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

bg bg at lysator.liu.se
Mon Aug 1 16:41:26 EDT 2016

```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 <tvb at LeapSecond.com> Date: 01/08/2016  20:01  (GMT+01:00) To: Discussion of precise time and frequency measurement <time-nuts at 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" <glennmaillist at bellsouth.net>
To: "Tom Van Baak" <tvb at leapsecond.com>; "Discussion of precise time and frequency measurement" <time-nuts at 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" <jim77742 at gmail.com>
> To: "Discussion of precise time and frequency measurement" <time-nuts at 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 <tvb at 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 at febo.com
>> To unsubscribe, go to
>> https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
>> and follow the instructions there.
>>
> _______________________________________________
> time-nuts mailing list -- time-nuts at febo.com
> To unsubscribe, go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> and follow the instructions there.
>
> ----- Original Message -----
> From: "Jim Palfreyman" <jim77742 at gmail.com>
> To: "Discussion of precise time and frequency measurement" <time-nuts at 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 <tvb at 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 at febo.com
>> To unsubscribe, go to
>> https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
>> and follow the instructions there.
>>
> _______________________________________________
> time-nuts mailing list -- time-nuts at febo.com
> To unsubscribe, go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts
> and follow the instructions there.
> _______________________________________________
> time-nuts mailing list -- time-nuts at 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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Glenn Little                ARRL Technical Specialist   QCWA  LM 28417
Amateur Callsign:  WB4UIV            wb4uiv at arrl.net    AMSAT LM 2178
QTH:  Goose Creek, SC USA (EM92xx)  USSVI LM   NRA LM   SBE ARRL TAPR
"It is not the class of license that the Amateur holds but the class
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