Paradoxes Resolved, Origins Illuminated - Requiem for Relativity
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Maurol

Argentina
37 Posts

Posted - 27 Jan 2009 :  16:18:40  Show Profile  Reply with Quote
quote:
Originally posted by nemesis

Thanks, Maurol. You describe the tropical year as a convenience - it sounds more like a fudge factor to me. I don't see the problem with the solstices changing calendar dates, especially over such a long time frame.



I suppose you mean equinoxes. Well, if I recall correctly, one of the Popes with a funny name starting with G didn't think the same. There must be an historical reason why the calendar is aligned with the equinoxes; you can surely google about it.

Oh, one "detail" that I got wrong: 13000 years in the future, the winter solstice will be around 18-20 of January. It was the other way around: shorter calendars will need more days to indicate the same moment.
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Joe Keller

USA
958 Posts

Posted - 27 Jan 2009 :  18:04:59  Show Profile  Reply with Quote
Synchronized Precession

Above, I note that the presumed Barbarossa/Frey objects show a binary orbit agreeing with Kepler's 2nd law to 0.125%. What about the changing aspect, of the binary orbit, due to the solar orbit? I used a kind of average ellipse which might compensate to first order, but the second order term is as large as (0.0965 radian)^2 * 0.5 = 0.47%. Maybe the phase was favorable. Anyway, there is another possible explanation: synchronized precession.

Let's compare the Sun/Earth/Luna system to the Sun/Barbarossa/Frey system. The precession period of the latter, would be 19yr
* 200^3 for weaker Sun tide
* (1/400)^2 for greater binary radius given same angular momentum
* sqrt(3000) for more massive central object
* sqrt(400) for the increase in angular momentum with binary radius
= 1,000,000 yr.

Let's replace the sun with a second, Saturn-mass moon, Freya, 2 AU from Barbarossa. Freya could precess Frey's orbit in 1,000,000 * (1/0.0003) * (2/200)^3 = 3000 yr. If this precession came to be synchronized with Barbarossa's solar orbit, the binary orbit's aspect would not change.
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Maurol

Argentina
37 Posts

Posted - 27 Jan 2009 :  18:59:06  Show Profile  Reply with Quote
quote:
Originally posted by Maurol

Oh, one "detail" that I got wrong: 13000 years in the future, the winter solstice will be around 18-20 of January. It was the other way around: shorter calendars will need more days to indicate the same moment.



And if we consider leap seconds, in 13000 years the calendar will fall behind between 8 and 12 days and fraction. We cannot know exactly in advance, because you cannot know with an advance bigger than 6 months if and when a leap second will be needed :-O
That certainly doesn't look as a very good way to keep time, don't you think?

So, considering leap seconds, the winter solstice will fall between January 6-10 in the year 13000.
But don't worry, leap seconds are here not to stay :-)
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Joe Keller

USA
958 Posts

Posted - 27 Jan 2009 :  20:30:10  Show Profile  Reply with Quote
REM program to fit Barbarossa/Frey orbits to 3 data points, predict others

REM initialize constants
PRINT : PRINT
REM GOTO 9000
pi# = 4 * ATN(1): pi180# = pi# / 180: pi2# = 2 * pi#: crit = 0
kk# = 0 - (pi2# / 365.25#) ^ 2 * 332447 / 332001: corr# = 1 / 1.01#
REM kk# is solar accel. const. in AU/day^2, including known planets
REM dimension variables; double precision throughout
REM positive integration weights for 1954, 1986, 1987:
DIM w(3) AS DOUBLE
w(1) = 1 / 4: w(2) = 3 / 8: w(3) = 3 / 8
DIM t(6) AS DOUBLE: DIM ra(6) AS DOUBLE: DIM decl(6) AS DOUBLE
DIM x(6) AS DOUBLE: DIM y(6) AS DOUBLE: DIM z(6) AS DOUBLE
DIM r(6) AS DOUBLE
DIM xe(6) AS DOUBLE: DIM ye(6) AS DOUBLE: DIM ze(6) AS DOUBLE
DIM re(6) AS DOUBLE
DIM rab(6) AS DOUBLE: DIM declb(6) AS DOUBLE: DIM raf(6) AS DOUBLE
DIM declf(6) AS DOUBLE
DIM nax(6) AS DOUBLE: DIM nay(6) AS DOUBLE: DIM naz(6) AS DOUBLE
DIM x0(6) AS DOUBLE: DIM y0(6) AS DOUBLE: DIM z0(6) AS DOUBLE
DIM theta(6, 6) AS DOUBLE

REM get times
GOSUB 1000
dt1# = 1 / (t(2) - t(1)): dt2# = 1 / (t(3) - t(2)): dt3# = 2 / (t(3) - t(1))
REM get coordinates
GOSUB 1100
REM get barycentric Earth positions
GOSUB 1200

REM PRINT "Done initializing; now searching..."
REM best mass ratio and radius
REM for circular orbit using A&A2, B&B3, C11&C, is q0#=.8747#; r00#=209.2#
REM and rp#=-.19 is best change from rp#=0
REM best using work and general orbit is q0#=.8770#; r00#=208.3#; rp0#=-25.9#
rel0# = 10 ^ 9
nn = 1

FOR i = 1 TO nn
q# = .877# + 0 * .0001# * i: p# = 1 - q#

FOR r0# = 208.3# TO 208.3# STEP .1

FOR rp# = -25.9 TO -25.9 STEP .1

rt# = rp# * pi2# / 3024 / 365
REM 3024 is an earlier approximation of the actual period; this isn't critical
REM getting c.o.m. coords. by weighted ave. of Barbarossa & Frey coords.,
REM causes < 0.5" error
FOR j = 1 TO 5
ra(j) = rab(j) * q# + raf(j) * p#
decl(j) = declb(j) * q# + declf(j) * p#
zz# = SIN(decl(j)): cs# = COS(decl(j))
xx# = cs# * COS(ra(j)): yy# = cs# * SIN(ra(j))
csthe# = (xx# * xe(j) + yy# * ye(j) + zz# * ze(j)) / re(j)
r1# = r0# + rt# * (t(j) - t(1))
REM cosine rule: rr# ^ 2 + 2 * rr# * re(j) * csthe# + re(j) ^ 2 - r1#^2 = 0
REM use quadratic equation for rr#
b# = 2 * re(j) * csthe#: c# = re(j) ^ 2 - r1# ^ 2
rr# = (0 - b# + SQR(b# ^ 2 - 4 * c#)) * .5#
x(j) = xe(j) + rr# * xx#
y(j) = ye(j) + rr# * yy#
z(j) = ze(j) + rr# * zz#: r(j) = r1#

NEXT j

REM one-time light time correction
IF crit = 0 THEN GOSUB 1050
crit = 1


REM Minimize difference between Newtonian accel.
REM and accel. of trial curve.
REM accel. of trial curve
GOSUB 2000
REM Newtonian accel.
GOSUB 2100
REM work done by difference
GOSUB 2200

IF rel# < rel0# THEN GOSUB 3000
NEXT rp#: NEXT r0#
REM IF i = 1 THEN PRINT i; "/"; nn; "done...";
REM IF i > 1 THEN PRINT i; "/"; nn; "...";
NEXT i

PRINT : PRINT "unexplained work, AU/day^2*AU:"; rel0#
PRINT "best mass ratio, initial radius, & rp# parameter:"
PRINT q0#, r00#, rp0#
PRINT "t(3) radius, midrange of t(1) & t(3)";
rt0# = rp0# * pi2# / 3024 / 365
rmid# = r00# + .5# * rt0# * (t(3) - t(1))
PRINT r00# + 2 * (rmid# - r00#), rmid#
PRINT "theoretical circular period"
PRINT rmid# ^ 1.5# * SQR(332000 / 332447)

FOR i = 2 TO 3
FOR j = 1 TO i - 1
cs# = (x0(i) * x0(j) + y0(i) * y0(j) + z0(i) * z0(j)) / r0(i) / r0(j)
theta(i, j) = ATN(SQR(1 - cs# ^ 2) / cs#)
PRINT "sector "; j; i; ": ";
PRINT theta(i, j)
PRINT "radians/day"; " ";
om# = theta(i, j) / (t(i) - t(j))

PRINT om#
IF i = 3 AND j = 1 THEN GOSUB 3100
NEXT j: NEXT i
PRINT theta(2, 1) + theta(3, 2) - theta(3, 1)

REM GOTO 990
PRINT "first order elliptical extrapolation"
xx# = x0(3) / r0(3): yy# = y0(3) / r0(3): zz# = z0(3) / r0(3)
dt# = t(3) - t(1)
x0# = x0(1) / r0(1): y0# = y0(1) / r0(1): z0# = z0(1) / r0(1)
cs# = xx# * x0# + yy# * y0# + zz# * z0#
xx# = xx# - cs# * x0#: yy# = yy# - cs# * y0#: zz# = zz# - cs# * z0#
rr# = SQR(xx# ^ 2 + yy# ^ 2 + zz# ^ 2)
xx# = xx# / rr#: yy# = yy# / rr#: zz# = zz# / rr#
th0# = ATN(SQR(1 - cs# ^ 2) / cs#)

FOR i = 4 TO 6
th# = th0# * (t(i) - t(1)) / dt#
th# = th# / (1 + rt0# * (t(i) - .5# * (t(3) + t(1))) / rmid#)
cs# = COS(th#): sn# = SIN(th#)
r# = r00# + rt0# * (t(i) - t(1))
xpred# = r# * (cs# * x0# + sn# * xx#)
ypred# = r# * (cs# * y0# + sn# * yy#)
zpred# = r# * (cs# * z0# + sn# * zz#)
REM PRINT "barycentric c.o.m. X, Y, Z for i-th time": PRINT
REM PRINT xpred#, ypred#, zpred#
REM PRINT "geocentric X, Y, Z"
xpred# = xpred# - xe(i): ypred# = ypred# - ye(i): zpred# = zpred# - ze(i)
REM PRINT xpred#, ypred#, zpred#
REM PRINT "J2000 celestial coords."
decl# = ATN(zpred# / SQR(xpred# ^ 2 + ypred# ^ 2)) / pi180#
ra# = ATN(ypred# / xpred#) / pi180# + 180
PRINT i; "-th point:"; ra#, decl#
PRINT "RA"; INT(ra# / 15); "h"; 4 * (ra# - INT(ra# / 15) * 15); "m"
PRINT "Decl"; -INT(-decl#); "deg"; 60 * (-decl# - INT(-decl#)); "arcmin"
NEXT i

990 END


REM Get Julian date - 2400000.
1000 t(1) = 34798.5# + 8 / 24 + 5 / 1440
t(2) = 46504.5# + 14 / 24 + 18 / 1440
REM Change time from B plate to C plate.
t(3) = t(2) + 365 - 28 - 31 + 16 + 3 / 24 - 16 / 1440
tbarbarossa3# = 54184.5# + 42 / 1440: tfrey3# = 54191.5# + 7 / 24 + 39 / 1440
REM Interpolate t(4) using est. of mass ratio.
t(4) = .877# * tbarbarossa3# + .123# * tfrey3#
REM Guess that 12/22/08 epoch, t(5), is approx. time on meridian;
REM Barbarossa was near stationarity anyway.
t(5) = 54822.5# + 12 / 24 + 47 / 1440
t(6) = 54851.5# + 9 / 24 + 48.5# / 1440
RETURN

