Moon around its own axis is the same as the sidereal lunar month.

As seen from the Earth it looks as if the Moon is being swung around on the end of a rope, the rope attached to the near side of the Moon. Another example is a pilot training centrifuge, where the capsule with the pilot seat is attached in such way that the capsule co-rotates with the arm. This particular situation of the Moon is very exceptional.


In general the relation of a Primary (heaviest celestial body of a system) with its satellites is described with the following mechanical model:
Imagine a wheel with teeth, like the gears of a bicycle. Take two of them, with the same amount of teeth, and wrap the bicycle chain around them to connect them.

When you swing around that contraption the swinging wheel does not "receive" the orbiting rotation rate as a given.
In celestial mechanics there is no such thing as automatically "inheriting" the a higher level orbiting rate into local rotation processes. (Not counting the small tidal effects for now (in fact, it was tidal effects that after millions of years of "massaging" locked the Moon into that 1:1 ratio)).


As far as I can tell the reasoning of the binary research institute is flawed from the very beginning, their story just does not add up.

If our Sun would have a companion at several lightyears distance, then that orbiting of the Solar system as a whole around a common center of mass of the Sun and its companion would not make an apreciable difference for the internal dynamics of the Solar system.

My judgement is that the reasoning of the binary research institute is based on lack of understanding of principles of dynamics of motion. My judgement is that they fail to properly visualise celestial motion. At some point they fail to take something into account, or they inadvertendly count it twice instead of once. --Cleon Teunissen | Talk 08:09, 22 August 2005 (UTC)

i can definitely appreciate your judgment, and i'm not here to change it. i'm just here to make sure a published alternative view on the causes of precession is represented on the page. npov suggests that all significant views be represented, and it strikes me that this view, being promoted by a number of independent organizations, deserves a brief mention and description. feel free, however, to add a referenced critique of the view to describe how it is flawed. Ungtss 18:45, 22 August 2005 (UTC)
<<If our Sun would have a companion at several lightyears distance, then that orbiting of the Solar system as a whole around a common center of mass of the Sun and its companion would not make an apreciable difference for the internal dynamics of the Solar system.>>
i don't see how anybody is claiming it does. the simple claim is that the location of objects within the solar system is calculated with reference to the tropical day and year, but the location of objects outside the solar system is calculated with reference to the sidereal year. if the precession of the earth were a result of a wobble in the earth's axis, all objects in the sky would appear to move as the axis moves. instead, all objects in the solar system move relative to the background stars. i can see no other reasonable interpretation of that fact than that the solar system is moving in a curved path through inertial space. Ungtss 18:45, 22 August 2005 (UTC)

You again avoid specifying whether you refer to gravitational calculations or calculations that consist of additions of calender dates and calender periods. The view you are promoting is an insignificant view, for it is in direct conflict with astronomical observation.

The earth's axis traces out a cone in inertial space. That is, the gyroscopic precession of the Earth is a process in which the angular momentum of the Earth in inertial space is changing. The orbits of all the planets of the solar system, including Earth, do not follow this change of angular momentum. In 13.000 years from now, the Earth's axis will be pointing in a direction with respect to the fixed stars that is about 46 degrees separated from where the Earth's axis is currently pointing. In 13.000 years from now, the planetery orbits of the solar system will be very closely in the same plane with resepect to the fixed stars as they are now. It is exclusively the Earth's axis that traces out that cone. --Cleon Teunissen | Talk 08:14, 26 August 2005 (UTC)

my friend, those conclusions you are drawing are all theory-dependent. indeed, in 13,000 years the axis will point a different direction in inertial space. but is that because the AXIS is moving, or because the SUN is moving?
you've said it's at odds with astronomical observations, but have not stated which.
You again avoid specifying whether you refer to gravitational calculations or calculations that consist of additions of calender dates and calender periods
i'm not exactly sure what you're looking for here. Ungtss 21:47, 28 August 2005 (UTC)

The precession rate of Mars[edit]

This site provides the following information:
http://www.nasa.gov/home/hqnews/2003/mar/HP_news_03094.html
The precession is the slow motion of the spin-pole of Mars as it moves along a cone in space (similar to a spinning top). For Mars it takes 170,000 years to complete one revolution.
[end quote]

--Cleon Teunissen | Talk 10:32, 26 August 2005 (UTC)

As i understand it, the estimates of mars's precession rate are not observed, but have been calculated based on the assumptions inherent in the static-sun newtonian model. however, if the assumption of a static sun is incorrect, those calculations will be incorrect. the calculated precession of mars doesn't mean anything when it's based on the very assumptions that are at issue. Ungtss 21:43, 28 August 2005 (UTC)


http://www.sciencemag.org/cgi/content/full/278/5344/1749
I quote from this source:

Doppler and range measurements to the Mars Pathfinder lander made using its radio communications system have been combined with similar measurements from the Viking landers to estimate improved values of the precession of Mars' pole of rotation and the variation in Mars' rotation rate. The observed precession of -7576 ± 35 milliarc seconds of angle per year implies a dense core and constrains possible models of interior composition. The estimated annual variation in rotation is in good agreement with a model of seasonal mass exchange of carbon dioxide between the atmosphere and ice caps.
[...]Accurate measurement of the precession is needed to determine the polar moment of inertia.

