Re: [time-nuts] Q/noise of Earth as an oscillator

[bg]

A bit short on the phone...
Both the ECI and ECEF frames are rotating at sidreal rate. The earth rate is (a 
small) part of the navigation equations. Thus its needed in inertial nav.
Glenn, look at the performance of the big land RLGs - NZ and German that were 
linked last week. How much better are those compared to your (also extremely 
good) sub gyros.--     Björn




Sent from my smartphone. Original message From: Tom Van Baak 
<t...@leapsecond.com> Date: 01/08/2016  20:01  (GMT+01:00) To: Discussion of 
precise time and frequency measurement <time-nuts@febo.com> Subject: Re: 
[time-nuts] Q/noise of Earth as an oscillator 
Hi Glenn,

Your 15.04 number rings a bell [1]. The conventional solar day is simply 86400 
seconds (24 hours). So each hour is 15 degrees, exactly.

But the actual (sidereal) earth rotation rate is about 86164 SI seconds (23h 
56m 4.091s). So each hour is 15.0411 degrees.

Someone who understands celestial mechanics or an ex-Navy person with sextant 
skills could explain better, but I bet that's where your 15.04 number comes 
from.

/tvb

[1] http://www.navy.mil/navydata/questions/bells.html ;-)

- Original Message - 
From: "Glenn Little WB4UIV" <glennmaill...@bellsouth.net>
To: "Tom Van Baak" <t...@leapsecond.com>; "Discussion of precise time and 
frequency measurement" <time-nuts@febo.com>
Sent: Monday, August 01, 2016 10:38 AM
Subject: Re: [time-nuts] Q/noise of Earth as an oscillator


In navigation we used the earth rate of 15.04 degrees per hour.
This was treated as a 'constant' even though it varied with wind, waves 
on the ocean and other things affecting the instantaneous rotational 
speed of the earth.

How does this factor into leap seconds, or, does it?

We accept that the day is 24 hours long, this would be for a earth 
rotational speed of 15.0 degrees per hour.

I am not a mathematician, but, I dis do electronic navigation on submarines.

73
Glenn
WB4UIV


On 8/1/2016 10:54 AM, Tom Van Baak wrote:
> Hi Jim.
>
>> You said: "you need energy; you need energy loss; you need cycles over which 
>> that loss repeatedly occurs."
>> With regard to the earth, where is the first one?
>
> By first one, do you mean where does the initial energy come from?
>
> For a pendulum clock, you supply energy with a lift or a push. For a lift to 
> the side, E = mgh, where h is the height above the base. For a push from 
> center, E = 1/2 mv^2. Either way, it takes all the potential or kinetic E you 
> provide and starts making time from there.
>
> For a rotating clock, you just give it a twist. In this case, E = 1/2 Iw^2, 
> where I is the moment of inertia and w (omega) is angular velocity. For earth 
> the total E is 2.1e29 J. That's the energy number you want, yes?
>
>> Sure it was there at the start when the solar system formed, but where is it 
>> now?
>
> I don't have data on where the initial swirl of solar system mass came from, 
> or how much of that rotational energy went into our planet and its pesky 
> moon, or Who or what gave that initial twist. The Q is pretty high so I 
> assume you could work backwards, but I leave that to astronomers and 
> cosmologists. I believe the 2 ms/day / century estimate we use is one such 
> measurement.
>
> For more on earth rotation rate, UTC and leap seconds see 
> https://www.ucolick.org/~sla/leapsecs/dutc.html
>
> Surely in the literature there is a pile of information or speculation 
> regarding all the rotational energy in the universe. It seems a common theme 
> everywhere you look; maybe it was as much Big Twist as Big Bang? Perhaps in 
> your Pulsar research you've run across some papers you could share. Off-list 
> is ok, unless you think it has general time-nuts appeal. We're running the 
> risk of spinning off-topic already.
>
> Thanks,
> /tvb
>
> ----- Original Message -
> From: "Jim Palfreyman" <jim77...@gmail.com>
> To: "Discussion of precise time and frequency measurement" 
> <time-nuts@febo.com>
> Sent: Sunday, July 31, 2016 7:34 PM
> Subject: Re: [time-nuts] Q/noise of Earth as an oscillator
>
>
> Hi Tom,
>
> You said: "you need energy; you need energy loss; you need cycles over
> which that loss repeatedly occurs."
>
> With regard to the earth, where is the first one? Sure it was there at the
> start when the solar system formed, but where is it now?
>
> Jim
>
>
> On 1 August 2016 at 12:16, Tom Van Baak <t...@leapsecond.com> wrote:
>
>> Hal:
>>> Is there a term other than Q that is used to describe the rate of energy
>> loss
>>> for things that aren't oscillators?
>>
>> Jim:
>>> cooling (as in hot things)
>>> discharge (as

Re: [time-nuts] Q/noise of Earth as an oscillator

[Hal Murray]

glennmaill...@bellsouth.net said:
> In navigation we used the earth rate of 15.04 degrees per hour. This was
> treated as a 'constant' even though it varied with wind, waves  on the ocean
> and other things affecting the instantaneous rotational  speed of the earth.

Were the wind and waves and whatever significant?   Measurable?

> How does this factor into leap seconds, or, does it?

Leap seconds are necessary to correct for the unpredictable things like waves 
and earthquakes and ice caps melting.

There is also tidal friction which is slowing down the rate of rotation.  So 
even if the wind and friends were insignificant we would need to adjust the 
clocks occasionally.


> We accept that the day is 24 hours long, this would be for a earth
> rotational speed of 15.0 degrees per hour.

The 0.04 is the difference between solar (sun) and sidereal (stars) time.

During a year, the earth goes around the sun.  If your clock runs on solar 
time but you are looking at the stars, that adds one more rotation per year.  
That's 360/365 degrees per day or 0.04 degrees per hour.


-- 
These are my opinions.  I hate spam.



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Re: [time-nuts] Q/noise of Earth as an oscillator

[Tom Van Baak]

Hi Glenn,

Your 15.04 number rings a bell [1]. The conventional solar day is simply 86400 
seconds (24 hours). So each hour is 15 degrees, exactly.

But the actual (sidereal) earth rotation rate is about 86164 SI seconds (23h 
56m 4.091s). So each hour is 15.0411 degrees.

Someone who understands celestial mechanics or an ex-Navy person with sextant 
skills could explain better, but I bet that's where your 15.04 number comes 
from.

/tvb

[1] http://www.navy.mil/navydata/questions/bells.html ;-)

- Original Message - 
From: "Glenn Little WB4UIV" <glennmaill...@bellsouth.net>
To: "Tom Van Baak" <t...@leapsecond.com>; "Discussion of precise time and 
frequency measurement" <time-nuts@febo.com>
Sent: Monday, August 01, 2016 10:38 AM
Subject: Re: [time-nuts] Q/noise of Earth as an oscillator


In navigation we used the earth rate of 15.04 degrees per hour.
This was treated as a 'constant' even though it varied with wind, waves 
on the ocean and other things affecting the instantaneous rotational 
speed of the earth.

How does this factor into leap seconds, or, does it?

We accept that the day is 24 hours long, this would be for a earth 
rotational speed of 15.0 degrees per hour.

I am not a mathematician, but, I dis do electronic navigation on submarines.

73
Glenn
WB4UIV


On 8/1/2016 10:54 AM, Tom Van Baak wrote:
> Hi Jim.
>
>> You said: "you need energy; you need energy loss; you need cycles over which 
>> that loss repeatedly occurs."
>> With regard to the earth, where is the first one?
>
> By first one, do you mean where does the initial energy come from?
>
> For a pendulum clock, you supply energy with a lift or a push. For a lift to 
> the side, E = mgh, where h is the height above the base. For a push from 
> center, E = 1/2 mv^2. Either way, it takes all the potential or kinetic E you 
> provide and starts making time from there.
>
> For a rotating clock, you just give it a twist. In this case, E = 1/2 Iw^2, 
> where I is the moment of inertia and w (omega) is angular velocity. For earth 
> the total E is 2.1e29 J. That's the energy number you want, yes?
>
>> Sure it was there at the start when the solar system formed, but where is it 
>> now?
>
> I don't have data on where the initial swirl of solar system mass came from, 
> or how much of that rotational energy went into our planet and its pesky 
> moon, or Who or what gave that initial twist. The Q is pretty high so I 
> assume you could work backwards, but I leave that to astronomers and 
> cosmologists. I believe the 2 ms/day / century estimate we use is one such 
> measurement.
>
> For more on earth rotation rate, UTC and leap seconds see 
> https://www.ucolick.org/~sla/leapsecs/dutc.html
>
> Surely in the literature there is a pile of information or speculation 
> regarding all the rotational energy in the universe. It seems a common theme 
> everywhere you look; maybe it was as much Big Twist as Big Bang? Perhaps in 
> your Pulsar research you've run across some papers you could share. Off-list 
> is ok, unless you think it has general time-nuts appeal. We're running the 
> risk of spinning off-topic already.
>
> Thanks,
> /tvb
>
> - Original Message -
> From: "Jim Palfreyman" <jim77...@gmail.com>
> To: "Discussion of precise time and frequency measurement" 
> <time-nuts@febo.com>
> Sent: Sunday, July 31, 2016 7:34 PM
> Subject: Re: [time-nuts] Q/noise of Earth as an oscillator
>
>
> Hi Tom,
>
> You said: "you need energy; you need energy loss; you need cycles over
> which that loss repeatedly occurs."
>
> With regard to the earth, where is the first one? Sure it was there at the
> start when the solar system formed, but where is it now?
>
> Jim
>
>
> On 1 August 2016 at 12:16, Tom Van Baak <t...@leapsecond.com> wrote:
>
>> Hal:
>>> Is there a term other than Q that is used to describe the rate of energy
>> loss
>>> for things that aren't oscillators?
>>
>> Jim:
>>> cooling (as in hot things)
>>> discharge (as in capacitors and batteries)
>>> leakage (as in pressure vessels)
>>> loss
>>
>> Scott:
>>> An irreversible process would be a better description versus energy loss.
>>> Like joule heating (resistance, friction).
>>
>> Notice that these are all energy losses over time; gradual processes with
>> perhaps an exponential time constant, but without cycles or periods. We
>> know not to apply Q in these scenarios.
>>
>> But when you have an oscillator, or a resonator, or (as I suggest) a
>> "rotator", it seems to make sense to use Q to describe the normalized rate
>&g

Re: [time-nuts] Q/noise of Earth as an oscillator

[Glenn Little WB4UIV]

In navigation we used the earth rate of 15.04 degrees per hour.
This was treated as a 'constant' even though it varied with wind, waves 
on the ocean and other things affecting the instantaneous rotational 
speed of the earth.


How does this factor into leap seconds, or, does it?

We accept that the day is 24 hours long, this would be for a earth 
rotational speed of 15.0 degrees per hour.


I am not a mathematician, but, I dis do electronic navigation on submarines.

73
Glenn
WB4UIV


On 8/1/2016 10:54 AM, Tom Van Baak wrote:

Hi Jim.


You said: "you need energy; you need energy loss; you need cycles over which that 
loss repeatedly occurs."
With regard to the earth, where is the first one?


By first one, do you mean where does the initial energy come from?

For a pendulum clock, you supply energy with a lift or a push. For a lift to 
the side, E = mgh, where h is the height above the base. For a push from 
center, E = 1/2 mv^2. Either way, it takes all the potential or kinetic E you 
provide and starts making time from there.

For a rotating clock, you just give it a twist. In this case, E = 1/2 Iw^2, 
where I is the moment of inertia and w (omega) is angular velocity. For earth 
the total E is 2.1e29 J. That's the energy number you want, yes?


Sure it was there at the start when the solar system formed, but where is it 
now?


I don't have data on where the initial swirl of solar system mass came from, or 
how much of that rotational energy went into our planet and its pesky moon, or 
Who or what gave that initial twist. The Q is pretty high so I assume you could 
work backwards, but I leave that to astronomers and cosmologists. I believe the 
2 ms/day / century estimate we use is one such measurement.

For more on earth rotation rate, UTC and leap seconds see 
https://www.ucolick.org/~sla/leapsecs/dutc.html

Surely in the literature there is a pile of information or speculation 
regarding all the rotational energy in the universe. It seems a common theme 
everywhere you look; maybe it was as much Big Twist as Big Bang? Perhaps in 
your Pulsar research you've run across some papers you could share. Off-list is 
ok, unless you think it has general time-nuts appeal. We're running the risk of 
spinning off-topic already.

Thanks,
/tvb

- Original Message -
From: "Jim Palfreyman" <jim77...@gmail.com>
To: "Discussion of precise time and frequency measurement" <time-nuts@febo.com>
Sent: Sunday, July 31, 2016 7:34 PM
Subject: Re: [time-nuts] Q/noise of Earth as an oscillator


Hi Tom,

You said: "you need energy; you need energy loss; you need cycles over
which that loss repeatedly occurs."

With regard to the earth, where is the first one? Sure it was there at the
start when the solar system formed, but where is it now?

Jim


On 1 August 2016 at 12:16, Tom Van Baak <t...@leapsecond.com> wrote:


Hal:

Is there a term other than Q that is used to describe the rate of energy

loss

for things that aren't oscillators?


Jim:

cooling (as in hot things)
discharge (as in capacitors and batteries)
leakage (as in pressure vessels)
loss


Scott:

An irreversible process would be a better description versus energy loss.
Like joule heating (resistance, friction).


Notice that these are all energy losses over time; gradual processes with
perhaps an exponential time constant, but without cycles or periods. We
know not to apply Q in these scenarios.

