Help me understand Foucault's pendulum?

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Date: 02 Sep 2006 12:25:41
From:
Subject: Help me understand Foucault's pendulum?

I always thought I understood why the Foucault pendulum rotates, but
after giving it some more thought recently I'm puzzled about something.

Explanations I've seen seem to rest either on the pendulum swinging in
the same direction as the Earth turns (it's easy to picture this at
either pole), or the Coriolis effect, which would explain why it turns
clockwise from a N-S swing, in the Northern Hemisphere. The Coriolis
Effect would also explain why it doesn't rotate at the equator: the
force works equally in both directions, for the downward and upward
halves of the swing respectively.

What I don't understand is this: Why, if it's the Coriolis effect, does
the midlatitude pendulum ever move past a due E/W swing. There is no
Coriolis effect then, because the relative speed of the Earth's
rotation is equal underfoot throughout the whole swing. Although
eastward winds (westerlies) will in fact curve south in the northern
hemisphere, it's because they are shunted over by the winds blowing
toward the northeast, which are affected by the Coriolis. But for the
pendulum, this wouldn't be true, and a N/S pushed pendulum would turn,
quickly at first, but then turn ever more slowly to stop at E/W.

But I know this isn't what happens!

And when I follow the always-swings-in-the-same-direction explanation,
I wonder why the speed of the pendulum's rotation isn't affected by the
angle of the original push of the pendulum, but is affected by
latitude? For example if I swung a pendulum due NE/SW at any
midlatitude, wouldn't it move to N/S when the Earth rotated 90 degrees,
then to NW/SE at 180 degrees, reverse to NS 270 and then back to NE/SW
at the full 360?? Sort of a 12-hour cycle? And if I shoved it just 10
degrees off of axis north, it would move to north, -10, north, +10
during a day -- at any midlatitude; the same 12 hour cycle but less
degree distance.

You could still tell time, but I know that's not what actually happens!
Beyond the polar and equatorial cases, I'm on shaky ground. Help?

Date: 02 Sep 2006 19:53:49
From: Alan French
Subject: Re: Help me understand Foucault's pendulum?

http://en.wikipedia.org/wiki/Foucault_pendulum

http://www.si.edu/resource/faq/nmah/pendulum.htm

Date: 03 Sep 2006 13:19:02
From: Rob Johnson
Subject: Re: Help me understand Foucault's pendulum?

In article <1157225141.524847.78590@i3g2000cwc.googlegroups.com >,
E-Howenstine@neiu.edu wrote:
>I always thought I understood why the Foucault pendulum rotates, but
>after giving it some more thought recently I'm puzzled about something.
>
>Explanations I've seen seem to rest either on the pendulum swinging in
>the same direction as the Earth turns (it's easy to picture this at
>either pole), or the Coriolis effect, which would explain why it turns
>clockwise from a N-S swing, in the Northern Hemisphere. The Coriolis
>Effect would also explain why it doesn't rotate at the equator: the
>force works equally in both directions, for the downward and upward
>halves of the swing respectively.
>
>What I don't understand is this: Why, if it's the Coriolis effect, does
>the midlatitude pendulum ever move past a due E/W swing. There is no
>Coriolis effect then, because the relative speed of the Earth's
>rotation is equal underfoot throughout the whole swing. Although
>eastward winds (westerlies) will in fact curve south in the northern
>hemisphere, it's because they are shunted over by the winds blowing
>toward the northeast, which are affected by the Coriolis. But for the
>pendulum, this wouldn't be true, and a N/S pushed pendulum would turn,
>quickly at first, but then turn ever more slowly to stop at E/W.

This is not true. The Coriolis effect does affect points separated
along a parallel of latitude as well as points separated along a line
of longitude (meridian). The point of confusion is that it is the
speed of the surface velocity that changes as a point moves along a
meridian; however, it is the direction of the surface velocity that
changes as a point moves along a parallel. Except at the equator, a
parallel is not a geodesic. Thus, in the northern hemisphere, the
surface velocity is constantly veering counterclockwise, while in the
southern hemisphere, it is veering clockwise. This is why a pendulum
does not get stuck swinging east-west.

>But I know this isn't what happens!
>
>And when I follow the always-swings-in-the-same-direction explanation,
>I wonder why the speed of the pendulum's rotation isn't affected by the
>angle of the original push of the pendulum, but is affected by
>latitude? For example if I swung a pendulum due NE/SW at any
>midlatitude, wouldn't it move to N/S when the Earth rotated 90 degrees,
>then to NW/SE at 180 degrees, reverse to NS 270 and then back to NE/SW
>at the full 360?? Sort of a 12-hour cycle? And if I shoved it just 10
>degrees off of axis north, it would move to north, -10, north, +10
>during a day -- at any midlatitude; the same 12 hour cycle but less
>degree distance.
>
>You could still tell time, but I know that's not what actually happens!
> Beyond the polar and equatorial cases, I'm on shaky ground. Help?

The whole point is that, relative to an inertial frame, the surface of
the Earth is rotating counterclockwise in the northern hemisphere and
clockwise in the southern hemisphere. Empirically, this rotation is
15 degrees per hour counterclockwise times the sine of the latitude.
At my latitude (34.2219 degrees north), that is 8.436 degrees per hour
counterclockwise, which is why a pendulum, being inertial, will appear
to rotate 8.436 degrees per hour clockwise.

Rob Johnson <rob@trash.whim.org >
take out the trash before replying
to view any ASCII art, display article in a monospaced font

Date: 03 Sep 2006 02:46:29
From: oriel36
Subject: Re: Help me understand Foucault's pendulum?

Alan French wrote:
> http://en.wikipedia.org/wiki/Foucault_pendulum
>
> http://www.si.edu/resource/faq/nmah/pendulum.htm

It always returns to the same thing,whether the Newtonian ballistic
agenda or its exotic 20th century extensions,the value to assign to
axial rotation.