According to this source, the precession rate had to be measured first, and subsequently an inference about the interior of Mars was bases on this measurement. Measuring the precession rate of Mars is far from straightforward, in that article it can be read why it is difficult. While it is difficult, it is possible. --Cleon Teunissen | Talk 22:21, 31 August 2005 (UTC)

i agree that a binary theory predicts similar precession affects across all the planets. however, again, as i understand it, the cited "estimate" of mars's precession rate is not reliable, because it is based on the assumption that it is Mars's axis that is wobbling, rather than the sun that is moving. If indeed the sun is moving, then the given "estimates" are meaningless. Ungtss 16:35, 3 September 2005 (UTC)
i guess the point is, those "estimates" are made based on the same assumptions about

What happened to the binary system hypothesis? I was searching for it recently and couldn't find it until I looked at Google's cache of this page, where it was...

Who dropped it, and why? Could I ask the moderators to have it back? Until I see a clear, unrefutable page or paper refutating completely each and every aspect of the binary hypothesis (as well as explaining each and every apparent conflict with current theory, as stated in page http://www.binaryresearchinstitute.org/evidence/lunarcycle.shtml and http://www.siriusresearchgroup.com/articles/index.shtml - in this last one pay attention specially to the Venus transits' problem), I believe it's a perfectly valid scientific hypothesis, worthable being paid attention and respect. I think the only way to settle this down definitely, once and for always, would be to place an observatory in the moon or in mars - if it's the whole solar system moving, it would have precession, and it's rate would be very similar to that of the earth... if it has much more less then it's just the moon's precession "natural" rate, and probe would've been found that the sun doesn't have a companion, and that it's just the earth that's "woobling"

Don't care, I've already replaced the information myself, slightly rephrased hoping to get a N.P.O.V. (jose pineda, same guy as last post)

http://www.binaryresearchinstitute.org/evidence/lunarcycle.shtml
I have had a good look at the info on the binaryresearchinstitute.org site, and I am inclined to think that that site is a hoax, designed to bamboozle people who have little experience in visualizing celestial mechanics.

I present a thought experiment, to illustrate some celestial mechanics. Two space probes are launched, one to Saturn, I call that one Castor, and one to Jupiter, I call that one Pollux. Both go into orbit around the planet of their destination, counterclockwise as seen from the north, both with a period of 10 days. The satellites are also given a rotation around their own axis, counterclockwise, with a period of 5 days. So for both Castor and Pollux their sidereal day is five Earth days. (The exact period is not crucial, the crucial bit is that it's the same for Castor and Pollux.)

At some point in time a fixed camera on the satellite looks straight at the planet. How long does it take to the next straight view? After 5 earth days of orbiting the satellites have made a full turn around their own axis. However, the camera now points exactly away from the planet, for in those 5 earth days they have covered 180 degrees of their orbit around their planet. So its another 5 Earth days before the planet is straight before the fixed camera.

The thing to note is that the period of orbit of Saturn around the Sun and the period of orbit of Jupiter around the sun does not enter into these considerations. Once you have thought the logic through it is very obvious that the orbits of Saturn and Jupiter do not matter for their satellites. The presentation of the binaryresearchinstitute.org site is aimed at people who are not aware that there is no such thing as some sort of "inheriting effect" of the orbit of the Primary. The shape and the periodicity of the orbit of the primary is inconsequential for the satellite.

The same logic applies for gyroscopic precession. Let Castor and Pollux be designed such that the gravitational attraction of the primary results in a torque on the satellite. Castor and Pollux have the same orbiting period, which means that the gravitational attraction is the same, which means that their gyroscopic precession will be the same. How much time will there be between two "equinoxes" of Castor's and Pollux' motion around their primary? For both satellitse that will be the same

Given the interconnectedness of the framework as a whole, there is no such thing as rejecting gyroscopic precession as described by newtonian dynamics, and at the same time retaining the rest of newtonian dynamics; it's a package deal.
To make the binary hypothes viable a new theory of motion would have to be developed from scratch, a theory of motion that equals newtonian dynamics in versatility and strength. The writers of the binaryresearchinstitute.org site seem unaware of these matters. That indicates that they either have no clue what they are dealing with, or the site is Hoax to see how many people can be fooled. --Cleon Teunissen | Talk 10:07, 1 September 2005 (UTC)

You are quite right Cleon. For a clearer understanding of the hoax see this site created by the author of the "theory". --Ian Pitchford 12:35, 1 September 2005 (UTC)
i challenge your assertion, mr. teunissen, that newtonian mechanics requires gyroscopic precession. On the contrary, mysterious forces of torque from the moon, sun, and other places are introduced into the equation in order to make them consistent with newtonian mechanics. newtonian mechanics can remain untouched, if those mysterious torque forces are not hypothesized. I don't see why this theory cannot be presented as an alternative view and then dispatched by facts to support your side. Ungtss 16:35, 3 September 2005 (UTC)

Torque-free precession and torque-induced precession[edit]

I propose to remove the remark about torque-free precession. Or if it is mentioned at all, then much further down in the article, for it is a very minor detail. --Cleon Teunissen | Talk 22:08, 31 August 2005 (UTC)

Currently the remark is just plain wrong - precession is not dependent on rotation about an axis of symmetry. TobyK 20:07, 9 December 2005 (UTC)

Direction of precession of a toy top[edit]

The article says: As a spinning object precesses, the tilt of its axis goes around in a circle in the opposite direction that the object is spinning. But it is not so for a toy top. I've just got one of these and started it rotating clockwise. I've clearly seen that the axis precessed clockwise too, apparently not counter-clockwise, as I expected from what I've read here. Only when it fell at last, it started moving counter-clockwise, due to friction with the surface.

