I hacked my units package to accept descending lists of target units, e.g., (c517) uconv(2*mph+inch/sec,[fps,furlong/fortnight]) 336 furlong (d517) ------------ + 3 fps 11 fortnight so you can have a pretty arbitrary mixed radix: (c520) uconv((furlong+inch)^2,[acres,feet,inch]) 2 2 (d520) inch + 110 feet + 10 acres The function anglicize contains a capricious (and negotiable) set of units lists: (c521) anglicize(13) (d521) bakersdozen (c523) anglicize(1000) (d523) thousand (c525) anglicize(1730) (d525) pair + greatgross (c560) anglicize(halfcrown) (d560) sixpence + 2 shilling (c526) anglicize(guinea-2*farthing) (d526) twopence + pound + ninepence + halfpenny (c535) anglicize(1.0*litre) (d535) 1.0 quart + 1.81388 fluidounce (c536) anglicize(mi^3-floz) 3 3 3 3 (d536) 166 yard + 63 rod + 3 quart + pint + 511 furlong + 10 foot 515 fluidounce 3 + -------------- + 999 chain 77 (c537) anglicize(d536+floz) 3 (d537) mile Now a question: my Webster's Unabridged gives sidereal day : the interval between two successive transits of the March equinox over the upper meridian of a place that is equal to 23 hours, 56 minutes, 4.09 seconds of mean solar time sidereal year : the time in which the earth completes one revolution in its orbit around the sun measured with respect to the fixed stars : 365 days, 6 hours, 9 minutes, and 9.54 seconds of solar time (c553) uconv(siderealyear,[days,hours,minutes,seconds]) 477 seconds (d553) ----------- + 9 minutes + 6 hours + 365 days 50 tropical year : the period occupied by the sun's center in passing from one equinox to the same again and having a mean length of 365 days, 5 hours, 48 minutes, 45.5 seconds (c551) anglicize(tropicalyear) 459747 second (d551) ------------- + 48 minute + 5 hour + 26 fortnight + day 10000 which differs by 10^-8 and is claimed to be better. I would think a sidereal day would be (c556) (1.d0/(1/siderealyear+1/day),%%=anglicize(%%)) 1.0d0 (d556) ------------------ = 4.09965986199637d0 second + 56.0d0 minute 1 1 ------------ + --- siderealyear day + 23.0d0 hour but (c555) (1.d0/(1/tropicalyear+1/day),%%=anglicize(%%)) 1.0d0 (d555) ------------------ = 4.09053820802342d0 second + 56.0d0 minute 1 1 ------------ + --- tropicalyear day + 23.0d0 hour agrees better with Webster. Is this because the March equinox is precessing? If so, what is the name of the interval required for the stars to complete one circle of Polaris? --rwg PS, the difference between the two notions of year (c557) anglicize(siderealyear-tropicalyear) 235653 second (d557) ------------- + 20 minute 10000 is called NECROPSIES of the equinoxes: tidal effects on Earth's nonsphericities hurry the seasons by 20+ minutes each orbit. Presently, the Northern Hemisphere is closer to the Sun in winter. In about (c558) canon(180.*days/d557) (d558) 12710.4 orbits, this ENCOPRESIS will switch this climatic moderation to the Southern Hemisphere, which also enjoys a higher %age of oceans. Go south, young carbophobe. (Sorry if I permuted a few letters, here.-) PPS, What a great language we have; "vulgar establishment" is a celestial term, and "celestial telescope" is a goldfish.
I would think a sidereal day would be 1/(1/siderealyear+1/day) but 1/(1/tropicalyear+1/day) agrees better with Webster. Is this because the March equinox is precessing? Yes. And I think that the sidereal day should have been called "tropical day".
If so, what is the name of the interval required for the stars to complete one circle of Polaris? Wikipedia calls that an "apparent sidereal day", and calls the other kind a "mean sidereal day". Alas, the mean of the "apparent"s isn't the "mean".
Presently, the Northern Hemisphere is closer to the Sun in winter. In about 12710.4 orbits, this ENCOPRESIS will switch this climatic moderation to the Southern Hemisphere, That is indeed when the Earth's axis will be tilted the other way, relative to the stars. But the perhelion moves around too, and you haven't accounted for that. See "anomalistic year".
4.09053820802342 second + 56 minute + 23 hour I wonder, how accurately has the sidereal day been measured, and how does it vary?