REM light time correction
1050 FOR j = 1 TO 6
t(j) = t(j) - rr# * 1.496 * 10 ^ 13 / (2.99793 * 10 ^ 10)
NEXT j
RETURN


REM get J2000 geocentric Barbarossa & Frey coordinates
REM "A" plate (1954); objects A2 & A
1100 raf(1) = pi180# * (11 * 15 + 2 / 4 + 25.16# / 240)
declf(1) = 0 - pi180# * (5 + 56 / 60 + 11.3# / 3600)
rab(1) = pi180# * (11 * 15 + 3 / 4 + 12.4# / 240)
declb(1) = 0 - pi180# * (5 + 58 / 60 + 9 / 3600)

REM "B" plate (1986); objects B3 & B
raf(2) = pi180# * (11 * 15 + 16 / 4 + 51.55# / 240)
declf(2) = 0 - pi180# * (7 + 49 / 60 + 41.1# / 3600)
rab(2) = pi180# * (11 * 15 + 16 / 4 + 56.07# / 240)
declb(2) = 0 - pi180# * (7 + 55 / 60 + 14.3# / 3600)

REM "C" plate (1987); objects C & C11
raf(3) = pi180# * (11 * 15 + 18 / 4 + 3.18# / 240)
declf(3) = 0 - pi180# * (7 + 58 / 60 + 46.1 / 3600)
rab(3) = pi180# * (11 * 15 + 18 / 4 + .41 / 240)
declb(3) = 0 - pi180# * (8 + 1 / 60 + 57.7 / 3600)

REM Genebriera March 25, 2007
rab(4) = pi180# * (11 * 15 + 26 / 4 + 22.2# / 240)
declb(4) = 0 - pi180# * (9 + 4 / 60 + 59 / 3600)

REM Riley April 1, 2007
raf(4) = pi180# * (11 * 15 + 26 / 4 + 25 / 240)
declf(4) = 0 - pi180# * (8 + 57 / 60 + 26 / 3600)

REM alternate coords. for Riley's "Frey" (see messageboard May 7, 2007)
REM raf(4) = pi180# * (11 * 15 + 26 / 4 + 24.6 / 240)
REM declf(4) = 0 - pi180# * (8 + 57 / 60 + 48.5 / 3600)

REM Genebriera's "Frey" (see messageboard May 15, 2007)
REM raf(4) = pi180# * (11 * 15 + 26 / 4 + 31.8 / 240)
REM declf(4) = 0 - pi180# * (9 + 0 / 60 + 11 / 3600)

REM Dec. 22, 2008 U. of Iowa photo
rab(5) = pi180# * (11 * 15 + 28 / 4 + 22.08# / 240)
declb(5) = 0 - pi180# * (9 + 16 / 60 + 6.4# / 3600)
raf(5) = pi180# * (11 * 15 + 29 / 4 + 4.66# / 240)
declf(5) = 0 - pi180# * (9 + 7 / 60 + 2.3# / 3600)
RETURN


REM Get J2000 barycentric Earth XYZ position (barycenter excludes Barbarossa).
REM In this program, "barycentric" = sun + traditional planets.
REM all but 1954, from Astronomical Almanac
1200 A# = 14 / 24 + 18 / 1440: i = 2
GOSUB 1220
A# = 17 / 24 + 2 / 1440: i = 3
GOSUB 1220
REM as in subroutine 1000, weighted time for 2007
A# = 42 / 1440 * .877# + (7 + 7 / 24 + 39 / 1440) * (1 - .877#): i = 4
GOSUB 1220
REM as in subroutine 1000, guess that 12/2008 epoch was meridian time
A# = 12 / 24 + 47 / 1440: i = 5
GOSUB 1220
A# = 9 / 24 + 48.5# / 1440: i = 6
GOSUB 1220
GOSUB 1250
RETURN

1220 b# = 1 - A#
READ xe1#, xe2#, ye1#, ye2#, ze1#, ze2#
xe(i) = b# * xe1# + A# * xe2#
ye(i) = b# * ye1# + A# * ye2#
ze(i) = b# * ze1# + A# * ze2#
re(i) = SQR(xe(i) ^ 2 + ye(i) ^ 2 + ze(i) ^ 2)
RETURN

REM find 1954 barycentric Earth position
1250 msun# = 332000: mjup# = 318.1#: msat# = 95.2#: mur# = 14.6#: mnep# = 17.2
me# = 1: mtot# = msun# + mjup# + msat# + mur# + mnep# + me#
sx# = 0: sy# = 0: sz# = 0
REM interpolate sun coordinates
A# = 8 / 24 + 5 / 1440: b# = 1 - A#
raa# = 22 * 15 + 30 / 4 + 33.54 / 240: rab# = 22 * 15 + 34 / 4 + 20.69 / 240
ra# = (raa# * b# + rab# * A#) * pi180#
decla# = 9 + 22 / 60 + 8.3 / 3600: declb# = 8 + 59 / 60 + 52 / 3600
decl# = 0 - (decla# * b# + declb# * A#) * pi180#
ra# = .9899423000000001#: rb# = .9901818999999999#
r# = ra# * b# + rb# * A#: w# = msun#
GOSUB 1270
REM planets interpolated to nearest RA second, or Decl arcminute
REM Jupiter
ra# = pi180# * (5 * 15 + 2 / 4 + 41 / 240)
decl# = pi180# * (22 + 29 / 60 + 51 / 3600)
r# = 4.846#: w# = mjup#
GOSUB 1270
REM Saturn
ra# = pi180# * (14 * 15 + 30 / 4 + 57 / 240)
decl = 0 - pi180# * (12 + 9 / 60 + 22 / 3600)
r# = 9.308#: w# = msat#
GOSUB 1270
REM Uranus
ra# = pi180# * (7 * 15 + 24 / 4 + 19 / 240)
decl# = pi180# * (22 + 31 / 60 + 46 / 3600)
r# = 18.001#: w# = mur#
GOSUB 1270
REM Neptune
ra# = pi180# * (13 * 15 + 38 / 4 + 22 / 240)
decl# = 0 - pi180# * (8 + 22 / 60 + 9 / 3600)
r# = 29.662#: w# = mnep#
GOSUB 1270
REM convert to 1950 ecliptic coords.
theta# = 23.45# * pi180#
GOSUB 1290
REM convert to 2000 ecliptic coords.
theta# = pi180# * 50.2619# / 3600 * 50
cs# = COS(theta#): sn# = SIN(theta#): sx0# = sx#
sx# = cs# * sx# - sn# * sy#
sy# = cs# * sy# + sn# * sx0#
REM convert to 2000 celestial coords.
theta# = 0 - 23.45# * pi180#
GOSUB 1290
xe(1) = 0 - sx# / mtot#
ye(1) = 0 - sy# / mtot#
ze(1) = 0 - sz# / mtot#
re(1) = SQR(xe(1) ^ 2 + ye(1) ^ 2 + ze(1) ^ 2)
RETURN

1270 z# = r# * SIN(decl#): cs# = COS(decl#)
x# = r# * cs# * COS(ra#): y# = r# * cs# * SIN(ra#)
sx# = sx# + w# * x#
sy# = sy# + w# * y#
sz# = sz# + w# * z#
RETURN

1290 cs# = COS(theta#): sn# = SIN(theta#): sz0# = sz#
sz# = cs# * sz# - sn# * sy#
sy# = cs# * sy# + sn# * sz0#
RETURN


REM find the accel. of the parabolas fitting 1st thru 3rd points
2000 vx1# = (x(2) - x(1)) * dt1#
vy1# = (y(2) - y(1)) * dt1#
vz1# = (z(2) - z(1)) * dt1#
vx2# = (x(3) - x(2)) * dt2#
vy2# = (y(3) - y(2)) * dt2#
vz2# = (z(3) - z(2)) * dt2#
REM correct accel. to fixed frame
ax# = (vx2# - vx1#) * dt3# * corr#
ay# = (vy2# - vy1#) * dt3# * corr#
az# = (vz2# - vz1#) * dt3# * corr#
RETURN

REM find Newtonian accel. at 1st thru 3rd points
2100 FOR j = 1 TO 3
dr# = kk# / r(j) ^ 3
nax(j) = x(j) * dr#
nay(j) = y(j) * dr#
naz(j) = z(j) * dr#
NEXT j
RETURN

REM weighted ave. sunward accel. & difference from observed
2200 avenax# = 0: avenay# = 0: avenaz# = 0
FOR j = 1 TO 3
avenax# = avenax# + w(j) * nax(j)
avenay# = avenay# + w(j) * nay(j)
avenaz# = avenaz# + w(j) * naz(j)
NEXT j
rel# = SQR((avenax# - ax#) ^ 2 + (avenay# - ay#) ^ 2 + (avenaz# - az#) ^ 2)
REM PRINT "subroutine 2200 --> Newtonian vs. apparent accel."
REM PRINT SQR(avenax# ^ 2 + avenay# ^ 2 + avenaz# ^ 2)
REM PRINT SQR(ax# ^ 2 + ay# ^ 2 + az# ^ 2)
REM work done by unexplained force
rel# = (avenax# - ax#) * (x(3) - x(1))
rel# = rel# + (avenay# - ay#) * (y(3) - y(1))
rel# = rel# + (avenaz# - az#) * (z(3) - z(1))
rel# = ABS(rel#)
RETURN


REM update best
3000 rel0# = rel#: i0 = i: q0# = q#: r00# = r0#: rp0# = rp#
FOR m = 1 TO 5
x0(m) = x(m): y0(m) = y(m): z0(m) = z(m): r0(m) = r(m)
NEXT m
RETURN

3100 PRINT "apparent circular period"
PRINT pi2# / om# / 365.25
RETURN

REM '86
DATA -.992473,-.99431,.097062,.081274,.042041,.035195
REM less accurate Earth XYZ for '87 "C" plate (see below)
REM DATA -.66451,-.66451,.6939,.6939,.30093,.30093
DATA -.647098,-.660316,.689355,.678874,.298895,.29435
REM '07
DATA -.992917,-.991887,-.057633,-.073429,-.025095,-.031943
REM '08
DATA -.008637,-.026129,.906672,.906369,.39304,.392909
REM '09
DATA -.493191,-.508316,.786488,.778431,.340941,.337449


REM data from April 2008 program
REM (times of plates or photos, Sun positions,
REM coords. of identified disappearing dots)
REM plate A
DATA 1954.154,.9900229, 22,34, 27.6,-8,-59,-7
REM object A2
DATA 11,2,25.47,-5,-56,-12
REM object A
DATA 11,3,12.4,-5,-58,-9

REM plate B
DATA 1986.201,.9945905,23,40,30.9,-2,-10,-49
REM object B3
DATA 11,16,51.55,-7,-49,-41.1
REM object B
DATA 11,16,56.07,-7,-55,-13.2

REM plate C
DATA 1987.08215,.9852006,20,53,6.9,-17,31,13
REM object C11 (fewer than 11 found; named for easy recall)
DATA 11,18,.41,-8,-1,-57.7
REM object C
DATA 11,18,3.18,-7,-58,-46.1

REM plate D (optical infrared)
REM DATA 1997.16711,.9914845,22,57,21.1,-6,-39,-54
REM DATA 11,22,32.9,-8,-26,-56
REM DATA 11,22,16.77,-8,-29,-32.6
REM DATA 11,22,1.33,-8,-34,-36.9
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Joe Keller

USA
958 Posts

Posted - 27 Jan 2009 :  20:46:14  Show Profile  Reply with Quote
The previous post is my new computer program for demonstrating the progress of the Barbarossa/Frey c.o.m. from 1954 through 1986 to 1987. It also allows predictions to be made for the geocentric c.o.m. position at any future time, if modified with appropriate Julian date and Earth XYZ coords.