I would agree that precession of a toy top seems to be in the same direction as the spin of the top, in the sense that (looking from above) the top and the ends of the axis seem to rotate in the same direction. When the top falls over the is usually fricitional slippage followed by a rolling rotation where the end of the axis seems to rotate in the opposite direction, but the top itself rotates in the original direction relative to the axis. --Henrygb 15:09, 8 February 2006 (UTC)
I agree that the article is misleading. The direction of precession depends on the direction of rotation and the direction of the applied force. In the case of a spinning top, gravity is trying to make it fall over. In the case of the Earth, the gravitational forces of the Sun and Moon on the spheroidal shape are trying to right it – hence the north pole precesses in retrograde motion. --Portnadler 18:30, 8 February 2006 (UTC)
I agree with Pordnadler; In the case of the Earth the gravitation of Sun and Moon exert a torque on the oblate spheroid that tends to align the Earth's equatorial plane with the plane of the Earth's orbit. (By the way, I find it pretty amazing that the gyroscopic precession has been preventing that alignment for as long as the Earth has been in existence.) A toy top that is suspended on string that is attached to the top end of the toy top's axis of rotation will precess like the Earth does. --Cleonis | Talk 19:10, 8 February 2006 (UTC)
I've simply removed the inaccurate sentence. If anyone can come up with a better explanation, please feel free. --Portnadler 11:25, 12 February 2006 (UTC)

Other types of precession[edit]

Weisstein has an article called free precession, but it says that there is constant torque, rather than no torque. Is this incorrect, the same thing or something different than torque-free precession? -- Kjkolb 14:34, 26 November 2005 (UTC)

I found some other types of precession in Weisstein. Be careful about using it as a source, as some of its information is technically incorrect, particularly about what terms are equivalent. You should check the Weisstein articles before you remove or redirect the topics below to make sure they are talking about exactly what you think they are talking about.

-- Kjkolb 07:53, 28 November 2005 (UTC)

General, Luni-Solar and Planetary Precession[edit]

My bible for these matters is Simon Newcomb's A Compendium of Spherical Astronomy, The Macmillan Company, 1906, republished by Dover, 1960. This makes the distinction between Luni-Solar Precession, which Newcomb describes as the motion of the celestial pole, and Planetary Precession, which is the long-term change of the plane of the ecliptic relative to a fixed frame of reference. These both have an effect on the motion of the equinoxes and Newcomb defines the combined effect as General precession. Planetary precession also causes the obliquity of the ecliptic to change over time.

Planetary Precession is not really precession. Perhaps we could split the article into Precession (the general physical/mathematical phenomenon associated with a rotating body) and Precession of the Equinoxes (the astronomical phenomenon).

--Portnadler 19:45, 5 December 2005 (UTC)

Response to applied torque is instantaneous[edit]

In the current version of the article it is stated:

If an external force pushes upon the rotating body, the body will resist the force by pushing back against it, but the reaction is delayed such that it occurs at a point 90 degrees later in its rotation.

I think it is incorrect to phrase it in terms of a delay. The response to the applied torque is instantaneous, but at a 90 degrees angle. I propose to borrow the expressions 'roll' and 'pitch' from aviation:
Vertical axis: spin axis
One horizonal axis: roll-axis
Other horizontal axis: pitch axis
If you have a gyroscope, spinning around a vertical axis, and you attempt to make it roll, it will immediately pitch . --Cleonis | Talk 11:55, 25 February 2006 (UTC)