The Wiki articles determines 23 hours 56 min for the poles while
another group correctly identifies the value as 24 hours or 15 degrees
of rotation per hour -

http://www.phys-astro.sonoma.edu/people/students/baker/SouthPoleFoucault.html

The astronomical principles only support 15 degrees per hour derived
from the two step process which renders the pre-Copernican equable 24
hour day derived from the noon Equation of Time correction into its
heliocentric adaption to the principle that the Earth has an
independent axial rotation.

The entire empirical framework supporting the hideous correlation
between the return of a star to a meridian and axial rotation is built
on a false premise -

"Flamsteed used the star Sirius as a timekeeper correcting the sidereal
time obtained from successive transits of the star into solar time, the
difference of course being due to the rotation of the Earth round the
Sun. Flamsteed wrote in a letter in 1677:-

... our clocks kept so good a correspondence with the Heavens that I
doubt it not but they would prove the revolutions of the Earth to be
isochronical... "

The correlation between clocks,terrestial longitudes and axial
rotation is based strictly between our parent star and the motions of
the Earth therefore appeals to the stellar background in determining
axial rotation prove to be sterile,barren and counter-productive.

Date: 03 Sep 2006 23:47:56
From: Alan French
Subject: Re: Help me understand Foucault's pendulum?

"oriel36" <geraldkelleher@yahoo.com > wrote in message
news:1157276789.704734.269890@e3g2000cwe.googlegroups.com...
>
> Alan French wrote:
> > http://en.wikipedia.org/wiki/Foucault_pendulum
> >
> > http://www.si.edu/resource/faq/nmah/pendulum.htm
>
> It always returns to the same thing,whether the Newtonian ballistic
> agenda or its exotic 20th century extensions,the value to assign to
> axial rotation.
>
> The Wiki articles determines 23 hours 56 min for the poles while
> another group correctly identifies the value as 24 hours or 15 degrees
> of rotation per hour -
>
>
http://www.phys-astro.sonoma.edu/people/students/baker/SouthPoleFoucault.html

I think they got it wrong, and their experiment certainly did not sound
rigorous enough to differentiate between a 24 hour day and a sidereal day.
The pendulum swings in a fixed plane relative to the stars, so the Wikipedia
value of a sidereal day should be correct.

Clear skies, Alan

Date: 04 Sep 2006 21:49:49
From: Oscar Lanzi III
Subject: Re: Help me understand Foucault's pendulum?

The true rotational period is in fact 23 hours 56 minutes 04 seconds
(give or take a bit). But because the Earth is also orbiting the Sun,
that motion must be comnbined with the rotational one to determine the
length of a daily "noon to noon" cycle as determined by observing the
Sun from Earth. The orbital motion is just enough to account for the
difference between 23:56:04 and 24:00:00.

Almost, that is. Earth's orbit is ELLIPICAL, which means that the
angular velocity of the orbital motion changes slightly. Therefore, the
time between one "noon" and the next oscillates slightly about its mean
of 24 hours. The effect is to alter sunrise and sunset times by several
minutes at some times of year.

--OL

Date: 05 Sep 2006 03:25:42
From: oriel36
Subject: Re: Help me understand Foucault's pendulum?

Oscar Lanzi III wrote:
> The true rotational period is in fact 23 hours 56 minutes 04 seconds
> (give or take a bit). But because the Earth is also orbiting the Sun,
> that motion must be comnbined with the rotational one to determine the
> length of a daily "noon to noon" cycle as determined by observing the
> Sun from Earth. The orbital motion is just enough to account for the
> difference between 23:56:04 and 24:00:00.
>

Never ,ever try to justify the return of a star to a terrestial
meridian in 23 hours 56 mion 04 sec using the axial and orbital motions
of the Earth.

http://astrosun2.astro.cornell.edu/academics/courses//astro201/images/sidereal_day.gif

Now sunshine,put the .986 degree orbital angle in an elliptical
framework and the Earth would travel faster along its orbital
circumference the further from the Sun it was.Of course Keplerian
orbital geometry states the opposite.

> Almost, that is. Earth's orbit is ELLIPICAL, which means that the
> angular velocity of the orbital motion changes slightly. Therefore, the
> time between one "noon" and the next oscillates slightly about its mean
> of 24 hours. The effect is to alter sunrise and sunset times by several
> minutes at some times of year.
>
> --OL

The actual astronomical principle which keeps the hand of a clock in
sync with axial rotation have nothing to do with the junk you and your
dynamicist friends have spouted for the last 3 centuries.

The astronomical principles are enjoyable,easy to understand and have
absolutely nothing whatsoever to do with the background stars and
celestial sphere geometry.It is a two step process and I request that
somebody else explain exactly how the pre-Copernican principle which
created the 24 hour day from the noon Equation of Time correction was
adapted by heliocentric astronomers to the principle that axial
rotation is an independent motion hence 15 degrees per hour and 24
hours/360 degrees exactly.

As far as I know,the only other people who have problems with the 24
hour day and natural phenomena are the creationists so drop the silly
23 hours 56 min 04 sec notion for axial rotation as quickely as you can
even though the entire Newtonian framework is based on that value.

Date: 06 Sep 2006 06:02:10
From: oriel36
Subject: Re: Help me understand Foucault's pendulum?

Richard Tobin wrote:
> In article <1157537137.994777.193140@m79g2000cwm.googlegroups.com>,
> oriel36 <geraldkelleher@yahoo.com> wrote:
>
> >It is hard to imagine that in the 21st century that I have to argue why
> >the Earth's axial rotation is exactly 15 degrees per hour .
>
> Do you have a special watch that runs fast, or do you have a special
> protractor marked in deviant degrees?
>
> What time is midnight in your world?
>
> -- Richard

The principles behind axial rotation and the 24 hour clock are there
for everyone to enjoy,being easy enough to understand and requiring
nothing more than the simple appreciation that the total length of the
natural day, using noon as the beginning and end of a cycle,is
unequal.

It does not require the background stars,it just requires the
admiration for our ancestors who created the equable 24 hour day by
applying a correction at each day to observation of natural noon by
adding or subtracting minutes and seconds thereby equalising the
variations to 24 hours exactly .One 24 hour day elapses seamlessly
into the next 24 hour day and from this important piece of information
the heliocentric astronomers set to work.

