I apologize for this being way off topic, but there appears to be some smart folks
here and I thought I would ask a few questions.
If a regular pendulum clock, which keeps quite accurate time, were taken from the
Earth to the Moon and the correct time set into the clock and the clock was
restarted, would the clock keep as accurate time as it did on Earth?
Would the clock's three weights have to be pulled up every 7 days as mine does?
Would the lower Lunar gravity make any difference in time keeping?
If so, which way and by how much?
If not, why not?
Hoyt W.
p.s. The Sun will rise in the East regardless of the above.
On Mon, 17 May 2004 15:04:14 -0500, Hoyt Weathers <[email protected]> wrote:
>Tom Veatch wrote:
>
>> On Mon, 17 May 2004 09:56:31 GMT, "Lawrence L'Hote" <[email protected]> wrote:
>>
>> >..now just wait a second, if you are at the north pole will the sun rise in
>> >the east?
>> >
>>
>> Nope. It'll rise in the South. Sets in the South, too.
>>
>> Tom Veatch
>> Wichita, KS USA
>
>South with respect to what, pray tell?
>
>Hoyt W.
>
Standing on the North Pole, look at any point on the horizon that you choose. That line of sight, extended indefinitely on the
surface of the earth, will pass through the South Pole. Therefore, that direction, by definition is South.
One of two singularities on the Earth's surface. The other, of course, is the South Pole where every direction you look is North.
There is an interesting little "trick" question that takes advantage of that.
Where on Earth can you go 1 mile South, 1 mile East, and 1 mile North and be back exactly where you started?
Tom Veatch
Wichita, KS USA
In article <[email protected]>,
Hoyt Weathers <[email protected]> wrote:
>Tom Veatch wrote:
>
>> On Mon, 17 May 2004 09:56:31 GMT, "Lawrence L'Hote" <[email protected]> wrote:
>>
>> >..now just wait a second, if you are at the north pole will the sun rise in
>> >the east?
>> >
>>
>> Nope. It'll rise in the South. Sets in the South, too.
>>
>> Tom Veatch
>> Wichita, KS USA
>
>South with respect to what, pray tell?
"Up", naturally. Take any map and look at it, there are 4 directions.
'East', 'West', 'South', and 'Up'. <snicker>
Seriously, at the North Pole, _every_ direction is South.
Thus, the entire horizon is 'South' of you. So _wherever_ the Sun comes over
the horizon -- or goes under it -- _is_ South.
Andy Dingley wrote:
>
> On Sun, 16 May 2004 16:29:18 -0500, Hoyt Weathers <[email protected]>
> wrote:
>
> >If a regular pendulum clock, which keeps quite accurate time, were taken from the
> >Earth to the Moon and the correct time set into the clock and the clock was
> >restarted, would the clock keep as accurate time as it did on Earth?
>
> Pendulum swing time= 2 * pi * sqrt (l) / g
>
> So the clock time will slow down in direct proportion to reduced
> gravity. A pendulum about 1/36 (or so) of the length would set it
> right though, as the time is also proportional to the square root of
> the length.
>
> It'll also stop working pretty quickly. The moon's atmosphere is a
> hard vacuum, and a dusty vacuum at that. Vacuum system lubrication is
> _difficult_ (many satellites lost because of it). Even low vapour
> pressure oils boil away in the vacuum and you're left with a sticky
> goo that jams mechanisms. Run the bearing clean and dry and you may
> get metallic welding instead. Then if you do find a bearing that
> works, the lunar dust is reputedly a real hazard for any long-term
> mechanism. With no air or moisture, there's not much to lay the dust
> and fine dust gets everywhere (Apollo suits started to show damage in
> just a few days)
Jeweled bearings, sealed case. It should run *more* accurately in a
vacuum. Remove air and you remove a whole set of variables.
> >Would the clock's three weights have to be pulled up every 7 days as mine does?
24 hours unwinds the same amount of cord.
> They'd descend at exactly the same rate (in clock hand-movement terms)
> as on Earth. They may also need to be made heavier, as their mass
> would stay the same but they'd only have 1/6th of the weight - is this
> still enough to drive the clock ?
How complex is the mechanism? Does it drive three hands, or one?
Or is it just a pendulum, which can by itself be considered a clock. You can
read time simply by counting the beats. Hands add nothing more than
convenience.
Father Haskell wrote:
> Andy Dingley wrote:
> >
> > On Sun, 16 May 2004 16:29:18 -0500, Hoyt Weathers <[email protected]>
> > wrote:
> >
> > >If a regular pendulum clock, which keeps quite accurate time, were taken from the
> > >Earth to the Moon and the correct time set into the clock and the clock was
> > >restarted, would the clock keep as accurate time as it did on Earth?
> >
> > Pendulum swing time= 2 * pi * sqrt (l) / g
> >
> > So the clock time will slow down in direct proportion to reduced
> > gravity. A pendulum about 1/36 (or so) of the length would set it
> > right though, as the time is also proportional to the square root of
> > the length.
> >
> > It'll also stop working pretty quickly. The moon's atmosphere is a
> > hard vacuum, and a dusty vacuum at that. Vacuum system lubrication is
> > _difficult_ (many satellites lost because of it). Even low vapour
> > pressure oils boil away in the vacuum and you're left with a sticky
> > goo that jams mechanisms. Run the bearing clean and dry and you may
> > get metallic welding instead. Then if you do find a bearing that
> > works, the lunar dust is reputedly a real hazard for any long-term
> > mechanism. With no air or moisture, there's not much to lay the dust
> > and fine dust gets everywhere (Apollo suits started to show damage in
> > just a few days)
>
> Jeweled bearings, sealed case. It should run *more* accurately in a
> vacuum. Remove air and you remove a whole set of variables.
>
> > >Would the clock's three weights have to be pulled up every 7 days as mine does?
>
> 24 hours unwinds the same amount of cord.
>
> > They'd descend at exactly the same rate (in clock hand-movement terms)
> > as on Earth. They may also need to be made heavier, as their mass
> > would stay the same but they'd only have 1/6th of the weight - is this
> > still enough to drive the clock ?
>
> How complex is the mechanism? Does it drive three hands, or one?
>
> Or is it just a pendulum, which can by itself be considered a clock. You can
> read time simply by counting the beats. Hands add nothing more than
> convenience.
Father Haskell, you raise several interesting points. I understand the dry bearing wear
problem in vacuum since I worked for Dr. von Braun for about 20 years before he was
forced to leave Huntsville.
With the lower gravity on the Moon, dust would settle in the near area, but more slowly
than on Earth, and not be blown about at all since there is no Lunar atmosphere of any
significance. There are no winds on the Moon. [well, the Solar Wind, but that is
another topic entirely] AFAIK, moisture has nothing to do with time keeping per se.
"24 hours unwinds the same amount of cord."
I assume you mean on the Moon as well as on the Earth. On the Moon, I now think the
clock would unwind 1/6 of the cable per a real 24 hours. so we seem not to agree on
that point.
IMHO, you have missed the point. I would agree with you IF the clock kept the same time
on the Moon as on Earth, but I now do not think that it would and other reply posters
seem to say that as well. I believe that it would take about 6 x 7 days for the clock
to run down, assuming that it now runs down in 7 days here on Earth - or in that neck
of the woods. In other words, the pendulum would move more slowly and there would be
fewer pendulum cycles, or ticks, per real minute on the Moon. Ergo, the clock would be
inaccurate and run slower and the hour and minute hands would correspondingly also move
more slowly.
I think it matters not how complex the mechanism is or how many hands the clock has.
The atmosphere and moisture are not factors because I had assumed the clock would
remain inside a manned Lunar Lander of some kind. I did not state that assumption and
that is my error. So forget about atmosphere and moisture - or lack thereof outside of
the Lander. The clock would not be exposed to the Lunar environment - other than the
lower gravity.
My Howard-Miller GF clock has 2 hands and 3 weights actually.
I thank you for your interesting reply.
Hoyt W.
Lawrence L'Hote wrote:
> "Robert Bonomi" <[email protected]> wrote in message
> news:[email protected]...
> > The weights have to be pulled up after the same number of cycles of the
> > pendulum. Which will take about 17 days on the moon.
>
> not that it matters a diddly twat, but a moon 'day' is quite different in
> length from an earth day.
>
> 'someone else' wrote
> > >p.s. The Sun will rise in the East regardless of the above.
>
> whadabout on the moon??
>
> Larry
I now fully agree with your first two points above.
Good question Larry. It depends entirely on the definition of where East is on the
Moon. If one assumes a certain direction for Lunar East then the the Sun should rise
there also every morning, though not as often as on Earth. If the Sun does not rise
in the Lunar East, then they screwed up again in determining East.
Drats!
This is further off the topic, but I assume the Moon "wobbles" in its Lunar orbit
just as the Earth does. Ergo, Lunar East would have to be specified on a particular
day in its Solar orbit. I could be in thick duck soup here, but I think East on the
Earth is applicable to a specific day of the year. I do not know what day that is.
It is interesting to look at a map of just the Lunar and Earth orbits around the Sun.
Some have argued that the Moon is a co-planet with Earth and not a satellite of the
Earth. I think there is some merit to that, but that is not my contention.
Hoyt W.