It should be noted that what does not happen is that the gyroscope pitches instead of rolling. It seems to me that the precession of a spinning top can be seen as a combination of pitching and rolling, with pitching and rolling present in equal measure.
That suggests to me the following interpretation: If you have a gyroscope, spinning around a vertical axis, and you attempt to make it roll, then the action is redistributed, and the gyroscope will in equal measure pitch and roll. --Cleonis | Talk 09:21, 4 March 2006 (UTC)
I believe (without having any proof) that precession IS NOT instantaneous. I think of it like this...
A toy top, spinning on a table, will remain stable with very little precession (wobble) if it is given a sufficient rotating speed, and a nearly vertical starting orientation. As the toy spins, friction of the tip on the table, and drag due to air will slow the rotating speed of the top. When this speed slows down sufficiently (for us to perceive), the top begins to precess more and more until eventually it topples over.
Now, it's not that there is no precession when the top is spinning fast, it's just that the delay it takes a unit mass to rotate 90 degrees is far too small for a human to perceive. Ex: 1 rotation per second (slow enough for us to perceive) means a .25 second delay. 10 revolutions per second means a .025 second delay and 100 revolutions per second means a .0025 second delay.
Here's how I think it works. Imagine a top, at 45 degrees to a surface and supported. The top is spun at low speed (whatever that is) and is let go. Since the top is spinning slowly, in the time it takes for the top to turn 90 degrees it will fall to an angle less than 45 degrees (obviously this quantity of time has to be smaller than the time it takes for the top to fall over). From that point on, as long as the rotational speed of the top is constant, the angle will not change. The same supported top spun at very high speed (whatever that is) and let go, will fall during the duration the top turns 90 degrees, but the amount of time will be so small that it will appear to remain at 45 degrees.
The way I see it, if precession was instantaneous, then a spinning top would never fall over until the top has stopped rotating completely. 141.119.184.10 14:38, 17 August 2007 (UTC)
If I understand you correctly, you are arguing that the time it takes for the rotating object to complete a quarter turn is a significant factor. Let's see how that works out for the Earth. The Earth has an equatorial bulge (the equator is 20 kilometers further away from the geometrical center than the poles). Since the Earth is not a perfect sphere the gravitational force exerted on the Earth by the Sun and the Moon has a net torque. If that torque would be the only factor then the plane of the equator would become aligned with the plane of the ecliptic. Since the Earth is rotating there is gyroscopic precession and as a consequence the angle between the plane of the equator and the plane of the ecliptic remains. The point is: if there would be a delay in the response to the torque, then the Earth would over time become aligned.
I am aware that the above reasoning is not complete. It is rather a plausibility argument that I am offering. The delay-hypothesis looks very counter-intuitive to me. In the case of the Earth the delay would have to be 6 hours, that's a huge delay.
The delayed-response hypothesis is offered to account for the fact that the response is at right angles to the applied torque. I find it very implausible that there would be some mechanism that stores the response to the torque for 6 hours, only to release it after the Earth has turned a quarter of a circle. Such delicate timing would require a sophisticated store/release mechanism. No such mechanism is in view.
By contrast, the instantaneous response depicted in Image:Gyroscopic_precession_256x256.png does not require the additional supposition of some timing/storing/releasing mechanism. --Cleonis | Talk 20:33, 17 August 2007 (UTC)
I agree, there is no such thing as a timing/storing/releasing mechanism. I attempted to explain that my way of understanding precession applies from the time a rotating object is first subjected to a torque until an element of mass has completed a 90 degree rotation. From that point on, there is a constant stream of elements of mass that complete their 90 degree rotation, so as long as the speed of rotation remains constant, then the precession remains constant. However, until that first element of mass has made its 90 degree rotation, then the object will not be under the influence of precession, and will have time to fall.
I really don't like thinking of precession using the Earth/Solar System as an example. I believe there are too many variables that can exert forces which may affect/enhance/deteriorate the precession to use it as a good example of "just" precession effects. However, to respond to your example, let me point out that the "first" 90 degree rotation occured a long long time ago. 141.119.184.10 13:29, 20 August 2007 (UTC)

Disambiguation?[edit]

There's also a webcomic http://precession.eviscerate.net/ called precession... does this need disambiguation? The webcomic in question doesn't have a wp page yet. —The preceding unsigned comment was added by 217.39.163.153 (talk) 16:03, 6 December 2006 (UTC).


Error in precession of the equinoxes?[edit]

Under precession of the equinoxes there is a statement that one degree of movement takes 180 years "from the point of view of the observer." This leads to a full cycle of 64,800 years. Shouldn't this be corrected to the 71.6 years per degree referenced in the Precession of the equinoxes page? --jb 15:52, 15 December 2006 (UTC)

  • Agree - but don't take my word for it. &#151; Xiutwel ♫☺♥♪ (talk) 07:31, 6 August 2007 (UTC)
The figure is correct. The celestial poles move 1 degree per 180 years. Since their circle is only 23.5 degrees in radius, 1 degree of that circle (which is 1/360 of a full cycle) takes just 71 years. The way, the truth, and the light 01:32, 11 August 2007 (UTC)

Strongly agree, in the article it says it takes about 25,800 years to be completed, this means about a degree every 71.66 years, which was already told even in ancient traditions. Horseshoekick 09:13, 6 November 2007 (UTC)

23,5 degrees[edit]

I am surprised to learn that the precession involves a 23,5 degree angle, exactly the same as the 23,5 degree angle of the Tropics. Is there an explanation for this? &#151; Xiutwel ♫☺♥♪ (talk) 07:31, 6 August 2007 (UTC)

They are the same angle - that between the Earth's axis and the ecliptic plane. The way, the truth, and the light 01:32, 11 August 2007 (UTC)

A disadvantage of precession ...[edit]

The paragraph goes on to discuss loosening of bicycle pedals. I have no doubt that this is called precession, but I'm sure that this is a completely unrelated phenomenon. It is a purely mechanical process and does not depend on inertia and is not inversely proportional to spin rate. I suggest it get its own article or at least a separate section that explains the difference. Any other thoughts? -AndrewDressel 14:46, 25 December 2006 (UTC)

Created separate section and explained the difference. -AndrewDressel 03:02, 27 December 2006 (UTC)

Torque-free precession[edit]

This sentence did not make sense:

Only moving objects can be in torque-free precession.

After all, only moving objects can be in precession at all. I have attempted to revise the section. Mark Foskey 13:19, 30 August 2007 (UTC)

I believe the equation connecting spin and precession frequency is incorrect. Equation (10) of Reference 1 is superficially similar, but alpha is measured relative to the total angular velocity, rather than the symmetry axis, and omega is the total angular velocity rather than the spin angular velocity. I think the equation you are looking for is omega_s = omega_p * cos(alpha) * ( I_p/I_s - 1 ) which can be derived starting from Equation (24) in this reference: [1] if you put omega_x = omega_p * sin(alpha) and omega_z = omega_s + omega_p * cos(alpha). Mgermaine8 (talk) 23:08, 11 June 2012 (UTC)

How can this statement be true?