But when you have an oscillator, or a resonator, or (as I suggest) a
"rotator", it seems to make sense to use Q to describe the normalized rate
of decay. So three keys to Q: you need energy; you need energy loss; you
need cycles over which that loss repeatedly occurs.

We use units of time (for example, SI seconds) when we describe a rate.
But here's why Q is unitless -- you normalize the energy (using E / dE)
*and* you also normalize the time (by cycle). No Joules. No seconds. So
having period is fundamental to Q. It's this unitless character of Q (in
both energy and time) that makes it portable from one branch of science to
another. And if you measure in radians you can even get rid of the 2*pi
factor ;-)

Without controversy, lots of articles define Q as 2*pi times {total
energy} / {energy lost per cycle}. To me, a slowly decaying spinning Earth
meets the three criteria. It appears to follow both the letter and the
spirit of Q.

Bob:

ummm…. Q is the general term of rate of energy loss and we just happen

to apply

it to oscillators in a very elegant fashion….


Oh, no. Now we have both quality factor and elegance factor!

/tvb
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Tom Van Baak]

Hi Jim.

> You said: "you need energy; you need energy loss; you need cycles over which 
> that loss repeatedly occurs."
> With regard to the earth, where is the first one?

By first one, do you mean where does the initial energy come from?

For a pendulum clock, you supply energy with a lift or a push. For a lift to 
the side, E = mgh, where h is the height above the base. For a push from 
center, E = 1/2 mv^2. Either way, it takes all the potential or kinetic E you 
provide and starts making time from there.

For a rotating clock, you just give it a twist. In this case, E = 1/2 Iw^2, 
where I is the moment of inertia and w (omega) is angular velocity. For earth 
the total E is 2.1e29 J. That's the energy number you want, yes?

> Sure it was there at the start when the solar system formed, but where is it 
> now?

I don't have data on where the initial swirl of solar system mass came from, or 
how much of that rotational energy went into our planet and its pesky moon, or 
Who or what gave that initial twist. The Q is pretty high so I assume you could 
work backwards, but I leave that to astronomers and cosmologists. I believe the 
2 ms/day / century estimate we use is one such measurement.

For more on earth rotation rate, UTC and leap seconds see 
https://www.ucolick.org/~sla/leapsecs/dutc.html

Surely in the literature there is a pile of information or speculation 
regarding all the rotational energy in the universe. It seems a common theme 
everywhere you look; maybe it was as much Big Twist as Big Bang? Perhaps in 
your Pulsar research you've run across some papers you could share. Off-list is 
ok, unless you think it has general time-nuts appeal. We're running the risk of 
spinning off-topic already.

Thanks,
/tvb

- Original Message - 
From: "Jim Palfreyman" <jim77...@gmail.com>
To: "Discussion of precise time and frequency measurement" <time-nuts@febo.com>
Sent: Sunday, July 31, 2016 7:34 PM
Subject: Re: [time-nuts] Q/noise of Earth as an oscillator


Hi Tom,

You said: "you need energy; you need energy loss; you need cycles over
which that loss repeatedly occurs."

With regard to the earth, where is the first one? Sure it was there at the
start when the solar system formed, but where is it now?

Jim


On 1 August 2016 at 12:16, Tom Van Baak <t...@leapsecond.com> wrote:

> Hal:
> > Is there a term other than Q that is used to describe the rate of energy
> loss
> > for things that aren't oscillators?
>
> Jim:
> > cooling (as in hot things)
> > discharge (as in capacitors and batteries)
> > leakage (as in pressure vessels)
> > loss
>
> Scott:
> > An irreversible process would be a better description versus energy loss.
> > Like joule heating (resistance, friction).
>
> Notice that these are all energy losses over time; gradual processes with
> perhaps an exponential time constant, but without cycles or periods. We
> know not to apply Q in these scenarios.
>
> But when you have an oscillator, or a resonator, or (as I suggest) a
> "rotator", it seems to make sense to use Q to describe the normalized rate
> of decay. So three keys to Q: you need energy; you need energy loss; you
> need cycles over which that loss repeatedly occurs.
>
> We use units of time (for example, SI seconds) when we describe a rate.
> But here's why Q is unitless -- you normalize the energy (using E / dE)
> *and* you also normalize the time (by cycle). No Joules. No seconds. So
> having period is fundamental to Q. It's this unitless character of Q (in
> both energy and time) that makes it portable from one branch of science to
> another. And if you measure in radians you can even get rid of the 2*pi
> factor ;-)
>
> Without controversy, lots of articles define Q as 2*pi times {total
> energy} / {energy lost per cycle}. To me, a slowly decaying spinning Earth
> meets the three criteria. It appears to follow both the letter and the
> spirit of Q.
>
> Bob:
> > ummm…. Q is the general term of rate of energy loss and we just happen
> to apply
> > it to oscillators in a very elegant fashion….
>
> Oh, no. Now we have both quality factor and elegance factor!
>
> /tvb
> ___
> time-nuts mailing list -- time-nuts@febo.com
> To unsubscribe, go to
> https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> and follow the instructions there.
>
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- Original Message - 
From: "Jim Palfreyman" <jim77...@gmail.com>
To: "Discussion of precise time and frequency measurement" <time-nuts@febo.com>
S

Re: [time-nuts] Q/noise of Earth as an oscillator

[Bill Byrom]

At the risk of boring everyone, here is an "Alice and Bob" thought
experiment concerning linear and circular movement:

Condition A: Let's say that Alice is in her spacecraft in inertial
straight line travel through nearly free space containing a thin gas
which creates a slight friction to movement. She is traveling at 628
meters per 24 hours adjacent to a distance scale with markers every 628
meters on the space highway. Passing a marker results in a tick. So she
can check her atomic clock every 24 hours at the tick. The thin gas
causes a slight reduction of her velocity (about a 2 millisecond
increase in tick interval every century).

Question: Does the movement of Alice exhibit Q? If so, how can we compute the 
value? If there a resonant frequency?

Condition B: Bob is in his spacecraft, which is identical to Alice's
model. A rod 200 meters long connects Alice's craft and Bob's craft. Bob
is moving at 628 meters per 24 hours in the opposite direction of Alice,
so that the midpoint of the rod is fixed and unmoving with regards to
the distant stars. Alice will return to the same point in space every 24
hours and pass a marker, generating a tick so she can check her atomic
clock once a day at the tick. Her linear motion has been constrained to
circular motion due to the stiff rod. The same thin gas is present,
resulting in a 2 millisecond increase in tick interval every century.

Question: Now does the movement of Alice exhibit Q? If so, how can we compute 
the value? If there a resonant frequency? 

Consider that in either Condition A or Condition B Alice (and Bob) can
increase or decrease their velocity to receive timing ticks at a
faster or slower rate. But there is no tendency for the velocity to
return to the one tick per day rate (628 meters per 24 hours), as
there would be with a harmonic oscillator. Due to Newton's First Law,
the velocity remains constant unless slowed by friction or affected by
external forces.

We could have started with a rod which was in the limit very long, so
the motion of Alice was only slightly diverted from a straight line
during a 24 hour interval. We could still measure the motion in 628
meter distance intervals on a circular or angular scale. Note that
nothing is fundamentally different if the rod is the exact length which
causes one rotation every 24 hours. The distance traveled and the
inertia resisting change in velocity is the same if the motion is linear
or circular, isn't it?
--
Bill Byrom N5BB


On Sun, Jul 31, 2016, at 10:27 PM, Bill Byrom wrote:
> I still claim that there is no natural frequency associated with the
> rotation of a body. The periodic nature of the rotating body motion is
> confusing you. The choice of a coordinate system is what is confusing.
>
> As I pointed out, what's the difference between an inertial
> body moving
> in a straight line and a rotating body? Let's say you take a
> body moving
> in free space with a small energy loss due to interactions with thin
> interstellar gas. You measure it on a ruler with marks every
> 628 meters
> (to use my example of a point on the Earth 100 meters from the
> pole as a
> comparison). That's the same as making an astronomical measurement on
> the Earth at a distant star which is overhead every 24 hours. In both
> cases the point moves 628 meters every 24 hours (measured with
> an atomic
> clock). We generate a tick when the Earth has rotated to the same
> relative position and when the body moving in a straight line reaches
> the next 628 meter mark. Both objects generate a tick every 24 hours,
> but the velocity is each case (angular or linear) is unconstrained by
> any periodic physical processes.
>
> What makes the rotating body (Earth) suitable for study as a harmonic
> oscillator with Q in this case? There is no energy transfer
> during each
> rotation. Should we establish a Q value for the body in free space
> straight line motion?  Both bodies have mass, inertia, a nearly
> constant
> velocity (linear and angular), and a slight loss. Each
> generates a tick
> every 24 hours (using the atomic clock as a reference). If we
> unwrap the
> polar coordinates and view the Earth rotation angle as increasing
> monotonically (or make marks every 628 meters on the scale
> measuring the
> free space body with straight line travel) they are identical.
>
> The geometry of the rotating body (Earth) is fooling you into thinking
> it's a periodic oscillator. Just because the position is similar after
> 24 hours doesn't mean anything, since there is no energy storage and
> transfer during each rotation. The Earth is reasonably symmetric (for
> this discussion), and it has no field which matters for this
> discussion
> which is rotating. It's just matter moving in a constrained circular
> fashion due to the geometry and constraints of a rigid body.
> Change the
> coordinate scale to linear (628 meters for each rotation at 100 meters
> from the axis) and compare it to the free space object moving in a
> straight 

Re: [time-nuts] Q/noise of Earth as an oscillator

[Hal Murray]

t...@radio.sent.com said:
> As I pointed out, what's the difference between an inertial body moving in a
> straight line and a rotating body?

The rotating body has a natural unit of time so there is a convenient way to 
make Q dimensionless.

For linear motion, the natural unit of decay would be the time constant.


-- 
These are my opinions.  I hate spam.



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Re: [time-nuts] Q/noise of Earth as an oscillator

[Bill Byrom]

I still claim that there is no natural frequency associated with the
rotation of a body. The periodic nature of the rotating body motion is
confusing you. The choice of a coordinate system is what is confusing.

As I pointed out, what's the difference between an inertial body moving
in a straight line and a rotating body? Let's say you take a body moving
in free space with a small energy loss due to interactions with thin
interstellar gas. You measure it on a ruler with marks every 628 meters
(to use my example of a point on the Earth 100 meters from the pole as a
comparison). That's the same as making an astronomical measurement on
the Earth at a distant star which is overhead every 24 hours. In both
cases the point moves 628 meters every 24 hours (measured with an atomic
clock). We generate a tick when the Earth has rotated to the same
relative position and when the body moving in a straight line reaches
the next 628 meter mark. Both objects generate a tick every 24 hours,
but the velocity is each case (angular or linear) is unconstrained by
any periodic physical processes.

What makes the rotating body (Earth) suitable for study as a harmonic
oscillator with Q in this case? There is no energy transfer during each
rotation. Should we establish a Q value for the body in free space
straight line motion?  Both bodies have mass, inertia, a nearly constant
velocity (linear and angular), and a slight loss. Each generates a tick
every 24 hours (using the atomic clock as a reference). If we unwrap the
polar coordinates and view the Earth rotation angle as increasing
monotonically (or make marks every 628 meters on the scale measuring the
free space body with straight line travel) they are identical.

The geometry of the rotating body (Earth) is fooling you into thinking
it's a periodic oscillator. Just because the position is similar after
24 hours doesn't mean anything, since there is no energy storage and
transfer during each rotation. The Earth is reasonably symmetric (for
this discussion), and it has no field which matters for this discussion
which is rotating. It's just matter moving in a constrained circular
fashion due to the geometry and constraints of a rigid body. Change the
coordinate scale to linear (628 meters for each rotation at 100 meters
from the axis) and compare it to the free space object moving in a
straight line. What's the difference?

--
Bill Byrom N5BB



On Sun, Jul 31, 2016, at 09:16 PM, Tom Van Baak wrote:
> Hal:
>> Is there a term other than Q that is used to describe the rate of
>> energy loss
>> for things that aren't oscillators?
>
> Jim:
>> cooling (as in hot things)
>> discharge (as in capacitors and batteries)
>> leakage (as in pressure vessels)
>> loss
>
> Scott:
>> An irreversible process would be a better description versus
>> energy loss.
>> Like joule heating (resistance, friction).
>
> Notice that these are all energy losses over time; gradual
> processes with
> perhaps an exponential time constant, but without cycles or
> periods. We
> know not to apply Q in these scenarios.
>
> But when you have an oscillator, or a resonator, or (as I suggest) a
> "rotator", it seems to make sense to use Q to describe the normalized
> rate of decay. So three keys to Q: you need energy; you need
> energy loss;
> you need cycles over which that loss repeatedly occurs.
>
> We use units of time (for example, SI seconds) when we describe a
> rate.
> But here's why Q is unitless -- you normalize the energy (using E /
> dE)
> **and** you also normalize the time (by cycle). No Joules. No
> seconds. So
> having period is fundamental to Q. It's this unitless character
> of Q (in
> both energy and time) that makes it portable from one branch of
> science
> to another. And if you measure in radians you can even get rid of the
> 2*pi factor ;-)
>
> Without controversy, lots of articles define Q as 2*pi times {total
> energy} / {energy lost per cycle}. To me, a slowly decaying spinning
> Earth meets the three criteria. It appears to follow both the
> letter and
> the spirit of Q.
>
> Bob:
>> ummm…. Q is the general term of rate of energy loss and we just
>> happen to apply
>> it to oscillators in a very elegant fashion….
>
> Oh, no. Now we have both quality factor and elegance factor!
>
> /tvb
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Jim Palfreyman]

Hi Tom,

You said: "you need energy; you need energy loss; you need cycles over
which that loss repeatedly occurs."