Mauro's computer instructions worked: after right-clicking on the top bar, I had to left-click edit->selectall; then left-click on the screen; then highlight the text on the screen to be copied; then left-click edit->copy. A few of the lines might become wrapped, but otherwise it looks OK. As it appears in the messageboard now, none of the lines are wrapped.
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nemesis

84 Posts

Posted - 27 Jan 2009 :  21:05:56  Show Profile  Reply with Quote
[/quote]

I suppose you mean equinoxes. Well, if I recall correctly, one of the Popes with a funny name starting with G didn't think the same. There must be an historical reason why the calendar is aligned with the equinoxes; you can surely google about it.

[/quote]
Is there any particular reason the equinoxes are preferred to the solstices? It's like the lunar cycle - a lunar month is traditionally reckoned from new moon to new moon but could just as well go from full moon to full moon. That makes more sense to me because it's easier to see when the moon is full.
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Maurol

Argentina
37 Posts

Posted - 28 Jan 2009 :  05:39:13  Show Profile  Reply with Quote
quote:
Originally posted by nemesis
Is there any particular reason the equinoxes are preferred to the solstices? It's like the lunar cycle - a lunar month is traditionally reckoned from new moon to new moon but could just as well go from full moon to full moon. That makes more sense to me because it's easier to see when the moon is full.



http://en.wikipedia.org/wiki/Gregorian_calendar
Read it. All of it.
Just joking. The answer to your question is under "Gregorian reform" in that article. Besides, the complete article is well written and worth reading.
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Stoat

United Kingdom
964 Posts

Posted - 28 Jan 2009 :  07:06:48  Show Profile  Reply with Quote
Hi Nemesis, hunter gatherers were hugely interested in the moon, then with the birth of a more sedentary agricultural society, the sun become more important. This was a major cultural crisis.

A bronze age wizard's hat was dug up in Germany, like all wizard's hats it had the moon and stars on it. The astonishing thing was that it showed the nineteen year cycle of the moon. That knowledge had to have taken a long time to establish. The new thinking about henges, is that they are constructed to show people that the setting sun and the rising moon align on a certain day. This restores harmony to the world.

Over the year, the sun rises north or south of east. So mark those two furthest points. In Oedipus Rex the shepherd explains when he found the dumped baby, by talking about the rising of Canupus. When that particular star rose just before the sun, he knew that it was time to bring the sheep down from the mountain pastures. Likewise the star Sirius was important in Egypt, as it brought with it the flooding of the Nile. A social contract with the gods.

People celebrated the equinox and the solstice and they knew the stars in a manner that modern man has totally lost. The newer religions had to fit their festivals into this world view. Problems arose with the Roman Empire. It was big and lasting! The calendar needed to be adjusted, by adding days to the year and setting up the leap year. Only the priests misinterpreted the new set up and had a leap year every three years. That was stopped and the julian calendar was adopted. However, July and August had to have the same number of days. Because Julius Caesar and Augustus Caesar had to have equal status. It worked though. Zoom on to the late middle ages, Easter is falling later and later. The church wants it close to the equinox. so the Gregorian calendar is brought in. Add ten days to the year and a new clause for working out leap years (a year divisible by four, except those that are divisible by 100 but not by 400; 1900 was not a leap year)

You've guessed it, it worked but again was rather messy. Britain and the colonies didn't join in until a hundred plus years later, and of course the general population thought it was a huge scam to rob them of their wages. The Russian October revolution wasn't in October, so rich Times readers had apoplexy over their brandy.

Anyway back to this twenty five thousand year cycle. There's something called the "argument of pericentre," if it is zero degrees, then the planet is closest to the sun in its orbit, at the ascending node of the orbit (for earth the spring equinox) if it's 180 degrees, then the planet is furthest away in its orbit at the ascending node. The argument for the earth is 95 degrees.

This doesn't change but if it did wander slightly over time, then at mid winter we could be a little further out in our slightly eccentric orbit. Let's say that the wander takes about 500 million years. Then we could think that the sun's output varies. Or we could think that the earth's orbit alters periodically in eccentricity. Or, that the tilt of the earth alters periodically. This does have relevance to the ice ages, as ice advanced in both hemispheres at the same time.

I once sent a letter to the New Musical Express, suggesting that we had a sliding rate of time. People's watches would be set by how much money they had. The richer they were the faster their watch would run. Pity it never caught on, as everyone would have learned calculus and no one would have a clue what the real time was.
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nemesis

84 Posts

Posted - 28 Jan 2009 :  08:15:48  Show Profile  Reply with Quote
For what it's worth, the Muslims use a pure lunar calendar and have no problem with their months rotating through the seasons. A lunisolar calendar had been used in pre-Islamic Arabia, but evidently Muhammad personally decreed an uncorrected lunar calendar.
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Maurol

Argentina
37 Posts

Posted - 29 Jan 2009 :  06:29:42  Show Profile  Reply with Quote
Hi,
I in no way want to turn this into a religious issue, but as Easter is the most important religious feast for the Christians, and as it is a moveable feast, that depends on the full moon and in the time of the Vernal equinox, for the Catholics and Christians it was important to fix the Vernal equinox.
They did't want the timing of a feast, which depends on astronomical events, to always advance with the passing of the years.
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Joe Keller

USA
958 Posts

Posted - 30 Jan 2009 :  12:13:40  Show Profile  Reply with Quote
REM program to fit Barbarossa/Frey orbits to 3 data points, predict others
REM and check binary orbit for Kepler's 2nd law

REM initialize constants
PRINT : PRINT
REM GOTO 9000
pi# = 4 * ATN(1): pi180# = pi# / 180: pi2# = 2 * pi#: crit = 0
kk# = 0 - (pi2# / 365.25#) ^ 2 * 332447 / 332001: corr# = 1 / 1.01#
REM kk# is solar accel. const. in AU/day^2, including known planets
REM dimension variables; double precision throughout
REM positive integration weights for 1954, 1986, 1987:
DIM w(3) AS DOUBLE
w(1) = 1 / 4: w(2) = 3 / 8: w(3) = 3 / 8
DIM t(7) AS DOUBLE: DIM ra(7) AS DOUBLE: DIM decl(7) AS DOUBLE
DIM x(7) AS DOUBLE: DIM y(7) AS DOUBLE: DIM z(7) AS DOUBLE
DIM r(7) AS DOUBLE: DIM rpl(7) AS DOUBLE: DIM rmi(7) AS DOUBLE
DIM drdth(7) AS DOUBLE
DIM xe(7) AS DOUBLE: DIM ye(7) AS DOUBLE: DIM ze(7) AS DOUBLE
DIM re(7) AS DOUBLE
DIM rab(7) AS DOUBLE: DIM declb(7) AS DOUBLE: DIM raf(7) AS DOUBLE
DIM declf(7) AS DOUBLE
DIM nax(7) AS DOUBLE: DIM nay(7) AS DOUBLE: DIM naz(7) AS DOUBLE
DIM x0(7) AS DOUBLE: DIM y0(7) AS DOUBLE: DIM z0(7) AS DOUBLE
DIM theta(7, 7) AS DOUBLE: DIM the(7) AS DOUBLE
DIM rapred(7) AS DOUBLE: DIM declpred(7) AS DOUBLE
DIM cc(5, 6) AS DOUBLE: DIM co(5) AS DOUBLE: DIM area(7) AS DOUBLE

REM get times
GOSUB 1000
dt1# = 1 / (t(2) - t(1)): dt2# = 1 / (t(3) - t(2)): dt3# = 2 / (t(3) - t(1))
REM get coordinates
GOSUB 1100
REM get barycentric Earth positions
GOSUB 1200

REM PRINT "Done initializing; now searching..."
REM best mass ratio, radius, dr/dtheta
REM using work and general orbit are q0#=.8898#; r00#=205.2#
REM rp0# = -1.0# rpp0#=-23.3#
nn = 1

FOR i = 1 TO nn
q# = .8898# + 0 * .0001# * i: p# = 1 - q#: rel0# = 10 ^ 9

FOR r0# = 205.2# TO 205.2# STEP .1#

REM dr/dtheta
FOR rp# = -1 TO -1 STEP .1#

REM d2r/dtheta2
FOR rpp# = -23.3# TO -23.3# STEP .1#

REM get dr/dt
rt# = rp# * pi2# / 3080 / 365.25#
REM get d2r/dt2
rtt# = rpp# * (pi2# / 3080 / 365.25#) ^ 2
REM getting c.o.m. coords. by weighted ave. of Barbarossa & Frey coords.,
REM causes < 0.5" error
FOR j = 1 TO 6
ra(j) = rab(j) * q# + raf(j) * p#
decl(j) = declb(j) * q# + declf(j) * p#
zz# = SIN(decl(j)): cs# = COS(decl(j))
xx# = cs# * COS(ra(j)): yy# = cs# * SIN(ra(j))
csthe# = (xx# * xe(j) + yy# * ye(j) + zz# * ze(j)) / re(j)
r1# = r0# + rt# * (t(j) - t(1)) + .5# * rtt# * (t(j) - t(1)) ^ 2
REM cosine rule: rr# ^ 2 + 2 * rr# * re(j) * csthe# + re(j) ^ 2 - r1#^2 = 0
REM use quadratic equation for rr#
b# = 2 * re(j) * csthe#: c# = re(j) ^ 2 - r1# ^ 2
rr# = (0 - b# + SQR(b# ^ 2 - 4 * c#)) * .5#
x(j) = xe(j) + rr# * xx#
y(j) = ye(j) + rr# * yy#
z(j) = ze(j) + rr# * zz#: r(j) = r1#
NEXT j

REM one-time light time correction
IF crit = 0 THEN GOSUB 1050
crit = 1


REM Minimize difference between Newtonian accel.
REM and accel. of trial curve.
REM accel. of trial curve
GOSUB 2000
REM Newtonian accel.
GOSUB 2100
REM difference^2
GOSUB 2200

IF rel# < rel0# THEN GOSUB 3000
NEXT rpp#: NEXT rp#: NEXT r0#
REM IF i = 1 THEN PRINT i; "/"; nn; "done...";
REM IF i > 1 THEN PRINT i; "/"; nn; "...";

