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
with respect to where you are standing, of course. all directions are due south from the north pole.....
Reminds me of The Pit and the Pendulum movie, or close enough to the actual title.
Hoyt W.
"'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?
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.
In article , Dan White wrote:
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".
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.)
"Up", naturally. Take any map and look at it, there are 4 directions. 'East', 'West', 'South', and 'Up'.
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.
There was a movie?
_Edgar Allan Poe_
scott
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
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
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
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!
No. The weighs do not fall continually, they fall a fixed amount at each "tick" as the paws engage and disengage.
correspondingly also move
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.
alt.astronomy alt.physics sci.physics
[snip]Nope. The ratio is sqrt(x).
-- Regards, Doug Miller (alphageek-at-milmac-dot-com)
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dust to remain in suspension and settle quite
actual effect, dust on the Moon falls much more
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.
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:
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.
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.
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.
We'll be sure to let you know, if you succeed.
Are your eyes brown?
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).
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).
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.
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.
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:
/* IAU (1976) System of Astronomical Constants
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.
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