With regard to the earth, where is the first one? Sure it was there at the
start when the solar system formed, but where is it now?

Jim


On 1 August 2016 at 12:16, Tom Van Baak  wrote:

> Hal:
> > Is there a term other than Q that is used to describe the rate of energy
> loss
> > for things that aren't oscillators?
>
> Jim:
> > cooling (as in hot things)
> > discharge (as in capacitors and batteries)
> > leakage (as in pressure vessels)
> > loss
>
> Scott:
> > An irreversible process would be a better description versus energy loss.
> > Like joule heating (resistance, friction).
>
> Notice that these are all energy losses over time; gradual processes with
> perhaps an exponential time constant, but without cycles or periods. We
> know not to apply Q in these scenarios.
>
> But when you have an oscillator, or a resonator, or (as I suggest) a
> "rotator", it seems to make sense to use Q to describe the normalized rate
> of decay. So three keys to Q: you need energy; you need energy loss; you
> need cycles over which that loss repeatedly occurs.
>
> We use units of time (for example, SI seconds) when we describe a rate.
> But here's why Q is unitless -- you normalize the energy (using E / dE)
> *and* you also normalize the time (by cycle). No Joules. No seconds. So
> having period is fundamental to Q. It's this unitless character of Q (in
> both energy and time) that makes it portable from one branch of science to
> another. And if you measure in radians you can even get rid of the 2*pi
> factor ;-)
>
> Without controversy, lots of articles define Q as 2*pi times {total
> energy} / {energy lost per cycle}. To me, a slowly decaying spinning Earth
> meets the three criteria. It appears to follow both the letter and the
> spirit of Q.
>
> Bob:
> > ummm…. Q is the general term of rate of energy loss and we just happen
> to apply
> > it to oscillators in a very elegant fashion….
>
> Oh, no. Now we have both quality factor and elegance factor!
>
> /tvb
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Tom Van Baak]

Hal:
> Is there a term other than Q that is used to describe the rate of energy loss 
> for things that aren't oscillators?

Jim:
> cooling (as in hot things)
> discharge (as in capacitors and batteries)
> leakage (as in pressure vessels)
> loss

Scott:
> An irreversible process would be a better description versus energy loss.
> Like joule heating (resistance, friction).

Notice that these are all energy losses over time; gradual processes with 
perhaps an exponential time constant, but without cycles or periods. We know 
not to apply Q in these scenarios.

But when you have an oscillator, or a resonator, or (as I suggest) a "rotator", 
it seems to make sense to use Q to describe the normalized rate of decay. So 
three keys to Q: you need energy; you need energy loss; you need cycles over 
which that loss repeatedly occurs.

We use units of time (for example, SI seconds) when we describe a rate. But 
here's why Q is unitless -- you normalize the energy (using E / dE) *and* you 
also normalize the time (by cycle). No Joules. No seconds. So having period is 
fundamental to Q. It's this unitless character of Q (in both energy and time) 
that makes it portable from one branch of science to another. And if you 
measure in radians you can even get rid of the 2*pi factor ;-)

Without controversy, lots of articles define Q as 2*pi times {total energy} / 
{energy lost per cycle}. To me, a slowly decaying spinning Earth meets the 
three criteria. It appears to follow both the letter and the spirit of Q.

Bob:
> ummm…. Q is the general term of rate of energy loss and we just happen to 
> apply 
> it to oscillators in a very elegant fashion….

Oh, no. Now we have both quality factor and elegance factor!

/tvb
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Gabs Ricalde]

On Sat, Jul 30, 2016 at 1:19 AM, Tom Van Baak wrote:
> The remaining question in this thread is if earth Q measurement has actual 
> meaning, that is, if the concept of Q is valid for a slowly decaying rotating 
> object, as it is for a slowly decaying simple harmonic oscillator. And that's 
> were get into the history and definition(s) and applicability of Q to non 
> harmonic oscillators, such as coils, capacitors, atomic clocks, planets, 
> pulsars, etc.
>
> /tvb
>

On Mon, Aug 1, 2016 at 6:50 AM, Bill Byrom wrote:
> My final argument is that the rotation frequency of the Earth is
> affected by tidal friction, but the amplitude of the motion of that
> point 100 meters from the axis is unaffected. The amplitude of a
> harmonic oscillator is directly affected by friction or other losses,
> but the effect on resonant frequency is tiny. So loss effects frequency
> in one situation and amplitude in the other. How can Q relate to both
> situations?
> --
> Bill Byrom N5BB
>

I'll try to answer both. Apologies for mistakes or shortcuts, I will
correct or provide details in future posts if needed.
If we consider this definition:

Q = 2*pi * energy stored / energy lost per cycle,

we can write it as

Q = 2*pi * E / (-dE/d(cycle))
  = E / (-dE/d(theta))
 => E = E0*exp(-theta/Q).

The energy decays exponentially if the frequency is constant. The
impulse response of an RLC circuit is an exponentially decaying sinusoid
with a fixed frequency, so Q naturally describes the decay of an RLC
circuit and other resonant systems that behave similarly.

Suppose we want to use Q for a rotating object that gradually slows
down. When a rotating object loses energy, the period increases. From
the definition, the energy of a rotating object with constant Q decays
exponentially when we are counting cycles. But each cycle gets stretched
over time, so the energy decay is slower than exponential over time.

Omitting the derivation, a rotating object with constant Q behaves like
this:

E0 = energy at t = 0
I = moment of inertia
k1 = sqrt(E0/I)
k2 = E0*I
theta = 2*Q*log(k1*t/(Q*sqrt(2)) + 1) rad
omega = k1*2*Q / (k1*t + Q*sqrt(2)) rad/s(1)
energy = k2*Q^2 / (E0/2*t^2 + sqrt(2*k2)*Q*t + I*Q^2)



Consider the Earth with an LOD increase of 2 ms/century (6.338e-13 s/s).
Suppose LOD = 0 at t = 0. Then

k3 = 6.338e-13/(86400 s)
omega0 = 72921151.467064e-12 rad/s
omega = omega0 * (1 - t*k3) rad/s.(2)
( from http://hpiers.obspm.fr/eop-pc/earthor/ut1lod/UT1.html )

The frequency linearly decreases over time, which is different from the
angular frequency of a rotating object with constant Q (1). Again
omitting the derivation, the changing Q of the Earth over time is

Q = omega0 * (1 - t*k3)^2 / (2*k3).(3)

The Q is around 4.97e12 and decreases very slowly to 4.74e12 after 100
million years, assuming a constant 2 ms/century for that duration.



If the period linearly increases over time, the angular frequency is

omega = 2*pi / (T0 + k4*t) rad/s(4)

where T0 is the period at t = 0 and k4 is the change in period per
second. This looks like the angular frequency with constant Q (1). Let's
make (4) look like (1):

omega = k1*2*(pi/k4) / (k1*t + T0*k1/k4)

so we set Q = pi/k4 and set k4 equal to the change in the period of (2)
at t = 0. Then we get

Q = omega0 / (2*k3)(5)

which is the same as (3) when t = 0. If we approximate omega0 =
2*pi/86400 (1 cycle = 1 solar day), (5) is the same as tvb's formula
pi * (86400 * 365 * 100 / 0.002).



To recap:
1. An oscillation with fixed frequency and exponential decay has constant Q.
2. A rotating object with linearly decreasing frequency (2) has
decreasing Q (3).
3. A rotating object with linearly increasing period (4) has constant Q (5).
4. Over short time scales or when Q is very large, (3) is almost equal to (5).
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Tom Van Baak]

Hal Murray wrote:
> It's energy loss in both cases.
> 
> Is there a term other than Q that is used to describe the rate of energy loss 
> for things that aren't oscillators?

I've seen "energy dissipation" or "energy decrement".

Before Q was used to describe line width of atomic and optical clocks,
before Q was used to describe performance of simple harmonic oscillators and 
quartz crystals,
before or while Q was used to describe the ratio of reactance to resistance in 
a coil,
horologists that worked with precision pendulum clocks needed a way to 
characterize the amount of energy lost per period.

This was very important because the energy lost each swing (due to all forms of 
friction) had to be replaced in order to keep the pendulum oscillator running. 
And the timekeeping performance seemed to be related to how small or how 
consistent this energy was. I'm sure "decrement" was used before then, but I'm 
not well-read enough in science history to have any idea. Still...

What I do know is this abstract from a 1938 paper [1], written by an early time 
nut:

THE DISSIPATION OF ENERGY BY A PENDULUM SWINGING IN AIR, By E. C. ATKINSON
--
ABSTRACT. The decrement of a pendulum falls slowly with the amplitude: 
hence the
need for determinations based on small changes of angle. The resulting 
errors of observation
lead to erratic values but not to systematic error. The result of 
measurements with a
seconds pendulum enclosed in a case is shown by a smoothed curve, the 
departure from
observed times being expressed by smoothing fractions, and a smoothing 
figure is a
measure of this departure for the whole or part of the experiment. From the 
decrement
the rate of loss of energy is calculated. This 7 kg. pendulum with 
amplitude 53' dissipates
a Board of Trade Unit (which serves a 70 w. lamp for 14 hours) in rather 
over 100,000
years. Experiments with different pendulums are described by which the 
component
losses due to suspension, rod, and bob are found. Suspension springs made 
from thin
strip clamped in chaps dissipate large and variable amounts of energy 
compared with
springs made from thick strip ground thin in the middle. The variable 
losses are associated
with variable rates of the pendulum. The cylindrical case adds considerably 
to the air
resistance. The measured loss due to a gravity impulse lever is little in 
excess of the
computed loss from collision with the pendulum: for a seconds pendulum 
1/2000 part of
the free pendulum loss.
--

This article is also a favorite of mine because it talks about a "root mean 
square of a series of such fractions may be called the smoothing figure for the 
series, and its smallness is an indication of the confidence which may be 
placed in the result." -- a 1938 precursor of the two-sample variance, or Allan 
deviation.

/tvb

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Re: [time-nuts] Q/noise of Earth as an oscillator

[Bob Camp]

Hi

> On Jul 31, 2016, at 8:19 PM, Hal Murray  wrote:
> 
> 
> t...@radio.sent.com said:
>> So loss effects frequency in one situation and amplitude in the other. How
>> can Q relate to both situations? 
> 
> It's energy loss in both cases.
> 
> Is there a term other than Q that is used to describe the rate of energy loss 
> for things that aren't oscillators?

ummm…. Q is the general term of rate of energy loss and we just happen to apply 
it to oscillators in a very elegant fashion….

Bob

> 
> 
> -- 
> These are my opinions.  I hate spam.
> 
> 
> 
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Scott Stobbe]

An irreversible process would be a better description versus energy loss.
Like joule heating (resistance, friction).

On Sunday, 31 July 2016, Hal Murray  wrote:

>
> t...@radio.sent.com  said:
> > So loss effects frequency in one situation and amplitude in the other.
> How
> > can Q relate to both situations?
>
> It's energy loss in both cases.
>
> Is there a term other than Q that is used to describe the rate of energy
> loss
> for things that aren't oscillators?
>
>
> --
> These are my opinions.  I hate spam.
>
>
>
> ___
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Re: [time-nuts] Q/noise of Earth as an oscillator

[jimlux]

On 7/31/16 5:19 PM, Hal Murray wrote:


t...@radio.sent.com said:

So loss effects frequency in one situation and amplitude in the other. How
can Q relate to both situations?


It's energy loss in both cases.

Is there a term other than Q that is used to describe the rate of energy loss
for things that aren't oscillators?


cooling (as in hot things)
discharge (as in capacitors and batteries)
leakage (as in pressure vessels)
loss





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Re: [time-nuts] Q/noise of Earth as an oscillator

[Hal Murray]

t...@radio.sent.com said:
> So loss effects frequency in one situation and amplitude in the other. How
> can Q relate to both situations? 

It's energy loss in both cases.

Is there a term other than Q that is used to describe the rate of energy loss 
for things that aren't oscillators?


-- 
These are my opinions.  I hate spam.



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Re: [time-nuts] Q/noise of Earth as an oscillator

[Tom Van Baak]

Ron Ott wrote:
> There might be two Qs: one relating to the axil rotation and another 
> concerning the volume behavior

Hi Ron, Chris, and now also Bill,

I was thinking this tangent wouldn't come up, but yes, in the fields of 
Seismology or Earth Science, you will also see "quality factor" and the letter 
Q used. In their case it refers to the attenuation of seismic waves traversing 
through the earth and bouncing back. It's a clever way to explore the 
composition of the earth, from the inner core outwards. It's an ironic (in the 
Fe sense) way to make large earthquakes a wonderful tool of science in addition 
to a dreadful threat to property and life.

Here's a few links that mention this type of Earth Q:

"Deep Earth Structure – Q of the Earth from Crust to Core"
in Treatise on Geophysics, Seismology and structure of the Earth, 2008
http://indico.ictp.it/event/a07174/session/116/contribution/64/material/0/0.pdf

"Anisotropy of Earth’s inner core intrinsic attenuation from seismic normal 
mode models"
in Earth and Planetary Science Letters, 2014
http://www.geo.uu.nl/~seismain/pdf/Earth_Planet_Sc_Lett_2014_Makinen.pdf

"Wide-band coupling of Earth’s normal modes due to anisotropic inner core 
structure"
in Geophysical Journal International, 2008
https://www.princeton.edu/geosciences/people/irving/publications/pdfs/Irving-2008-GJI-WideBand.pdf

"Tidal dissipation compared to seismic dissipation: in small bodies, in earths, 
and in superearths"
in The Astrophysical Journal, 2012
https://arxiv.org/pdf/1105.3936.pdf


Any you're right. The "vibrating planet jello" Q is unrelated to the "rotating 
planet timekeeping" Q that I mentioned. The jello Q is a couple of hundred. The 
rotation/clock Q is a couple of trillion. That's why we define the second from 
the rotation of the earth and not the sound of the earth.