The torque-free precession rate of an object with an axis of symmetry, such as a disk, spinning about an axis not aligned with that axis of symmetry can be calculated as follows:
[Equation showing relation between precession rate and spin rate which has no dependence on coning angle]

We KNOW that in the limit of rotation purely about one of the Ip axes there is no spin at all. the (unprintable by me) equation CAN'T be right. What's up? Dan Watts (talk) 12:07, 6 May 2014 (UTC)

I found a reference that allows one to get the (new) equation. Done? Dan Watts (talk) 13:45, 8 May 2014 (UTC)

Magnitude of Precession[edit]

I have not seen in any paper/books/internet any discussion/analysis on the magnitude of the precession angle and how it varies with I or omega and the force or impulse tending to pull it off axis. Is there a simple formula for this. -Coli.white (talk) 21:44, 14 August 2008 (UTC)

Hmmm. Not sure what you mean by magnitude, but this article does provide an expression for the precession rate based on moment of inertia, spin rate, and applied torque. -AndrewDressel (talk) 23:34, 14 August 2008 (UTC)

OK - when a gyroscope is spinning and aligned with the gravitional field and assumed not precessing. It is then given a little knock (or a force is applied) in a direction at 90deg to the gravitational field, the gyro will start precessing with a period as described. BUT, what about the size of the precessing angle? This is obviously related to the size of the knock - the bigger the knock the greater the angle although the precession period is obvioulsy unaffected by the size of the knock. For your interest, I am writing about the stabilizing effects of gyroscopic behaviour on discuses and frisbees, also the rifling of gun barrels. Thank you for your consideration of this problem, Colin 148.197.179.190 (talk) 10:35, 15 August 2008 (UTC)

I believe you are asking about the nutation angle. Unfortunately, calculating that requires the general equations of motion for a gyroscope, and they consist of three, coupled, second-order, nonlinear, differential equations that do not have a general closed-form solution. Special cases can be examined, for example if nutation angle, spin rate, and precession rate are constant, as is done in the current article, and which doesn't help you. Or, they can be solved numerically, most likely with a computer and a software package, such as MATLAB, that performs forward integration. The latter wouldn't be the hardest thing in the world to do, if you have the tools, because the equations are already well known, and the situation you wish to model it pretty well isolated and defined.
Most college level engineering dynamics texts can get you started on the topic: at least provide you with the necessary equations. To write this reply, I checked in Hibbeler's Engineering Mechanics: Dynamics, McGill and King's Engineering Mechanics: An Introduction to Dynamics, and Meriam and Kraige's Engineering Mechanics: Dynamics. I liked the treatment in the last one the best. Luckily for you, old editions, which most students avoid because the homework problems won't match the current syllabus, will be perfectly suitable and probably cheap as dirt. -AndrewDressel (talk) 14:36, 15 August 2008 (UTC)

Thank you so much for your advice. I will eagerly attempt to dig these texts out on Monday. I thought nutation was a small amplitude 'wobble' superimposed on the steady precessing motion but I will look into this also78.149.200.84 (talk) 14:46, 16 August 2008 (UTC).

Yes, the current nutation article doesn't mention it as one of the names of the Euler rotations, but does at least link to Euler angles in the "See also" section. -AndrewDressel (talk) 15:24, 16 August 2008 (UTC)

Another little misunderstanding ...I think. I was of the opinion that the period of precession only depends on the moment of inertia and the angular velocity of the flywheel (and the value of g) -ie not dependent on the angle of precession. I thought it was a bit like a pendulum whose period also does not depend on the angle of swing (ok, so long as the angle stays small) but if the period of precession is dependent on the torque then that must vary with angle. So is the period, in fact dependent on the angle; the precession period being greater the greater the angle of precession? Perhaps we can make small angle assumptions like the pendulum and thereby cancel out the angular dependence.91.111.30.74 (talk) 01:00, 19 August 2008 (UTC)

Since the precession rate is proportional to the torque, and the torque due to gravity acting on the center of mass is proportional to the distance perpendicular to the gravity vector between the center of mass and the support point, and that perpendicular distance is proportional to the sine of the nutation angle, then the precession rate is proportional to the sine of the nutation angle and the period of precession is inversely proportional to the sine of the nutation angle. If that angle is very close to 90 degrees, then its sine can be approximated by 1 and the precession rate is effectively independent of the nutation angle. -AndrewDressel (talk) 03:36, 19 August 2008 (UTC)

Yes. This is as I thought. Thank you. Coli.white (talk) 01:02, 20 August 2008 (UTC)


Discus/Frisbee Wobble[edit]

When a frisbee or discus wobbles as it flies due to, say, errors in throw, I have thought that this is equivalent to precession in a gyroscope but with the axis not fixed at either end. So we could calculate the period of wobble using the gyroscope precession equation. However for this to be true, the centre of mass of the discus must lie some distance displaced horizontally from the spin axis (that distance is one variable in the precession period equation). However, a discus wobble I've always envisioned to be wobbling around the centre of mass point, so the c of m does not actually deviate from the trajectory path. Or, to put another way, the locus of the spin axis traces out a double cone joined at the points. This cannot be true if it is a form of precession with the axis not fixed at either end. If the wobble is a form of true precession, there must be a 'virtual point' below the discus which represents the fixed end of the virtual axle so the discus then forms the flywheel of a virtual gyroscope. Would really appreciate people's thoughts on this. Thanks Coli.white (talk) 22:05, 12 September 2008 (UTC)