Tom Veatch wrote:
> On Mon, 17 May 2004 09:56:31 GMT, "Lawrence L'Hote" <[email protected]> wrote:
>
> >..now just wait a second, if you are at the north pole will the sun rise in
> >the east?
> >
>
> Nope. It'll rise in the South. Sets in the South, too.
>
> Tom Veatch
> Wichita, KS USA
South with respect to what, pray tell?
Hoyt W.
Father Haskell wrote:
<snip, snip>
> Related point -- check out record producer George Martin's autobiography. Seargent
> Pepper was cut with a pendulum-regulated turntable. Martin knew this was something
> special, and wanted every detail done perfectly.
Reminds me of The Pit and the Pendulum movie, or close enough to the actual title.
Hoyt W.
Hoyt Weathers wrote:
>
> Father Haskell wrote:
>
> > Andy Dingley wrote:
> > >
> > > On Sun, 16 May 2004 16:29:18 -0500, Hoyt Weathers <[email protected]>
> > > wrote:
> > >
> > > >If a regular pendulum clock, which keeps quite accurate time, were taken from the
> > > >Earth to the Moon and the correct time set into the clock and the clock was
> > > >restarted, would the clock keep as accurate time as it did on Earth?
> > >
> > > Pendulum swing time= 2 * pi * sqrt (l) / g
> > >
> > > So the clock time will slow down in direct proportion to reduced
> > > gravity. A pendulum about 1/36 (or so) of the length would set it
> > > right though, as the time is also proportional to the square root of
> > > the length.
> > >
> > > It'll also stop working pretty quickly. The moon's atmosphere is a
> > > hard vacuum, and a dusty vacuum at that. Vacuum system lubrication is
> > > _difficult_ (many satellites lost because of it). Even low vapour
> > > pressure oils boil away in the vacuum and you're left with a sticky
> > > goo that jams mechanisms. Run the bearing clean and dry and you may
> > > get metallic welding instead. Then if you do find a bearing that
> > > works, the lunar dust is reputedly a real hazard for any long-term
> > > mechanism. With no air or moisture, there's not much to lay the dust
> > > and fine dust gets everywhere (Apollo suits started to show damage in
> > > just a few days)
> >
> > Jeweled bearings, sealed case. It should run *more* accurately in a
> > vacuum. Remove air and you remove a whole set of variables.
> >
> > > >Would the clock's three weights have to be pulled up every 7 days as mine does?
> >
> > 24 hours unwinds the same amount of cord.
> >
> > > They'd descend at exactly the same rate (in clock hand-movement terms)
> > > as on Earth. They may also need to be made heavier, as their mass
> > > would stay the same but they'd only have 1/6th of the weight - is this
> > > still enough to drive the clock ?
> >
> > How complex is the mechanism? Does it drive three hands, or one?
> >
> > Or is it just a pendulum, which can by itself be considered a clock. You can
> > read time simply by counting the beats. Hands add nothing more than
> > convenience.
>
> Father Haskell, you raise several interesting points. I understand the dry bearing wear
> problem in vacuum since I worked for Dr. von Braun for about 20 years before he was
> forced to leave Huntsville.
"'Vonce ze rockets go up, who cares vere zey come down,
dat's not my department,' says Werner Von Braun." (Lehrer)
The tower shot of the Apollo 11 liftoff still leaves me speechless, regardless. My
favorite arg to the moon hoaxers is, well, where did this monster end up?
> With the lower gravity on the Moon, dust would settle in the near area, but more slowly
> than on Earth, and not be blown about at all since there is no Lunar atmosphere of any
> significance. There are no winds on the Moon. [well, the Solar Wind, but that is
> another topic entirely] AFAIK, moisture has nothing to do with time keeping per se.
>
> "24 hours unwinds the same amount of cord."
>
> I assume you mean on the Moon as well as on the Earth. On the Moon, I now think the
> clock would unwind 1/6 of the cable per a real 24 hours. so we seem not to agree on
> that point.
>
> IMHO, you have missed the point. I would agree with you IF the clock kept the same time
> on the Moon as on Earth, but I now do not think that it would and other reply posters
> seem to say that as well. I believe that it would take about 6 x 7 days for the clock
> to run down, assuming that it now runs down in 7 days here on Earth - or in that neck
> of the woods. In other words, the pendulum would move more slowly and there would be
> fewer pendulum cycles, or ticks, per real minute on the Moon. Ergo, the clock would be
> inaccurate and run slower and the hour and minute hands would correspondingly also move
> more slowly.
>
> I think it matters not how complex the mechanism is or how many hands the clock has.
> The atmosphere and moisture are not factors because I had assumed the clock would
> remain inside a manned Lunar Lander of some kind. I did not state that assumption and
> that is my error. So forget about atmosphere and moisture - or lack thereof outside of
> the Lander. The clock would not be exposed to the Lunar environment - other than the
> lower gravity.
>
> My Howard-Miller GF clock has 2 hands and 3 weights actually.
Related point -- check out record producer George Martin's autobiography. Seargent
Pepper was cut with a pendulum-regulated turntable. Martin knew this was something
special, and wanted every detail done perfectly.
Tom Veatch wrote:
<snip, snip>
> Hoyt,
>
> Minor quibble having little or no effect on the discussion.
>
> Although Moon gravity is appx. 1/6 Earth gravity, the Earth's atmosphere causes dust to remain in suspension and settle quite
> slowly. No atmosphere means dust on the Moon literally drops like a rock. So, in actual effect, dust on the Moon falls much more
> rapidly than on the Earth.
>
> Tom Veatch
> Wichita, KS USA
Quite correct Tom. IIRC, there was one Apollo flight where an astronaut, on video, dropped a feather and a more massive and compact
object at the same time. Both the feather and the other object fell at the same rate and hit the Lunar surface at the same time.
The reason for that, of course, is that there is no atmosphere on the Moon. I think we fully agree on this. You just stated it
differently than I did and quite well I think. TKX
Hoyt W.
Scott Lurndal wrote:
> Hoyt Weathers <[email protected]> writes:
> <snip>
> >Reminds me of The Pit and the Pendulum movie, or close enough to the actual title.
> >
>
> There was a movie?
>
> _Edgar Allan Poe_ <http://bau2.uibk.ac.at/sg/poe/works/pit_pend.html>
>
> scott
Oh yes - and now there is a DVD of it. The ending was different from Poe's novel IIRC.
Vincent Price starred in it.
Here is one URL for the DVD:
http://www.uln.com/cgi-bin/vlink/027616862884IE.html?ptitle=Pit-and-the-Pendulum-DVD
Hoyt W.
"Hoyt Weathers" <[email protected]> wrote in message
news:[email protected]...
> Father Haskell wrote:
>
> > Andy Dingley wrote:
> > >
> > > On Sun, 16 May 2004 16:29:18 -0500, Hoyt Weathers <[email protected]>
> > > wrote:
> > >
> > > >If a regular pendulum clock, which keeps quite accurate time, were
taken from the
> > > >Earth to the Moon and the correct time set into the clock and the
clock was
> > > >restarted, would the clock keep as accurate time as it did on Earth?
> > >
> > > Pendulum swing time= 2 * pi * sqrt (l) / g
> > >
> > > So the clock time will slow down in direct proportion to reduced
> > > gravity. A pendulum about 1/36 (or so) of the length would set it
> > > right though, as the time is also proportional to the square root of
> > > the length.
> > >
> > > It'll also stop working pretty quickly. The moon's atmosphere is a
> > > hard vacuum, and a dusty vacuum at that. Vacuum system lubrication is
> > > _difficult_ (many satellites lost because of it). Even low vapour
> > > pressure oils boil away in the vacuum and you're left with a sticky
> > > goo that jams mechanisms. Run the bearing clean and dry and you may
> > > get metallic welding instead. Then if you do find a bearing that
> > > works, the lunar dust is reputedly a real hazard for any long-term
> > > mechanism. With no air or moisture, there's not much to lay the dust
> > > and fine dust gets everywhere (Apollo suits started to show damage in
> > > just a few days)
> >
> > Jeweled bearings, sealed case. It should run *more* accurately in a
> > vacuum. Remove air and you remove a whole set of variables.
> >
> > > >Would the clock's three weights have to be pulled up every 7 days as
mine does?
> >
> > 24 hours unwinds the same amount of cord.
No.
The weighs do not fall continually, they fall a fixed amount at each "tick"
as the paws engage and disengage.
> > > They'd descend at exactly the same rate (in clock hand-movement terms)
> > > as on Earth. They may also need to be made heavier, as their mass
> > > would stay the same but they'd only have 1/6th of the weight - is this
> > > still enough to drive the clock ?
> >
> > How complex is the mechanism? Does it drive three hands, or one?
> >
> > Or is it just a pendulum, which can by itself be considered a clock.
You can
> > read time simply by counting the beats. Hands add nothing more than
> > convenience.
>
> Father Haskell, you raise several interesting points. I understand the dry
bearing wear
> problem in vacuum since I worked for Dr. von Braun for about 20 years
before he was
> forced to leave Huntsville.
>
> With the lower gravity on the Moon, dust would settle in the near area,
but more slowly
> than on Earth, and not be blown about at all since there is no Lunar
atmosphere of any
> significance. There are no winds on the Moon. [well, the Solar Wind, but
that is
> another topic entirely] AFAIK, moisture has nothing to do with time
keeping per se.