Just in case readers think there can't ever be more than one Q, I refer you to 
pendulum clocks. The main Q, the one that is related to timekeeping, is derived 
from the periodic decay of the swing of the pendulum. Especially for precision 
pendulum clocks, there are other Q's as well: the rod/bob combination lends 
itself to many unwanted modes of physical vibration, up/down, front/back, 
left/right, twist, "violin modes", etc. Each of these modes have amplitude, 
period, and decay. They all interact with each other and with the main swinging 
of the pendulum in nasty ways. And yes, they all have their own Q.

/tvb

- Original Message - 
From: "Ron Ott" <ron...@sbcglobal.net>
To: <ch...@chriscaudle.org>; "Discussion of precise time and frequency 
measurement" <time-nuts@febo.com>
Sent: Wednesday, July 27, 2016 9:57 AM
Subject: Re: [time-nuts] Q/noise of Earth as an oscillator


There might be two Qs: one relating to the axil rotation and another concerning 
the volume behavior of the earth as a giant bowl of Jello. But you'd have to 
figure out how to really slam the planet to excite the entire volume. 
Earthquakes are probably too wimpy.
Ron


      From: Chris Caudle <ch...@chriscaudle.org>
 To: time-nuts@febo.com 
 Sent: Wednesday, July 27, 2016 8:50 AM
 Subject: Re: [time-nuts] Q/noise of Earth as an oscillator
   
On Wed, July 27, 2016 10:33 am, Chris Caudle wrote:
> Does that imply that this value is not constant:
>>> And if you take the classic definition
>>> Q = 2 pi * total energy /energy lost per cycle
>>> then it would seem earth has a Q factor.

After re-reading "The Story of Q" I agree that Q of a rotating body could
be non-constant, but also consistent with the original definition of Q as
the ratio of reactance to resistance of an inductor, which of course would
vary almost completely linearly over a wide frequency range where the
resistive dissipation was not frequency dependent (i.e. where skin effect
was negligible).

Perhaps a more useful question is whether that is still a useful
definition compared to how the term is more typically used now to refer to
resonance bandwidth.

-- 
Chris Caudle



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Re: [time-nuts] Q/noise of Earth as an oscillator

[Bill Byrom]

 inertia at a
specific moment in our history, and is not controlled by physical
properties (density, Young's modulus, mass, etc.).

The situation is completely different for parameters of objects which
exhibit Q. The Earth has a resonant vibration frequency due to material
properties, and this causes a characteristic Q due to material
properties (as described in the paper referenced above). But this has
only a tiny effect on the rotation of the Earth, which can speed up or
slow down largely irrespective of the Q. A harmonic oscillator is
constrained to a specific resonant frequency, which changes only
slightly with material properties sensitive to temperature and other
effect. If you add energy to a harmonic oscillator the amplitude of
oscillation increases but not the frequency (which is constrained by the
Q to be centered on a natural resonance frequency).

My final argument is that the rotation frequency of the Earth is
affected by tidal friction, but the amplitude of the motion of that
point 100 meters from the axis is unaffected. The amplitude of a
harmonic oscillator is directly affected by friction or other losses,
but the effect on resonant frequency is tiny. So loss effects frequency
in one situation and amplitude in the other. How can Q relate to both
situations?
--
Bill Byrom N5BB


On Wed, Jul 27, 2016, at 08:42 AM, Tom Van Baak wrote:
> Hi Michael,
>
> I sympathize with both your and Attila's comments and would
> like to dig
> deeper for the truth on this.
>
> Clearly both the earth and a pendulum (and many other periodic
> systems)
> exhibit a decay of energy, when you remove the periodic
> restoring force.
> And if you take the classic definition Q = 2 pi * total energy
> / energy
> lost per cycle then it would seem earth has a Q factor.
>
> In fact, if you use real energy numbers you get:
>
> - total rotational energy of earth is 2.14e29 J
> - energy lost per cycle (day) is 2.7e17 J
> - so Q = 2pi * 2.14e29 / 2.7e17 = 5e12, the same 5 trillion as my
>   earlier
> calculation.
>
> But your point about resonance is a good one and it has always
> intrigued
> me. Is this one difference between a pendulum and the earth as
> timekeepers?
>
> On the other hand, if you swept the earth with an external powerful
> frequency in the range well below to well above 1.16e-5 Hz (1/86164 s)
> would you not see a resonance peak right at the center? Given
> the mass of
> the planet and its pre-existing rotational energy, it seems like
> there is
> a "resonance", a preference to remain at its current frequency. Plus
> it
> has a slow decay due to internal friction. This sounds like any other
> timing system with Q to me.
>
> Or imagine a planet the same size as earth made from a Mylar balloon.
> Much less mass. Give it the same rotational speed. Much easier to
> increase or decrease its energy by applying external force. Far lower
> Q
> than earth, yes?
>
> It might also be useful at this point, to:
>
> read the history Q and its definition:
> http://www.collinsaudio.com/Prosound_Workshop/The_story_of_Q.pdf
>
> and read the patent in which Q first appeared:
> http://leapsecond.com/pages/Q/1927-US1628983.pdf
>
> or view the first paragraph in which Q appeared:
> http://leapsecond.com/pages/Q/1927-Q-patent-600x300.gif
>
> /tvb
>
> - Original Message -----
> From: "Michael Wouters" <michaeljwout...@gmail.com>
> To: "Discussion of precise time and frequency measurement"
> <time-nuts@febo.com>; <att...@kinali.ch>
> Sent: Wednesday, July 27, 2016 5:43 AM
> Subject: Re: [time-nuts] Q/noise of Earth as an oscillator
>
>
>> On Wed, Jul 27, 2016 at 8:08 AM, Attila Kinali
>> <att...@kinali.ch> wrote:
>>
>> "I am not sure you can apply this definition of Q onto earth."
>>
>> It  doesn't make sense to me either.
>>
>> If you mark a point on the surface of a sphere then you can observe
>> that point as the sphere
>> rotates and count rotations to make a clock. If you think of just a
>> circle, then a point on it viewed in a rectilinear coordinate system
>> executes simple harmonic motion so the motion of that point
>> looks like
>> an oscillator, so that much is OK.
>>
>> But unlike the LCR circuit, the pendulum and quartz crystal, the
>> sphere's rotational motion does not have a
>> resonant frequency. Another way of characterizing the Q of an
>> oscillator, the relative width of the resonance, makes
>> no sense in this context.
>>
>> It seems to me that the model of the earth as an oscillator is
>> misapplied and that the 'Q' is not a meaningful number.
>> I think the confusion arises here because of a conflation of a

Re: [time-nuts] Q/noise of Earth as an oscillator

[Scott Stobbe]

Thanks Tom, I would agree LIGOs efforts are beyond heroic, I will try to
find some of their phase noise plots.

Regarding Q of the earth, I would agree one could compute an UNloaded Q for
the earth as if it were a mass element in some form of a mechanical
oscillator. The first sticky point is which Q, a loaded Q (QL) would assume
it's oscillating (which as many have outlined) it's not, an unloaded Q
(1/DF, or X/R) would be a reasonable value to commute for some specific
frequency. It is interesting (although expected) I have arrived to similar
Q as you, though through different reasoning, but similar assumptions.

Taking inspiration from the many brilliant controls and analog Engineers
from the analog computing days, we can create a circuit equivalent model
for the rotational dynamics of the earth.

Some physical parameters:

 omega_e = 7.3E-5 rad/s (angular rate of the earth)
 alpha_e = -6.3E-22 rad/s^2 (angular deceleration of the earth)
 J_e = 8E37 kg m^2(Inertia of the earth)

Mapping the torque-(angular displacement) space to the volt-coulomb space

Capacitor - Torsional Spring
Q = C V,  tau = k theta

Resistor - Damper / Friction
Qdot = V/R,tau = B thetadot

Inductor - Mass
Qdotdot = V/L,tau = J thetadotdot

In this circuit equivalent model the inertia of the earth would be
represented by an inductor of inductance
L = 8E37 kg m^2

An approximate rotational friction coefficient (tidal friction, doubt it's
a first order relation, but for the sake of Q, assume it is, and other
losses) can be found from the net angular deceleration.
B = J * |alpha_e|/omega_e = 6.9E20 Nm s

Solving as an equivalent resistance yields,
Rs = 6.9E20 Nm s

Finally the unloaded Q at 11 and change uHz,
Q = XL/Rs = (7.3E-5)(8E37)/(6.9E20) = 8.5E12

8.5 Trillion, not bad for an inductor at 11 uHz... Now if you really wanted
an 11 uHz oscillator you could ram a torsional spring up the earth's south
pole.

>From a circuit perspective, the earth's rotation looks like a monster near
superconducting inductor that at some point and somehow was precharged to a
current Io, and then had its terminals crowbarred. Our solar time is like
watching a reference electron run round and round a coil.

Björn and Dave, thanks for the gyro reference I will take a look.

On Fri, Jul 29, 2016 at 1:19 PM, Tom Van Baak  wrote:

> Scott Stobbe wrote:
> > I believe a phase noise plot deep into the uHz or lower would apply to
> the
> > rotation rate of the earth.
>
> Yup. You'll see lots of uHz to Hz noise plots by people working with
> seismic noise, for example. My introduction to the subject were the many
> plots and papers that describe the heroic effort LIGO goes through to
> measure tidal and seismic noise in order to keep their gravity wave servos
> locked. Short-term, the surface of the earth is a very noisy place. But if
> you can model or measure it before it hits the mirrors, you can mostly back
> it out.
>
> One can use PN and ADEV statistics on earth rotation, just like any other
> clock or oscillator. And it seems we can also compute Q for the earth, as
> if it were a mechanical oscillator.
>
> The remaining question in this thread is if earth Q measurement has actual
> meaning, that is, if the concept of Q is valid for a slowly decaying
> rotating object, as it is for a slowly decaying simple harmonic oscillator.
> And that's were get into the history and definition(s) and applicability of
> Q to non harmonic oscillators, such as coils, capacitors, atomic clocks,
> planets, pulsars, etc.
>
> /tvb
>
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Jim Palfreyman]

> What about an ADEV/TDEV plot of the pulsar J0437-4715?

Very boring. It's a straight line from top left to bottom right. :-)

See page 5 of this: http://arxiv.org/pdf/1004.0115.pdf

Jim Palfreyman


On 29 July 2016 at 17:33, Azelio Boriani  wrote:

> What about an ADEV/TDEV plot of the pulsar J0437-4715?
>
> On Fri, Jul 29, 2016 at 12:39 AM, Jim Palfreyman 
> wrote:
> > Hi All,
> >
> > Tom gave me a nudge to look here - I hadn't been following this thread.
> >
> > For those that don't know, I study pulsars and so the way we measure what
> > pulsars do could be relevant to this discussion.
> >
> > First, I have never heard of a Q measure when referencing a pulsar. I
> think
> > the key here is that it's not resonating as such. Rotating yes,
> resonating
> > no.
> >
> > Pulsars spin and slow down due to giving off energy (magnetic dipole
> > radiation). So in the pulsar world we mainly refer to spin frequency (F0)
> > and frequency derivative (F1). With some of the younger and more
> "erratic"
> > pulsars, F2 (and further) can be modelled.
> >
> > Here's some data on the Vela pulsar (hot off the presses - measured just
> > now):
> >
> > F0  11.1867488542579
> > F1  -1.55859177352837e-11
> > F2  1.23776878287221e-21
> >
> > Vela is young and erratic. Millisecond pulsars are outstanding clocks.
> > Here's the data for J0437-4715 - one of the most stable pulsars we know
> > about:
> >
> > F0  173.6879458121843
> > F1  -1.728361E-15
> >
> > I'm sure the "Q" of Vela would be pretty decent - but I can tell you now,
> > as a time-keeper, she's useless.
> >
> >
> > Jim Palfreyman
> >
> >
> >
> > On 28 July 2016 at 20:50, Tony Finch  wrote:
> >
> >> Neville Michie  wrote:
> >>
> >> > The conical pendulum has a simple form of a weight on a string,
> instead
> >> > of oscillating in one plane as a conventional pendulum, it swings
> around
> >> > in a circular orbit in the horizontal plane. It has a definite
> resonant
> >> > frequency.
> >>
> >> I don't think it does have a resonant frequency, any more than the Earth
> >> does: the angular velocity of the pendulum is sqrt(g/h) where h is the
> >> height of the pendulum; give it more energy, it swings higher, so h is
> >> smaller, so the frequency is higher.
> >>
> >> Tony.
> >> --
> >> f.anthony.n.finch    http://dotat.at/  -  I xn--zr8h
> >> punycode
> >> South Thames, Dover: Southwesterly 5 or 6. Slight or moderate. Rain or
> >> showers. Good, occasionally poor.
> >> ___
> >> time-nuts mailing list -- time-nuts@febo.com
> >> To unsubscribe, go to
> >> https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> >> and follow the instructions there.
> >>
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> > To unsubscribe, go to
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> > and follow the instructions there.
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Dave Brown]

The main issue has been deemed to be safety- or lack of it, due to the ring 
lasers location in an underground cavern that cannot continue to be used 
because of a high risk of additional ground/rock failure. For the present 
there has been no further ring laser work at canterbury university since the 
2010/2011 quakes and cave access is not permitted.