PRINT : PRINT "unexplained accel., AU/day^2:"; SQR(rel0#)
PRINT "best mass ratio, initial radius, dr/dtheta & d2r/dtheta2:"
PRINT q0#, r00#, rp0#, rpp0#
PRINT "t(3) radius, midrange of t(1) & t(3) radii";
rmid# = .5# * (r0(1) + r0(3))
PRINT r0(3), rmid#
PRINT "circular period at this distance"
PRINT rmid# ^ 1.5# * SQR(332001 / 332447)

FOR i = 2 TO 3
FOR j = 1 TO i - 1
cs# = (x0(i) * x0(j) + y0(i) * y0(j) + z0(i) * z0(j)) / r0(i) / r0(j)
theta(i, j) = ATN(SQR(1 - cs# ^ 2) / cs#)
PRINT "sector "; j; i; "in radians: ";
PRINT theta(i, j)
PRINT "radians/day"; " ";
om# = theta(i, j) / (t(i) - t(j))
PRINT om#
IF i = 3 AND j = 1 THEN GOSUB 3100
NEXT j: NEXT i
PRINT "excess travel"
PRINT theta(2, 1) + theta(3, 2) - theta(3, 1)
NEXT i

PRINT "first order elliptical extrapolation"
xx# = x0(3) / r0(3): yy# = y0(3) / r0(3): zz# = z0(3) / r0(3)
x0# = x0(1) / r0(1): y0# = y0(1) / r0(1): z0# = z0(1) / r0(1)
cs# = xx# * x0# + yy# * y0# + zz# * z0#
xx# = xx# - cs# * x0#: yy# = yy# - cs# * y0#: zz# = zz# - cs# * z0#
rr# = SQR(xx# ^ 2 + yy# ^ 2 + zz# ^ 2)
xx# = xx# / rr#: yy# = yy# / rr#: zz# = zz# / rr#
th0# = ATN(SQR(1 - cs# ^ 2) / cs#)

REM To get all of them, let i=4 to 7.
FOR i = 6 TO 6
th# = th0# * (t(i) - t(1)) / (t(3) - t(1))
ri# = r00# + .5# * rt0# * (t(i) - t(1)) + .1667# * rtt0# * (t(i) - t(1)) ^ 2
r3# = r00# + .5# * rt0# * (t(3) - t(1)) + .1667# * rtt0# * (t(3) - t(1)) ^ 2
th# = th# * (r3# / ri#) ^ 2
cs# = COS(th#): sn# = SIN(th#)
r# = r00# + rt0# * (t(i) - t(1)) + .5# * rtt0# * (t(i) - t(1)) ^ 2
xpred# = r# * (cs# * x0# + sn# * xx#)
ypred# = r# * (cs# * y0# + sn# * yy#)
zpred# = r# * (cs# * z0# + sn# * zz#)
REM PRINT "barycentric c.o.m. X, Y, Z for i-th time": PRINT
REM PRINT xpred#, ypred#, zpred#
REM PRINT "geocentric X, Y, Z"
xpred# = xpred# - xe(i): ypred# = ypred# - ye(i): zpred# = zpred# - ze(i)
REM PRINT xpred#, ypred#, zpred#
REM PRINT "J2000 celestial coords."
decl# = ATN(zpred# / SQR(xpred# ^ 2 + ypred# ^ 2)) / pi180#
ra# = ATN(ypred# / xpred#) / pi180# + 180
rapred(i) = ra# * pi180#: declpred(i) = decl# * pi180#
PRINT i; "-th point:"; ra#, decl#
PRINT "RA"; INT(ra# / 15); "h"; 4 * (ra# - INT(ra# / 15) * 15); "m"
PRINT "Decl"; -INT(-decl#); "deg"; 60 * (-decl# - INT(-decl#)); "arcmin"
PRINT
NEXT i

600 PRINT "PART 2: binary orbit - A, B, C; and Dec. 2008 U. of Iowa photo"
FOR i = 1 TO 3
y(i) = declf(i) - declb(i): th# = declf(i) * .1102# + declb(i) * .8898#
x(i) = (raf(i) - rab(i)) * COS(th#)
y(i) = y(i) * r0(i): x(i) = x(i) * r0(i)
NEXT i
y(4) = (declf(6) - declpred(6)) / .8898#
x(4) = (raf(6) - rapred(6)) / .8898# * COS(declpred(6))
r# = r00# + rt0# * (t(6) - t(1)) + .5# * rtt0# * (t(6) - t(1)) ^ 2
x(4) = x(4) * r#: y(4) = y(4) * r#
t(4) = t(6)

REM try to correct for precession
REM slope of 1986-1987 line is 54deg15'+180
dth# = .159#
REM precession correction estimated graphically is .157#
REM this is the aspect change expected from orbital motion
cs# = .81157#: sn# = .58425#
x(1) = x(1) + cs# * dth#
x(4) = x(4) - cs# * dth#
y(1) = y(1) + sn# * dth#
y(4) = y(4) - sn# * dth#

FOR i = 1 TO 4
PRINT x(i), y(i)
NEXT i
PRINT
PRINT (y(3) - y(2)) / (x(3) - x(2)), (y(4) - y(3)) / (x(4) - x(3))
PRINT
REM vary 5th point & fit ellipse
y(5) = 0: rel0# = 10 ^ 9

FOR x0# = .355# TO .355# STEP .001#

x(5) = -x0#
FOR i = 1 TO 5
cc(i, 1) = x(i) ^ 2: cc(i, 2) = x(i) * y(i): cc(i, 3) = y(i) ^ 2
cc(i, 4) = x(i): cc(i, 5) = y(i): cc(i, 6) = 1
NEXT i
REM solve for ellipse coeffs.
FOR i = 1 TO 5
FOR j = 1 TO 5
IF j = i GOTO 650
f# = cc(j, i) / cc(i, i)
FOR k = 1 TO 6
cc(j, k) = cc(j, k) - cc(i, k) * f#
NEXT k
650 NEXT j
NEXT i
FOR i = 1 TO 5
co(i) = cc(i, 6) / cc(i, i)
NEXT i
discrim# = co(2) ^ 2 - 4 * co(1) * co(3)
IF discrim# > 0 GOTO 690
REM PRINT x0#, discrim#
REM find relative standard deviation of areal speed
GOSUB 4000
690 NEXT x0#
PRINT "5th x-intercept, best relative std. dev. of areal speed"
PRINT -x00#, rel0#


990 END


REM Get Julian date - 2400000.
1000 t(1) = 34798.5# + 8 / 24 + 5 / 1440
t(2) = 46504.5# + 14 / 24 + 18 / 1440
t(3) = t(2) + 365 - 28 - 31 + 16 + 3 / 24 - 16 / 1440
t(4) = t(3) + 10 * 365 + 3 + 28 + 3 - 4 / 24 + 57 / 1440
tfrey3# = 54184.5# + 42 / 1440
tbarbarossa3# = 54191.5# + 7 / 24 + 39 / 1440
REM Interpolate t(5) using est. of mass ratio.
t(5) = .8743# * tbarbarossa3# + .1257# * tfrey3#
REM Guess that 12/22/08 epoch, t(6), is approx. time on meridian;
REM Barbarossa was near stationarity anyway.
t(6) = 54822.5# + 12 / 24 + 47 / 1440
t(7) = 54851.5# + 9 / 24 + 48.5# / 1440
RETURN

REM light time correction
1050 FOR j = 1 TO 7
t(j) = t(j) - rr# * 1.496 * 10 ^ 13 / (2.99793 * 10 ^ 10)
NEXT j
dt1# = 1 / (t(2) - t(1)): dt2# = 1 / (t(3) - t(2)): dt3# = 2 / (t(3) - t(1))
RETURN


REM get J2000 geocentric Barbarossa & Frey coordinates
REM "A" plate (1954); objects A2 & A
1100 raf(1) = pi180# * (11 * 15 + 2 / 4 + 25.16# / 240)
declf(1) = 0 - pi180# * (5 + 56 / 60 + 11.3# / 3600)
rab(1) = pi180# * (11 * 15 + 3 / 4 + 12.4# / 240)
declb(1) = 0 - pi180# * (5 + 58 / 60 + 9 / 3600)

REM "B" plate (1986); objects B3 & B
raf(2) = pi180# * (11 * 15 + 16 / 4 + 51.67# / 240)
declf(2) = 0 - pi180# * (7 + 49 / 60 + 40.4# / 3600)
rab(2) = pi180# * (11 * 15 + 16 / 4 + 55.78# / 240)
declb(2) = 0 - pi180# * (7 + 55 / 60 + 14# / 3600)

REM "C" plate (1987); objects C & C11
raf(3) = pi180# * (11 * 15 + 18 / 4 + 3.18# / 240)
declf(3) = 0 - pi180# * (7 + 58 / 60 + 46.1 / 3600)
rab(3) = pi180# * (11 * 15 + 18 / 4 + .41 / 240)
declb(3) = 0 - pi180# * (8 + 1 / 60 + 57.7 / 3600)

REM "D" plate (1997); objects uncertain

REM Genebriera March 25, 2007
raf(5) = pi180# * (11 * 15 + 26 / 4 + 22.2# / 240)
declf(5) = 0 - pi180# * (9 + 4 / 60 + 59 / 3600)

REM Riley April 1, 2007
rab(5) = pi180# * (11 * 15 + 26 / 4 + 25 / 240)
declb(5) = 0 - pi180# * (8 + 57 / 60 + 26 / 3600)

REM alternate coords. for Riley's "Barbarossa"
REM (see messageboard May 7, 2007; called "Frey" there)
REM rab(5) = pi180# * (11 * 15 + 26 / 4 + 24.6 / 240)
REM declb(5) = 0 - pi180# * (8 + 57 / 60 + 48.5 / 3600)

REM Genebriera's "Barbarossa"
REM (see messageboard May 15, 2007; called "Frey" there)
REM rab(5) = pi180# * (11 * 15 + 26 / 4 + 31.8 / 240)
REM declb(5) = 0 - pi180# * (9 + 0 / 60 + 11 / 3600)

REM Dec. 22, 2008 U. of Iowa photo
raf(6) = pi180# * (11 * 15 + 27 / 4 + 30.17# / 240)
declf(6) = 0 - pi180# * (9 + 21 / 60 + 48.6# / 3600)
REM (Barbarossa would have been off west edge)
REM objects identified for Dec. 22 in earlier theory
REM rab(6) = pi180# * (11 * 15 + 28 / 4 + 22.08# / 240)
REM declb(6) = 0 - pi180# * (9 + 16 / 60 + 6.4# / 3600)
REM raf(6) = pi180# * (11 * 15 + 29 / 4 + 4.66# / 240)
REM declf(6) = 0 - pi180# * (9 + 7 / 60 + 2.3# / 3600)
RETURN