It is not clear what you mean by "errors in throw". Do you mean that the spin axis upon release is not aligned with an axis of symmetry, or that aerodynamics forces apply a torque? Which "gyroscope precession equation" do you mean? The one given in this article does not include a distance.
It would seem that a frisbee could be subject to both types of precession: torque-free and torque-induced. It would experience torque-free precession because the spin axis does not align with an axis of symmetry, and it would experience torque-induced precession because of aerodynamic forces. In neither case is there any requirement for the center of mass to be displaced from the spin axis nor for there to be a "fixed end", virtual or real, to the spin axis. -AndrewDressel (talk) 14:19, 14 September 2008 (UTC)

I was referring to the 'Classical (Newtonian) section and the equation for precession period. Reference also the diagram in that section which show that, for stable precession to occur, the centre of mass moves around the fixed point of the axle with a component at some horizontal distance from it. This provides a downward torque which then moves around the fixed point someway below the C of M.

Equating this to a discus which maybe is given a slight push downwards on the rim at the moment of release, it will then wobble as it spins through the air. I was thinking, in such a case, the spin axis will precess around the axis of symmetry, but the axis, although precessing will, at all times cut the C of G (unlike the case of a precessing gyroscope). I am considering how to calculate the period of the wobble. Given, say, the M of I of the discus, the degree of wobble (ie the offset angle of spin axis to symmetry axis) and the angular velocity of the discus, can the period of wobble be calculated using said equation. It could just be that a wobbling discus with a spin axis which only moves around the C of M but always cuts it(thereby not creating a torque), is a mechanical system unrelated to gyroscopic precession and I can't use the equation to calculate the wobble period. —Preceding unsigned comment added by Coli.white (talkcontribs) 23:16, 14 September 2008 (UTC)

Below the diagram, the caption says"The torque caused by the two opposing forces Fg and -Fg causes a change in the angular momentum L in the direction of that torque. This causes the top to precess." But in my scheme both forces will be aligned and there will be no reaction force from the ground. This is very rambly - hope it makes sense. Thanks so much for your consideration. Coli.white (talk) 23:20, 14 September 2008 (UTC)

In the equation given:
in which Is is the moment of inertia, is the angular velocity of spin about the spin axis, and Q is the torque, there is no length. In the diagram, the length is only used to calculate the torque. In the case of a flying frisbee, the torque due to aerodynamic forces would also be calculated with a length, probably from the center of mass to the center of pressure.
However, the wobble you describe, of a frisbee release with a spin axis that is not aligned with an axis of symmetry, is torque free precession. To calculate it, you would need to use the technique described in the second paragraph of that section:
"...the vector speed (w) will have to evolve in time so that their cross product L = I x w remains constant." -AndrewDressel (talk) 01:50, 15 September 2008 (UTC)

Thank you. Yes. I understand the torque induced precession. It's the torque-free version I struggle with. I see the cross-product formula L = I x w, but do you think it is possible to calculate the period of wobble given I and w for the discus (assuming w is the angular velocity of the discus spin rather than precession (ie the wobble) angular velocity). And L will be tha angular momentum of the discus. Basically I'm after an expression for wp for the torque free case. Again, thanks. Coli.white (talk) 21:52, 16 September 2008 (UTC)

David Boal of Simon Fraser University gives a nice derivation of an expression for the torque-free precession rate of an object with an axis of symmetry, such as a disk, spinning about an axis not aligned with that axis of symmetry can be calculated as follows:
where is the precession rate, is the spin rate about the axis of symmetry, is the angle between the axis of symmetry and the axis about which it precess, is the moment of inertia about the axis of symmetry, and is moment of inertia about either of the other two perpendicular principal axes. They should be the same, due to the symmetry of the disk. -AndrewDressel (talk) 18:58, 17 September 2008 (UTC)

That is brilliant. Exactly what I need! Thank you. Just to clarify, you weren't suggesting that Ip and Is were the same, were you? I guess you meant that Ip would be the same for any angle at 90deg to the axis of symmetry. Finally,do you have an academic reference for this equation? Otherwise, again I am so grateful for your support on this matter. Coli.white (talk) 23:10, 17 September 2008 (UTC)

No, Ip and Is are not the same. Is is the moment of inertia about the axis of symmetry. The other two, in the case of a disk or football, are identical, perpendicular to Is and each other, hence they are just labeled Ip in Boal's derivation. In his lecture notes, he cites Analytical Mechanics, 6th Edition, by G.R. Fowles and G. R. Cassiday (Saunders, 1999). -AndrewDressel (talk) 00:33, 18 September 2008 (UTC)

Thank you so much. I now have all the answers I need!

Classical (Newtonian)[edit]

I have a question about the discussion of the person on a rotating platform with a spinning bicycle wheel. I have seen this demonstration before and struggle to understand it and the explanation here does not help me.

Now I am new to this whole area but the way I understand it (correct me if I'm wrong) is that precession is just an effect of conservation of angular momentum. It would seem to me that the case of the person on the platform would be a clear demonstration of this, or, more specifically, torque-free precession. I don't understand where the torque comes from in this situation as is discussed in the article. If the person is holding the wheel in place by the axle at an angle to the horizontal then the wheel is from the point of view of the person, in static equilibrium. Hence, no torque?