DaveB,
Christchurch, NZ


- Original Message - 
From: ""Björn Gabrielsson"" <b...@lysator.liu.se>
To: "Discussion of precise time and frequency measurement" 
<time-nuts@febo.com>

Sent: Saturday, July 30, 2016 8:48 AM
Subject: Re: [time-nuts] Q/noise of Earth as an oscillator



I'am  not sure how the big ring lasers have progressed over the past
years. It seems the big New Zeeland earthquake messed up the nice ring
lasers over there.

http://www.fs.wettzell.de/LKREISEL/G/LaserGyros.html
http://www.phys.canterbury.ac.nz/ringlaser/about_us.shtml

--

   Björn

I believe a phase noise plot deep into the uHz or lower would apply to 
the

rotation rate of the earth.

On Saturday, 23 July 2016, Hal Murray <hmur...@megapathdsl.net> wrote:



t...@leapsecond.com said:
> Earth is a very noisy, wandering, drifting,
incredibly-expensive-to-measure,
> low-precision (though high-Q) clock.

What is the Q of the Earth?  It might be on one of your web pages, but I
don't remember seeing it.  Google found a few mentions, but I didn't
find a
number.

I did find an interesting list of damping mechanisms in a geology book.
Geology-nuts are as nutty as time-nuts.  Many were discussing damping of
seismic waves rather than rotation.

I've seen mention that the rotation rate of the Earth changed by a few
microseconds per day as a result of the 2011 earthquake in Japan.  Does
that
show up in any data?  Your recent graph doesn't go back that far and
it's
got
a full scale of 2000 microseconds so a few is going to be hard to see.



--
These are my opinions.  I hate spam.



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Re: [time-nuts] Q/noise of Earth as an oscillator

[Björn Gabrielsson]

I'am  not sure how the big ring lasers have progressed over the past
years. It seems the big New Zeeland earthquake messed up the nice ring
lasers over there.

http://www.fs.wettzell.de/LKREISEL/G/LaserGyros.html
http://www.phys.canterbury.ac.nz/ringlaser/about_us.shtml

--

Björn

> I believe a phase noise plot deep into the uHz or lower would apply to the
> rotation rate of the earth.
>
> On Saturday, 23 July 2016, Hal Murray  wrote:
>
>>
>> t...@leapsecond.com said:
>> > Earth is a very noisy, wandering, drifting,
>> incredibly-expensive-to-measure,
>> > low-precision (though high-Q) clock.
>>
>> What is the Q of the Earth?  It might be on one of your web pages, but I
>> don't remember seeing it.  Google found a few mentions, but I didn't
>> find a
>> number.
>>
>> I did find an interesting list of damping mechanisms in a geology book.
>> Geology-nuts are as nutty as time-nuts.  Many were discussing damping of
>> seismic waves rather than rotation.
>>
>> I've seen mention that the rotation rate of the Earth changed by a few
>> microseconds per day as a result of the 2011 earthquake in Japan.  Does
>> that
>> show up in any data?  Your recent graph doesn't go back that far and
>> it's
>> got
>> a full scale of 2000 microseconds so a few is going to be hard to see.
>>
>>
>>
>> --
>> These are my opinions.  I hate spam.
>>
>>
>>
>> ___
>> time-nuts mailing list -- time-nuts@febo.com 
>> To unsubscribe, go to
>> https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
>> and follow the instructions there.
>>
> ___
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> https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> and follow the instructions there.
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Tom Van Baak]

Scott Stobbe wrote:
> I believe a phase noise plot deep into the uHz or lower would apply to the
> rotation rate of the earth.

Yup. You'll see lots of uHz to Hz noise plots by people working with seismic 
noise, for example. My introduction to the subject were the many plots and 
papers that describe the heroic effort LIGO goes through to measure tidal and 
seismic noise in order to keep their gravity wave servos locked. Short-term, 
the surface of the earth is a very noisy place. But if you can model or measure 
it before it hits the mirrors, you can mostly back it out.

One can use PN and ADEV statistics on earth rotation, just like any other clock 
or oscillator. And it seems we can also compute Q for the earth, as if it were 
a mechanical oscillator.

The remaining question in this thread is if earth Q measurement has actual 
meaning, that is, if the concept of Q is valid for a slowly decaying rotating 
object, as it is for a slowly decaying simple harmonic oscillator. And that's 
were get into the history and definition(s) and applicability of Q to non 
harmonic oscillators, such as coils, capacitors, atomic clocks, planets, 
pulsars, etc.

/tvb

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Re: [time-nuts] Q/noise of Earth as an oscillator

[Alex Pummer]

the Q factor could be derived from the modulation bandwidth of an 
oscillator [ the "old way" of measuring the Q of resonator of the 
running oscillator's ], therefore if we look the fluctuation spectrum of 
the frequency of an oscillator we could determine the Q. Any circular 
movement could be seen as the source of a harmonic oscillation.


73
KJ6UHN
Alex



On 7/29/2016 9:28 AM, Attila Kinali wrote:

On Fri, 29 Jul 2016 03:29:27 -0500
David  wrote:


Capacitors and inductors have an associated Q while lacking a resonate
frequency except for parasitic elements.  Their Q increases with
frequency up to a point; does that apply to a spinning body?  I guess
it depends on the loss mechanism.

The Q of an inductor (or capacitor) is defined at a specific frequency.
You can see it as the Q factor that would be achieved, if the inductor
(capacitor) would be paired up with an ideal capacitor (inductor) with
a value such, that it would result in the specified frequency.

Hence, if you increase the frequency, the Q factor increases for an inductor. 
Conversly, the Q factor of an capacitor decreases with increasing  frequency.

See also:
https://en.wikipedia.org/wiki/Inductor#Q_factor
https://en.wikipedia.org/wiki/Capacitor#Q_factor


Attila Kinali



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Re: [time-nuts] Q/noise of Earth as an oscillator

[Scott Stobbe]

I believe a phase noise plot deep into the uHz or lower would apply to the
rotation rate of the earth.

On Saturday, 23 July 2016, Hal Murray  wrote:

>
> t...@leapsecond.com said:
> > Earth is a very noisy, wandering, drifting,
> incredibly-expensive-to-measure,
> > low-precision (though high-Q) clock.
>
> What is the Q of the Earth?  It might be on one of your web pages, but I
> don't remember seeing it.  Google found a few mentions, but I didn't find a
> number.
>
> I did find an interesting list of damping mechanisms in a geology book.
> Geology-nuts are as nutty as time-nuts.  Many were discussing damping of
> seismic waves rather than rotation.
>
> I've seen mention that the rotation rate of the Earth changed by a few
> microseconds per day as a result of the 2011 earthquake in Japan.  Does
> that
> show up in any data?  Your recent graph doesn't go back that far and it's
> got
> a full scale of 2000 microseconds so a few is going to be hard to see.
>
>
>
> --
> These are my opinions.  I hate spam.
>
>
>
> ___
> time-nuts mailing list -- time-nuts@febo.com 
> To unsubscribe, go to
> https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> and follow the instructions there.
>
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Attila Kinali]

On Fri, 29 Jul 2016 03:29:27 -0500
David  wrote:

> Capacitors and inductors have an associated Q while lacking a resonate
> frequency except for parasitic elements.  Their Q increases with
> frequency up to a point; does that apply to a spinning body?  I guess
> it depends on the loss mechanism.

The Q of an inductor (or capacitor) is defined at a specific frequency.
You can see it as the Q factor that would be achieved, if the inductor
(capacitor) would be paired up with an ideal capacitor (inductor) with
a value such, that it would result in the specified frequency.

Hence, if you increase the frequency, the Q factor increases for an inductor. 
Conversly, the Q factor of an capacitor decreases with increasing  frequency.

See also:
https://en.wikipedia.org/wiki/Inductor#Q_factor
https://en.wikipedia.org/wiki/Capacitor#Q_factor


Attila Kinali

-- 
It is upon moral qualities that a society is ultimately founded. All 
the prosperity and technological sophistication in the world is of no 
use without that foundation.
 -- Miss Matheson, The Diamond Age, Neil Stephenson
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Azelio Boriani]

What about an ADEV/TDEV plot of the pulsar J0437-4715?

On Fri, Jul 29, 2016 at 12:39 AM, Jim Palfreyman  wrote:
> Hi All,
>
> Tom gave me a nudge to look here - I hadn't been following this thread.
>
> For those that don't know, I study pulsars and so the way we measure what
> pulsars do could be relevant to this discussion.
>
> First, I have never heard of a Q measure when referencing a pulsar. I think
> the key here is that it's not resonating as such. Rotating yes, resonating
> no.
>
> Pulsars spin and slow down due to giving off energy (magnetic dipole
> radiation). So in the pulsar world we mainly refer to spin frequency (F0)
> and frequency derivative (F1). With some of the younger and more "erratic"
> pulsars, F2 (and further) can be modelled.
>
> Here's some data on the Vela pulsar (hot off the presses - measured just
> now):
>
> F0  11.1867488542579
> F1  -1.55859177352837e-11
> F2  1.23776878287221e-21
>
> Vela is young and erratic. Millisecond pulsars are outstanding clocks.
> Here's the data for J0437-4715 - one of the most stable pulsars we know
> about:
>
> F0  173.6879458121843
> F1  -1.728361E-15
>
> I'm sure the "Q" of Vela would be pretty decent - but I can tell you now,
> as a time-keeper, she's useless.
>
>
> Jim Palfreyman
>
>
>
> On 28 July 2016 at 20:50, Tony Finch  wrote:
>
>> Neville Michie  wrote:
>>
>> > The conical pendulum has a simple form of a weight on a string, instead
>> > of oscillating in one plane as a conventional pendulum, it swings around
>> > in a circular orbit in the horizontal plane. It has a definite resonant
>> > frequency.
>>
>> I don't think it does have a resonant frequency, any more than the Earth
>> does: the angular velocity of the pendulum is sqrt(g/h) where h is the
>> height of the pendulum; give it more energy, it swings higher, so h is
>> smaller, so the frequency is higher.
>>
>> Tony.
>> --
>> f.anthony.n.finch    http://dotat.at/  -  I xn--zr8h
>> punycode
>> South Thames, Dover: Southwesterly 5 or 6. Slight or moderate. Rain or
>> showers. Good, occasionally poor.
>> ___
>> time-nuts mailing list -- time-nuts@febo.com
>> To unsubscribe, go to
>> https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
>> and follow the instructions there.
>>
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Re: [time-nuts] Q/noise of Earth as an oscillator

[David]

On Wed, 27 Jul 2016 15:15:48 -0400, you wrote:

>On 7/27/2016 10:04 AM, time-nuts-requ...@febo.com wrote:
>
>> Exciting the Earth with a new frequency (and an adeguate amount of
>> energy) sets a new rotational speed: you cannot retune a (for example)
>> quartz crystal in the same way...
>
>Sure you can. Spin it at 100 RPM, or 1000, RPM or even 25000  RPM... :)
>
>Interesting conversation. I tend to agree the earth is not a classic 
>harmonic oscillator. Energy is not exchanged between different storage 
>mechanisms. It's rotational period has no natural harmonic frequency. 
>i.e. rotational period could be anything.
>
>However I also agree it exhibits characteristics of other items that Q 
>can be calculated for. Rate of slowing, loss of energy per cycle, etc.
>And since the definition of Q is varied and used quite widely, it seems 
>Q is also appropriate here.
>
>Maybe Earth is a special case since after all it DID give us the second, 
>and we DO set our atomic clocks to IT every 6 to 12 months...
>
>...now I'll be thinking of this all night...  ...I think Tom is just 
>toying with us now...
>
>Dan

Capacitors and inductors have an associated Q while lacking a resonate
frequency except for parasitic elements.  Their Q increases with
frequency up to a point; does that apply to a spinning body?  I guess
it depends on the loss mechanism.

If you used the Earth's rotation as part of a harmonic oscillator,
what would limit the Q?  All the sloshing fluid, physical
displacement, and mechanical hysteresis add up to energy lost per
cycle.

This seems like one of those fun physics problems where you start with
a bunch of seemingly unrelated pieces of numerical data and calculate
the mass of the Milky Way.
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Jim Palfreyman]

Hi All,

Tom gave me a nudge to look here - I hadn't been following this thread.

For those that don't know, I study pulsars and so the way we measure what
pulsars do could be relevant to this discussion.

First, I have never heard of a Q measure when referencing a pulsar. I think
the key here is that it's not resonating as such. Rotating yes, resonating
no.

Pulsars spin and slow down due to giving off energy (magnetic dipole
radiation). So in the pulsar world we mainly refer to spin frequency (F0)
and frequency derivative (F1). With some of the younger and more "erratic"
pulsars, F2 (and further) can be modelled.

Here's some data on the Vela pulsar (hot off the presses - measured just
now):

F0  11.1867488542579
F1  -1.55859177352837e-11
F2  1.23776878287221e-21

Vela is young and erratic. Millisecond pulsars are outstanding clocks.
Here's the data for J0437-4715 - one of the most stable pulsars we know
about:

F0  173.6879458121843
F1  -1.728361E-15

I'm sure the "Q" of Vela would be pretty decent - but I can tell you now,
as a time-keeper, she's useless.