As one 24 hour day elapses into the next 24 hour cycle ,the
heliocentric astronomers (the earliest appears to be Gemma Frisius in
1545 ) being familiar with the idea that the Earth has an independent
and constant axial rotation,adapted the 24 hour to terrestial
lomgitudes and axial rotation at 15 degrees per hour.They never needed
an external reference to infer that axial rotation was independent for
they already knew that Copernicus had justified the orbital motion of
the Earth as an independent motion.You can see the orbital motion of
the Earth in an inner orbital circuit overtaking the slower moving
Jupiter and Saturn to affirm this -

http://antwrp.gsfc.nasa.gov/apod/image/0112/JuSa2000_tezel.gif

So,what an incredible and pragmatic principle the correlation between
clocks,terrestial longitudes and axial rotation at 15 degres per hour
is and 24 hours/360 degrees in total .It never requires an external
justification like orbital motion does hence the ridiculous attempt by
celestial sphere geometers to justify axial rotation through the return
of a star in 23 hours 56 min 04 sec.

Why bother sticking with celestial sphere geometry when future
generations are going to shake their heads in wonder how a
civilisatuion became so stupid when they had modern imaging before
them.

In any case, the Earth rotational correlation with the pace of a clock
hand is 24 hours/360 degrees ,any other value is astronomical
creationism.

Date: 06 Sep 2006 03:05:38
From: oriel36
Subject: Re: Help me understand Foucault's pendulum?

John Carruthers wrote:
> Oscar Lanzi III wrote:
> > The true rotational period is in fact 23 hours 56 minutes 04 seconds
> > (give or take a bit). But because the Earth is also orbiting the Sun,
> > that motion must be comnbined with the rotational one to determine the
> > length of a daily "noon to noon" cycle as determined by observing the
> > Sun from Earth. The orbital motion is just enough to account for the
> > difference between 23:56:04 and 24:00:00.
> >
> > Almost, that is. Earth's orbit is ELLIPICAL, which means that the
> > angular velocity of the orbital motion changes slightly. Therefore, the
> > time between one "noon" and the next oscillates slightly about its mean
> > of 24 hours. The effect is to alter sunrise and sunset times by several
> > minutes at some times of year.
> >
> > --OL
>
> You are wasting your breath on this one Oscar. I engage in light troll
> baiting but serious education is wasted here. He's never even built a
> sundial let alone an accurate Orrery or a telescope drive :-)
> jc

It is hard to imagine that in the 21st century that I have to argue why
the Earth's axial rotation is exactly 15 degrees per hour .

If you want to say it is 15.041 degrees per hour like Rob here then
fine,whatever people did to deserve this nonsense I do not know,that
they accept it is altogether another matter.I have to consider that Rob
is 10 years old along with the Harry Potter fan Brian,but I assume most
here are adults who can freely decide for themselves the geometry which
creates the 24 hour day and how it applies to axial rotation.

Never,ever justify the return of a star to a terrestial meridian in 23
hours 56 min 04 sec using the axial and orbital motions of the Earth..

http://astrosun2.astro.cornell.edu/academics/courses//astro201/images/sidereal_day.gif

Date: 06 Sep 2006 12:21:04
From: Richard Tobin
Subject: Re: Help me understand Foucault's pendulum?

In article <1157537137.994777.193140@m79g2000cwm.googlegroups.com >,
oriel36 <geraldkelleher@yahoo.com > wrote:

>It is hard to imagine that in the 21st century that I have to argue why
>the Earth's axial rotation is exactly 15 degrees per hour .

Do you have a special watch that runs fast, or do you have a special
protractor marked in deviant degrees?

What time is midnight in your world?

-- Richard

Date: 06 Sep 2006 02:54:02
From: oriel36
Subject: Re: Help me understand Foucault's pendulum?

Eugene Griessel wrote:
> "John Carruthers" <joncarruthers@hotmail.com> wrote:
>
> >
> >Oscar Lanzi III wrote:
> >> The true rotational period is in fact 23 hours 56 minutes 04 seconds
> >> (give or take a bit). But because the Earth is also orbiting the Sun,
> >> that motion must be comnbined with the rotational one to determine the
> >> length of a daily "noon to noon" cycle as determined by observing the
> >> Sun from Earth. The orbital motion is just enough to account for the
> >> difference between 23:56:04 and 24:00:00.
> >>
> >> Almost, that is. Earth's orbit is ELLIPICAL, which means that the
> >> angular velocity of the orbital motion changes slightly. Therefore, the
> >> time between one "noon" and the next oscillates slightly about its mean
> >> of 24 hours. The effect is to alter sunrise and sunset times by several
> >> minutes at some times of year.
> >>
> >> --OL
> >
> >You are wasting your breath on this one Oscar. I engage in light troll
> >baiting but serious education is wasted here. He's never even built a
> >sundial let alone an accurate Orrery or a telescope drive :-)
>
> Or used a sextant to navigate with - a challenge I have issued to him
> repeatedly.
>
> Eugene L Griessel
>
> A critic is a man who knows the way, but cannot drive the car.

Direct him to your website where you explain the relationship betwen
clocks, terrestial longitudes and axial rotation at 15 degrees per hour
making 24 hours/360 degrees in total -

Longitude
"Imagine the earth as a large orange. Slice the peel into gores as if
you were about to peel it. These "slices" equate to the lines of
longitude. Longitude is a "man made" measure insofar that the line from
which all longitude is measured is the line that runs through the old
naval observatory in Greenwich. Longitudinal distances are related to
time and every 15 is the distance the earth rotates in one hour. The
distance of a degree of longitude varies with the place on earth it is
measured. "

http://www.dynagen.co.za/eugene/where/index.html

Saying things like this is not going to get you invited to star parties
Gene but it happens to be correct. The working principles for the rate
of axial rotation beneath Foucault's pendulum will be 24 hours/360
degrees at the surface geographical location of the rotational axis
of the Earth.