>
> "24 hours unwinds the same amount of cord."
>
> I assume you mean on the Moon as well as on the Earth. On the Moon, I now
think the
> clock would unwind 1/6 of the cable per a real 24 hours. so we seem not to
agree on
> that point.
>
> IMHO, you have missed the point. I would agree with you IF the clock kept
the same time
> on the Moon as on Earth, but I now do not think that it would and other
reply posters
> seem to say that as well. I believe that it would take about 6 x 7 days
for the clock
> to run down, assuming that it now runs down in 7 days here on Earth - or
in that neck
> of the woods. In other words, the pendulum would move more slowly and
there would be
> fewer pendulum cycles, or ticks, per real minute on the Moon. Ergo, the
clock would be
> inaccurate and run slower and the hour and minute hands would
correspondingly also move
> more slowly.
>
> I think it matters not how complex the mechanism is or how many hands the
clock has.
> The atmosphere and moisture are not factors because I had assumed the
clock would
> remain inside a manned Lunar Lander of some kind. I did not state that
assumption and
> that is my error. So forget about atmosphere and moisture - or lack
thereof outside of
> the Lander. The clock would not be exposed to the Lunar environment -
other than the
> lower gravity.
>
> My Howard-Miller GF clock has 2 hands and 3 weights actually.
Most likely two of the weights drive the pendulum and one the chimes.
The pendulum is driven at the end of swing going both directions. In
many clocks this is redundant and as long as one or the other weight
is would it will run just fine.
Having messed up the night before and not getting this message sent to
the news group as I intended, I'm reposting now (slightly modified).
> JMartin957 wrote:
>
> > >
> > >The _stability_ of the time tick will be unchanged.
> > >
> > >The _frequency_ will be lower. By a factor of about 2.42.
> >
> > Actually, it will be much less accurate.
> >
> > Think tides. They would be much greater on the moon, due to the greater
> > gravitational attraction of the Earth. The gravitational attraction of
the Sun
> > would have a greater effect as well - while it would be the same on the
Moon as
> > on the Earth, it would be greater proportionally to the Moon's own
weaker
> > gravitational force.
Umm, no on the "tides" influence.
The moon does not spin in relation to the earth and thus the tidal
forces are fixed: if the seas on the moon were full of sea water
there would be no tides coming in and going out.
If the moon was covered with water it wouldn't be round, it would
bulge out on the sides facing and turned away from the earth.
In regards to the sun: it's too far away to have near the effect the earth
would have (assuming the moon spun). Also, it's the different distances
from two points on a body to the other body that creates tidal forces,
the moon is smaller so the distances are nearer the same as thus tidal
forces are smaller. A gas planet will experience greater tidal forces
than a rock the same mass in the same orbit.
> > Other than those "outside influences", the pendulum should be just as
accurate
> > on the Moon as on the Earth. Slower, but just as accurate.
> >
> > John Martin
>
> John, I was agreeing with you up top that it would be less accurate, then
at the
> bottom you said, in effect, that it would be just as accurate.
>
> Did your fingers stumble or am I not understanding you? How could it be
just as
> accurate if the pendulum has a slower cycle rate on the Moon? I am not
trying to pick
> a fuss, but I fail to understand your point. And I do want to understand.
It would be just as accurate, just ticking at a slower rate.
Well, I guess that depends on what you mean by "accurate": the
clock would mark time at a stable rate but the rate would be
much slower than here on earth.
Nobody ever set a fixed swing rate for pendulums.
In rec.woodworking
Hoyt Weathers <[email protected]> wrote:
>If a regular pendulum clock, which keeps quite accurate time, were taken from the
>Earth to the Moon and the correct time set into the clock and the clock was
>restarted, would the clock keep as accurate time as it did on Earth?
>
>Would the clock's three weights have to be pulled up every 7 days as mine does?
>
>Would the lower Lunar gravity make any difference in time keeping?
>
>If so, which way and by how much?
>
>If not, why not?
I'm surprised with over 40 posts, that no one has mentioned the effect
gravity has on time. While the pendulum will have the largest effect for
this experiment, EVERY clock, regardless of type be it atomic, spring,
quartz, etc, will run faster on the moon due to the lower gravity.
Einstein's general theory of relativity.
In rec.woodworking
[email protected] (Robert Bonomi) wrote:
>>I'm surprised with over 40 posts, that no one has mentioned the effect
>>gravity has on time. While the pendulum will have the largest effect for
>>this experiment, EVERY clock, regardless of type be it atomic, spring,
>>quartz, etc, will run faster on the moon due to the lower gravity.
>>Einstein's general theory of relativity.
>>
>
>But, but,... The moon is moving -faster- than the earth, whenever
>the moon is more than 1/2 full, time is -slower- on the Moon, due to
>velocity effects. To an observer outside the Earth-Moon, that is.
Granted, the velocity of the moon has to be considered but it is an
additional factor and all relatavistic effects have to be taken into
account, as it is in the GPS satellite system.
General relativity in the global positioning system
http://www.phys.lsu.edu/mog/mog9/node9.html
On Mon, 17 May 2004 01:14:32 +0100, Andy Dingley
<[email protected]> wrote:
>Pendulum swing time= 2 * pi * sqrt (l) / g
>So the clock time will slow down in direct proportion to reduced
>gravity. A pendulum about 1/36 (or so) of the length would set it
>right though, as the time is also proportional to the square root of
>the length.
Doh ! My memory must be going - that should of course be
2 * pi * sqrt (l / g)
(which considering the dimensions would have told me anyway)
So you'll run slow at about 1/2.5, or a 1/6 pendulum rod would fix it.
In article <[email protected]>,
Sysiphus <[email protected]> wrote:
>Hoyt Weathers thought it a good use of my time to say:
>
>
>Wow! I wish there was a newsgroup where this was on-topic! More "space
>math" than I have seen in a long time. The way I see it, though,(like you
>really want to know) is that the issue has been "under simplified". The
>clock works through the use of (a) falling weight(s). This weight has its
>"weight" based on the total force required to overcome the
>friction/inertia/gearing needed to move Mickey's hands around the clockface.
>
>If the moon's gravity is .XXX times the earth's gravity, then that clock
>will move .XXX times slower on the moon. AFAIK, there is no other
>weight-save the pendulum(which is "driven" by the rate of the falling
>weight), and therefore there is no additional 'gravitational' requirement,
>only the torque needed to spin the gears.
>
>But, I could be full of it, and not know!
Are your eyes brown? <grin>
The pendulum regulates how fast the clock 'ticks'.
The energy of the 'falling' weight is just there to make up for the
friction losses in the gear chain, and the entropy loss in the pendulum
itself. This is why the weight falls _so_much_ slower than it would
if it were 'free falling'. It is only _allowed_ to fall enough to provide
the 'make-up' energy.
The period of oscillation of a pendulum is (a) proportional to the square-root
of the length of the pendulum and (b) inversely proportional to the square-root
of the local gravitational constant.
Thus, if all else remains the same, and the gravitational constant is
lowered to .xxx, the clock will run slower, by a factor of 1/sqrt(.xxx).
On Sun, 16 May 2004 16:29:18 -0500, Hoyt Weathers <[email protected]>
wrote:
>If a regular pendulum clock, which keeps quite accurate time, were taken from the
>Earth to the Moon and the correct time set into the clock and the clock was
>restarted, would the clock keep as accurate time as it did on Earth?
Pendulum swing time= 2 * pi * sqrt (l) / g
So the clock time will slow down in direct proportion to reduced
gravity. A pendulum about 1/36 (or so) of the length would set it
right though, as the time is also proportional to the square root of
the length.
It'll also stop working pretty quickly. The moon's atmosphere is a
hard vacuum, and a dusty vacuum at that. Vacuum system lubrication is
_difficult_ (many satellites lost because of it). Even low vapour
pressure oils boil away in the vacuum and you're left with a sticky
goo that jams mechanisms. Run the bearing clean and dry and you may
get metallic welding instead. Then if you do find a bearing that
works, the lunar dust is reputedly a real hazard for any long-term
mechanism. With no air or moisture, there's not much to lay the dust
and fine dust gets everywhere (Apollo suits started to show damage in
just a few days)
>Would the clock's three weights have to be pulled up every 7 days as mine does?
They'd descend at exactly the same rate (in clock hand-movement terms)
as on Earth. They may also need to be made heavier, as their mass
would stay the same but they'd only have 1/6th of the weight - is this
still enough to drive the clock ?
--
Smert' spamionam
By Sun, 16 May 2004 16:29:18 -0500, Hoyt Weathers <[email protected]>
decided to post "Waaay OT- Lunar physics question" to
rec.woodworking:
>I apologize for this being way off topic, but there appears to be some smart folks
>here and I thought I would ask a few questions.
>
>If a regular pendulum clock, which keeps quite accurate time, were taken from the
>Earth to the Moon and the correct time set into the clock and the clock was
>restarted, would the clock keep as accurate time as it did on Earth?
>
>Would the clock's three weights have to be pulled up every 7 days as mine does?
>
>Would the lower Lunar gravity make any difference in time keeping?
>
>If so, which way and by how much?