REM Get J2000 barycentric Earth XYZ position (excludes Barbarossa).
REM In this program, "barycentric" = sun + traditional planets.
REM all but 1954, from Astronomical Almanac
1200 a# = 14 / 24 + 18 / 1440: i = 2
GOSUB 1220
a# = 17 / 24 + 2 / 1440: i = 3
GOSUB 1220
a# = 13 / 24 + 59 / 1440: i = 4
GOSUB 1220
REM as in subroutine 1000, weighted time for 2007
REM warning: need to consult Almanac to accurize
a# = 42 / 1440 * .1102# + (7 + 7 / 24 + 39 / 1440) * .8898#: i = 5
GOSUB 1220
REM as in subroutine 1000, guess that 12/2008 epoch was meridian
a# = 12 / 24 + 47 / 1440: i = 6
GOSUB 1220
a# = 9 / 24 + 48.5# / 1440: i = 7
GOSUB 1220
GOSUB 1250
RETURN

1220 b# = 1 - a#
READ xe1#, xe2#, ye1#, ye2#, ze1#, ze2#
xe(i) = b# * xe1# + a# * xe2#
ye(i) = b# * ye1# + a# * ye2#
ze(i) = b# * ze1# + a# * ze2#
re(i) = SQR(xe(i) ^ 2 + ye(i) ^ 2 + ze(i) ^ 2)
RETURN

REM find 1954 barycentric Earth position
1250 msun# = 332000: mjup# = 318.1#
msat# = 95.2#: mur# = 14.6#: mnep# = 17.2#
me# = 1: mtot# = msun# + mjup# + msat# + mur# + mnep# + me#
sx# = 0: sy# = 0: sz# = 0
REM interpolate sun coordinates
a# = 8 / 24 + 5 / 1440: b# = 1 - a#
raa# = 22 * 15 + 30 / 4 + 33.54# / 240
rab# = 22 * 15 + 34 / 4 + 20.69# / 240
ra# = (raa# * b# + rab# * a#) * pi180#
decla# = 9 + 22 / 60 + 8.300000000000001# / 3600
declb# = 8 + 59 / 60 + 52 / 3600
decl# = 0 - (decla# * b# + declb# * a#) * pi180#
ra# = .9899423000000001#: rb# = .9901818999999999#
r# = ra# * b# + rb# * a#: w# = msun#
GOSUB 1270
REM planets interpolated to nearest RA second, or Decl arcminute
REM Jupiter
ra# = pi180# * (5 * 15 + 2 / 4 + 41 / 240)
decl# = pi180# * (22 + 29 / 60 + 51 / 3600)
r# = 4.846#: w# = mjup#
GOSUB 1270
REM Saturn
ra# = pi180# * (14 * 15 + 30 / 4 + 57 / 240)
decl = 0 - pi180# * (12 + 9 / 60 + 22 / 3600)
r# = 9.308#: w# = msat#
GOSUB 1270
REM Uranus
ra# = pi180# * (7 * 15 + 24 / 4 + 19 / 240)
decl# = pi180# * (22 + 31 / 60 + 46 / 3600)
r# = 18.001#: w# = mur#
GOSUB 1270
REM Neptune
ra# = pi180# * (13 * 15 + 38 / 4 + 22 / 240)
decl# = 0 - pi180# * (8 + 22 / 60 + 9 / 3600)
r# = 29.662#: w# = mnep#
GOSUB 1270
REM convert to 1950 ecliptic coords.
theta# = 23.45# * pi180#
GOSUB 1290
REM convert to 2000 ecliptic coords.
theta# = pi180# * 50.2619# / 3600 * 50
cs# = COS(theta#): sn# = SIN(theta#): sx0# = sx#
sx# = cs# * sx# - sn# * sy#
sy# = cs# * sy# + sn# * sx0#
REM convert to 2000 celestial coords.
theta# = 0 - 23.45# * pi180#
GOSUB 1290
xe(1) = 0 - sx# / mtot#
ye(1) = 0 - sy# / mtot#
ze(1) = 0 - sz# / mtot#
re(1) = SQR(xe(1) ^ 2 + ye(1) ^ 2 + ze(1) ^ 2)
RETURN

1270 z# = r# * SIN(decl#): cs# = COS(decl#)
x# = r# * cs# * COS(ra#): y# = r# * cs# * SIN(ra#)
sx# = sx# + w# * x#
sy# = sy# + w# * y#
sz# = sz# + w# * z#
RETURN

1290 cs# = COS(theta#): sn# = SIN(theta#): sz0# = sz#
sz# = cs# * sz# - sn# * sy#
sy# = cs# * sy# + sn# * sz0#
RETURN


REM find the accel. of the parabolas fitting 1st thru 3rd points
2000 vx1# = (x(2) - x(1)) * dt1#
vy1# = (y(2) - y(1)) * dt1#
vz1# = (z(2) - z(1)) * dt1#
vx2# = (x(3) - x(2)) * dt2#
vy2# = (y(3) - y(2)) * dt2#
vz2# = (z(3) - z(2)) * dt2#
REM accel., corrected to fixed frame
ax# = (vx2# - vx1#) * dt3# * corr#
ay# = (vy2# - vy1#) * dt3# * corr#
az# = (vz2# - vz1#) * dt3# * corr#
RETURN

REM find Newtonian accel. at 1st thru 3rd points
2100 FOR j = 1 TO 3
dr# = kk# / r(j) ^ 3
nax(j) = x(j) * dr#
nay(j) = y(j) * dr#
naz(j) = z(j) * dr#
NEXT j
RETURN

REM weighted ave. Newtonian accel., and work done
2200 avenax# = 0: avenay# = 0: avenaz# = 0
FOR j = 1 TO 3
avenax# = avenax# + w(j) * nax(j)
avenay# = avenay# + w(j) * nay(j)
avenaz# = avenaz# + w(j) * naz(j)
NEXT j
rel# = (avenax# - ax#) ^ 2 + (avenay# - ay#) ^ 2 + (avenaz# - az#) ^ 2
RETURN

REM update best
3000 rel0# = rel#: i0 = i: q0# = q#
r00# = r0#: rp0# = rp#: rt0# = rt#: rpp0# = rpp#: rtt0# = rtt#
FOR m = 1 TO 6
x0(m) = x(m): y0(m) = y(m): z0(m) = z(m): r0(m) = r(m)
NEXT m
RETURN

3100 PRINT "apparent circular period"
PRINT pi2# / om# / 365.25#
RETURN

REM find areas of the four sectors
REM PRINT "radii"
4000 FOR i = 1 TO 4
the(i) = ATN(-y(i) / x(i))
IF x(i) > 0 THEN GOSUB 4100
IF the(i) < 0 THEN GOSUB 4120
r(i) = SQR(x(i) ^ 2 + y(i) ^ 2)
REM PRINT r(i),
NEXT i
REM PRINT
r(5) = r(1): the(5) = the(1) + pi2#
REM PRINT "drdth"
FOR i = 1 TO 4
u# = 2 * co(1) * x(i) + co(2) * y(i) + co(4)
w# = co(2) * x(i) + 2 * co(3) * y(i) + co(5)
sn# = 0 - (u# * y(i) - w# * x(i)) / SQR(u# ^ 2 + w# ^ 2)
drdth(i) = sn# / SQR(1 - sn# ^ 2)
REM PRINT drdth(i),
NEXT i
REM PRINT
drdth(5) = drdth(1)
PRINT "interpolated radii"
REM interpolate r quadratically
FOR i = 1 TO 5
IF i < 5 THEN dthpl# = the(i + 1) - the(i)
IF i < 5 THEN drpl# = r(i + 1) - r(i)
IF i > 1 THEN dthmi# = the(i) - the(i - 1)
IF i > 1 THEN drmi# = r(i) - r(i - 1)
IF i < 5 THEN rpl(i) = r(i) + 2 / 9 * drdth(i) * dthpl# + 1 / 9 * drpl#
IF i > 1 THEN rmi(i) = r(i) - 2 / 9 * drdth(i) * dthmi# - 1 / 9 * drmi#
PRINT rpl(i), rmi(i)
NEXT i
REM Cotes' 3/8 rule
REM PRINT "areas"
FOR i = 1 TO 4
r2m# = r(i) ^ 2 + r(i + 1) ^ 2 + 3 * (rpl(i) ^ 2 + rmi(i + 1) ^ 2)
area(i) = r2m# / 8 * (the(i + 1) - the(i)) / 2
REM PRINT area(i),
NEXT i
REM PRINT
are# = area(1) + area(2) + area(3) + area(4)
s# = 0: ss# = 0
PRINT "rates"
FOR i = 1 TO 3
IF i = 1 THEN lap# = 1
IF i = 2 THEN lap# = 0
IF i = 3 THEN lap# = 1
rate# = (area(i) + lap# * are#) / (t(i + 1) - t(i))
PRINT rate#; " ";
s# = s# + rate#: ss# = ss# + rate# ^ 2
NEXT i
rel# = SQR((ss# - s# ^ 2 / 3) / 2) / (s# / 3)
IF rel# < rel0# THEN GOSUB 4200
RETURN

4100 the(i) = the(i) + pi#
RETURN
4120 the(i) = the(i) + pi2#
RETURN

4200 rel0# = rel#: x00# = x0#
RETURN


REM Astronomical Almanac barycentric Earth XYZ '86
DATA -.992473#,-.99431#,.097062#,.081274#,.042041#,.035195#
REM '87
DATA -.647098#,-.660316#,.689355#,.678874#,.298895#,.29435#
REM '97
DATA -.951971#,-.957296#,.278791#,.263632#,.121036#,.114464#
REM '07
DATA -.992917#,-.991887#,-.057633#,-.073429#,-.025095#,-.031943#
REM '08
DATA -.008637#,-.026129#,.906672#,.906369#,.39304#,.392909#
REM '09
DATA -.493191#,-.508316#,.786488#,.778431#,.340941#,.337449#


REM (times of plates, Sun positions,
REM coords. of identified disappearing dots)
REM plate A (Red)
REM DATA 1954.154,.9900229, 22,34, 27.6,-8,-59,-7
REM object A2
REM DATA 11,2,25.47,-5,-56,-12
REM object A
REM DATA 11,3,12.4,-5,-58,-9

REM plate B (Red)
REM DATA 1986.201,.9945905,23,40,30.9,-2,-10,-49
REM object B3
REM DATA 11,16,51.67,-7,-49,-40.4
REM object B
REM DATA 11,16,55.78,-7,-55,-14

REM plate C (Red)
REM DATA 1987.08215,.9852006,20,53,6.9,-17,31,13
REM object C11 (named for easy recall, like "CII")
REM DATA 11,18,.41,-8,-1,-57.7
REM object C
REM DATA 11,18,3.18,-7,-58,-46.1

REM plate D (optical infrared)
REM DATA 1997.16711,.9914845,22,57,21.1,-6,-39,-54
REM DATA 11,22,32.9,-8,-26,-56
REM DATA 11,22,16.77,-8,-29,-32.6
REM DATA 11,22,1.33,-8,-34,-36.9
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Joe Keller

USA
958 Posts

Posted - 30 Jan 2009 :  13:06:25  Show Profile  Reply with Quote
Hi *********, and Prof. *********!