Date: 06 Sep 2006 01:20:19
From: John Carruthers
Subject: Re: Help me understand Foucault's pendulum?

Date: 06 Sep 2006 08:24:47
From: Eugene Griessel
Subject: Re: Help me understand Foucault's pendulum?

"John Carruthers" <joncarruthers@hotmail.com > wrote:

>
>Oscar Lanzi III wrote:
>> The true rotational period is in fact 23 hours 56 minutes 04 seconds
>> (give or take a bit). But because the Earth is also orbiting the Sun,
>> that motion must be comnbined with the rotational one to determine the
>> length of a daily "noon to noon" cycle as determined by observing the
>> Sun from Earth. The orbital motion is just enough to account for the
>> difference between 23:56:04 and 24:00:00.
>>
>> Almost, that is. Earth's orbit is ELLIPICAL, which means that the
>> angular velocity of the orbital motion changes slightly. Therefore, the
>> time between one "noon" and the next oscillates slightly about its mean
>> of 24 hours. The effect is to alter sunrise and sunset times by several
>> minutes at some times of year.
>>
>> --OL
>
>You are wasting your breath on this one Oscar. I engage in light troll
>baiting but serious education is wasted here. He's never even built a
>sundial let alone an accurate Orrery or a telescope drive :-)

Or used a sextant to navigate with - a challenge I have issued to him
repeatedly.

Eugene L Griessel

A critic is a man who knows the way, but cannot drive the car.

Date: 03 Sep 2006 13:39:20
From: oriel36
Subject: Re: Help me understand Foucault's pendulum?

I loved the Wiki article on this and the deviation of the Earth from a
perfect sphere arising from the dead center of the Earth

http://en.wikipedia.org/wiki/Foucault_pendulum

The actual mechanism is differential rotation along the entire length
of the Earth's axis in the molten/flexible interior.If you have
problems wondering how the flexible/molten interior behaves then here
latitude dependent differential rotation visible on the external plasma
of the Sun as a guide -

http://www.astronomynotes.com/starsun/sun-rotation.gif

The bonus is that the rotational dynamics of the molten interior also
provide the mechanism for the motion of the fractured surface crust.

This is what real astronomy is all about,but you have to get the value
for axial rotation right.Why you insist in referencing it off celestial
sphere geometry is anyone's guess but time to grow up.

Rob Johnson wrote:
> In article <1157225141.524847.78590@i3g2000cwc.googlegroups.com>,
> E-Howenstine@neiu.edu wrote:
> >I always thought I understood why the Foucault pendulum rotates, but
> >after giving it some more thought recently I'm puzzled about something.
> >
> >Explanations I've seen seem to rest either on the pendulum swinging in
> >the same direction as the Earth turns (it's easy to picture this at
> >either pole), or the Coriolis effect, which would explain why it turns
> >clockwise from a N-S swing, in the Northern Hemisphere. The Coriolis
> >Effect would also explain why it doesn't rotate at the equator: the
> >force works equally in both directions, for the downward and upward
> >halves of the swing respectively.
> >
> >What I don't understand is this: Why, if it's the Coriolis effect, does
> >the midlatitude pendulum ever move past a due E/W swing. There is no
> >Coriolis effect then, because the relative speed of the Earth's
> >rotation is equal underfoot throughout the whole swing. Although
> >eastward winds (westerlies) will in fact curve south in the northern
> >hemisphere, it's because they are shunted over by the winds blowing
> >toward the northeast, which are affected by the Coriolis. But for the
> >pendulum, this wouldn't be true, and a N/S pushed pendulum would turn,
> >quickly at first, but then turn ever more slowly to stop at E/W.
>
> This is not true. The Coriolis effect does affect points separated
> along a parallel of latitude as well as points separated along a line
> of longitude (meridian). The point of confusion is that it is the
> speed of the surface velocity that changes as a point moves along a
> meridian; however, it is the direction of the surface velocity that
> changes as a point moves along a parallel. Except at the equator, a
> parallel is not a geodesic. Thus, in the northern hemisphere, the
> surface velocity is constantly veering counterclockwise, while in the
> southern hemisphere, it is veering clockwise. This is why a pendulum
> does not get stuck swinging east-west.
>
> >But I know this isn't what happens!
> >
> >And when I follow the always-swings-in-the-same-direction explanation,
> >I wonder why the speed of the pendulum's rotation isn't affected by the
> >angle of the original push of the pendulum, but is affected by
> >latitude? For example if I swung a pendulum due NE/SW at any
> >midlatitude, wouldn't it move to N/S when the Earth rotated 90 degrees,
> >then to NW/SE at 180 degrees, reverse to NS 270 and then back to NE/SW
> >at the full 360?? Sort of a 12-hour cycle? And if I shoved it just 10
> >degrees off of axis north, it would move to north, -10, north, +10
> >during a day -- at any midlatitude; the same 12 hour cycle but less
> >degree distance.
> >
> >You could still tell time, but I know that's not what actually happens!
> > Beyond the polar and equatorial cases, I'm on shaky ground. Help?
>
> The whole point is that, relative to an inertial frame, the surface of
> the Earth is rotating counterclockwise in the northern hemisphere and
> clockwise in the southern hemisphere. Empirically, this rotation is
> 15 degrees per hour counterclockwise times the sine of the latitude.
> At my latitude (34.2219 degrees north), that is 8.436 degrees per hour
> counterclockwise, which is why a pendulum, being inertial, will appear
> to rotate 8.436 degrees per hour clockwise.
>
> Rob Johnson <rob@trash.whim.org>
> take out the trash before replying
> to view any ASCII art, display article in a monospaced font

Date: 04 Sep 2006 05:56:12
From: oriel36
Subject: Re: Help me understand Foucault's pendulum?