>
>If not, why not?
>
>Hoyt W.
>
>p.s. The Sun will rise in the East regardless of the above.
Hoyt,
In deciding on using a division of the physical properties of the earth to
help create a universal (almost) system of measurement, of about
1/10,000,000 of 1/4 of half of a complete meridian passing through France
-- which they messed up anyway because of several factors, not the least of
which was failing to recognize completely the true oddity of the shape of
this ball we live on -- the French Academy rejected a measurement based on
moment of pendulums of given arc at a certain latitude mostly because they
determined that local distortions of gravity, such as provided by
mountains, would fail to provide the exactitude which these astronomers
wished for their standard.
Such a measurement taken over a fixed period of time, at one given
location, would not necessarily match that of another at another location.
Other factors, such as measuring the length of the pendulum, determining
the timing of the swing, atmospheric density and the quality and
repeatability of the observations, to name but a few, all proved that a
pendulum (eg clock) was not really that accurate. Given all of this, I
think that if you wish to measure time on the moon, you would do better
with a Swatch or Timex, at the very least.
/ts
--
find / -iname "*gw*" -exec rm -r{} \;
In heaven, there is no beer,
That's why we drink it here,
And when we're all gone from here,
Our friends will be drinking all the beer!
-- Famous old Czech song about beer --
In article <[email protected]>,
Bruce <[email protected]> wrote:
>In rec.woodworking
>Hoyt Weathers <[email protected]> wrote:
>
>>If a regular pendulum clock, which keeps quite accurate time, were
>taken from the
>>Earth to the Moon and the correct time set into the clock and the clock was
>>restarted, would the clock keep as accurate time as it did on Earth?
>>
>>Would the clock's three weights have to be pulled up every 7 days as mine does?
>>
>>Would the lower Lunar gravity make any difference in time keeping?
>>
>>If so, which way and by how much?
>>
>>If not, why not?
>
>I'm surprised with over 40 posts, that no one has mentioned the effect
>gravity has on time. While the pendulum will have the largest effect for
>this experiment, EVERY clock, regardless of type be it atomic, spring,
>quartz, etc, will run faster on the moon due to the lower gravity.
>Einstein's general theory of relativity.
>
But, but,... The moon is moving -faster- than the earth, whenever
the moon is more than 1/2 full, time is -slower- on the Moon, due to
velocity effects. To an observer outside the Earth-Moon, that is.
"Hoyt Weathers" <[email protected]> wrote in message
news:[email protected]...
> I apologize for this being way off topic, but there appears to be some
smart folks
> here and I thought I would ask a few questions.
>
> If a regular pendulum clock, which keeps quite accurate time, were taken
from the
> Earth to the Moon and the correct time set into the clock and the clock
was
> restarted, would the clock keep as accurate time as it did on Earth?
ABSOLUTELY NOT... The clocks TRUNNION would not survive the trip to the
moon.. ;~)
Hoyt Weathers <[email protected]> wrote in
news:[email protected]:
> I apologize for this being way off topic, but there appears to be some
> smart folks here and I thought I would ask a few questions.
>
> If a regular pendulum clock, which keeps quite accurate time, were
> taken from the Earth to the Moon and the correct time set into the
> clock and the clock was restarted, would the clock keep as accurate
> time as it did on Earth?
A pendulum clock will run slower on the moon.
http://scienceworld.wolfram.com/physics/Pendulum.html
Hoyt Weathers <[email protected]> writes:
>Father Haskell wrote:
>
><snip, snip>
>
>> Related point -- check out record producer George Martin's autobiography. Seargent
>> Pepper was cut with a pendulum-regulated turntable. Martin knew this was something
>> special, and wanted every detail done perfectly.
>
>Reminds me of The Pit and the Pendulum movie, or close enough to the actual title.
>
There was a movie?
_Edgar Allan Poe_ <http://bau2.uibk.ac.at/sg/poe/works/pit_pend.html>
scott
In article <[email protected]>,
Hoyt Weathers <[email protected]> wrote:
>I apologize for this being way off topic, but there appears to be some
>smart folks here and I thought I would ask a few questions.
OK, but _who's_ homework assignment is this?
This usually shows up in about an 8th-grade science class.
>If a regular pendulum clock, which keeps quite accurate time, were taken
>from the Earth to the Moon and the correct time set into the clock and the
>clock was >restarted, would the clock keep as accurate time as it did on
>Earth?
The _stability_ of the time tick will be unchanged.
The _frequency_ will be lower. By a factor of about 2.42.
>
>Would the clock's three weights have to be pulled up every 7 days as mine
>does?
The weights have to be pulled up after the same number of cycles of the
pendulum. Which will take about 17 days on the moon.
>Would the lower Lunar gravity make any difference in time keeping?
Yes.
>If so, which way and by how much?
The frequency of a pendulum is proportional to the square root of the
local gravitational constant.
Thus, since Lunar gravity is _less_ than that of Earth, the frequency
will be lower. And the clock will run slower.
"How much?" The factor is the square root of the ratio of the gravitational
field strengths. A quantitative answer is left as an exercise for the
student. <Big sh*t-eating grin>
>
>If not, why not?
not applicable.
>
>Hoyt W.
>
>p.s. The Sun will rise in the East regardless of the above.
>
>
>The _stability_ of the time tick will be unchanged.
>
>The _frequency_ will be lower. By a factor of about 2.42.
>>
Actually, it will be much less accurate.
Think tides. They would be much greater on the moon, due to the greater
gravitational attraction of the Earth. The gravitational attraction of the Sun
would have a greater effect as well - while it would be the same on the Moon as
on the Earth, it would be greater proportionally to the Moon's own weaker
gravitational force.
Other than those "outside influences", the pendulum should be just as accurate
on the Moon as on the Earth. Slower, but just as accurate.
John Martin
In article <[email protected]>, BruceR <[email protected]> wrote:
>Hoyt Weathers wrote:
>> BruceR wrote:
>>
>>
>>>[snip]
>>>
>>>At work here we need to include gnat fart constants. Here is a sample of
>>>some constants to better pin down how accurate that darn moon clock is:
>>>
>>
>>
>> <big snip>
>>
>> Gee whillerkers Bruce, how does "global Love Number H" and "global
>Love Number L"
>> affect whether the clock will run slower or not? I have always been
>wanting to know
>> and you are the first one to bring it up.
>
>Gee silly! Of course you have to consider horizontal displacement of the
> earths crust as the tidal forces move about, being different depending
>on the density of the body in question....
>
>BTW, would a moon clock have a inset face that shows the phases of the
>earth?
Ask the Addams family what they have on their moondial. <grin>
In article <[email protected]>,
Tom Veatch <[email protected]> wrote:
>On Mon, 17 May 2004 21:33:59 +0000, [email protected] (Robert
>Bonomi) wrote:
>
><snip>
>> at a
>> fixed location on the Moon, the effect of the Earth's gravitational pull
>> is a constant -- both in magnitude and direction -- because the moon
>> is in a 'tidal lock' with the same face constantly towards the earth.
><snip>
>
>Strictly in the interest of precision, you are almost correct saying the
>Earth's gravitational pull is a constant vector in
>the"lunacentric" reference system, and would be precisely correct if the
>Moon's orbit was a perfect circle and both the Earth and
>the Moon were perfect homogeneous spheres. But, the orbit of the Moon
>about the Earth is Oh-So-Slightly Elliptical and both bodies
>are just a tad off from being perfectly spherical. Oh, hell, let's
>ignore variations in mass concentrations. I did mention
>"precision", didn't I? 8^)
The fact that the bodies are not perfectly spherical,and not perfectly
homogeneous, and/or the mass concentration variations are NOT a factor.
Two reasons -- first, the axis of rotation is the 'center of mass', and
second, for gravitation calculations, _at_or_above_ the surface of the
object, it can be treated as a 'point source', with the entire mass existent
at the putative center of mass.
Eccentricity of the Lunar orbit does contribute some variance.
its about +/- 0.015% in the gravitational constant. Which translates
to a variance of about 74 parts in a million in the pendulum period.
Over the short term, this introduces a maximum effect of about 6 seconds
per (earth) day.
Over a complete _lunar_ day, on the other hand, the effects cancel out,
completely.
>As a result, there is a slight "back and forth" precession of the
>surface of the Moon as seen from Earth and an equivalent back and
>forth precession of the position of the Earth as seen from the Moon.
>This causes the direction of the vector to oscillate back and
>fourth with a period matching the Moon's orbital period. Likewise, the
>eccentricity of the orbit causes the distance between the
>centers of mass of the Earth and the Moon to oscillate with that same
>period. Ergo, the magnitude of the vector is also non
>constant.
I don't have a quantitative figure on the precession. However, *IF*
it is +/- 1 degree. then the effect is almost exactly the same as the
orbital eccentricity. +/- 0.015%
Which raises the _very_ interesting situation that if the effects are 180
degrees out-of-phase, relative to each other, then they nullify each other
almost perfectly.
>
>Very much a higher order effect that can be, and rightfully was, ignored
>in your analysis. However, you might want to take it into
>account the next time you plan a Lunar Landing Mission.