Here's a guide star (too dim to be a real guide star)(at present, Jan. 30, Frey is about 6 arcminutes west of it, mainly due to change in Earth parallax since Dec. 22, and Barbarossa about 3.5 arcmin west of Frey; so, it would be best to ask those astronomers in Boston to center the photo (2*6+3.5)/2 = 7.75 arcminutes west of this star; i.e., 31 "seconds of Right Ascension" west of it) and 5.5*0.4 arcminutes north of it:

USNO-B 0806-0230017
USNO-B Red1 mag +19.61, USNO-B Red2 mag +19.21
Star's coordinates:
RA 11:27:32.00
Decl -9:21:48.3

Center of desired photo:
RA 11:27:01
decl -9:20

(Subtract about one second of RA, and add about 6" of Decl., that is, aim farther west and north, for each day past Jan. 30.)

I happen to know about this star, because its pixel appearance is very similar to a nearby "disappearing dot" (not in 1987 Red sky survey) not too near the west edge of the Dec. 22, 2008 U. of Iowa photo, at

RA 11:27:30.17
Decl -9:21:48.6

This object not only is very similar in appearance to the above Red mag +19.4 star, it also accords perfectly with my revised, corrected and accurized theory of the solar and binary Barbarossa/Frey orbits. This theory is completely contained within the BASIC program I posted to Dr. Van Flandern's messageboard about an hour ago.

There is now so little doubt in my mind of the reality of these objects, that I'm hardly in a rush to get photos. "They'll keep."

Your *******,
Joe
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Maurol

Argentina
37 Posts

Posted - 30 Jan 2009 :  13:49:01  Show Profile  Reply with Quote
quote:
Originally posted by Maurol

quote:
Originally posted by nemesis

I'll rephrase my precession question a little differently. Currently, at the winter solstice in late December the Earth is near perigee. In roughly 13,000 years, halfway through the precession cycle, at the winter solstice, still late December by the tropical year, will the Earth be near apogee?



Hi nemesis,
Short answer: No.
What encompasses the points of perihelion and aphelion is the length of the sidereal year. The tropical year is 20 minutes shorter, to align the calendar with the seasons, that is, to maintain the equinocces(marks of the seasons) in relatively the same days, due to the earth axis precessing, and then changing the occurrence of the equinocces. The tropical year is a convenience.
In 13.000 years, as the difference is cumulative, you'll have a (significant) difference of 13000 * 20 = 26000 minutes, around 18 days, between a calendar based in the sidereal year, and one based in the tropical year.
So in late December, the Earth will still be near perihelion, although in a calendar based in the tropical year perihelion will really occur around December 13.



This is incorrect! As the seasons depend on the amount and angle of incidence of light on each hemisphere, and as that depends on the direction of the inclination of the Earth axis of rotation, in 13000 years, having the direction of the axis preceded 180 degrees, winter will occur near aphelion, as nemesis says.
That moment will be called December 21, if our calendar is still in use, but the Earth will be then in the opposite side of its orbit.

I was confused by the proclivity to associate closeness to perihelion and occurrence of the summer/winter seasons, common to our time.

All this holds if, as current astronomy says, perihelion is stable and precedes only a small number of arc secs/century.
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Joe Keller

USA
958 Posts

Posted - 30 Jan 2009 :  15:38:12  Show Profile  Reply with Quote
The above program uses objects A & A2 from 1954, B & B3 from 1986, and C11 & C from 1987. All three are online scans of Red sky survey plates. The coordinates and dates are to be found in the program. The former object of each pair is the one I now identify as Barbarossa, the more massive member of the binary; and the latter is the one I now identify as Frey.

The designation "C11" was chosen for ease of remembering (like "CII"), not because I found so many disappearing dots. The object "C" is the very first "disappearing dot" I ever announced; I thought it was Barbarossa but now realize it is Frey.

USNO-B Blue magnitudes of stars comparable to these, if given at all, usually are near the faintness limit of accuracy, +21. On the other hand, there are many comparison stars with reproducible Red magnitudes (i.e., R1 = R2). Looking almost perpendicularly out of the galaxy, most stars this faint, are red dwarfs (also, many objects this faint are galaxies). According to the Red magnitudes of nearby similar stars, my estimates of these objects' magnitudes range from about +17.5 to +19.5.

The USNO-B Red magnitudes of comparable stars are consistently about 1.5 mag brighter than the photometric magnitudes on the Jan. 20, 2009, U. of Iowa photo. That is, a typical comparable star has USNO-B Red magnitude +18.5 but is said to be +20.0 Visual magnitude on the U. of Iowa photo. This is reasonable for red dwarfs, which most of them must be, at this high galactic latitude.

Objects A2 and B3, both Frey, have one or more dimmer companions (other disappearing dots) within about an arcminute or less. These might be sub-moons of Frey.

One other object I use in my program, is located at RA 11:27:30.17, Decl -9:21:48.6 on the Dec. 22, 2008 U. of Iowa CCD photo, taken with their 15 inch robotic telescope in southern Arizona and median stacked by Professor Mutel. This object has a starlike appearance; it is the most starlike of any candidate image on any CCD photo I have examined. It is similar to the nearby USNO-B 0806-0230017, whose Red magnitudes are +19.4 +/- 0.2. The object is near the west edge of the photo but well outside the obviously affected edge region. It is Frey; the theoretical position of Barbarossa is outside the photo, because I changed my orbital theory.

Other "Barbarossas" and "Freys" identified on this and other prospective CCD photos, are not used in the program, because although such "disappearing dots" are important discoveries, they do not fit my revised, corrected and accurized theory of the orbits. Even with the new predicted coordinates, the other photos are, more or less, of the correct patch of sky, but I have not had time to check them. Someone else might check them using the above program, which gives the center of mass prediction, given the Julian date. The positions of Barbarossa and Frey then can be found from the center of mass. For Frey, subtract the predicted Dec. 22, 2008, c.o.m. coordinates from Frey's actual Dec. 22 position. The binary period is about 22 yr, so this difference still can be used for Frey now. The mass ratio is Barbarossa::Frey = 0.8898::0.1102, so to find Barbarossa, multiply Frey's lever arm by -0.1102/0.8898. The Jan. 20, 2009, U. of Iowa photo, made according to my earlier theory, is too far east to contain either Barbarossa or Frey.

The program has two parts. The first part uses the three sky survey dot pairs, to get the c.o.m. solar orbit. The mass ratio, orbital radius, and first and second derivatives of the orbital radius, are varied to find the trajectory which minimizes the difference between Newtonian gravitation and the theoretical orbit's acceleration. This difference can be made very small for these three dot pairs, but not for any other sets of three pairs tried. The result is the mass ratio Barbarossa::Frey = 0.8898::0.1102, initial 1954 radius 205.2 AU, initial 1954 dr/dt = -1.0 AU per radian of orbit, and d2r/dt2 = -23.3 AU per radian per radian. So, Barbarossa is near aphelion, and its solar orbit has eccentricity e = 0.114. When parameters are varied enough to change predictions by an arcsecond, the unexplained acceleration becomes much larger, so, the prediction might have arcsecond accuracy. However, the path is so short, that the eccentricity cannot be known very accurately.

The program's second part analyzes the Barbarossa/Frey binary orbit. A fourth binary orbital point is achieved by predicting Barbarossa's position, from the Dec. 22, 2008 U. of Iowa Frey observation, and the program's predicted c.o.m. position. Yesterday I graphed all these, noticing that the 1986, 1987, and 2008 Frey positions, relative to Barbarossa, are almost collinear. There is only a narrow range of possible fifth points such that an ellipse can be fitted.

Frey's 2008 position (relative to Barbarossa) is only a little past its 1986 and 1987 positions, on an almost straight line, so by conservation of angular momentum, the speed on this segment must be almost constant, and the time of travel on this segment can be estimated. I also drew a straight line through Frey's 1954 position, and the origin, to get an estimate of the position 1/2 orbit past 1954; this happens to be between the 1987 & 2008 positions. Assuming between one and two orbits during 1954-1987, and again during 1987-2008, and a mass Barbarossa+Frey = 0.01 solar mass (estimated from my theory of outer solar system precession resonances, and in other ways) these two constructions, with straight edge and ruler on graph paper, imply semimajor axes of 1.61 and 1.73 AU, resp. The former figure should be more accurate for a noncircular orbit, because it employs a whole orbit, not an orbit and a half.

If the binary orbital eccentricity is small, then the foci lie near the center, and the tangents to the apparent binary orbit ellipse, near 1987 and near 1954, are parallel because these points lie near the opposite ends of the "diameter" drawn through, the focus, which approximates the center. If these two parallel tangent lines are near eclipse (suggested by the small area of the apparent ellipse) then points on them appear to move perpendicular to the lines, but also to stream along the lines.

The three collinear Freys, slope 54.25deg NW to SE. Early in my research I had noted that Barbarossa's solar orbit slopes 27 deg on average during this time.
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Joe Keller

USA
958 Posts

Posted - 01 Feb 2009 :  12:54:34  Show Profile  Reply with Quote
(email response sent 5 min. ago)

Hi Prof *********!

The "royal road" to Planet X / Barbarossa, is Paul Wesson's article in Physical Review D, in 1981, on "tax day", April 15. Wesson noticed that (angular momentum)/(mass^2) is roughly constant in astronomical systems of all sizes, and thought this phenomenon might reflect an underlying universal physical constant. I would augment Wesson's article three ways:

1. Some systems should be disregarded because it's fairly certain that their angular momentum value is due to special effects. These systems include Mercury, Venus, and Luna rotation (tidal drag, spin-orbit locking), double planets (Earth-Luna, Pluto-Charon)(capture?) and binary stars (capture?)(undetected distant member?). The most pristine systems are the rotations of the four giant planets. These all have J/M^2 = Wesson "p" factor within a factor of two of each other.

2. I didn't see this mentioned in Wesson's article; maybe I missed it (I looked up the article in the library downstairs the day that I visited the U. of Iowa Astronomy Dept.) but the "p" factor for the giant planets, differs from the "p" factor for a spinning electron, by a factor equal, to within an order of magnitude, to the "electric-to-gravity ratio". Wesson's "p", seems to be the gravitational counterpart of electron spin.

3. It's a good guess that the proto-sun, and hence the solar system, has the same "p" factor as the giant planets. One might ask, does it have enough angular momentum? Is another planet needed to bring the "p" factor up to the norm? How big a planet, how far out? It turns out that the solar system's angular momentum is at least 20x too small, by comparison with the giant planets. Addition of Barbarossa at my estimated mass and distance, makes the angular momentum about right, no more than 2x too large.

This is the "royal road". It's something that can be understood and checked quickly. It doesn't prove that my "dots" are the planet, but it's strong circumstantial evidence for such a planet, somewhere.

>...Peter Goldreich is probably the most respected solar system dynamical theorist in the world. He is not known for making sloppy approximations. ...