Alan French wrote:
> "oriel36" <geraldkelleher@yahoo.com> wrote in message
> news:1157276789.704734.269890@e3g2000cwe.googlegroups.com...
> >
> > Alan French wrote:
> > > http://en.wikipedia.org/wiki/Foucault_pendulum
> > >
> > > http://www.si.edu/resource/faq/nmah/pendulum.htm
> >
> > It always returns to the same thing,whether the Newtonian ballistic
> > agenda or its exotic 20th century extensions,the value to assign to
> > axial rotation.
> >
> > The Wiki articles determines 23 hours 56 min for the poles while
> > another group correctly identifies the value as 24 hours or 15 degrees
> > of rotation per hour -
> >
> >
> http://www.phys-astro.sonoma.edu/people/students/baker/SouthPoleFoucault.html
>
> I think they got it wrong, and their experiment certainly did not sound
> rigorous enough to differentiate between a 24 hour day and a sidereal day.
> The pendulum swings in a fixed plane relative to the stars, so the Wikipedia
> value of a sidereal day should be correct.
>
> Clear skies, Alan

There is nothing dull or dreary about the founding principles which
keep the clock hand in sync with axial rotation at 15 degrees per
hour,it just takes a sharper mind to recognise that it is a two stage
process which links the whole thing together.One of the greatest tools
to emerge is Goggle Books,things such as the Equation of Time can be
put into historical context,even where Flamsteed went astray is
assigning a cause for the natural inequality -

http://books.google.com/books?vid=0e8wj0K_xQ2xlXO4iX&id=INpRAJ0dxGYC&pg=PA11&lpg=PA12&dq=equation+of+time+longitude

The astronomical nuances can be oh so subtle,the extraction of the 24
hours day from the natural unequal day and the application of the
Equation of Time to keep one 24 hour day elapsing seamlessly into the
next 24 hour day.This turning from one 24 hour day to the next is vital
for the heliocentric adaption which spread the 24 hours across the
globe as terrestial longitudes and treated axial rotation as an
independent motion.No references were ever required to justify axial
rotation at 15 degrees per hour much less the two that the sidereal
justification requires.

The people at the South pole,just like John Harrison and his
watch,worked off the principle that axial rotation is 15 degrees per
hour and 24 hours/360 degrees in total.How long people wish to remain
with the sidereal justification is anyone's guess but it affects
everything ahead of it where the Earth's axial and orbital motions are
required.

Never,ever try to justify the return of a star in 23 hours 56 min 04
sec through the axial and orbital motion of the Earth is probably the
harshest judgement I can offer but otherwise it is time to recover
productive astronomical working principles based on an accurate look at
axial and orbital motions.

Date: 04 Sep 2006 02:57:28
From: oriel36
Subject: Re: Help me understand Foucault's pendulum?

Date: 04 Sep 2006 13:53:52
From: oriel36
Subject: Re: Help me understand Foucault's pendulum?

Rob Johnson wrote:
> In article <1157225141.524847.78590@i3g2000cwc.googlegroups.com>,

> The whole point is that, relative to an inertial frame, the surface of
> the Earth is rotating counterclockwise in the northern hemisphere and
> clockwise in the southern hemisphere.

Goodness me !!,how did I miss this one.

This is incredibly funny,I know I may have to enjoy it all by myself
but given that the same dynamicists as this guy removed Pluto from
astrophotographers,I wonder how long it will be before somebody else
realises just what a nightmare of a situation exists.

I am really beginning to enjoy this comical situation unfold,while
astronomical methods and principles are temporarily buried under 3
centuries of empirical junk,they can be restored easily enough.In the
meantime,this stuff is priceless for its comical content.

Date: 04 Sep 2006 12:41:12
From:
Subject: Re: Help me understand Foucault's pendulum?

I can see that, on a non-rotating planet, a pendulum starting a swing
due East in the N hemisphere would bend slightly southeast (from the
surface vantage) because the midlatitude parallel does not follow a
great arc, while a pendulum does. On the backswing, I would think it
would follow the same arc and therefore go back and forth in place.
Add the Earth's spin, and in theory at least you would have a little
Coriolis effect because there is a little N/S motion of any pendulum,
as none can follow any midlatitude parallel. But the Coriolis effect on
that swing would be miniscule, wouldn't it? And if the Coriolis is
miniscule near E/W'ish trajectory, although the pendulum would not be
stuck there, it would move through that area very very slowly. But in
fact, it precedes at a fixed rate. ?

Rob Johnson wrote:

... The Coriolis effect does affect points separated
along a parallel of latitude as well as points separated along a line
of longitude (meridian). The point of confusion is that it is the
speed of the surface velocity that changes as a point moves along a
meridian; however, it is the direction of the surface velocity that
changes as a point moves along a parallel. Except at the equator, a
parallel is not a geodesic. Thus, in the northern hemisphere, the
surface velocity is constantly veering counterclockwise, while in the
southern hemisphere, it is veering clockwise. This is why a pendulum
does not get stuck swinging east-west.

Date: 04 Sep 2006 12:41:06
From:
Subject: Re: Help me understand Foucault's pendulum?

Date: 05 Sep 2006 16:16:30
From: Rob Johnson
Subject: Re: Help me understand Foucault's pendulum?

In article <1157398866.013860.268050@p79g2000cwp.googlegroups.com >,
E-Howenstine@neiu.edu wrote:
>I can see that, on a non-rotating planet, a pendulum starting a swing
>due East in the N hemisphere would bend slightly southeast (from the
>surface vantage) because the midlatitude parallel does not follow a
>great arc, while a pendulum does. On the backswing, I would think it
>would follow the same arc and therefore go back and forth in place.
>
>Add the Earth's spin, and in theory at least you would have a little
>Coriolis effect because there is a little N/S motion of any pendulum,
>as none can follow any midlatitude parallel. But the Coriolis effect on
>that swing would be miniscule, wouldn't it? And if the Coriolis is
>miniscule near E/W'ish trajectory, although the pendulum would not be
>stuck there, it would move through that area very very slowly. But in
>fact, it precedes at a fixed rate. ?

Let's compute the difference in tangential velocity between two points
on a spinning planet, with period p and radius r. To simplify the
formulas in the following, we will measure longitude and latitude in
radians.

