<p><strong>In theory</strong>, if the Earth was a solid object, the Chandler wobble would be easy to understand. It's simply a result of the polhode rolling without slipping on the herpolhode lying in the invariable plane. In other words, $\mathrm I \dot \omega + \omega \times (\mathrm I \omega) = 0$, where $\mathrm I$ is the Earth's inertia tensor and $\omega$ is the Earth's angular velocity. The Chandler wobble is simply the consequence of the Earth not having a spherical mass distribution and the Earth not having it's angular momentum and angular velocity coaligned. </p>

<p><strong>In practice</strong>, the Earth is not a solid object. The theory behind the Earth's orientation is rather (putting it mildly) complex. The best model involves over a thousand parameters. Unfortunately, this best model of the Earth's precession, nutation, and rotation does not explain the Earth's orientation. The Chandler wobble, along with a number of lesser phenomena, is an after-the-fact discrepancy between the very best analytic model and the very best observational results.</p>

<p>Knowing the Earth's orientation is extremely important to some sciences, high resolution astronomy in particular. The US Naval Observatory (USNO) publishes its estimates of the deviation of the Earth's true orientation versus that of the best analytic models on a daily basis. The USNO is a key member of the International Earth Rotation and Reference Systems (IERS, for short, and yes, the acronym doesn't work. It used to work.)</p>

CC BY-SA 3.0
2015-03-03T19:37:39.147Z
2015-03-03T19:37:39.147Z
David Hammen
<p>My knowledge of science is very basic, but after some research I have some knowledge, but I am still not clear with so many things regarding to this earth. </p>

<p>I have read <a href="http://www.livescience.com/17619-lasers-measure-earth-rotation-wobble.html">Laser's Measure the Earth's Rotation and Wobble</a> but would appreciate it if anyone who knows science well, could explain how the rotation of the Earth is measured and how it is affected by the Chandler wobble and the annual wobble? </p>

CC BY-SA 3.0
2015-03-03T05:26:05.477Z
What do you mean by "measure the cycles of earth moving"?  I think we have a language understanding problem.
gerrit
CC BY-SA 3.0
2015-03-03T06:22:33.623Z
Hi both sorry for poor with English writing :), what I mean is cycle of earth roll or rotation. here is what I mean http://www.livescience.com/17619-lasers-measure-earth-rotation-wobble.html  But I don't know what is the measurement to proof.
pitoukhmer
CC BY-SA 3.0
2015-03-03T07:04:21.040Z
Thank @Fred for your correcting me, that really appreciated
pitoukhmer
<earth-history><earth-rotation><earth-observation>
How to measure the cycles of earth moving/rotation?
pitoukhmer
# How to measure the cycles of earth moving/rotation?

(Asked by: username_0 on 2015-03-03)

My knowledge of science is very basic, but after some research I have some knowledge, but I am still not clear with so many things regarding to this earth.

I have read [Laser's Measure the Earth's Rotation and Wobble](http://www.livescience.com/17619-lasers-measure-earth-rotation-wobble.html) but would appreciate it if anyone who knows science well, could explain how the rotation of the Earth is measured and how it is affected by the Chandler wobble and the annual wobble?

**Tags:** earth-history, earth-rotation, earth-observation

> **Comments:**
> **username_1**: What do you mean by "measure the cycles of earth moving"?  I think we have a language understanding problem.  
> **username_0**: Hi both sorry for poor with English writing :), what I mean is cycle of earth roll or rotation. here is what I mean http://www.livescience.com/17619-lasers-measure-earth-rotation-wobble.html  But I don't know what is the measurement to proof.  
> **username_0**: Thank @Fred for your correcting me, that really appreciated  

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## 1 answers

### Answer by username_2 on 2015-03-03. Score: 8

**In theory**, if the Earth was a solid object, the Chandler wobble would be easy to understand. It's simply a result of the polhode rolling without slipping on the herpolhode lying in the invariable plane. In other words, $\mathrm I \dot \omega + \omega \times (\mathrm I \omega) = 0$, where $\mathrm I$ is the Earth's inertia tensor and $\omega$ is the Earth's angular velocity. The Chandler wobble is simply the consequence of the Earth not having a spherical mass distribution and the Earth not having it's angular momentum and angular velocity coaligned.

**In practice**, the Earth is not a solid object. The theory behind the Earth's orientation is rather (putting it mildly) complex. The best model involves over a thousand parameters. Unfortunately, this best model of the Earth's precession, nutation, and rotation does not explain the Earth's orientation. The Chandler wobble, along with a number of lesser phenomena, is an after-the-fact discrepancy between the very best analytic model and the very best observational results.

Knowing the Earth's orientation is extremely important to some sciences, high resolution astronomy in particular. The US Naval Observatory (USNO) publishes its estimates of the deviation of the Earth's true orientation versus that of the best analytic models on a daily basis. The USNO is a key member of the International Earth Rotation and Reference Systems (IERS, for short, and yes, the acronym doesn't work. It used to work.)


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