Jim Palfreyman



On 28 July 2016 at 20:50, Tony Finch  wrote:

> Neville Michie  wrote:
>
> > The conical pendulum has a simple form of a weight on a string, instead
> > of oscillating in one plane as a conventional pendulum, it swings around
> > in a circular orbit in the horizontal plane. It has a definite resonant
> > frequency.
>
> I don't think it does have a resonant frequency, any more than the Earth
> does: the angular velocity of the pendulum is sqrt(g/h) where h is the
> height of the pendulum; give it more energy, it swings higher, so h is
> smaller, so the frequency is higher.
>
> Tony.
> --
> f.anthony.n.finch    http://dotat.at/  -  I xn--zr8h
> punycode
> South Thames, Dover: Southwesterly 5 or 6. Slight or moderate. Rain or
> showers. Good, occasionally poor.
> ___
> time-nuts mailing list -- time-nuts@febo.com
> To unsubscribe, go to
> https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> and follow the instructions there.
>
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Bob Camp]

Hi


> On Jul 27, 2016, at 12:57 PM, Ron Ott <ron...@sbcglobal.net> wrote:
> 
> There might be two Qs: one relating to the axil rotation and another 
> concerning the volume behavior of the earth as a giant bowl of Jello.  But 
> you'd have to figure out how to really slam the planet to excite the entire 
> volume. Earthquakes are probably too wimpy.

Run into a bit smaller Earth with an object somewhat larger than the moon?
Give us all a bit of a warning before you run the experiment so we can book 
that flight to Mars ...

Bob

> Ron
> 
> 
>  From: Chris Caudle <ch...@chriscaudle.org>
> To: time-nuts@febo.com 
> Sent: Wednesday, July 27, 2016 8:50 AM
> Subject: Re: [time-nuts] Q/noise of Earth as an oscillator
> 
> On Wed, July 27, 2016 10:33 am, Chris Caudle wrote:
>> Does that imply that this value is not constant:
>>>> And if you take the classic definition
>>>> Q = 2 pi * total energy /energy lost per cycle
>>>> then it would seem earth has a Q factor.
> 
> After re-reading "The Story of Q" I agree that Q of a rotating body could
> be non-constant, but also consistent with the original definition of Q as
> the ratio of reactance to resistance of an inductor, which of course would
> vary almost completely linearly over a wide frequency range where the
> resistive dissipation was not frequency dependent (i.e. where skin effect
> was negligible).
> 
> Perhaps a more useful question is whether that is still a useful
> definition compared to how the term is more typically used now to refer to
> resonance bandwidth.
> 
> -- 
> Chris Caudle
> 
> 
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> 
> 
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Tony Finch]

Neville Michie  wrote:

> The conical pendulum has a simple form of a weight on a string, instead
> of oscillating in one plane as a conventional pendulum, it swings around
> in a circular orbit in the horizontal plane. It has a definite resonant
> frequency.

I don't think it does have a resonant frequency, any more than the Earth
does: the angular velocity of the pendulum is sqrt(g/h) where h is the
height of the pendulum; give it more energy, it swings higher, so h is
smaller, so the frequency is higher.

Tony.
-- 
f.anthony.n.finch    http://dotat.at/  -  I xn--zr8h punycode
South Thames, Dover: Southwesterly 5 or 6. Slight or moderate. Rain or
showers. Good, occasionally poor.
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Ron Ott]

There might be two Qs: one relating to the axil rotation and another concerning 
the volume behavior of the earth as a giant bowl of Jello.  But you'd have to 
figure out how to really slam the planet to excite the entire volume. 
Earthquakes are probably too wimpy.
Ron


  From: Chris Caudle <ch...@chriscaudle.org>
 To: time-nuts@febo.com 
 Sent: Wednesday, July 27, 2016 8:50 AM
 Subject: Re: [time-nuts] Q/noise of Earth as an oscillator
   
On Wed, July 27, 2016 10:33 am, Chris Caudle wrote:
> Does that imply that this value is not constant:
>>> And if you take the classic definition
>>> Q = 2 pi * total energy /energy lost per cycle
>>> then it would seem earth has a Q factor.

After re-reading "The Story of Q" I agree that Q of a rotating body could
be non-constant, but also consistent with the original definition of Q as
the ratio of reactance to resistance of an inductor, which of course would
vary almost completely linearly over a wide frequency range where the
resistive dissipation was not frequency dependent (i.e. where skin effect
was negligible).

Perhaps a more useful question is whether that is still a useful
definition compared to how the term is more typically used now to refer to
resonance bandwidth.

-- 
Chris Caudle


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Re: [time-nuts] Q/noise of Earth as an oscillator

[Alex Pummer]

in case the Earth, the circular frequency of the Earth is equal of a 
sinusoidal wave's frequency, which is a projection of the Earth circular 
movement in a plain, which is perpendicular to the axis of the Earth. 
The Q of the Earth in interpreted as the quotient of the phase speed of 
the  Earth's  circular rotation [or the projected wave's] and the 
fluctuation of the phase speed.


73
KJ6UHN
Alex

On 7/27/2016 8:10 PM, Neville Michie wrote:

The discussion of Earth as a system with Q, but which is not resonant,
is a more extreme case than the CONICAL PENDULUM.

The conical pendulum has a simple form of a weight on a string, instead of 
oscillating
in one plane as a conventional pendulum, it swings around in a circular orbit
in the horizontal plane. It has a definite resonant frequency.
Now a simple pendulum oscillates kinetic energy and potential energy, but a 
conical pendulum
has constant potential energy and oscillates its energy from North-South energy 
to East-West
energy.
I believe that a conical pendulum still has the circular error associated with 
amplitude.

But will it be as good a time keeper as the simple pendulum?

A curious fact about the conical pendulum is that whereas the simple pendulum 
has earth
rotation forces that show in the Foucault Pendulum, the conical pendulum has a 
different
period depending whether is swings clockwise or anticlockwise due to the earths 
rotation.

What do you think?

Cheers,
Neville Michie
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-
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Neville Michie]

The discussion of Earth as a system with Q, but which is not resonant,
is a more extreme case than the CONICAL PENDULUM.

The conical pendulum has a simple form of a weight on a string, instead of 
oscillating 
in one plane as a conventional pendulum, it swings around in a circular orbit 
in the horizontal plane. It has a definite resonant frequency.
Now a simple pendulum oscillates kinetic energy and potential energy, but a 
conical pendulum
has constant potential energy and oscillates its energy from North-South energy 
to East-West 
energy.
I believe that a conical pendulum still has the circular error associated with 
amplitude.

But will it be as good a time keeper as the simple pendulum?

A curious fact about the conical pendulum is that whereas the simple pendulum 
has earth 
rotation forces that show in the Foucault Pendulum, the conical pendulum has a 
different 
period depending whether is swings clockwise or anticlockwise due to the earths 
rotation.

What do you think?

Cheers, 
Neville Michie
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Dan Kemppainen]

On 7/27/2016 10:04 AM, time-nuts-requ...@febo.com wrote:

Exciting the Earth with a new frequency (and an adeguate amount of
energy) sets a new rotational speed: you cannot retune a (for example)
quartz crystal in the same way...


Sure you can. Spin it at 100 RPM, or 1000, RPM or even 25000  RPM... :)

Interesting conversation. I tend to agree the earth is not a classic 
harmonic oscillator. Energy is not exchanged between different storage 
mechanisms. It's rotational period has no natural harmonic frequency. 
i.e. rotational period could be anything.


However I also agree it exhibits characteristics of other items that Q 
can be calculated for. Rate of slowing, loss of energy per cycle, etc.
And since the definition of Q is varied and used quite widely, it seems 
Q is also appropriate here.


Maybe Earth is a special case since after all it DID give us the second, 
and we DO set our atomic clocks to IT every 6 to 12 months...


...now I'll be thinking of this all night...  ...I think Tom is just 
toying with us now...


Dan




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Re: [time-nuts] Q/noise of Earth as an oscillator

[Jim Harman]

On Wed, Jul 27, 2016 at 11:33 AM, Chris Caudle 
wrote:

> Who has a globe on magnetic bearings in a vacuum chamber and will run the
> experiment for us?
>

The superconducting gyroscopes in the Gravity Probe B satellite did an
extraordinary job of eliminating frictional and other losses in a spinning
object, with a spin-down time constant of 15,000 years.

https://einstein.stanford.edu/TECH/technology1.html


-- 

--Jim Harman
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Chris Caudle]

On Wed, July 27, 2016 10:33 am, Chris Caudle wrote:
> Does that imply that this value is not constant:
>>> And if you take the classic definition
>>> Q = 2 pi * total energy /energy lost per cycle
>>> then it would seem earth has a Q factor.

After re-reading "The Story of Q" I agree that Q of a rotating body could
be non-constant, but also consistent with the original definition of Q as
the ratio of reactance to resistance of an inductor, which of course would
vary almost completely linearly over a wide frequency range where the
resistive dissipation was not frequency dependent (i.e. where skin effect
was negligible).

Perhaps a more useful question is whether that is still a useful
definition compared to how the term is more typically used now to refer to
resonance bandwidth.

-- 
Chris Caudle


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Re: [time-nuts] Q/noise of Earth as an oscillator

[Chris Caudle]

On Wed, July 27, 2016 8:58 am, Azelio Boriani wrote:
> Exciting the Earth with a new frequency (and an adeguate amount of
> energy) sets a new rotational speed: you cannot retune a (for example)
> quartz crystal in the same way...

Does that imply that this value is not constant:

> On Wed, Jul 27, 2016 at 3:42 PM, Tom Van Baak  wrote:
>> And if you take the classic definition Q = 2 pi * total energy / energy
>> lost per cycle then it would seem earth has a Q factor.
>>
>> In fact, if you use real energy numbers you get:
>>
>> - total rotational energy of earth is 2.14e29 J
>> - energy lost per cycle (day) is 2.7e17 J
>> - so Q = 2pi * 2.14e29 / 2.7e17 = 5e12, the same 5 trillion as my
>> earlier calculation.

My first intuition is that energy lost per cycle would not increase as the
square of angular velocity, but angular kinetic energy does increase as
the square of angular velocity.  The energy losses are from a complicated
interaction of tidal forces and fluid friction between crust and
atmosphere/ocean and mantle and outer core, so I don't trust that my
intuition is correct at all.  If however that turns out to be correct in a
rough sense then adding energy to speed up the rotation would change the
period and the Q, which isn't a property of most things with high Q that
we use as stable clocks.

Who has a globe on magnetic bearings in a vacuum chamber and will run the
experiment for us?

-- 
Chris Caudle


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Re: [time-nuts] Q/noise of Earth as an oscillator

[Peter Reilley]

If you consider viewing earth from above the equator at a long distance 
and imagine a
spot on the surface of the earth.   That spot will appear to have a 
sinusoidal motion.
The frequency of the sinusoid exhibits a decay.   That decay can be 
considered as the Q

of the earths rotation.

Pete.



On 7/27/2016 9:00 AM, jimlux wrote:

On 7/27/16 5:43 AM, Michael Wouters wrote:

On Wed, Jul 27, 2016 at 8:08 AM, Attila Kinali  wrote:

"I am not sure you can apply this definition of Q onto earth."

It  doesn't make sense to me either.

If you mark a point on the surface of a sphere then you can observe
that point as the sphere
rotates and count rotations to make a clock. If you think of just a
circle, then a point on it viewed in a rectilinear coordinate system
executes simple harmonic motion so the motion of that point looks like
an oscillator, so that much is OK.

But unlike the LCR circuit, the pendulum and quartz crystal, the
sphere's rotational motion does not have a
resonant frequency. Another way of characterizing the Q of an
oscillator, the relative width of the resonance, makes
no sense in this context.



There's also the thing that "things that resonate" typically have 
energy transferring back and forth between modes or components: E 
field and H field for an antenna; kinetic vs potential energy for 
pendulums and weight/spring; charge and current (C & L, really E 
field/H field again).


Spinning earth is more of an "rotational inertia and loss" thing, with 
zero frequency, just the exponential decay term.


If you think of a single measurand in any of these scenarios you have 
at the core some sort of exp(-kt)*cos(omega*t+phi) and we're relating 
Q to the coefficient k.



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Re: [time-nuts] Q/noise of Earth as an oscillator

[David]

We have discussed quartz crystals which despite their high Q, may
suffer from periodic "jumps" in frequency do to what I assume are
imperfection that hardly affect Q.  If they did affect Q, then that
would have been a good way to grade them for this behavior.

Temperature coefficient is also independent of Q.  If I built a
pendulum out of a material with a high thermal coefficient of
expansion, it will not be very stable but the Q will be unaffected.

On Wed, 27 Jul 2016 22:43:04 +1000, you wrote:

>On Wed, Jul 27, 2016 at 8:08 AM, Attila Kinali  wrote:
>
>...
>
>Something else that indicates that the model is suspect is that the
>apparently high 'Q' implies a stability which the earth does not have,
>as Tom observes. Viewed another way, this suggests that the model is
>inappropriate because it leads to an incorrect conclusion.
>
>Time for bed. I'll probably lie awake thinking about this now :-)
>
>Cheers
>Michael
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Azelio Boriani]

Exciting the Earth with a new frequency (and an adeguate amount of
energy) sets a new rotational speed: you cannot retune a (for example)
quartz crystal in the same way...