Help? —Preceding unsigned comment added by 203.161.75.108 (talk) 01:48, 8 January 2009 (UTC)

I don't quite get that explanation either. Another way to look at it is:
  1. When the wheel is held with its axis in the horizontal position and spun, there is no angular momentum about the vertical axis.
  2. When the wheel is tilted so that its axis is vertical, there are not external torques so there should still be no angular momentum about the vertical axis. The only way to cancel the angular moment of the spinning wheel about the vertical axis is for the demonstrator to begin to rotate in the opposite direction.
  3. When the wheel is tilted back so that its axis is horizontal, the demonstrator must also stop rotating so that there is still no angular momentum about the vertical axis.
-AndrewDressel (talk) 03:27, 8 January 2009 (UTC)

Cheers, that was my understanding too. The article confused me. —Preceding unsigned comment added by 121.221.165.185 (talk) 14:11, 8 January 2009 (UTC)

Actually, I watched an MIT physics lecture on gyroscopes on youtube that helped explain it.

http://au.youtube.com/watch?v=zLy0IQT8ssk

Maybe this should be added to external links? I don't edit pages so I don't know the guidelines regarding what can and can't be an external link, but thats my suggestion anyway. —Preceding unsigned comment added by 203.161.75.108 (talk) 03:12, 9 January 2009 (UTC)

That MIT open courseware video ( http://au.youtube.com/watch?v=zLy0IQT8ssk ) is lovely to watch. However, I notice that the narrator fails to deliver a crucial bit of information. Proceeding as he does he is suggesting that the torque that is applied to the spinning mass causes the precession. What happens in fact is as follows: the weel is rolling, and then a torque is applied that would pitch down the wheel if it wouldn't be spinning. As the torque is applied the wheel does pitch down, but only a tiny amount. The pitching down, unnoticable as it is, induces a swivel (as per gyroscopic precession). Next in the causal sequence: when a rolling wheel swivels it tends to pitch up. But now it can't pitch up, because of that torque that is being applied. So rather than pitching up the swivel of the wheel is sustained.
The torque is not a continuous cause of the gyroscopic precesssion. During the sustained precession the torque keeps the wheel from pitching up, that is what it does on a sustained basis.
Another way of setting up a state of sustained precession is to first start a swiveling motion, and then, as the spinning wheel responds with starting to pitch up, quickly add precisely the torque that keeps the wheel from pitching up.
The counterintuitive part is: when you try to pitch a rolling wheel it responds with swivel; the response is at right angles to the input. The mechanics of that is explained in the article section 'torque induced'. --Cleonis | Talk 19:59, 9 January 2009 (UTC)

In November 2012, Espotom added "a non-mathematical way of visualizing the cause of gyroscopic precession". It's good, but I think there's an even clearer way, so I'm hereby asking for critiques of the following explanation; and also whether Espotom would mind if I accordingly replace or add to what he or she wrote. Here's my version: "You observe a spinning bicycle wheel edge on, with its nearest edge moving downwards before your eyes. Marker lines are visible; horizontal across the rim of the wheel. The left-hand end of the axle, as you face it, is suspended somehow; but gravity is 'inclined' to pull the right-hand end of the axle downward. From your viewpoint, gravity could be regarded as trying to turn the axle of the wheel clockwise. Also at this moment, the marker line that's closest to you is mid-way on its descent from the top point of the wheel to the bottom. In effect, gravity is trying to turn this marker line in a clockwise direction; the same as it's trying to do with the axle. The right-hand end of this marker line correspondingly has more downward force upon it than has the left-hand end; though both are already moving downward. In addition, the wheel has lots of spokes. So, and roughly speaking will do, spoke R, say, joins the right-hand end of the marker line to the axle and some spoke L joins the left-hand end of the marker line to the axle. Since the wheel is undergoing no motion apart from spinning, spoke R must transmit more tension to the rim than spoke L does. This tension is in the form of centripetal acceleration; not only that involved in the wheel spinning, but also the greater downward force resulting from gravity as per its action upon the right-hand end of the axle. Where spoke L joins the rim, this downward force is smaller, as the left-hand end of the axle has some compensating force which is keeping it suspended. (A mirror-image scenario is taking place on the far edge of the wheel, which is moving upwards.) The greater tension in spoke R will logically make the wheel begin to precess in a clockwise direction as viewed from below. That is, the wheel will tend to pivot around the left-hand end of the axle, which continues to be suspended. The tension difference between spokes R and L is repeated to less degree among all left-right pairs of spokes. The width of the rim doesn't matter because a wide rimmed wheel can be viewed as simply a combination of any variety of thinner rimmed wheels. However, why do the L and R spoke tensions equalize once the wheel is precessing with a certain angular velocity? Well, change the scenario by supposing the wheel to be already precessing. Hence the right-hand end of the axle (in the moment when it is still orthogonal to your line of sight) is moving away from you, and the left-hand end is moving toward you. However, relative to the axle's rotating frame of reference, the force of gravity acting upon the right-hand end of the axle is being directed not vertically downward, but downward and slightly toward you, while the force of gravity on the left-hand end is directed downward and slightly away from you. So the tension in spoke R is decreased while that in spoke L is increased. Therefore, with precession happening, the tension in spoke R can equalize with the tension in spoke L, so that the wheel enters a state of equilibrium. Though this explanation talks about bicycle wheels and spokes, it may be fully generalized to account for any instance of precession. In fact, it applies just as readily to a wheel which isn't spinning at all, although that situation would be far less likely to involve the wheel remaining upright." Subtendant (talk) 21:56, 5 December 2013 (UTC)