I never called him sloppy or said he wasn't respected. Goldreich himself expressed at least some doubt, in his article, of the validity of at least one of his approximations. Making a nonrigorous calculation doesn't damn him; a rigorous calculation was impossible, as far as anyone knows. This is the infamous "many body problem" with at least 1+4+1=6 massive bodies. Someone said that even the wanderings of the Jupiter-Saturn resonance were analytically intractable, and that's only a three-body problem. Also, respected solar system dynamical theorists can disagree. Harrington said on his deathbed that Standish's correction to Neptune's mass, didn't remove the residuals. Eckert found a residual so big that Standish's correction couldn't possibly remove it.

There's a tendency to escape uncomfortable ambiguity by a "rush to judgment". To say, "Whew, Goldreich says I can stop thinking about this, what a relief!" when really "stop thinking about it" is only the "executive summary" of what Goldreich really said, and other respected dynamical theorists, Eckert and Harrington, have made claims which seem to contradict Goldreich and Standish.

The Kimura phenomenon provides a paradigm. Basically, Kimura noticed that accurate determinations of declinations of bright stars near the north ecliptic pole, made from northern observatories, showed systematic seasonal change not explained by any known correction. Soon afterwards Chandler endorsed, not exactly Kimura's conclusion, but at least Kimura's reputation. One would think that the Kimura phenomenon, with at least Chandler's limited endorsement, would be taken seriously, and apparently it was for about a generation.

Then in Astronomische Nachrichten (I accidentally discovered this article at Drake University) someone said, in a short letter, basically, "Kimura didn't know how to average data correctly; it's all just an artifact of that." I, for one, could not decipher, let alone verify, the objection to Kimura's method of averaging data. Furthermore I could not imagine how any erroneous method of averaging data, could cause a seasonal change. I suppose everyone was tired of looking at the unsightly clutter of Kimura's phenomenon, and unconsciously wanted to throw it in the trash so we could all feel better. I think it's good to have a sense of humor about this, but I'm not belittling anyone in particular; these are human failings we all share.

A careful, peer-reviewed theoretical estimate of the albedo of one particular class of cool brown dwarf (indirectly supported by some observations) is, < 1%. I've posted all this material to Dr. Van Flandern's messageboard during the last two years.

Sincerely,
Joe Keller
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Joe Keller

USA
958 Posts

Posted - 01 Feb 2009 :  13:06:20  Show Profile  Reply with Quote
(previous email, sent last night)

Hi Prof. *********!

I've become something of a specialist in the literature on this!

>...the perturbing effect of such an object on the orbits of the outer planets...

I read Goldreich's article about that, in the Pacific astronomy journal. I counted that Goldreich makes four explicit mathematical simplifying assumptions in his lengthy calculation. It was a nonrigorous calculation. There's a lot that can go wrong with that kind of calculation.

For the distance and mass I propose, the orbital precession rates of Neptune, the plutinos, and the classical Edgeworth-Kuiper belt objects, stand in 3::2::1 resonance. Furthermore the torque at the classical Kuiper belt due to Barbarossa (the outward part of the solar system), equals the torque there due to Jupiter, Saturn, Uranus & Neptune (the inward part of the solar system). That's solving (3-1)+1 = 3 rather fundamental equations with, really, one adjustable parameter (Barbarossa's mass / distance^3). Why should that be, unless there is something happening, dynamically, that Goldreich didn't know?

>...the size and therefore the apparent magnitude of such an object...

Nobody knows what size it would be, or what temperature it would be. I saw a quantum mechanical calculation from Physical Review in the 1990s, concluding that every mass between Jupiter and a hot brown dwarf, gives about that same 120,000 km diameter, and that even a 5 billion year old 0.01 solar mass hyperjovian that never burned deuterium, should be room temperature or so. But that's just a quantum mechanical calculation, and a pretty big extrapolation from the lab! What if there is an unforeseen mechanism of cooling by convection? What if formation is by accretion so that all the gravitational energy is released at the surface where it can radiate away quickly? What if, somehow, it's pulsar-like (nuclear density) material?

There's an article saying that at least some brown dwarfs ("black smokers") have albedo 1% or even less - something about alkali metals at the surface. We know little about very cool brown dwarfs, distant from any star, because there's no way to observe them directly. A recent article estimates that, accounting for observation bias, the cooler known brown dwarf spectral type outnumbers the hotter type, almost 10:1. What if there's an even cooler type that outnumbers it 100:1? Is that our dark matter?

I don't think that surprisingly high density or surprisingly low albedo have been ruled out. There's a tendency to be overconfident in current theory, therefore not to look at contradictory data, and to miss discoveries. Why couldn't dark matter come in big blobs?

Sincerely,
Joe Keller
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Joe Keller

USA
958 Posts

Posted - 03 Feb 2009 :  14:41:15  Show Profile  Reply with Quote
The four Freys I've identified above (the Dec. 22, 2008, U. of Iowa photo's Frey, referred to the center of mass extrapolated using my program; and the 1954, 1986 & 1987 Red sky survey Freys referred to the centers of mass calculated using their and Barbarossa's positions) do not, in the most straightforward way, fit an orbital ellipse. I investigated this using my program above, and a supplemental program based on formulas appearing in, inter alia, Love's Analytic Geometry, especially ch. VIII, sec. 69.

The satisfaction of Kepler's second law by the adjustment in the program as posted above, is suggestive, but the adjustment made differs from textbook reality in three ways:

1. The sector area is calculated only to quadratic accuracy.
2. Each side of the orbit is given the full displacement, instead of dividing the displacement between the sides according to their foreground or background distance.
3. The component of displacement parallel to the binary orbit is ignored.

The change in aspect of the Barbarossa-Frey orbit, from 1954 to 2008, due to Barbarossa's orbital motion, amounts to a precession of 0.112 radians about an axis at 27deg counterclockwise of North (Barbarossa's orbit slopes about 27deg south and east). To adjust for this, the 1954 and 2008 Freys must be moved parallel to Barbarossa's orbit. The distances to be moved, are unknown because they depend on the size and shape of the binary orbit. However, the 2008 point must be moved westward, because it is almost collinear with 1986 & 1987, so with eastward motion, the orbit quickly becomes hyperbolic and then fails even to be convex (no solution).

I tried many plausible pairs of distances by which to adjust these points. For each pair of distances, I used my program to find the "fifth point" of the ellipse (that is, the western x-intercept) satisfying Kepler's 2nd law best. Because there are 4-1=3 sectors, there are two Kepler's 2nd law equations to satisfy. Generally only one member of the pair of adjustment distances, was independent, and often there was no solution. (I corrected the accuracy of the sector areas to a cubic approximation, which should suffice.)

I found only orbits that are too small or too eccentric (e.g., e = 0.99). None of them implied the parallax displacements assumed.

[Digression about believable eccentricities:

[The six binary pulsars in Taylor's 1995 catalog (VizieR) with e>0.3, have 0.61<e<0.87, median 0.74. (The binary orbit part of the catalog has 45 entries; eccentricities are given for 30.) So, Frey's eccentricity is typical of the eccentric 20% of binary pulsars. Worley's 1983 visual binary catalog (N=933) contains 79 systems with 0.3<e<0.35; this drops off smoothly to 49 with 0.85<e<0.9, then drops faster to 31 with 0.9<e<0.95 and 11 with 0.95<e<0.99999. Neptune's moon Nereid has eccentricity almost 0.8.

[Digression about method:

[Following the basic methods in Love's Analytic Geometry, and using a short supplemental computer program, I found the real orbit from the apparent one, by searching (with an IBM 486 desktop computer) all axes and angles by which to rotate the apparent orbit: first I transformed the axes by theta, to the trial axis of rotation, and in the same stroke transformed one axis to expand the orbit perpendicular to the trial axis, by sec(alpha). I then put the transformed orbital equation into standard form (using another axis rotation to eliminate the cross term) and saw whether or not the center of mass (origin) equalled a focus. (For the correct real orbit, this will be so. The process amounts to solving two equations - that (0,0) is a focus - by varying two parameters - the direction of the unit normal to the real orbit.)
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Joe Keller

USA
958 Posts

Posted - 03 Feb 2009 :  15:42:00  Show Profile  Reply with Quote
Hi Prof. *********!

>...tidal effect of a 0.01 solar mass body at 200 AU on the Sun...

Goldreich and I refer to the tide on the outer solar system (e.g., Planet X acceleration of Neptune, minus Planet X acceleration of Sun) which Goldreich and I agree would not be negligible in its effect on solar system structure. This effect is difficult if not impossible to calculate rigorously in a many-body system. It might express itself most obviously as a small-integer resonance between precession periods.

>The Sun's spin angular momentum...

Paul Wesson and I refer to the solar system's angular momentum (i.e., proto-sun's angular momentum), not the small part retained by the sun today.

>...darker than every other known solar system object...

Stranger things have happened. Also, we don't know its albedo because we really don't know its size. Anyway, the true object might be brighter but undetected (the binary orbit could fool detection algorithms).

>...extraordinary claims require extraordinary evidence.

Extraordinary evidence will not be found without looking (sometimes looking quite a bit). To motivate the photographic effort, I make the case for finding something.

>I looked at your 'detection' in the Dec 22 image - it was so faint that I could not determine a point-spread function...

The third, latest and best detection (11:27:30.17, -9:21:48.6 on the Dec. 22, 2008 U. of Iowa photo) closely resembles a nearby cataloged USNO-B object, with R1=19.6, R2=19.2. It conforms to my revised, corrected and accurized orbital theory (see computer program posted to Dr. Van Flandern's messageboard).

Sincerely,
Joe Keller
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Joe Keller

USA
958 Posts

Posted - 05 Feb 2009 :  20:42:19  Show Profile  Reply with Quote
Frey is a Low-Mass, Black, Iron Dwarf

The term, "Iron Dwarf", usually refers to a white dwarf containing mainly elements near iron. The term "Black Dwarf" refers to a white dwarf which has lost enough heat that its surface has become cold. The partial collapse (partial electron degeneracy) that would be seen in an iron object of Frey's mass, gives Frey a size consistent with Frey's observed Visual and Red magnitude, assuming the usual albedo for outer solar system objects, and a surface resembling a red class Kuiper belt object.

Only one certain photograph of Frey (and none of Barbarossa) were taken with clear, not red, filter. This is the Dec. 22, 2008, U. of Iowa photo. Let's assume that this CCD response approximates the Visual band.

Thirty arcseconds from Frey in the U. of Iowa photo, is a USNO-B catalog star with Red magnitude (average of R1 & R2) +19.41 +/- 0.20, Blue magnitude +20.55, and Optical Infrared mag +18.45. According to Straizys, "Multicolor Stellar Photometry", Table 27, R-I for this star corresponds to spectral type K2V, and B-R corresponds to type G7V. The catalog documentation warns that Blue magnitudes > +21 are invalid, so I'll estimate this star closer to the former type, and say K0V; Straizys gives V-R=+0.52 for this type, so the Visual magnitude of this comparison star is +19.93 +/- 0.20. Frey seems slightly brighter, so I'll say Frey has Visual magnitude +19.8.

