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Author Topic: Thoughts on"Q" or Quality Factor  (Read 975 times)
Group: Moderator
Hero Member
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Posts: 3537
It's turtles all the way down
Here we will try to look at and discuss several things, like the spin of the earth and determine "Q"

For example, the earth slows it's rotation by 1.4 milliseconds per hundred years. That's 1400 uSec / 100 years or 14 uSec / year.

One cycle would be 24 hours.

Q is defined as energy per cycle divided by energy lost per cycle. We can also look at this in the time domain.

I encourage some of you folks to do the math using either energy or time. The results are truly amazing considering a viscous damped oscillator such as the earth / moon system.

Everyone knows that a hard boiled egg will spin much longer than a raw egg. Considering the gravitational pull of the moon and the amount of energy created in the frictional heating of the waves, why is the earth not slowing down faster than it appears to be?

Lets see who can come up with the correct figure for the "Q" of the earth spin, then let's compare it to other  frictionally damped viscous systems.

Let's also compare it to the "Q" of atomic oscillators vs normal mechanical or electrical systems.

Anyone want to play?


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"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
Group: Guest
why is the earth not slowing down faster than it appears to be?

Same reason the Moon's orbit is increasing very slowly over time.

Mass is added to the Earth on a continuous basis. Since adding mass to a spinning object will tend to cause a decrease in the spin rate.

I'm assuming the net change in spin rate is zero from the impact of objects.

The Moon's orbital distance from the Earth increases for the same reason. Mass is increasing while the orbital time remains almost the same. With this scenario the orbit must increase.

The complexity begins when you figure this oscillating system is no simple oscillator. There are a few entrainment values for calculation. After all, this system is a Van der Pol unforced oscillator and energy is lost and gained ;)

Anyway, that is my mid-morning four cups of coffee answer  :D
   
Group: Moderator
Hero Member
*****

Posts: 3537
It's turtles all the way down
Have another cup of coffee and see if you can come up with a Q for earth spin. A first cut approximate answer will do.


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"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
Group: Guest
First cut?

It varies between 100 and an ungodly high embarrassing number. The majority of cycles will be around 400.

I'll have to finish my open-source super computer before I can supply a fairly reasonable answer. Every cycle is slightly different.

My point was... a fairly accurate answer for any one cycle within - just an Earth year - has many variables. The result is this value of Q varies wildly because this dirt-ball oscillator is not a linear oscillator.

Are we to generalize or should we pick a particular period within this millennium  :D


   
Group: Moderator
Hero Member
*****

Posts: 3537
It's turtles all the way down
WW said:

Quote
Are we to generalize or should we pick a particular period within this millennium.

Lets just take the average slowdown of the last 100 years, 1.4 milliseconds, and apply that to one cycle i.e 24 hour period of rotation.

e.g. a spinning top will slowdown at a certain rate. We can calculate the cycle by cycle (360 degrees of rotation) and come up with a reasonable value for the Q.

So averaging we have 1400 microseconds per 100 years, or 1.4 uSec per year.

Now divide that by 365 days / year to get the slowdown per day or 24 hours. (one cycle)

I'll let you finish from here. :D


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
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