On Wed, Jul 27, 2016 at 3:42 PM, Tom Van Baak <t...@leapsecond.com> wrote:
> Hi Michael,
>
> I sympathize with both your and Attila's comments and would like to dig 
> deeper for the truth on this.
>
> Clearly both the earth and a pendulum (and many other periodic systems) 
> exhibit a decay of energy, when you remove the periodic restoring force. And 
> if you take the classic definition Q = 2 pi * total energy / energy lost per 
> cycle then it would seem earth has a Q factor.
>
> In fact, if you use real energy numbers you get:
>
> - total rotational energy of earth is 2.14e29 J
> - energy lost per cycle (day) is 2.7e17 J
> - so Q = 2pi * 2.14e29 / 2.7e17 = 5e12, the same 5 trillion as my earlier 
> calculation.
>
> But your point about resonance is a good one and it has always intrigued me. 
> Is this one difference between a pendulum and the earth as timekeepers?
>
> On the other hand, if you swept the earth with an external powerful frequency 
> in the range well below to well above 1.16e-5 Hz (1/86164 s) would you not 
> see a resonance peak right at the center? Given the mass of the planet and 
> its pre-existing rotational energy, it seems like there is a "resonance", a 
> preference to remain at its current frequency. Plus it has a slow decay due 
> to internal friction. This sounds like any other timing system with Q to me.
>
> Or imagine a planet the same size as earth made from a Mylar balloon. Much 
> less mass. Give it the same rotational speed. Much easier to increase or 
> decrease its energy by applying external force. Far lower Q than earth, yes?
>
> It might also be useful at this point, to:
>
> read the history Q and its definition:
> http://www.collinsaudio.com/Prosound_Workshop/The_story_of_Q.pdf
>
> and read the patent in which Q first appeared:
> http://leapsecond.com/pages/Q/1927-US1628983.pdf
>
> or view the first paragraph in which Q appeared:
> http://leapsecond.com/pages/Q/1927-Q-patent-600x300.gif
>
> /tvb
>
> - Original Message -
> From: "Michael Wouters" <michaeljwout...@gmail.com>
> To: "Discussion of precise time and frequency measurement" 
> <time-nuts@febo.com>; <att...@kinali.ch>
> Sent: Wednesday, July 27, 2016 5:43 AM
> Subject: Re: [time-nuts] Q/noise of Earth as an oscillator
>
>
>> On Wed, Jul 27, 2016 at 8:08 AM, Attila Kinali <att...@kinali.ch> wrote:
>>
>> "I am not sure you can apply this definition of Q onto earth."
>>
>> It  doesn't make sense to me either.
>>
>> If you mark a point on the surface of a sphere then you can observe
>> that point as the sphere
>> rotates and count rotations to make a clock. If you think of just a
>> circle, then a point on it viewed in a rectilinear coordinate system
>> executes simple harmonic motion so the motion of that point looks like
>> an oscillator, so that much is OK.
>>
>> But unlike the LCR circuit, the pendulum and quartz crystal, the
>> sphere's rotational motion does not have a
>> resonant frequency. Another way of characterizing the Q of an
>> oscillator, the relative width of the resonance, makes
>> no sense in this context.
>>
>> It seems to me that the model of the earth as an oscillator is
>> misapplied and that the 'Q' is not a meaningful number.
>> I think the confusion arises here because of a conflation of a
>> rotation of the sphere (which marks out a time interval) with an
>> oscillation. Both can be used to define an energy lost per unit time
>> but the former doesn't have anything to do with the properties of an
>> oscillator.
>>
>> Something else that indicates that the model is suspect is that the
>> apparently high 'Q' implies a stability which the earth does not have,
>> as Tom observes. Viewed another way, this suggests that the model is
>> inappropriate because it leads to an incorrect conclusion.
>>
>> Time for bed. I'll probably lie awake thinking about this now :-)
>>
>> Cheers
>> Michael
>>
>> On Wed, Jul 27, 2016 at 8:08 AM, Attila Kinali <att...@kinali.ch> wrote:
>>> Hoi Tom,
>>>
>>> On Tue, 26 Jul 2016 12:36:37 -0700
>>> "Tom Van Baak" <t...@leapsecond.com> wrote:
>>>
>>>> Among other things, the quality-factor, or Q is a measure of how slowly a
>>>> free-running oscillator runs down. There are lots of examples of periodic 
>>>> or
>>>> damped oscillat

Re: [time-nuts] Q/noise of Earth as an oscillator

[Tom Van Baak]

Hi Michael,

I sympathize with both your and Attila's comments and would like to dig deeper 
for the truth on this.

Clearly both the earth and a pendulum (and many other periodic systems) exhibit 
a decay of energy, when you remove the periodic restoring force. And if you 
take the classic definition Q = 2 pi * total energy / energy lost per cycle 
then it would seem earth has a Q factor.

In fact, if you use real energy numbers you get:

- total rotational energy of earth is 2.14e29 J
- energy lost per cycle (day) is 2.7e17 J
- so Q = 2pi * 2.14e29 / 2.7e17 = 5e12, the same 5 trillion as my earlier 
calculation.

But your point about resonance is a good one and it has always intrigued me. Is 
this one difference between a pendulum and the earth as timekeepers?

On the other hand, if you swept the earth with an external powerful frequency 
in the range well below to well above 1.16e-5 Hz (1/86164 s) would you not see 
a resonance peak right at the center? Given the mass of the planet and its 
pre-existing rotational energy, it seems like there is a "resonance", a 
preference to remain at its current frequency. Plus it has a slow decay due to 
internal friction. This sounds like any other timing system with Q to me.

Or imagine a planet the same size as earth made from a Mylar balloon. Much less 
mass. Give it the same rotational speed. Much easier to increase or decrease 
its energy by applying external force. Far lower Q than earth, yes?

It might also be useful at this point, to:

read the history Q and its definition:
http://www.collinsaudio.com/Prosound_Workshop/The_story_of_Q.pdf

and read the patent in which Q first appeared:
http://leapsecond.com/pages/Q/1927-US1628983.pdf

or view the first paragraph in which Q appeared:
http://leapsecond.com/pages/Q/1927-Q-patent-600x300.gif

/tvb

- Original Message - 
From: "Michael Wouters" <michaeljwout...@gmail.com>
To: "Discussion of precise time and frequency measurement" 
<time-nuts@febo.com>; <att...@kinali.ch>
Sent: Wednesday, July 27, 2016 5:43 AM
Subject: Re: [time-nuts] Q/noise of Earth as an oscillator


> On Wed, Jul 27, 2016 at 8:08 AM, Attila Kinali <att...@kinali.ch> wrote:
> 
> "I am not sure you can apply this definition of Q onto earth."
> 
> It  doesn't make sense to me either.
> 
> If you mark a point on the surface of a sphere then you can observe
> that point as the sphere
> rotates and count rotations to make a clock. If you think of just a
> circle, then a point on it viewed in a rectilinear coordinate system
> executes simple harmonic motion so the motion of that point looks like
> an oscillator, so that much is OK.
> 
> But unlike the LCR circuit, the pendulum and quartz crystal, the
> sphere's rotational motion does not have a
> resonant frequency. Another way of characterizing the Q of an
> oscillator, the relative width of the resonance, makes
> no sense in this context.
> 
> It seems to me that the model of the earth as an oscillator is
> misapplied and that the 'Q' is not a meaningful number.
> I think the confusion arises here because of a conflation of a
> rotation of the sphere (which marks out a time interval) with an
> oscillation. Both can be used to define an energy lost per unit time
> but the former doesn't have anything to do with the properties of an
> oscillator.
> 
> Something else that indicates that the model is suspect is that the
> apparently high 'Q' implies a stability which the earth does not have,
> as Tom observes. Viewed another way, this suggests that the model is
> inappropriate because it leads to an incorrect conclusion.
> 
> Time for bed. I'll probably lie awake thinking about this now :-)
> 
> Cheers
> Michael
> 
> On Wed, Jul 27, 2016 at 8:08 AM, Attila Kinali <att...@kinali.ch> wrote:
>> Hoi Tom,
>>
>> On Tue, 26 Jul 2016 12:36:37 -0700
>> "Tom Van Baak" <t...@leapsecond.com> wrote:
>>
>>> Among other things, the quality-factor, or Q is a measure of how slowly a
>>> free-running oscillator runs down. There are lots of examples of periodic or
>>> damped oscillatory motion that have Q -- RC or LC circuit, tuning fork,
>>> pendulum, vibrating quartz; yes, even a rotating planet in space.
>>
>> I am not sure you can apply this definition of Q onto earth. Q is defined
>> for harmonic oscillators (or oscillators that can be approximated by an
>> harmonic oscillator) but the earth isn't oscillating, it's rotating.
>> While, for time keeping purposes, similar in nature, the physical
>> description of both are different.
>>
>> Attila Kinali
>>
>> --
>> Malek's Law:
>> Any simple idea will be worded in the most complicated way.
>&g

Re: [time-nuts] Q/noise of Earth as an oscillator

[jimlux]

On 7/27/16 5:43 AM, Michael Wouters wrote:

On Wed, Jul 27, 2016 at 8:08 AM, Attila Kinali  wrote:

"I am not sure you can apply this definition of Q onto earth."

It  doesn't make sense to me either.

If you mark a point on the surface of a sphere then you can observe
that point as the sphere
rotates and count rotations to make a clock. If you think of just a
circle, then a point on it viewed in a rectilinear coordinate system
executes simple harmonic motion so the motion of that point looks like
an oscillator, so that much is OK.

But unlike the LCR circuit, the pendulum and quartz crystal, the
sphere's rotational motion does not have a
resonant frequency. Another way of characterizing the Q of an
oscillator, the relative width of the resonance, makes
no sense in this context.



There's also the thing that "things that resonate" typically have energy 
transferring back and forth between modes or components: E field and H 
field for an antenna; kinetic vs potential energy for pendulums and 
weight/spring; charge and current (C & L, really E field/H field again).


Spinning earth is more of an "rotational inertia and loss" thing, with 
zero frequency, just the exponential decay term.


If you think of a single measurand in any of these scenarios you have at 
the core some sort of exp(-kt)*cos(omega*t+phi) and we're relating Q to 
the coefficient k.



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Re: [time-nuts] Q/noise of Earth as an oscillator

[Michael Wouters]

On Wed, Jul 27, 2016 at 8:08 AM, Attila Kinali  wrote:

"I am not sure you can apply this definition of Q onto earth."

It  doesn't make sense to me either.

If you mark a point on the surface of a sphere then you can observe
that point as the sphere
rotates and count rotations to make a clock. If you think of just a
circle, then a point on it viewed in a rectilinear coordinate system
executes simple harmonic motion so the motion of that point looks like
an oscillator, so that much is OK.

But unlike the LCR circuit, the pendulum and quartz crystal, the
sphere's rotational motion does not have a
resonant frequency. Another way of characterizing the Q of an
oscillator, the relative width of the resonance, makes
no sense in this context.

It seems to me that the model of the earth as an oscillator is
misapplied and that the 'Q' is not a meaningful number.
I think the confusion arises here because of a conflation of a
rotation of the sphere (which marks out a time interval) with an
oscillation. Both can be used to define an energy lost per unit time
but the former doesn't have anything to do with the properties of an
oscillator.

Something else that indicates that the model is suspect is that the
apparently high 'Q' implies a stability which the earth does not have,
as Tom observes. Viewed another way, this suggests that the model is
inappropriate because it leads to an incorrect conclusion.

Time for bed. I'll probably lie awake thinking about this now :-)

Cheers
Michael

On Wed, Jul 27, 2016 at 8:08 AM, Attila Kinali  wrote:
> Hoi Tom,
>
> On Tue, 26 Jul 2016 12:36:37 -0700
> "Tom Van Baak"  wrote:
>
>> Among other things, the quality-factor, or Q is a measure of how slowly a
>> free-running oscillator runs down. There are lots of examples of periodic or
>> damped oscillatory motion that have Q -- RC or LC circuit, tuning fork,
>> pendulum, vibrating quartz; yes, even a rotating planet in space.
>
> I am not sure you can apply this definition of Q onto earth. Q is defined
> for harmonic oscillators (or oscillators that can be approximated by an
> harmonic oscillator) but the earth isn't oscillating, it's rotating.
> While, for time keeping purposes, similar in nature, the physical
> description of both are different.
>
> Attila Kinali
>
> --
> Malek's Law:
> Any simple idea will be worded in the most complicated way.
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Re: [time-nuts] Q/noise of Earth as an oscillator

[bg]

http://www.prc68.com/I/Sensors.shtml#Gyroscopic
"A note on gyros.  When located in a dynamic platform, like an airplane, the 
gyro needs to be located at the center of mass of the vehicle.  This way when 
the plane banks, pitches or yaws the gyro will only respond to the angles.  But 
if it's located anywhere else it will be in error.  This applies to models as 
well as to full size vehicles."
This is not true. You can - and people do - mount gyros (interial systems) 
anywhere in the rigid part of the body - wingtip mount might be a bad idea...
If the gyro is away from the center of rotation it will experience the same 
rotation and also an acceleration. The chosen gyro should be good enough at 
separating rotation from acceleration. This has not been a problem for a very 
long time - 40+ years - for typical aviation nav systems.
For specialized applications - say spinning munitions - where your performance 
is really limited by acceleration - you should keep your sensors at the center. 
The same could be said for accelerometers, oscillators and many other sensors 
that might have errors amplified by the extra acc.
--      Björn




Sent from my smartphone. Original message From: Brooke Clarke 
<bro...@pacific.net> Date: 27/07/2016  01:00  (GMT+01:00) To: Discussion of 
precise time and frequency measurement <time-nuts@febo.com> Subject: Re: 
[time-nuts] Q/noise of Earth as an oscillator 
Hi Hal:

I resemble that remark.

Momentum and drift.  It's interesting that the drift rate depends on the 
physical volume.  See table at: 
http://www.prc68.com/I/Sensors.shtml#Gyroscopic
http://www.prc68.com/I/Gyroscopes.html

-- 
Have Fun,

Brooke Clarke
http://www.PRC68.com
http://www.end2partygovernment.com/2012Issues.html
The lesser of evils is still evil.