For _that_, I d*mn well need better data than 4 sig-fig accuracy on the
Lunar gravitational constant, 3 sig-fig accuracy on the mean Sun-Earth
distance, and the Earth-Moon distance, and 2 sig-figs on the diameter of
the Earth and the Moon.
I _do_ have 'pi' memorized to 20 decimal places, so precision is not
restricted on _that_ basis. <grin>
>
>Just trying to be "precise" (with maybe just a little "wise-assedness"
>thrown in). ;-)
We'll be sure to let you know, if you succeed. <snicker>
JMartin957 wrote:
> >
> >The _stability_ of the time tick will be unchanged.
> >
> >The _frequency_ will be lower. By a factor of about 2.42.
> >>
>
> Actually, it will be much less accurate.
>
> Think tides. They would be much greater on the moon, due to the greater
> gravitational attraction of the Earth. The gravitational attraction of the Sun
> would have a greater effect as well - while it would be the same on the Moon as
> on the Earth, it would be greater proportionally to the Moon's own weaker
> gravitational force.
>
> Other than those "outside influences", the pendulum should be just as accurate
> on the Moon as on the Earth. Slower, but just as accurate.
>
> John Martin
John, I was agreeing with you up top that it would be less accurate, then at the
bottom you said, in effect, that it would be just as accurate.
Did your fingers stumble or am I not understanding you? How could it be just as
accurate if the pendulum has a slower cycle rate on the Moon? I am not trying to pick
a fuss, but I fail to understand your point. And I do want to understand.
Hoyt W.
Doug Miller wrote:
> In article <[email protected]>, [email protected] wrote:
> >JMartin957 wrote:
> >
> >> >
> >> >The _stability_ of the time tick will be unchanged.
> >> >
> >> >The _frequency_ will be lower. By a factor of about 2.42.
> >> >>
> >>
> >> Actually, it will be much less accurate.
> >>
> >> Think tides. They would be much greater on the moon, due to the greater
> >> gravitational attraction of the Earth. The gravitational attraction of the
> > Sun
> >> would have a greater effect as well - while it would be the same on the Moon
> > as
> >> on the Earth, it would be greater proportionally to the Moon's own weaker
> >> gravitational force.
> >>
> >> Other than those "outside influences", the pendulum should be just as
> > accurate
> >> on the Moon as on the Earth. Slower, but just as accurate.
> >>
> >> John Martin
> >
> >John, I was agreeing with you up top that it would be less accurate, then at
> > the
> >bottom you said, in effect, that it would be just as accurate.
> >
> >Did your fingers stumble or am I not understanding you? How could it be just as
> >accurate if the pendulum has a slower cycle rate on the Moon? I am not trying
> > to pick
> >a fuss, but I fail to understand your point. And I do want to understand.
> >
>
> It's both. :-)
>
> Accurate in the sense that it will be self-consistent; that is, an hour
> measured this week will be the same length of time as an hour measured last
> week, next week, or next year (assuming that you keep the clock wound).
>
> Inaccurate, in the sense that each hour measured by the clock on the moon will
> actually be about 144 minutes long, and thus the deviation from "correct" time
> will grow continuously.
>
> So a pendulum clock designed for use on Earth, moved to the moon, will keep
> time consistently, and could be considered accurate for lunar timekeeping, as
> long as those who use it are willing to agree on a new definition of "hour".
> Problems occur only when comparing the time shown by such a clock to the time
> shown by other timekeeping devices such as an Earth-based pendulum clock or a
> battery-powered digital wristwatch.
>
> --
> Regards,
> Doug Miller (alphageek-at-milmac-dot-com)
>
> For a copy of my TrollFilter for NewsProxy/Nfilter,
> send email to autoresponder at filterinfo-at-milmac-dot-com
> You must use your REAL email address to get a response.
Thanks for clearing that up Doug. Consistency is a good clarification, STS, to the
use of the single word of accuracy. I learn something new every day - or try to.
Hoyt W.
Dan White wrote:
> "Doug Miller" <[email protected]> wrote in message
> news:[email protected]...
> > In article <[email protected]>, [email protected] wrote:
> > >JMartin957 wrote:
> > >
> > >> >
> > >> >The _stability_ of the time tick will be unchanged.
> > >> >
> > >> >The _frequency_ will be lower. By a factor of about 2.42.
> > >> >>
> > >>
> > >> Actually, it will be much less accurate.
> > >>
> > >> Think tides. They would be much greater on the moon, due to the
> greater
> > >> gravitational attraction of the Earth. The gravitational attraction of
> the
> > > Sun
> > >> would have a greater effect as well - while it would be the same on the
> Moon
> > > as
> > >> on the Earth, it would be greater proportionally to the Moon's own
> weaker
> > >> gravitational force.
> > >>
> > >> Other than those "outside influences", the pendulum should be just as
> > > accurate
> > >> on the Moon as on the Earth. Slower, but just as accurate.
> > >>
> > >> John Martin
> > >
> > >John, I was agreeing with you up top that it would be less accurate, then
> at
> > > the
> > >bottom you said, in effect, that it would be just as accurate.
> > >
> > >Did your fingers stumble or am I not understanding you? How could it be
> just as
> > >accurate if the pendulum has a slower cycle rate on the Moon? I am not
> trying
> > > to pick
> > >a fuss, but I fail to understand your point. And I do want to understand.
> > >
> >
> > It's both. :-)
> >
> > Accurate in the sense that it will be self-consistent; that is, an hour
> > measured this week will be the same length of time as an hour measured
> last
> > week, next week, or next year (assuming that you keep the clock wound).
> >
> > Inaccurate, in the sense that each hour measured by the clock on the moon
> will
> > actually be about 144 minutes long, and thus the deviation from "correct"
> time
> > will grow continuously.
> >
>
> I think maybe the terms "accuracy" and "precision" would be helpful here.
> You could say the clock has excellent precision, in that it will measure an
> hour almost the same exact way every time. However, its accuracy is poor,
> because it will not measure an Earth hour very well at all. Precision
> signifies how repeatable or closely matched measurements are to an average.
> Accuracy signifies how close those measurements are to the true value.
>
> dwhite
I like consistency and precision equally, each in their own way. But I vote
Republican anyway!
Hoyt W.
BruceR wrote:
> [snip]
>
> At work here we need to include gnat fart constants. Here is a sample of
> some constants to better pin down how accurate that darn moon clock is:
>
<big snip>
Gee whillerkers Bruce, how does "global Love Number H" and "global Love Number L"
affect whether the clock will run slower or not? I have always been wanting to know
and you are the first one to bring it up.
Hoyt W.
BruceR wrote:
> Hoyt Weathers wrote:
> > BruceR wrote:
> >
> >
> >>[snip]
> >>
> >>At work here we need to include gnat fart constants. Here is a sample of
> >>some constants to better pin down how accurate that darn moon clock is:
> >>
> >
> >
> > <big snip>
> >
> > Gee whillerkers Bruce, how does "global Love Number H" and "global Love Number L"
> > affect whether the clock will run slower or not? I have always been wanting to know
> > and you are the first one to bring it up.
>
> Gee silly! Of course you have to consider horizontal displacement of the
> earths crust as the tidal forces move about, being different depending
> on the density of the body in question....
>
> BTW, would a moon clock have a inset face that shows the phases of the
> earth?
>
> -Bruce
>
> >
> > Hoyt W.
> >
> >
> >
> >
>
> -----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
> http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
> -----== Over 100,000 Newsgroups - 19 Different Servers! =-----
The clock in question does not have an inset face, but that was not a "given" within my
original questions. Therefore your question is N/A. Interesting though.
-Hoyt W.
[snip]
At work here we need to include gnat fart constants. Here is a sample of
some constants to better pin down how accurate that darn moon clock is:
>
>
> For _that_, I d*mn well need better data than 4 sig-fig accuracy on the
> Lunar gravitational constant, 3 sig-fig accuracy on the mean Sun-Earth
> distance, and the Earth-Moon distance, and 2 sig-figs on the diameter of
> the Earth and the Moon.