First, we will look at the case which seems more easily understood:
two points along a line of longitude. Suppose we are standing on a
parallel of latitude, b, and we are observing a point on the surface
and the same meridian at latitude b+db. The directions that we and
the observed point are moving are the same; the only difference is
their magnitude (speed). Our speed is 2 pi r/p cos(b) eastward while
the speed of the observed point is 2 pi r/p cos(b+db) eastward, which
is approximately 2 pi r/p (cos(b) - sin(b) db). Thus, the relative
velocity of the observed point is -2 pi r/p sin(b) db east. When the
pendulum is over the observed point, it would seem to be moving with
a relative velocity tangential to the surface of 2 pi r/p sin(b) db
eastward. On the surface, dy = r db, so the relative velocity is
2 pi sin(b)/p dy east

Next, consider the case which seems to be less easily understood:
two points along a parallel of latitude. Suppose we are standing on
a parallel of latitude, b, at east longitude a, and we are observing
a point on the surface at the same latitude, but at longitude a+da.
Our speed is the same as that of the observed point, but its direction
is different. Since the motion of both is in a plane perpindicular to
the axis of rotation, so must be their difference. For small da, the
difference in velocity is approximately the common speed times the
change in direction: 2 pi r/p cos(b) da. Furthermore, this difference
in velocity is almost perpindicular to our velocity and, as mentioned
above, perpindicular to the axis of rotation. Thus, the change in
velocity is directly toward the axis, however, we want the tangential
change in velocity. Since the change in velocity is directly toward
the axis, the magnitude of its tangential component is sin(b) times
that of the full difference 2 pi r/p cos(b) da north. On the surface,
dx = r cos(b) da, so the relative velocity is 2 pi sin(b)/p dx north.
When the pendulum is over the observed point, it would seem to be
moving with a relative velocity tangential to the surface of
2 pi sin(b)/p dx south.

Summarizing, swinging a distance dy north makes the pendulum move
with a relative eastward velocity of 2 pi sin(b)/p dy and swinging
east a distance dx makes the pendulum move with a relative southward
velocity of 2 pi sin(b)/p dx. Thus, swinging east-west is similar to
swinging north-south. They both produce a relative clockwise rotation
with angular velocity 2 pi sin(b)/p.

This can also be verified with very sensitive accelerometers. They
tell us that we are all spinning on the surface of the Earth with an
angular velocity 2 pi sin(b)/p. Luckily, this is not nearly enough
to make us dizzy, even at the poles where the rotation is a bit more
than 15 degrees per hour (approximately 15.041 degrees per hour).

>Rob Johnson wrote:
>
>... The Coriolis effect does affect points separated
>along a parallel of latitude as well as points separated along a line
>of longitude (meridian). The point of confusion is that it is the
>speed of the surface velocity that changes as a point moves along a
>meridian; however, it is the direction of the surface velocity that
>changes as a point moves along a parallel. Except at the equator, a
>parallel is not a geodesic. Thus, in the northern hemisphere, the
>surface velocity is constantly veering counterclockwise, while in the
>southern hemisphere, it is veering clockwise. This is why a pendulum
>does not get stuck swinging east-west.

I hope the explanation above makes things a bit clearer. If not, at
least it provides some mathematical support.

Rob Johnson <rob@trash.whim.org >
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Date: 05 Sep 2006 14:36:45
From: Brian Tung
Subject: Re: Help me understand Foucault's pendulum?

> You could still tell time, but I know that's not what actually happens!
> Beyond the polar and equatorial cases, I'm on shaky ground. Help?

It sounds like you've got it, but I'll take this opportunity to say that
it originally didn't make sense to me, either. (You can even find my
posts on it on SAA several years ago!) Especially the explanation that
the pendulum continues to swing in the same plane. That makes sense at
the poles, but it doesn't really make much sense at the equator. Also,
it seems to suggest that after a day, the pendulum should appear to be
swinging in the same direction (relative to the ground), which it does
not. It only does that at the poles and at the equator (well, and at
30 degrees latitude, for reasons I'll explain in a bit).

Here's one way to visualize it that helped me. The first insight is to
recognize that the effect would be unchanged if the Earth were a big
cone (with gravity continuing to pull downward, of course). If you live
at, say, 50 degrees north latitude, then at your latitude, the pendulum
reacts pretty much as it would if the surface of the Earth were a big
cone with a (half) angle of 50 degrees. (That is, a big cone pointed
north, where the angle at the tip is 100 degrees.)

You can cut that cone from the base to the tip and unfurl it flat, so
that it looks a bit like a Pac-Man. It's not a complete circle because
the parallel at 50 degrees latitude is not as long all the way around as
the equator at 0 degrees latitude is. As a result, if you were to mark
points of 0, 90, 180, and 270 degrees longitude along the outer edge of
the Pac-Man, they would not be 90 degrees apart. Instead, they would
be separated by 90 degrees times the sine of 50 degrees, or about 69
degrees.

Suppose we lay this Pac-Man out so that the "mouth" points toward the
right, and we put a penny on the upper "jaw," with the head on the penny
pointing to the lower left, toward the center--the north pole of the
cone. This penny prepresents a pendulum swinging north-south. One
quarter of a day later, it has moved 69 degrees counter-clockwise around
the edge of the Pac-Man. If you slide the penny over so that it still
points to the lower left, it is of course no longer pointed toward the
center. Instead, it is "off" by 69 degrees, indicating that in that
quarter of a day, the pendulum rotates, relative to the ground around
it, about 69 degrees, and therefore about four times that--276 degrees--
in an entire day. At the end of that day the penny is now at the lower
"jaw" of the Pac-Man.

There's nothing special about 50 degrees latitude--I could have used any
latitude at all. So the angle through which the pendulum rotates will
be 360 degrees times the sine of the latitude. That's why the pendulum
appears to be swinging in the same direction after one day at the
equator (since the sine of 0 degrees is 0) and at the poles (since the
sine of 90 degrees is 1). It also swings in the same direction at 30
degrees latitude, because the sine of 30 degrees is 1/2. In this case,
the Pac-Man has opened its mouth so widely that it's a semi-circle, with
the upper and lower "jaws" forming a straight line. After one day, the
pendulum is swinging south-north instead of north-south--but that looks
exactly the same.