Where does the work go?[edit]

I'm sorry if this question has already been answered before in the article or the talk page, in that case I missed it. I'm just curious about this, when we apply a torque to create the precession, we are doing real work, because we apply a force over distance (and of course torque is just angular work). Still I wonder, where exactly does that spent energy end up? Does it reduce the speed of the wheel somehow, has it gained some potential or is it just missing ;D ???? I'm not trying to act dumb, I just don't see where that energy goes....--Nabo0o (talk) 12:25, 18 November 2009 (UTC)

Well? I really hope that the people writing this article, explaining what precession is, knows the subject well enough to say how torque interacts with the angular momentum. If not, then it should be cleared out. --Nabo0o (talk) 22:33, 11 December 2009 (UTC)

Precession of a ball (sphere)[edit]

I'm very sorry to admit I don't understand the cause of gyroscopic precession at all. Anyway, I wonder whether there can be precession (the gyroscopic or torque-induced type) in a rotating perfectly round ball (supposed it is possible to exert torque on it). I had thought precession of the Earth (or Mars) was an example, but Axial_precession says that this is due to gravitaonal forces and would not occur if the Earth were perfectly round. I'm not sure about Mars here. Can anyone show me a real-world example of a precessing, near-perfect ball? And what about precession in rotating black holes? Would that be produce a specific type of gravitational waves? Thank you very much. Illegal604 (talk) 11:11, 22 April 2010 (UTC)

Precession of a Gyroscope illustration[edit]

I may be overlooking something, but it seems to me that the rotational axis of the gyroscope illustrated in the first picture in the article, is a constant vertical axis through the bottom pin of the gyroscope. Since precession is defined as "a change in the orientation of the rotational axis of a rotating body," I wouldn't think this is an example of precession. In any case, it isn't in general the case that one side of the precessing object is always faced with the z-axis, so this is a poor illustration of the general phenomenon. — Preceding unsigned comment added by 94.224.49.243 (talk) 07:15, 2 January 2013 (UTC)

Malware Link Removal[edit]

--Gary Dee 18:32, 21 July 2013 (UTC)

heisenberg uncertainty principle[edit]

Earth precession is told to be around 26000 years. I would like to know how exactly this has been determined. One argument involves accurate measurement. But as far as i know, it is impossible, no matter how accurate you measure, to determine this with any reasonable certainty if the measuring time is so short. This follows from Heisenberg. If you have less than one period measured, uncertainty about the exact period will increase. With less than 1/10th of a period, one can at best fit a 1st degree polynomial through the data, but never a sinuoid. You can create artificially a perfect sinusoid (e.g.with matlab/octave) from which you know the period, but I was never able to recover the exact period from the datapoints if not a complete period is in the signal. Not with harmonic decomposition (which wil at best lead to a probability distribution) and also not with more exotic datafitting methods such as fminbound, fninsearch, fmincon, fminunc. Please show me a good article that explains how earth precession is determined — Preceding unsigned comment added by 86.93.56.137 (talk) 18:40, 30 April 2014 (UTC)

"it is impossible, no matter how accurate you measure, to determine this with any reasonable certainty if the measuring time is so short" - yes, and that's why the Earth's precession is said to be around 26,000 years. The three zeroes at the end suggests it's only accurate to within about 1000 years. 212.9.31.12 (talk) 12:34, 28 January 2016 (UTC)

Disappointingly Soupy technical writing.[edit]

My response to this article, "Is this supposed to be comprehensible?"

...at least the first quarter, after which I quit.

I'm thinking, of all the dark secrets of technology, maybe they want to still keep this one a secret by making it seem incomprehensible?

Or maybe our poor experts just aren't talented at technical writing for comprehension?

If I have to learn this subject, and I may because it's on the trailing edge of cutting edge technology across so many areas ...

I will rewrite the entire thing - without any non-essential waste of anyone's time.

Xgenei (talk) 01:57, 4 November 2015 (UTC)

On the simplified explanation with the bicycle wheel[edit]

Okay, I know this isn't a physics forum, but something just doesn't make sense about the explanation given in the "Classical (Newtonian)" section.

Actually, the force attempts to displace them some amount horizontally at that moment, but the ostensible component of that motion, attributed to the horizontal force, never occurs, as it would if the wheel was not spinning.

How can the force "never occur"? If you push (instantaneously) on the rim of a wheel, spinning or not, that point must move, surely? Even if it only moves a little bit and immediately moves back. It can't offer infinite resistance to any force.

The force can't be instantly "teleported" to the 6 and 12 o'clock positions. That obviously doesn't happen when the wheel is stationary (the point pushed will definitely move in that case), and it doesn't noticeably happen if the wheel is spinning slowly. There isn't some cut-off speed where suddenly the force takes effect elsewhere.

Right? 212.9.31.12 (talk) 12:32, 28 January 2016 (UTC)

  1. ^ http://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec29.pdf