>
> I _do_ have 'pi' memorized to 20 decimal places, so precision is not
> restricted on _that_ basis. <grin>
>
>>Just trying to be "precise" (with maybe just a little "wise-assedness"
>>thrown in). ;-)
>
>
> We'll be sure to let you know, if you succeed. <snicker>
>
/* IAU (1976) System of Astronomical Constants
* SOURCE: USNO Circular # 163 (1981dec10)
* ALL ITEMS ARE DEFINED IN THE SI (MKS) SYSTEM OF UNITS
*
*/
#define GAUSS_GRAV 0.01720209895 /* Gaussian gravitational constant */
#define C_LIGHT 299792458. /* Speed of light; m/s */
#define TAU_A 499.004782 /* Light time for one a.u.; sec */
#define E_EQ_RADIUS 6378137. /* Earth's Equatorial Radius, meters
(IUGG value) */
#define E_FORM_FCTR 0.00108263 /* Earth's dynamical form factor */
#define GRAV_GEO 3.986005e14 /* Geocentric gravitational constant;
(m^3)(s^-2) */
#define GRAV_CONST 6.672e-11 /* Constant of gravitation;
(m^3)(kg^-1)(s^-2) */
#define LMASS_RATIO 0.01230002 /* Ratio of mass of Moon to mass of
Earth */
#define PRECESS 5029.0966 /* General precession in longitude;
arcsec per Julian century
at standard epoch J2000 */
#define OBLIQUITY 84381.448 /* Obliquity of the ecliptic at
epoch J2000; arcsec */
#define NUTATE 9.2025 /* Constant of nutation at
epoch J2000; arcsec */
#define ASTR_UNIT 1.49597870e11 /* Astronomical unit; meters */
#define SOL_PRLX 8.794148 /* Solar parallax; arcsec */
#define ABERRATE 20.49552 /* Constant of aberration at
epoch J2000; arcsec */
#define E_FLAT_FCTR 0.00335281 /* Earth's flattening factor */
#define GRAV_HELIO 1.32712438e20 /* Heliocentric gravitational
constant
(m^3)(s^-2) */
#define S_E_RATIO 332946.0 /* Ratio of mass of Sun to mass of
Earth */
#define S_EMOON_RATIO 328900.5 /* Ratio of mass of sun to
mass of Earth plus Moon */
#define SOLAR_MASS 1.9891e30 /* Mass of Sun; kg */
#define JD_J2000 2451545.0 /* Julian Day Number of 2000jan1.5 */
#define BES_YEAR 365.242198781 /* Length of Besselian Year in days
at B1900.0 (JD 2415020.31352)
*/#define SOLAR_SID 0.997269566329084 /* Ratio of Solar time interval to
Sidereal time interval at
J2000 */
#define SID_SOLAR 1.002737909350795 /* Ratio of Sidereal time interval
to Solar time interval at
J2000 */
#define ACCEL_GRV 9.78031846 /* acceleration of gravity at the
* earth's surface (m)(s^-2) */
#define GRAV_MOON 4.90279750e12 /* Lunar-centric gravitational
constant
(m^3)(s^-2) */
#define ETIDE_LAG 0.0 /* Earth tides: lag angle
(radians) */
#define LOVE_H 0.60967 /* Earth tides: global Love Number H,
IERS value (unitless) */
#define LOVE_L 0.0852 /* Earth tides: global Love Number L,
IERS value (unitless) */
-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----
Hoyt Weathers wrote:
> BruceR wrote:
>
>
>>[snip]
>>
>>At work here we need to include gnat fart constants. Here is a sample of
>>some constants to better pin down how accurate that darn moon clock is:
>>
>
>
> <big snip>
>
> Gee whillerkers Bruce, how does "global Love Number H" and "global Love Number L"
> affect whether the clock will run slower or not? I have always been wanting to know
> and you are the first one to bring it up.
Gee silly! Of course you have to consider horizontal displacement of the
earths crust as the tidal forces move about, being different depending
on the density of the body in question....
BTW, would a moon clock have a inset face that shows the phases of the
earth?
-Bruce
>
> Hoyt W.
>
>
>
>
-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----
"Doug Miller" <[email protected]> wrote in message
news:[email protected]...
> In article <[email protected]>, [email protected] wrote:
> >JMartin957 wrote:
> >
> >> >
> >> >The _stability_ of the time tick will be unchanged.
> >> >
> >> >The _frequency_ will be lower. By a factor of about 2.42.
> >> >>
> >>
> >> Actually, it will be much less accurate.
> >>
> >> Think tides. They would be much greater on the moon, due to the
greater
> >> gravitational attraction of the Earth. The gravitational attraction of
the
> > Sun
> >> would have a greater effect as well - while it would be the same on the
Moon
> > as
> >> on the Earth, it would be greater proportionally to the Moon's own
weaker
> >> gravitational force.
> >>
> >> Other than those "outside influences", the pendulum should be just as
> > accurate
> >> on the Moon as on the Earth. Slower, but just as accurate.
> >>
> >> John Martin
> >
> >John, I was agreeing with you up top that it would be less accurate, then
at
> > the
> >bottom you said, in effect, that it would be just as accurate.
> >
> >Did your fingers stumble or am I not understanding you? How could it be
just as
> >accurate if the pendulum has a slower cycle rate on the Moon? I am not
trying
> > to pick
> >a fuss, but I fail to understand your point. And I do want to understand.
> >
>
> It's both. :-)
>
> Accurate in the sense that it will be self-consistent; that is, an hour
> measured this week will be the same length of time as an hour measured
last
> week, next week, or next year (assuming that you keep the clock wound).
>
> Inaccurate, in the sense that each hour measured by the clock on the moon
will
> actually be about 144 minutes long, and thus the deviation from "correct"
time
> will grow continuously.
>
I think maybe the terms "accuracy" and "precision" would be helpful here.
You could say the clock has excellent precision, in that it will measure an
hour almost the same exact way every time. However, its accuracy is poor,
because it will not measure an Earth hour very well at all. Precision
signifies how repeatable or closely matched measurements are to an average.
Accuracy signifies how close those measurements are to the true value.
dwhite
In article <[email protected]>,
Dan White <[email protected]> wrote:
>
>"Doug Miller" <[email protected]> wrote in message
>news:[email protected]...
>> In article <[email protected]>, [email protected] wrote:
>> >JMartin957 wrote:
>> >
>> >> >
>> >> >The _stability_ of the time tick will be unchanged.
>> >> >
>> >> >The _frequency_ will be lower. By a factor of about 2.42.
>> >> >>
>> >>
>> >> Actually, it will be much less accurate.
>> >>
>> >> Think tides. They would be much greater on the moon, due to the
>greater
>> >> gravitational attraction of the Earth. The gravitational attraction of
>the
>> > Sun
>> >> would have a greater effect as well - while it would be the same on the
>Moon
>> > as
>> >> on the Earth, it would be greater proportionally to the Moon's own
>weaker
>> >> gravitational force.
>> >>
>> >> Other than those "outside influences", the pendulum should be just as
>> > accurate
>> >> on the Moon as on the Earth. Slower, but just as accurate.
>> >>
>> >> John Martin
>> >
>> >John, I was agreeing with you up top that it would be less accurate, then
>at
>> > the
>> >bottom you said, in effect, that it would be just as accurate.
>> >
>> >Did your fingers stumble or am I not understanding you? How could it be
>just as
>> >accurate if the pendulum has a slower cycle rate on the Moon? I am not
>trying
>> > to pick
>> >a fuss, but I fail to understand your point. And I do want to understand.
>> >
>>
>> It's both. :-)
>>
>> Accurate in the sense that it will be self-consistent; that is, an hour
>> measured this week will be the same length of time as an hour measured
>last
>> week, next week, or next year (assuming that you keep the clock wound).
>>
>> Inaccurate, in the sense that each hour measured by the clock on the moon
>will
>> actually be about 144 minutes long, and thus the deviation from "correct"
>time
>> will grow continuously.
>>
>
>I think maybe the terms "accuracy" and "precision" would be helpful here.
>You could say the clock has excellent precision, in that it will measure an
>hour almost the same exact way every time. However, its accuracy is poor,
>because it will not measure an Earth hour very well at all. Precision
>signifies how repeatable or closely matched measurements are to an average.
>Accuracy signifies how close those measurements are to the true value.
You know, there was a *reason* I described things the way I did, in the
original posting.
I'll repeat:
1) The 'stability', aka 'repeatability', is essentially unchanged.
John Martin contests that , claiming that 'outside influences'-- claimed
to be comparatively larger on the moon -- beyond the local gravitational
constant, will degrade the stability. I disagree. Because: (a) at a
fixed location on the Moon, the effect of the Earth's gravitational pull
is a constant -- both in magnitude and direction -- because the moon
is in a 'tidal lock' with the same face constantly towards the earth.
and (b) the gravitational effect of any other solar body is essentially
identical, as the distance from Earth, or the Moon, to that solar
body is 'for all practical purposes' identical. (gravitational
attraction is inversely proportional to the square of the distance
between the bodies. considering the Sun, as felt from the Earth
and the Moon, there is a maximum difference of about 2.688/1000ths of
the Earth-Sun distance. which means the relative difference in
gravitational effect is 1/(1.002688^2), since period of a pendulum
is proportional to the sqrt of the gravitational constant,
you've got a variance +/- 0.2688% of the Sun-Earth gravitational effect
at the Moon. On Earth, The Sun's gravitational effect varies the
local gravitational constant every 24 hours. (maximum it the
middle of the night, when it adds to the earth effect, and minimum
at noon, when it is in the opposite direction). Exactly the same
thing occurs on the Moon, albeit on a circa 28-earthday cycle.
Solar gravitational constant, at the moon's surface, is about
1/12,750,000th of the moons gravity. On the Earth's surface,
the solar effect is about 1/4,900,000th the Earth's gravity.
This *is* counter-intuitive, but true, nonetheless -- it works out
that way because the Moon is much smaller in diameter than the Earth,
and thus, the surface is closer to the 'center of mass'.
What all these gyrations show is that the 'outside influence' effect of
other solar bodies, measured on the surface of the Moon, will be _less_
than the effect from the same source, measured on the surface of the Earth.
And, as mentioned, the effect of the Earth does _not_ affect the
stability of the tick because it is constant in both magnitude and
direction.