--
Brian Tung <brian@isi.edu >
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.html

Date: 06 Sep 2006 01:19:06
From: Rob Johnson
Subject: Re: Help me understand Foucault's pendulum?

In article <edkqld$mi6$1@praesepe.isi.edu >,
brian@isi.edu (Brian Tung) wrote:
>> You could still tell time, but I know that's not what actually happens!
>> Beyond the polar and equatorial cases, I'm on shaky ground. Help?
>
>It sounds like you've got it, but I'll take this opportunity to say that
>it originally didn't make sense to me, either. (You can even find my
>posts on it on SAA several years ago!) Especially the explanation that
>the pendulum continues to swing in the same plane. That makes sense at
>the poles, but it doesn't really make much sense at the equator. Also,
>it seems to suggest that after a day, the pendulum should appear to be
>swinging in the same direction (relative to the ground), which it does
>not. It only does that at the poles and at the equator (well, and at
>30 degrees latitude, for reasons I'll explain in a bit).
>
>Here's one way to visualize it that helped me. The first insight is to
>recognize that the effect would be unchanged if the Earth were a big
>cone (with gravity continuing to pull downward, of course). If you live
>at, say, 50 degrees north latitude, then at your latitude, the pendulum
>reacts pretty much as it would if the surface of the Earth were a big
>cone with a (half) angle of 50 degrees. (That is, a big cone pointed
>north, where the angle at the tip is 100 degrees.)
>
>You can cut that cone from the base to the tip and unfurl it flat, so
>that it looks a bit like a Pac-Man. It's not a complete circle because
>the parallel at 50 degrees latitude is not as long all the way around as
>the equator at 0 degrees latitude is. As a result, if you were to mark
>points of 0, 90, 180, and 270 degrees longitude along the outer edge of
>the Pac-Man, they would not be 90 degrees apart. Instead, they would
>be separated by 90 degrees times the sine of 50 degrees, or about 69
>degrees.
>
>Suppose we lay this Pac-Man out so that the "mouth" points toward the
>right, and we put a penny on the upper "jaw," with the head on the penny
>pointing to the lower left, toward the center--the north pole of the
>cone. This penny prepresents a pendulum swinging north-south. One
>quarter of a day later, it has moved 69 degrees counter-clockwise around
>the edge of the Pac-Man. If you slide the penny over so that it still
>points to the lower left, it is of course no longer pointed toward the
>center. Instead, it is "off" by 69 degrees, indicating that in that
>quarter of a day, the pendulum rotates, relative to the ground around
>it, about 69 degrees, and therefore about four times that--276 degrees--
>in an entire day. At the end of that day the penny is now at the lower
>"jaw" of the Pac-Man.
>
>There's nothing special about 50 degrees latitude--I could have used any
>latitude at all. So the angle through which the pendulum rotates will
>be 360 degrees times the sine of the latitude. That's why the pendulum
>appears to be swinging in the same direction after one day at the
>equator (since the sine of 0 degrees is 0) and at the poles (since the
>sine of 90 degrees is 1). It also swings in the same direction at 30
>degrees latitude, because the sine of 30 degrees is 1/2. In this case,
>the Pac-Man has opened its mouth so widely that it's a semi-circle, with
>the upper and lower "jaws" forming a straight line. After one day, the
>pendulum is swinging south-north instead of north-south--but that looks
>exactly the same.

That is one of my favorite explanations, and I was thinking of writing
it up, but I wanted to show that the north-south and the east-west
swings produce the same deflection, and I didn't see a way to do that
without more math. Of course, I worry that the math in my explanation
may induce more shut-eye than understanding.

Rob Johnson <rob@trash.whim.org >
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Date: 05 Sep 2006 11:51:43
From: oriel36
Subject: Re: Help me understand Foucault's pendulum?

Rob Johnson wrote:
> In article <1157398866.013860.268050@p79g2000cwp.googlegroups.com>,

> This can also be verified with very sensitive accelerometers. They
> tell us that we are all spinning on the surface of the Earth with an
> angular velocity 2 pi sin(b)/p. Luckily, this is not nearly enough
> to make us dizzy, even at the poles where the rotation is a bit more
> than 15 degrees per hour (approximately 15.041 degrees per hour).
>

I love this !!.

15.041 degrees per hour at the poles making 360.986 degrees per 24 hour
day !.

Personally I liked the Wiki explanation which has 23 hours 56 min 04
sec which "Wraps Around" at 24 hours -

http://en.wikipedia.org/wiki/Sidereal_time

Make you dizzy ! ,the reasoning is incredibly funny when you do not
know the actual exquisite principles which keeps the clock hand in sync
with axial rotation without a need for an external reference.

Apparently astrophotographers will tolerate anything by the dynamicists
who would much prefer that planets not exist and they could discuss
what exists inbetween.

So who is going to be the first the news to Rob here that rotation is
and always will be 15 degrees per hour and 24 hours/360 degrees
exactly.

Date: 05 Sep 2006 10:58:11
From:
Subject: Aha -- thanks very much...

Rob - thanks for that explanation. That helps a lot. I'll summarize
my observations without any math, what do you think...

Imagine a pendulum at 45 degrees north latitude, swinging east.
Keep in mind these things...

1. The pendulum's pivot is at a fixed distance from the planet, and
moves with the Earth's rotation
2. The pendulum always cuts a great arc, in essence cutting the Earth
exactly in half with each swing
3. The platform is always flat on the earth's surface, so it swivels as
the Earth turns
4. The platform does not rotate with respect to latitude and longitude
-- the same side is always facing north

You can replicate this with the palm of your hand, fingers always
pointing north, placed on a large imaginary globe, spinning to the
right (east).