This is consistent with the Red magnitudes, +17.5 to +19.5, I've estimated for Barbarossa and Frey from Red sky surveys. Wickramasinghe & Hoyle, Ap & Space Sci 268:369, 1999, say that for typical "red class" Kuiper Belt objects, reflectance at 6500A is 2.50 times (1.0 magnitude) that at 4500A. Those wavelengths are near the "central wavelengths" of the Blue & Red bands, 4350A & 6800A, resp. (www.elvis.rowan.edu). So, with a small correction for the larger actual difference (6800-4350)/(6500-4500)=1.2, red class Kuiper belt objects have B-R=1.2+0.97=2.2 (the second term arises because Vega, which reaches the zenith at midnorthern latitudes, was chosen as a convenient color photometric standard; so, G2V stars like the sun, have B-R=0.97 in the Cousins VRI system according to Straizys' Table 27). For G2V stars (see Straizys) (V-R)/(B-R)=0.35/0.97=0.36, which times 2.2 is 0.79. So if Frey's surface resembles a typical red class Kuiper object, it has Red mag +19.8 - 0.8 = +19.0, consistent enough with my estimates on Red sky surveys.

Iron-nickel meteorites are common; some contain mainly nickel (specific gravity under ordinary conditions, 8.90). If Frey were nickel, its specific gravity certainly would be higher than this, because the pressure in Frey's interior, times the area of an atom, is comparable to atomic electrical force (the atom would be compressed as though Z, the atomic number, were increased but the number of electrons held the same). Suppose Frey has exactly the mass of Jupiter but Frey's density is 11, i.e., 8x Jupiter's. Suppose also that Frey's albedo is 0.04 (common, some say typical, for outer solar system objects, which usually are fairly black; Luna's average albedo is 0.07, facing us). Then Frey's Visual magnitude would be * +18.3 *. This still is slightly too bright, because the Red magnitude would be +17.5 (if Frey is as red as red class Kuiper objects) which is brighter than any of the comparison Red magnitudes I've found. The Visual magnitude would be 1.5 mag brighter than the one Visual magnitude I have (above). This is assuming 100% nickel "ordinary matter" composition with some compression but no collapse.

Following Stein & Cameron, "Stellar Evolution", 1966, pp. 72-73, I find that if composed of iron, with mass equal to Jupiter, Frey is already semi-collapsed, almost like a white dwarf. Here is the approximate calculation:

It's about right, to say that a pure hydrogen body of the mass of Jupiter, is on the verge of collapse. That is, the gravitational potential energy difference for a hydrogen atom, between Jupiter's surface and center, about equals the ionization energy of the hydrogen atom. Let's consider an iron atom near the center of this body. The gravitational potential for the iron atom is 56x greater, enough to ionize it to about Fe IX (see Lange's Handbook of Chemistry, 13th ed., or recent edition CRC handbooks, for atomic ionization potentials). For a conservative estimate, suppose it is Fe VI. In Fe VI, the 4s shell electrons detect 7.5 net positive charge units, vs. 1.5 for the 4s shell of calcium, so should be 5x smaller, i.e., 1.9/5=0.38A (this is tolerably consistent with the Fe II & III ionic crystal radii on www.webelements.com ) instead of 1.5A. So, Fe VI is about (?) 1/4 the diameter of neutral Fe. The free, degenerate electrons also fill space, but their size is estimated from the nonrelativistic version of Stein & Cameron eqn. 5.8, to be roughly 5x the Compton radius, negligible vs. the Fe ions. If Frey is all iron, it shrinks even more, because it starts smaller with bigger gravitational potential, ionizing the iron more.

Frey shrinks by a factor of better than 4, so its brightness decreases by more than 3 magnitudes, to dimmer than +18.3 + 3 = +21.3. This is consistent with the Visual +19.8 mag observed, if the albedo is higher than 0.04 * 4 = 0.16.

Barbarossa, if iron, would be so collapsed, that rings, a nearby moon, high albedo or faint self-luminosity, would be needed to explain its brightness. On the other hand, Barbarossa might be composed of lighter atoms than Frey. Even a hydrogen composition for Barbarossa is possible, if present estimates of hyperjovian diameter turn out to be inaccurate. Such inaccuracy might arise either from assuming too much retained heat, or from approximations made in solving the integral equations involved.
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Joe Keller

USA
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Posted - 06 Feb 2009 :  22:44:52  Show Profile  Reply with Quote
(Feb. 6, 2009)

Please Try This Yourself at Home


My (first) Feb. 3 post, explains how I could not find a real binary orbit satisfying Kepler's second law. Today I realized that luckily, the binary orbit can be solved graphically to good accuracy, if Kepler's second law is set aside for the short 1986-1987 interval (the rationale for this will be discussed in a later post).

You can do this yourself at home with graph paper and a ruler. A short table of trig functions is optional. I think this is the best opportunity there ever will be, to see for yourself that these objects are real, before many photos become available.

First, let's review Barbarossa's solar orbit. In 2007, I discovered that three pairs of "disappearing dots", on 1954, 1986 and 1987 Red sky surveys (online scans) have a center of mass which makes a straight, constant-speed arc across the celestial sphere, if the mass ratio is chosen right. One adjustable parameter, the mass ratio, solves two equations: for straightness and for constant speed. The mass ratio (most accurate, from my recent program) is Barbarossa::Frey = 0.89::0.11. The orbit (corrected for Earth parallax) cuts Declination lines at about 27 degrees southward to the east.

With the program above, I got accurate enough fitting of these three center of mass points, to an orbit accelerated only by the Sun, to make arcsecond-accuracy predictions. It turned out that a starlike "disappearing dot" of comparison Visual magnitude +19.8, on the Dec. 22, 2008 (call it 2009, since the others all were late winter or early spring) U. of Iowa photo, is in a position consistent with Frey, but the corresponding Barbarossa, assuming the predicted center of mass, is off the west edge of the photo.

Now, there are four observations of the Frey-Barbarossa binary orbit. I converted radians on the celestial sphere, to A.U., by multiplying by Barbarossa's distance. Frey's coordinates with the center of mass (predicted, for 2008; actual, for the others) at the origin, are:

1954 -0.701, +0.117
1986 -0.061, +0.332
1987 +0.041, +0.190
2008 +0.223, -0.063

The first number is how many A.U. Frey is east of the c.o.m., and the second number is how many A.U. north.

When you plot these, you'll notice that three of these "Frey" dots are almost perfectly collinear (draw a line through them; it will be sloped about 54 degrees south and to the east). This a clue that the dots are not random accidents. The orbit is roughly circular but very slanted, almost edge-on to our line of sight. The 1986, 1987 and 2008 Freys are all at about the same distance from us, on the nearer side.

The 1954 Frey is on the farther side of the orbit (thus the orbit is retrograde). As Barbarossa revolves around the Sun, that orbital position seems to move, by parallax, westward parallel to Barbarossa's solar orbit. If 1986-1987 is the reference time, then the precession correction for the 2008 Frey is small, mainly because it isn't much nearer us than the origin is. On the other hand, the precession correction for the 1954 Frey is almost (assuming a 1.4 AU Frey/c.o.m. orbit, which will be justified later) 2*1.4*(1986.5-1954)/(2009-1954)*0.112 radians of orbital travel = 0.185 AU, of which, multiplying by sin(54-27), 0.09 AU is perpendicular to the 1986-1987-2008 Frey line. On your graph, make a new "corrected 1954" Frey, 0.185 AU from the original Frey, moving along a ray sloped 27deg north and west (27deg = arctan 0.5). The corrected Frey is to the west, because that is where Frey would have been if the orbit had been in its apparent position of ~1986.5. The rearward limb of the orbit moves west due to parallax.

Draw two more lines parallel to the line through the "three Freys". One of these lines, draw through the origin, and the other, draw through the "corrected 1954 Frey". Note that the line through the corrected 1954 Frey, is four times as far from the line through the origin, as is the line through the "three amigos" Freys. This is because the 1954 Frey is near apogee, and the "three amigos" Freys near perigee. In this arrangement, perspective doesn't change the apogee::perigee ratio, which is 4::1, corresponding to eccentricity 0.6 ( (1+0.6)/(1-0.6) = 4 ).

How long is the semimajor axis, "a"? Call the angle between our line of sight, and the normal to the binary orbital plane, "i". From the graph (draw a line perpendicular to the "three amigos" line, and through the "corrected 1954 Frey") 2 * a * cos(i) = 0.74 AU. From 1954 to 1987, Barbarossa moved 1/15 radian in its solar orbit. To get 0.185 AU parallax, the 1954 Barbarossa must have been 0.185 * 15 AU closer to us, than the others. So, the semimajor axis must satisfy

2 * a * sin(i) = 2.775 AU

So, tan(i) = 2.775/0.74 and i = 75deg, and a = 2.775/2/sin(75) = 1.44 AU.

Now let's find the orbital period and investigate Kepler's second law. Draw a line from the "corrected 1954 Frey" through the origin. This line happens to go through the 2008 Frey. This shows that from 1954 to about 1987 (33 yr) is a period and a half, and from about 1987 to 2008 (really more like 2009, 22 yr), another period, for two and a half total. This satisfies Kepler's second law for the long time intervals, and shows that the period is about 22 yr.

Another way to find the period, is to recall that the three amigos are collinear. By conservation of angular momentum, their speed along that line must be constant. The time from the 1987 point (really, one orbit ahead of 1987) to the 12/22/2008 point, is the distance, divided by the speed from 3/15/1986 to 1/31/1987. Subtract this extra time from (12/22/2008 minus 1/31/1987); I found 20.4 yr. However, if for some reason Kepler's second law (conservation of angular momentum) fails for the 1986-1987 interval, thus the speed is not constant (for example, Frey might have a moon of its own), this would not be so accurate. In fact, the true Frey of 1987 probably is only 40% as far from the 2008 Frey position, as shown on the graph, and moving 3x as fast; so, my best estimate of the true period is 21.7 yr.

Knowing a = 1.44 and e = 0.6, I find (see chapter on ellipses, in any analytic geometry book) b = 1.15, and the apparent ellipse area pi*a*b*cos(i) = 1.35 sq AU. The area of the triangle formed by the origin and the 1986 & 1987 Freys should be 1.35*(0.87yr / 21.7 yr) = 0.054 sq AU. Finding the area of the small origin-1986-1987 triangle on my graph, I find only 0.013 sq AU. So, this short interval does not conform to Kepler's second law. This is hardly surprising, since the 1954 and 1986 sky surveys show additional fainter "disappearing dots" near Frey, more than ten times closer to Frey than to Barbarossa (i.e., within the Jacobi, or other, stability limits), which moons could displace Frey enough to give big deviations from Kepler's second law, in short time intervals. The demand that Kepler's second law be satisfied, even for the short 1986-1987 interval, might be what made it impossible for me to find a consistent binary orbit.

From hypothetical outer solar system precession resonances and other theories, I've estimated the Barbarossa+Frey system as 0.01 solar mass. The a = 1.44, is for the Frey-c.o.m. distance; the Frey-Barbarossa distance is a = 1.44/0.89 = 1.62 AU. The 21.7 yr orbit and 0.01 solar mass imply 1.68 AU, fair agreement for this rough graphical method.
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