Overall, the clock tick will be _more_ stable on the Moon than it
is on Earth. By _maybe_ "one part in a ten million". <grin>
2) The _frequency_ of the tick -- also known as the 'period' of the
pendulum -- *IS* lower. By the square-root of the ratio of the local
gravitational constant. On the surface of the Moon, it is 0.1645g
(I looked it up! :) this means that the pendulum will be slower by
a factor of 2.4655. Or it will take 2 hrs, 27 minutes 55+ seconds for
the clock to show the passage of one hour.
And, finally, it will take a little over 17-1/4 'earth' days for that 7-day
mechanism to get to the point of requiring raising the weights. (which _will_
still be at the passage of '7 days' as "indicated* by the clock.)
Robert Bonomi wrote:
> In article <[email protected]>,
> Tom Veatch <[email protected]> wrote:
>>On Mon, 17 May 2004 21:33:59 +0000, [email protected] (Robert
>>Bonomi) wrote:
>>
>><snip>
>>> at a
>>> fixed location on the Moon, the effect of the Earth's gravitational
>>> pull is a constant -- both in magnitude and direction -- because the
>>> moon is in a 'tidal lock' with the same face constantly towards the
>>> earth.
>><snip>
>>
>>Strictly in the interest of precision, you are almost correct saying the
>>Earth's gravitational pull is a constant vector in
>>the"lunacentric" reference system, and would be precisely correct if the
>>Moon's orbit was a perfect circle and both the Earth and
>>the Moon were perfect homogeneous spheres. But, the orbit of the Moon
>>about the Earth is Oh-So-Slightly Elliptical and both bodies
>>are just a tad off from being perfectly spherical. Oh, hell, let's
>>ignore variations in mass concentrations. I did mention
>>"precision", didn't I? 8^)
>
> The fact that the bodies are not perfectly spherical,and not perfectly
> homogeneous, and/or the mass concentration variations are NOT a factor.
> Two reasons -- first, the axis of rotation is the 'center of mass', and
> second, for gravitation calculations, _at_or_above_ the surface of the
> object, it can be treated as a 'point source', with the entire mass
> existent at the putative center of mass.
That's actually true only if the body is homogeneous. The Moon is
not--Lunar mass concentrations were discovered by their effect on the
motion of the Lunar Orbiter satellite, which orbit would not have been
perturbed by them if your contention was correct.
However, if our clock is in a fixed location on the lunar surface then the
effect of any nearby mass concentration would be accounted for when the
clock was regulated--that would only be an issue if the clock was moved
around, and even then I suspect it would be a very small effect.
Libration would have some effect--objects on the surface of a macroscopic
body in orbit about the other (or to be pedantic about their common center
of mass) are subject to tidal stresses which would affect the rate of a
pendulum clock, however most of that would againb be accounted for when the
clock was regulated, with libration being the only significant variable in
that regard. But again I suspect that that would be a very small effect.
> Eccentricity of the Lunar orbit does contribute some variance.
> its about +/- 0.015% in the gravitational constant. Which translates
> to a variance of about 74 parts in a million in the pendulum period.
>
> Over the short term, this introduces a maximum effect of about 6 seconds
> per (earth) day.
>
> Over a complete _lunar_ day, on the other hand, the effects cancel out,
> completely.
>
>>As a result, there is a slight "back and forth" precession of the
>>surface of the Moon as seen from Earth and an equivalent back and
>>forth precession of the position of the Earth as seen from the Moon.
>>This causes the direction of the vector to oscillate back and
>>fourth with a period matching the Moon's orbital period. Likewise, the
>>eccentricity of the orbit causes the distance between the
>>centers of mass of the Earth and the Moon to oscillate with that same
>>period. Ergo, the magnitude of the vector is also non
>>constant.
>
> I don't have a quantitative figure on the precession. However, *IF*
> it is +/- 1 degree. then the effect is almost exactly the same as the
> orbital eccentricity. +/- 0.015%
>
> Which raises the _very_ interesting situation that if the effects are 180
> degrees out-of-phase, relative to each other, then they nullify each other
> almost perfectly.
>
>>
>>Very much a higher order effect that can be, and rightfully was, ignored
>>in your analysis. However, you might want to take it into
>>account the next time you plan a Lunar Landing Mission.
>
> For _that_, I d*mn well need better data than 4 sig-fig accuracy on the
> Lunar gravitational constant, 3 sig-fig accuracy on the mean Sun-Earth
> distance, and the Earth-Moon distance, and 2 sig-figs on the diameter of
> the Earth and the Moon.
>
> I _do_ have 'pi' memorized to 20 decimal places, so precision is not
> restricted on _that_ basis. <grin>
>>
>>Just trying to be "precise" (with maybe just a little "wise-assedness"
>>thrown in). ;-)
>
> We'll be sure to let you know, if you succeed. <snicker>
--
--John
Reply to jclarke at ae tee tee global dot net
(was jclarke at eye bee em dot net)
"Hoyt Weathers" <[email protected]> wrote in message
news:[email protected]...
> Dan White wrote:
>
> > "Doug Miller" <[email protected]> wrote in message
> > news:[email protected]...
> > > In article <[email protected]>, [email protected] wrote:
> > > It's both. :-)
> > >
> > > Accurate in the sense that it will be self-consistent; that is, an
hour
> > > measured this week will be the same length of time as an hour measured
> > last
> > > week, next week, or next year (assuming that you keep the clock
wound).
> > >
> > > Inaccurate, in the sense that each hour measured by the clock on the
moon
> > will
> > > actually be about 144 minutes long, and thus the deviation from
"correct"
> > time
> > > will grow continuously.
> > >
> >
> > I think maybe the terms "accuracy" and "precision" would be helpful
here.
> > You could say the clock has excellent precision, in that it will measure
an
> > hour almost the same exact way every time. However, its accuracy is
poor,
> > because it will not measure an Earth hour very well at all. Precision
> > signifies how repeatable or closely matched measurements are to an
average.
> > Accuracy signifies how close those measurements are to the true value.
> >
> > dwhite
>
> I like consistency and precision equally, each in their own way. But I
vote
> Republican anyway!
>
> Hoyt W.
>
Ditto here, Hoyt, but the terms "accuracy" and "precision" are more than
just my idea. These are the scientifically "accepted" terms to describe how
close data are to each other (precision), and to the true value (accuracy).
dwhite
On Mon, 17 May 2004 21:33:59 +0000, [email protected] (Robert Bonomi) wrote:
<snip>
> at a
> fixed location on the Moon, the effect of the Earth's gravitational pull
> is a constant -- both in magnitude and direction -- because the moon
> is in a 'tidal lock' with the same face constantly towards the earth.
<snip>
Strictly in the interest of precision, you are almost correct saying the Earth's gravitational pull is a constant vector in
the"lunacentric" reference system, and would be precisely correct if the Moon's orbit was a perfect circle and both the Earth and
the Moon were perfect homogeneous spheres. But, the orbit of the Moon about the Earth is Oh-So-Slightly Elliptical and both bodies
are just a tad off from being perfectly spherical. Oh, hell, let's ignore variations in mass concentrations. I did mention
"precision", didn't I? 8^)
As a result, there is a slight "back and forth" precession of the surface of the Moon as seen from Earth and an equivalent back and
forth precession of the position of the Earth as seen from the Moon. This causes the direction of the vector to oscillate back and
fourth with a period matching the Moon's orbital period. Likewise, the eccentricity of the orbit causes the distance between the
centers of mass of the Earth and the Moon to oscillate with that same period. Ergo, the magnitude of the vector is also non
constant.
Very much a higher order effect that can be, and rightfully was, ignored in your analysis. However, you might want to take it into
account the next time you plan a Lunar Landing Mission.
Just trying to be "precise" (with maybe just a little "wise-assedness" thrown in). ;-)
Tom Veatch
Wichita, KS USA
In article <[email protected]>, [email protected] wrote:
>JMartin957 wrote:
>
>> >
>> >The _stability_ of the time tick will be unchanged.
>> >
>> >The _frequency_ will be lower. By a factor of about 2.42.
>> >>
>>
>> Actually, it will be much less accurate.
>>
>> Think tides. They would be much greater on the moon, due to the greater
>> gravitational attraction of the Earth. The gravitational attraction of the
> Sun
>> would have a greater effect as well - while it would be the same on the Moon
> as
>> on the Earth, it would be greater proportionally to the Moon's own weaker
>> gravitational force.
>>
>> Other than those "outside influences", the pendulum should be just as
> accurate
>> on the Moon as on the Earth. Slower, but just as accurate.
>>
>> John Martin
>
>John, I was agreeing with you up top that it would be less accurate, then at
> the
>bottom you said, in effect, that it would be just as accurate.
>
>Did your fingers stumble or am I not understanding you? How could it be just as
>accurate if the pendulum has a slower cycle rate on the Moon? I am not trying
> to pick
>a fuss, but I fail to understand your point. And I do want to understand.
>
It's both. :-)
Accurate in the sense that it will be self-consistent; that is, an hour
measured this week will be the same length of time as an hour measured last
week, next week, or next year (assuming that you keep the clock wound).
Inaccurate, in the sense that each hour measured by the clock on the moon will
actually be about 144 minutes long, and thus the deviation from "correct" time
will grow continuously.
So a pendulum clock designed for use on Earth, moved to the moon, will keep
time consistently, and could be considered accurate for lunar timekeeping, as
long as those who use it are willing to agree on a new definition of "hour".