Let's say a swing of the pendulum is so slow that the Earth takes a
quarter turn under it. The platform has swiveled beneath the swinging
pendulum in a curious way. It has done three things: (1) fallen
eastward over the side of the Earth, because the planet has turned that
way (2) maintained its tilt toward the pole, because the platform is on
the curved surface of the Earth, and (3) rotated a bit counter
clockwise. This last bit is the most important. The equatorial side
of the platform has moved faster and farther than the polar side. It
has swept around the polar side of the platform causing the platform to
rotate -- not in respect to latitude and longitude, but in respect to
its original position. And in respect to the pendulum. The pendulum
starts an eastward swing and the platform is torqued a bit under it.
This is easily seen with your palm and imaginary globe, but is even
more clear if you tape a cardboard disk to a real globe. If you use a
pencil as a pendulum over a cardboard disk and real globe, it's easy to
see that it sketches a line that bends to the right on the swing and to
the right again on the return. Near the equator the effect is less to
none, and at the pole -- where the platform doesn't tilt at all -- the
sketched line is bent the most.

How's that?

Date: 06 Sep 2006 02:22:29
From: Rob Johnson
Subject: Re: Aha -- thanks very much...

In article <1157479091.122889.192510@i3g2000cwc.googlegroups.com >,
E-Howenstine@neiu.edu wrote:
>Rob - thanks for that explanation. That helps a lot. I'll summarize
>my observations without any math, what do you think...
>
>
>Imagine a pendulum at 45 degrees north latitude, swinging east.
>Keep in mind these things...
>
>1. The pendulum's pivot is at a fixed distance from the planet, and
>moves with the Earth's rotation
>2. The pendulum always cuts a great arc, in essence cutting the Earth
>exactly in half with each swing
>3. The platform is always flat on the earth's surface, so it swivels as
>the Earth turns
>4. The platform does not rotate with respect to latitude and longitude
>-- the same side is always facing north
>
>You can replicate this with the palm of your hand, fingers always
>pointing north, placed on a large imaginary globe, spinning to the
>right (east).
>
>Let's say a swing of the pendulum is so slow that the Earth takes a
>quarter turn under it. The platform has swiveled beneath the swinging
>pendulum in a curious way. It has done three things: (1) fallen
>eastward over the side of the Earth, because the planet has turned that
>way (2) maintained its tilt toward the pole, because the platform is on
>the curved surface of the Earth, and (3) rotated a bit counter
>clockwise. This last bit is the most important. The equatorial side
>of the platform has moved faster and farther than the polar side. It
>has swept around the polar side of the platform causing the platform to
>rotate -- not in respect to latitude and longitude, but in respect to
>its original position. And in respect to the pendulum. The pendulum
>starts an eastward swing and the platform is torqued a bit under it.
>This is easily seen with your palm and imaginary globe, but is even
>more clear if you tape a cardboard disk to a real globe. If you use a
>pencil as a pendulum over a cardboard disk and real globe, it's easy to
>see that it sketches a line that bends to the right on the swing and to
>the right again on the return. Near the equator the effect is less to
>none, and at the pole -- where the platform doesn't tilt at all -- the
>sketched line is bent the most.
>
>How's that?

This sounds to be very close to the idea of Brian Tung's explanation.
It sounds as if you now have a good grasp of how the Foucault Pendulum
works.

Rob Johnson <rob@trash.whim.org >
take out the trash before replying
to view any ASCII art, display article in a monospaced font

Date: 05 Sep 2006 22:20:27
From: Richard Adams
Subject: Re: Aha -- thanks very much...

Rob Johnson wrote:

> In article <1157479091.122889.192510@i3g2000cwc.googlegroups.com>,
> E-Howenstine@neiu.edu wrote:
>
>>Rob - thanks for that explanation. That helps a lot. I'll summarize
>>my observations without any math, what do you think...
>>
>>
>>Imagine a pendulum at 45 degrees north latitude, swinging east.
>>Keep in mind these things...
>>
>>1. The pendulum's pivot is at a fixed distance from the planet, and
>>moves with the Earth's rotation
>>2. The pendulum always cuts a great arc, in essence cutting the Earth
>>exactly in half with each swing
>>3. The platform is always flat on the earth's surface, so it swivels as
>>the Earth turns
>>4. The platform does not rotate with respect to latitude and longitude
>>-- the same side is always facing north
>>
>>You can replicate this with the palm of your hand, fingers always
>>pointing north, placed on a large imaginary globe, spinning to the
>>right (east).
>>
>>Let's say a swing of the pendulum is so slow that the Earth takes a
>>quarter turn under it. The platform has swiveled beneath the swinging
>>pendulum in a curious way. It has done three things: (1) fallen
>>eastward over the side of the Earth, because the planet has turned that
>>way (2) maintained its tilt toward the pole, because the platform is on
>>the curved surface of the Earth, and (3) rotated a bit counter
>>clockwise. This last bit is the most important. The equatorial side
>>of the platform has moved faster and farther than the polar side. It
>>has swept around the polar side of the platform causing the platform to
>>rotate -- not in respect to latitude and longitude, but in respect to
>>its original position. And in respect to the pendulum. The pendulum
>>starts an eastward swing and the platform is torqued a bit under it.
>>This is easily seen with your palm and imaginary globe, but is even
>>more clear if you tape a cardboard disk to a real globe. If you use a
>>pencil as a pendulum over a cardboard disk and real globe, it's easy to
>>see that it sketches a line that bends to the right on the swing and to
>>the right again on the return. Near the equator the effect is less to
>>none, and at the pole -- where the platform doesn't tilt at all -- the
>>sketched line is bent the most.
>>
>>How's that?
>
>
> This sounds to be very close to the idea of Brian Tung's explanation.
> It sounds as if you now have a good grasp of how the Foucault Pendulum
> works.
>

I might've (accurately or not) suggested as an example examining the
behaviour of a gyroscope (or a bicycle wheel) spinning and then turning
yourself around in a circle. It's motion in each direction, though 180
degrees opposite, I think it's pretty much the same net result and easy
for the layman to see, if he/she considers themselves the earth and the
gyroscope/wheel as a motion similar to the pendulum.