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.