Problems occur only when comparing the time shown by such a clock to the time
shown by other timekeeping devices such as an Earth-based pendulum clock or a
battery-powered digital wristwatch.
--
Regards,
Doug Miller (alphageek-at-milmac-dot-com)
For a copy of my TrollFilter for NewsProxy/Nfilter,
send email to autoresponder at filterinfo-at-milmac-dot-com
You must use your REAL email address to get a response.
"Robert Bonomi" <[email protected]> wrote in message
news:[email protected]...
> The weights have to be pulled up after the same number of cycles of the
> pendulum. Which will take about 17 days on the moon.
not that it matters a diddly twat, but a moon 'day' is quite different in
length from an earth day.
'someone else' wrote
> >p.s. The Sun will rise in the East regardless of the above.
whadabout on the moon??
Larry
"Lawrence L'Hote" <[email protected]> wrote in message
news:j%%pc.13787$qA.1744579@attbi_s51...
<snip>
> ..now just wait a second, if you are at the north pole will the sun rise
in
> the east?
Only if you are facing south.
Greg
On Wed, 19 May 2004 02:56:09 -0400, "John Keeney" <[email protected]>
wrote:
>
>Umm, no on the "tides" influence.
>The moon does not spin in relation to the earth and thus the tidal
>forces are fixed: if the seas on the moon were full of sea water
>there would be no tides coming in and going out.
it's all a bit fuzzy now but I seem to remember reading sometime in
the last few years an article about the moon and it's orbital
relationship with Terra. yes, one side always faces the earth, sorta.
the moon wobbles quite a bit. the "back" side does have a bit of area
that has never been visible to earth, but it's a lot smaller than you
might think.
>If the moon was covered with water it wouldn't be round, it would
>bulge out on the sides facing and turned away from the earth.
yup.
>In regards to the sun: it's too far away to have near the effect the earth
>would have (assuming the moon spun).
but in relation to the sun the moon *does* spin. the effects of
gravity obey the square of the distance bit, and the sun is a long way
away, but it's also freakin' huge. nothing compared to the grav effect
of Terra, but non-trivial.
> Also, it's the different distances
>from two points on a body to the other body that creates tidal forces,
>the moon is smaller so the distances are nearer the same as thus tidal
>forces are smaller. A gas planet will experience greater tidal forces
>than a rock the same mass in the same orbit.
and the moon is basically a rock, though there is some recent data
indicating that it may have a squishier core than previously thought.
On Mon, 17 May 2004 09:56:31 GMT, "Lawrence L'Hote" <[email protected]> wrote:
>..now just wait a second, if you are at the north pole will the sun rise in
>the east?
>
Nope. It'll rise in the South. Sets in the South, too.
Tom Veatch
Wichita, KS USA
"Lawrence L'Hote" <[email protected]> wrote in message
news:bJ%pc.12699$gr.1169536@attbi_s52...
>
> "Robert Bonomi" <[email protected]> wrote in message
> news:[email protected]...
> > The weights have to be pulled up after the same number of cycles of the
> > pendulum. Which will take about 17 days on the moon.
> not that it matters a diddly twat, but a moon 'day' is quite different in
> length from an earth day.
>
> 'someone else' wrote
> > >p.s. The Sun will rise in the East regardless of the above.
>
> whadabout on the moon??
>
..now just wait a second, if you are at the north pole will the sun rise in
the east?
"Hoyt Weathers" <[email protected]> wrote in message
news:[email protected]...
> I apologize for this being way off topic, but there appears to be some
smart folks
> here and I thought I would ask a few questions.
>
> If a regular pendulum clock, which keeps quite accurate time, were taken
from the
> Earth to the Moon and the correct time set into the clock and the clock
was
> restarted, would the clock keep as accurate time as it did on Earth?
Looks like your question is how gravity affects a pendulum. A pendulum's
movement is caused by gravity. Since a pendulum is just another falling
object it should fall with the same acceleration as a free falling object
(neglecting friction). An object on Earth will fall at 32 ft/sec^2. An
object on the moon will fall with (I think) with 1/6 that acceleration, and
1/6 as fast. (An object in space, without gravity, will not fall at all).
So a pendulum on the moon will keep time more slowly than on Earth. It
seems like it should be 1/6 as fast, but I don't know exactly how the clock
works, so maybe 1/6 isn't right.
dwhite <--- at least that's how it looks to me
In article <[email protected]>, Sysiphus <[email protected]> wrote:
>Hoyt Weathers thought it a good use of my time to say:
>
>
>Wow! I wish there was a newsgroup where this was on-topic!
alt.astronomy
alt.physics
sci.physics
[snip]
>If the moon's gravity is .XXX times the earth's gravity, then that clock
>will move .XXX times slower on the moon.
Nope. The ratio is sqrt(x).
--
Regards,
Doug Miller (alphageek-at-milmac-dot-com)
For a copy of my TrollFilter for NewsProxy/Nfilter,
send email to autoresponder at filterinfo-at-milmac-dot-com
You must use your REAL email address to get a response.
Hoyt Weathers thought it a good use of my time to say:
Wow! I wish there was a newsgroup where this was on-topic! More "space
math" than I have seen in a long time. The way I see it, though,(like you
really want to know) is that the issue has been "under simplified". The
clock works through the use of (a) falling weight(s). This weight has its
"weight" based on the total force required to overcome the
friction/inertia/gearing needed to move Mickey's hands around the clockface.
If the moon's gravity is .XXX times the earth's gravity, then that clock
will move .XXX times slower on the moon. AFAIK, there is no other
weight-save the pendulum(which is "driven" by the rate of the falling
weight), and therefore there is no additional 'gravitational' requirement,
only the torque needed to spin the gears.
But, I could be full of it, and not know!
On Sun, 16 May 2004 16:29:18 -0500, Hoyt Weathers <[email protected]>
wrote:
>If a regular pendulum clock, which keeps quite accurate time, were taken from the
>Earth to the Moon and the correct time set into the clock and the clock was
>restarted, would the clock keep as accurate time as it did on Earth?
The others are good references, but this might be simpler and to the
point: Look at the period being proportional to the root of (l/g).
This means that it is proportional to the length [gets longer as the
root of the length increases, not important here since that won't
change] and inversely proportional to the root of gravitational
acceleration, g, as measured on that lump of dirt.
http://physics.about.com/cs/dimensional/a/230603b.htm
So the period will increase [slower clock] as the gravity decreases,
and is measured by the root of that value [about 1/6th Earth's
gravity.].
Look at it this way: It takes longer [for the pendulum bob] to fall
the same distance on the moon.
So, was this your son's weekend homework assignment? :-)
Bill.
On Mon, 17 May 2004 15:04:14 -0500, Hoyt Weathers <[email protected]>
wrote:
>Tom Veatch wrote:
>
>> On Mon, 17 May 2004 09:56:31 GMT, "Lawrence L'Hote" <[email protected]> wrote:
>>
>> >..now just wait a second, if you are at the north pole will the sun rise in
>> >the east?
>> >
>>
>> Nope. It'll rise in the South. Sets in the South, too.
>>
>> Tom Veatch
>> Wichita, KS USA
>
>South with respect to what, pray tell?
>
>Hoyt W.
>
with respect to where you are standing, of course. all directions are
due south from the north pole.....
On Mon, 17 May 2004 10:53:03 -0500, Hoyt Weathers <[email protected]> wrote:
<snip>
>With the lower gravity on the Moon, dust would settle in the near area, but more slowly
>than on Earth, and not be blown about at all since there is no Lunar atmosphere of any
>significance.
<snip>
Hoyt,
Minor quibble having little or no effect on the discussion.
Although Moon gravity is appx. 1/6 Earth gravity, the Earth's atmosphere causes dust to remain in suspension and settle quite
slowly. No atmosphere means dust on the Moon literally drops like a rock. So, in actual effect, dust on the Moon falls much more
rapidly than on the Earth.
Tom Veatch
Wichita, KS USA
Just a little searching would get you this:
http://online.cctt.org/physicslab/content/Phy1HON/lessonnotes/pendulums/period.asp
Period goes as 1/sqrt g, so the clock would run slow by 1/sqrt 6. I'm not
sure of the six...look up lunar gravity. The clock would need different
gears!
If the clock is running more slowly, then the weights must be falling more
slowly.
Here are clock pages:
http://home.howstuffworks.com/clock.htm
http://www.britannica.com/clockworks/t_pendulum.html
Obviously one can make a gravimeter (device to measure the strength of
gravity) using a pendulum, given the equation.
Use your google.
Wilson
"Hoyt Weathers" <[email protected]> wrote in message
news:[email protected]...
> I apologize for this being way off topic, but there appears to be some
smart folks
> here and I thought I would ask a few questions.
>
> If a regular pendulum clock, which keeps quite accurate time, were taken
from the
> Earth to the Moon and the correct time set into the clock and the clock
was
> restarted, would the clock keep as accurate time as it did on Earth?
>
> Would the clock's three weights have to be pulled up every 7 days as mine
does?
>
> Would the lower Lunar gravity make any difference in time keeping?
>
> If so, which way and by how much?
>
> If not, why not?
>
> Hoyt W.
>
> p.s. The Sun will rise in the East regardless of the above.
>