My HP 48GX calculator does unit conversions: it seems to use the Bessel year, when converting years to days. -- Don Reble djr@nk.ca
djr>That is indeed when the Earth's axis will be tilted the other way, relative to the stars. But the perhelion moves around too, and you haven't accounted for that. See "anomalistic year". Yow, the perihelion is *delayed* almost 5 min/orbit, while the solstice advances > 20 min, so anglicize(anomalisticyear-solaryear) gives 25 min, 7.1 sec, and canon(180.*days/%) gives 10319 orbits for a "full tilt". I thought perihelion precession was only detectable for Mercury.
4.09053820802342 second + 56 minute + 23 hour djr>I wonder, how accurately has the sidereal day been measured, Of course, everything past the 4.09 in the above figure is rubbish. djr>and how does it vary? Besides tidal slowdown, we have those Hawaiian volcanoes piling up inertial moment. Until someday they pile too high ...-( djr>My HP 48GX calculator does unit conversions: it seems to use the Bessel year, news to Webster, but not the Web djr> when converting years to days. I think most people want calendar year.
Suppose a large comet hit Mercury. Could the dust cloud shade Earth noticeably? Ice age theory 2: The solar system passes through occasional galactic dust clouds. --rwg PS, http://www.nature.com/news/2005/050117/full/433181a.html wants you to pay money for a story entitled Titan team claims just deserts as probe hits moon of creme brulee. I thought they said it was *wet*! It would be just like the British to make creme brulee from a mix. Conspiracy theory: the final 350 images were not lost, but are being suppressed because they show indignant blancmanges menacing the lander. "Hyperbole is the sinh qua non of exponents."
It's the general-relativity-induced perihelion precession that's only significant for Mercury. Most of the perihelion precession is due to the gravitational influence of the other planets. R. William Gosper wrote:
djr>That is indeed when the Earth's axis will be tilted the other way, relative to the stars. But the perhelion moves around too, and you haven't accounted for that. See "anomalistic year".
Yow, the perihelion is *delayed* almost 5 min/orbit, while the solstice advances > 20 min, so anglicize(anomalisticyear-solaryear) gives 25 min, 7.1 sec, and canon(180.*days/%) gives 10319 orbits for a "full tilt". I thought perihelion precession was only detectable for Mercury.
4.09053820802342 second + 56 minute + 23 hour
djr>I wonder, how accurately has the sidereal day been measured, Of course, everything past the 4.09 in the above figure is rubbish. djr>and how does it vary? Besides tidal slowdown, we have those Hawaiian volcanoes piling up inertial moment. Until someday they pile too high ...-( djr>My HP 48GX calculator does unit conversions: it seems to use the Bessel year, news to Webster, but not the Web djr> when converting years to days. I think most people want calendar year.
Suppose a large comet hit Mercury. Could the dust cloud shade Earth noticeably? Ice age theory 2: The solar system passes through occasional galactic dust clouds. --rwg PS, http://www.nature.com/news/2005/050117/full/433181a.html wants you to pay money for a story entitled Titan team claims just deserts as probe hits moon of creme brulee. I thought they said it was *wet*! It would be just like the British to make creme brulee from a mix. Conspiracy theory: the final 350 images were not lost, but are being suppressed because they show indignant blancmanges menacing the lander. "Hyperbole is the sinh qua non of exponents."
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If so, what is the name of the interval required for the stars to complete one circle of Polaris? --rwg
This sounds like an unambiguous question, but it isn't because of a couple of complicating factors. First, the stars don't exactly circle polaris, or any other fixed point. They circle a point that circles the zodiacal pole every 25752 years. In 10k years, the pole star will be Vega. This 25752 number takes into account both the precession of the pole (25784 years) and the slower precession of earth's orbit. Thus the sidereal day, is (today) about 86164.09 seconds whereas the period of rotation of the earth with respect to the fixed stars is more like 86164.10 seconds. I say "today" because they both slows down with time, somewhat unpredictably, by about 2.5 milliseconds a century.
..., what is the name of the interval required for the stars to complete one circle of Polaris?
This sounds like an unambiguous question,
More ambiguous than we imagined! All I wanted was a term like Sidereal Day (which isnt quite correct) for the 24 hrish period it takes to make a full turn wrt the stars. --Bill
participants (4)
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Danny Hillis -
Don Reble -
Mike Speciner -
R. William Gosper