 Original Message 
> att...@kinali.ch said:
>> I am not sure you can apply this definition of Q onto earth. Q is defined
>> for harmonic oscillators (or oscillators that can be approximated by an
>> harmonic oscillator) but the earth isn't oscillating, it's rotating. While,
>> for time keeping purposes, similar in nature, the physical description of
>> both are different.
> What do gyroscope-nuts use to describe the quality of their toys?
>
>

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Re: [time-nuts] Q/noise of Earth as an oscillator

[Will Kimber]

Around mid 1978 a scientific magazine had an item on the shift of axis 
of the earth due to earthquakes.  Sorry but searching has not found the 
item. My, not to be  relied on memory, thinks that it was from the 
French Bureau International de l'Heure in Science or Scientific American


Cheers,
Will

On 07/27/2016 10:45 AM, Tom Van Baak wrote:

Hal Murray wrote:

I've seen mention that the rotation rate of the Earth changed by a few
microseconds per day as a result of the 2011 earthquake in Japan.  Does that
show up in any data?  Your recent graph doesn't go back that far and it's got
a full scale of 2000 microseconds so a few is going to be hard to see.


Right. The IERS graphs I posted are real measurements of earth rotation. The 
earthquake / tsunami numbers are theoretical calculations only; numbers far 
smaller than what can be measured.

A couple of guys at NASA have carefully modeled all of this and the predictions 
are quite interesting. It's just that the official NASA/JPL press releases, 
once filtered by the popular press, give the impression that these are 
worrisome or real measurable effects on the earth's rotation rate or axis tilt. 
Instead they are order(s) of magnitude below what can be measured, given the 
level of VLBI resolution or earth rotation noise.

Here are three recent press releases:

2005 "NASA Details Earthquake Effects on the Earth"
http://www.jpl.nasa.gov/news/news.php?feature=716

2010 "Chilean Quake May Have Shortened Earth Days"
http://www.nasa.gov/topics/earth/features/earth-20100301.html

2011 "Japan Quake May Have Shortened Earth Days, Moved Axis"
http://www.nasa.gov/topics/earth/features/japanquake/earth20110314.html

Google for Richard Gross and Benjamin Fong Chao for more information.

/tvb

- Original Message -
From: "Hal Murray" 
To: "Tom Van Baak" ; "Discussion of precise time and frequency 
measurement" 
Cc: 
Sent: Saturday, July 23, 2016 5:59 PM
Subject: Q/noise of Earth as an oscillator




t...@leapsecond.com said:

Earth is a very noisy, wandering, drifting, incredibly-expensive-to-measure,
low-precision (though high-Q) clock.


What is the Q of the Earth?  It might be on one of your web pages, but I
don't remember seeing it.  Google found a few mentions, but I didn't find a
number.

I did find an interesting list of damping mechanisms in a geology book.
Geology-nuts are as nutty as time-nuts.  Many were discussing damping of
seismic waves rather than rotation.

I've seen mention that the rotation rate of the Earth changed by a few
microseconds per day as a result of the 2011 earthquake in Japan.  Does that
show up in any data?  Your recent graph doesn't go back that far and it's got
a full scale of 2000 microseconds so a few is going to be hard to see.



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Re: [time-nuts] Q/noise of Earth as an oscillator

[Brooke Clarke]

Hi Hal:

I resemble that remark.

Momentum and drift.  It's interesting that the drift rate depends on the physical volume.  See table at: 
http://www.prc68.com/I/Sensors.shtml#Gyroscopic

http://www.prc68.com/I/Gyroscopes.html

--
Have Fun,

Brooke Clarke
http://www.PRC68.com
http://www.end2partygovernment.com/2012Issues.html
The lesser of evils is still evil.

 Original Message 

att...@kinali.ch said:

I am not sure you can apply this definition of Q onto earth. Q is defined
for harmonic oscillators (or oscillators that can be approximated by an
harmonic oscillator) but the earth isn't oscillating, it's rotating. While,
for time keeping purposes, similar in nature, the physical description of
both are different.

What do gyroscope-nuts use to describe the quality of their toys?




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Re: [time-nuts] Q/noise of Earth as an oscillator

[Tom Van Baak]

Hal Murray wrote: 
> I've seen mention that the rotation rate of the Earth changed by a few 
> microseconds per day as a result of the 2011 earthquake in Japan.  Does that 
> show up in any data?  Your recent graph doesn't go back that far and it's got 
> a full scale of 2000 microseconds so a few is going to be hard to see.

Right. The IERS graphs I posted are real measurements of earth rotation. The 
earthquake / tsunami numbers are theoretical calculations only; numbers far 
smaller than what can be measured.

A couple of guys at NASA have carefully modeled all of this and the predictions 
are quite interesting. It's just that the official NASA/JPL press releases, 
once filtered by the popular press, give the impression that these are 
worrisome or real measurable effects on the earth's rotation rate or axis tilt. 
Instead they are order(s) of magnitude below what can be measured, given the 
level of VLBI resolution or earth rotation noise.

Here are three recent press releases:

2005 "NASA Details Earthquake Effects on the Earth"
http://www.jpl.nasa.gov/news/news.php?feature=716

2010 "Chilean Quake May Have Shortened Earth Days"
http://www.nasa.gov/topics/earth/features/earth-20100301.html

2011 "Japan Quake May Have Shortened Earth Days, Moved Axis"
http://www.nasa.gov/topics/earth/features/japanquake/earth20110314.html

Google for Richard Gross and Benjamin Fong Chao for more information.

/tvb

- Original Message - 
From: "Hal Murray" 
To: "Tom Van Baak" ; "Discussion of precise time and 
frequency measurement" 
Cc: 
Sent: Saturday, July 23, 2016 5:59 PM
Subject: Q/noise of Earth as an oscillator


> 
> t...@leapsecond.com said:
>> Earth is a very noisy, wandering, drifting, incredibly-expensive-to-measure,
>> low-precision (though high-Q) clock.
> 
> What is the Q of the Earth?  It might be on one of your web pages, but I 
> don't remember seeing it.  Google found a few mentions, but I didn't find a 
> number.
> 
> I did find an interesting list of damping mechanisms in a geology book.  
> Geology-nuts are as nutty as time-nuts.  Many were discussing damping of 
> seismic waves rather than rotation.
> 
> I've seen mention that the rotation rate of the Earth changed by a few 
> microseconds per day as a result of the 2011 earthquake in Japan.  Does that 
> show up in any data?  Your recent graph doesn't go back that far and it's got 
> a full scale of 2000 microseconds so a few is going to be hard to see.
> 

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Re: [time-nuts] Q/noise of Earth as an oscillator

[David]

On Wed, 27 Jul 2016 00:08:47 +0200, you wrote:

>Hoi Tom,
>
>On Tue, 26 Jul 2016 12:36:37 -0700
>"Tom Van Baak"  wrote:
>
>> Among other things, the quality-factor, or Q is a measure of how slowly a 
>> free-running oscillator runs down. There are lots of examples of periodic or 
>> damped oscillatory motion that have Q -- RC or LC circuit, tuning fork, 
>> pendulum, vibrating quartz; yes, even a rotating planet in space.
>
>I am not sure you can apply this definition of Q onto earth. Q is defined
>for harmonic oscillators (or oscillators that can be approximated by an
>harmonic oscillator) but the earth isn't oscillating, it's rotating.
>While, for time keeping purposes, similar in nature, the physical
>description of both are different.
>
>   Attila Kinali

It seems reasonable to me.  The calculation returns a dimensionless
number which does represent the run-down time.  Any reference duration
like hours, months, or years would have returned the same result.
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Hal Murray]

att...@kinali.ch said:
> I am not sure you can apply this definition of Q onto earth. Q is defined
> for harmonic oscillators (or oscillators that can be approximated by an
> harmonic oscillator) but the earth isn't oscillating, it's rotating. While,
> for time keeping purposes, similar in nature, the physical description of
> both are different. 

What do gyroscope-nuts use to describe the quality of their toys?


-- 
These are my opinions.  I hate spam.



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Re: [time-nuts] Q/noise of Earth as an oscillator

[Attila Kinali]

Hoi Tom,

On Tue, 26 Jul 2016 12:36:37 -0700
"Tom Van Baak"  wrote:

> Among other things, the quality-factor, or Q is a measure of how slowly a 
> free-running oscillator runs down. There are lots of examples of periodic or 
> damped oscillatory motion that have Q -- RC or LC circuit, tuning fork, 
> pendulum, vibrating quartz; yes, even a rotating planet in space.

I am not sure you can apply this definition of Q onto earth. Q is defined
for harmonic oscillators (or oscillators that can be approximated by an
harmonic oscillator) but the earth isn't oscillating, it's rotating.
While, for time keeping purposes, similar in nature, the physical
description of both are different.

Attila Kinali

-- 
Malek's Law:
Any simple idea will be worded in the most complicated way.
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Re: [time-nuts] Q/noise of Earth as an oscillator

[Tom Van Baak]

Hal Murray wrote: 
> What is the Q of the Earth?  It might be on one of your web pages, but I 
> don't remember seeing it.  Google found a few mentions, but I didn't find a 
> number.

3.1415 * 86400 * 365 * 100 / 0.002 = 5e12, or 5 trillion.  Here's why:

(1)
Among other things, the quality-factor, or Q is a measure of how slowly a 
free-running oscillator runs down. There are lots of examples of periodic or 
damped oscillatory motion that have Q -- RC or LC circuit, tuning fork, 
pendulum, vibrating quartz; yes, even a rotating planet in space.

In the high vacuum of space you'd think the Earth would rotate forever, but the 
moon affects earth's rotation; it causes "tidal friction". Astronomers have 
measured the slowing as roughly "2 ms per day per century".

For example, the average length of a day in 1900 was about 86400.000 seconds 
and the average length of a day in 2000 was about 86400.002 seconds. That's a 
deceleration, or slight loss in rate, or slight gain in period, of 2 ms/day per 
century. Since the spinning earth oscillator is slowing down, you can calculate 
its Q factor.

(2)
The common definition of Q is 2pi times the number of periods it takes for the 
energy to fall to 37% (1/e). Although correct and simple, it is sometimes 
awkward to use this definition. I mean, it can take forever to lose 63% of your 
energy if Q is really high or the period is really long.

Another definition of Q is 2pi over dE/E, the latter term being called the 
"decrement", which is the relative energy loss per period. For example, if your 
wine glass oscillator loses 1% of its energy each period, then its Q = 2pi/1%, 
or 628.

This formula for Q is more convenient because you can do it with data from a 
single period. For earth the period is 1 day. So what's the decrement, the 
relative energy loss for one day?

(3)
Rotational energy goes as w^2, where w (omega) is the rotation rate. This means 
energy drops twice as fast as rotation rate drops.

And how fast does the earth rotation rate drop? That would be 2 ms per day per 
century, which is 2.3e-8 per century, or 2.3e-10 per year, or 6.3e-13 per day. 
So the relative loss of energy per period (day) is twice that: 1.3e-12. Plug 
that into the decrement definition of Q and your calculator gives 5e12.

So the Q of the earth is 5 trillion.

https://www.google.com/#q=3.1415+*+86400+*+365+*+100+/+0.002

/tvb

p.s.

This Q is really high, and you might expect a similarly superb Allan deviation. 
But no, not at all:

http://leapsecond.com/museum/earth/

The reason the ADEV is so poor compared to the Q is that Earth, as a 
timekeeper, is an unstable mess inside and out. There are all sorts of massive 
perturbing factors from the liquid core to the swirling atmosphere to the 
annual redistribution of water and biomass; these all affect the short-term 
performance.


- Original Message - 
From: "Hal Murray" 
To: "Tom Van Baak" ; "Discussion of precise time and 
frequency measurement" 
Cc: 
Sent: Saturday, July 23, 2016 5:59 PM
Subject: Q/noise of Earth as an oscillator


> 
> t...@leapsecond.com said:
>> Earth is a very noisy, wandering, drifting, incredibly-expensive-to-measure,
>> low-precision (though high-Q) clock.
> 
> What is the Q of the Earth?  It might be on one of your web pages, but I 
> don't remember seeing it.  Google found a few mentions, but I didn't find a 
> number.
> 
> I did find an interesting list of damping mechanisms in a geology book.  
> Geology-nuts are as nutty as time-nuts.  Many were discussing damping of 
> seismic waves rather than rotation.
> 
> I've seen mention that the rotation rate of the Earth changed by a few 
> microseconds per day as a result of the 2011 earthquake in Japan.  Does that 
> show up in any data?  Your recent graph doesn't go back that far and it's got 
> a full scale of 2000 microseconds so a few is going to be hard to see.
> 

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Re: [time-nuts] Q/noise of Earth as an oscillator

[Graham / KE9H]

42

On Sat, Jul 23, 2016 at 7:59 PM, Hal Murray  wrote:

>
> t...@leapsecond.com said:
> > Earth is a very noisy, wandering, drifting,
> incredibly-expensive-to-measure,
> > low-precision (though high-Q) clock.
>
> What is the Q of the Earth?  It might be on one of your web pages, but I
> don't remember seeing it.  Google found a few mentions, but I didn't find a
> number.
>
> I did find an interesting list of damping mechanisms in a geology book.
> Geology-nuts are as nutty as time-nuts.  Many were discussing damping of
> seismic waves rather than rotation.
>
> I've seen mention that the rotation rate of the Earth changed by a few
> microseconds per day as a result of the 2011 earthquake in Japan.  Does
> that
> show up in any data?  Your recent graph doesn't go back that far and it's
> got
> a full scale of 2000 microseconds so a few is going to be hard to see.
>
>
>
> --
> These are my opinions.  I hate spam.
>
>
>
> ___
> time-nuts mailing list -- time-nuts@febo.com
> To unsubscribe, go to
> https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> and follow the instructions there.
>
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