Requiem for Relativity

(March 9, 2009; corrected March 16)

Here's an ephemeris:

From my heretofore best prediction program (see Feb. 28 post),

0h Feb. 23, 2009: Barbarossa/Frey c.o.m. (geocentric)
11:26:35.37, -9:15:29.8

7h Feb. 16, 2009: Barbarossa/Frey c.o.m. (geocentric)
11:26:41.67, -9:15:58.3

(but likely Barbarossa detections are ~1' S of these predictions)

These predictions can be extrapolated linearly through March with practically no loss of accuracy. Lately, my various prediction strategies for 2009, starting from 1997, differ only about 0.5'. These predictions are for the c.o.m., but because I now think that the mass ratio Barbarossa::Frey is 40::1, Barbarossa now is, I think, only ~10" NW of the c.o.m. Recently I've detected objects (see above) consistent with Barbarossa and Frey's estimated mutual postion, but with Barbarossa generally ~1' S of the c.o.m., however.

My newest prediction program (see below) gives

0h Feb. 23, 2009: Barbarossa/Frey c.o.m. (geocentric)
11:26:34.55, -9:15:27.6

One could use this for the base, and the earlier pair for the rate of change. Most of the apparent motion now is due to Earth parallax.

*******

My newest prediction program is more rigorous, yet simpler. With its help, I know much more about Barbarossa's orbit. It seems that on March 3, 1997, Barbarossa was 89.22deg past perihelion, near the outgoing latus rectum of a typical eccentric orbit, with e = 0.6106, semimajor axis a = 343.8 AU, and Period = 6340 yr.

My newest method is to approximate Barbarossa's barycentric X,Y,Z (the barycenter is that of the known solar system, i.e. excluding Barbarossa) as sinusoidal functions of time, all with period equal to 2*pi/(Barbarossa's average heliocentric angular speed during the period of observation). Approximately, they're first order Fourier expansions in the angle.

This gives only three unknowns (the amplitude & phase, i.e. sine & cosine amplitudes, and a constant term) to solve four equations (for the 1954, 1986, 1987 & 1997 positions). As a temporary convenience, I include another term, an unknown coefficient times a "step function" (a kind of "wavelet") of +/- 1 AU, in X, Y, or Z, for 1986 & 1987. A 4x4 system of linear equations for four coefficients, is uniquely solved, so that the estimator exactly fits each of the four sky survey c.o.m. positions.

To get the heliocentric XYZ coordinates, I guess the mass ratio, initial orbital radius, and the first time derivative of the radius. Searching on these three parameters (even a 486 desktop is fast enough), I find those that make the step function coefficient practically zero. That is, I find the mass ratio, radius, and dr/dt (assuming, at first, that all higher derivatives of r are zero) that allow pure sine waves in theta, to fit all three of XYZ, for all four sky surveys: the unique perfect fit.

The search result is Barbarossa::Frey = 0.9791::0.0209, 1997 radius = 213.87 AU, 1997 dr/dtheta = +129.5 AU/radian. The second time derivative of r is very small here near the latus rectum, and is considered constant at its midrange value, +0.00003 AU/yr^2.

Knowing r, dr/dtheta, and dtheta/dt, suffices to solve for the major axis, eccentricity, and phase of the orbit. I used the Sun's Kepler "k" constant adjusted for Barbarossa's reduced mass and for the gravity of the giant planets.

Often, adjusting three parameters can solve three equations; so, it's only weak evidence for Barbarossa's existence, that adjusting the mass ratio, radius and dr/dt, places the fourth observed c.o.m. point, on the unique time-parameterized 2936yr sinusoidal curve through the other three. Likewise, often dr/dt will be small enough to give an elliptical orbit rather than a hyperbola.

The main proof that these starlike images are Barbarossa & Frey, is that, according to the program below, the unique orbit fitting all four c.o.m. points, is planar, and satisfies Kepler's second law. The excess arc traveled is consistent (for a worst-case example) with a 0.9 degree deflection of the path, as viewed from the sun, on the 1986-1987 segment only.

As for "equal areas in equal times", none of the three segments differs from the other by more than 1.1% in areal speed, and the 1986-1987 areal speed roughly interpolates the other two. This departure from Kepler's second law could be caused by a second, more distant, smaller moon, Freya. Such a short-period 1% perturbation of areal speed could cause a 3% error in the estimated period.


REM program to fit Barbarossa-Frey c.o.m. orbit to 4 data points
REM & predict others

REM initialize constants
PRINT : PRINT
pi# = 4 * ATN(1): pi180# = pi# / 180: pi2# = 2 * pi#: crit = 0
REM kk# is reduced solar accel. const. in AU/day^2, including known planets
kk# = (pi2# / 365.25#) ^ 2 * 332447 / 332001 * 1.01# / (332002 / 332001)
REM graphical estimate of period=2960yr
ww# = pi2# / 2936.42# / 365.25#: w2# = ww# ^ 2
REM dimension variables; double precision throughout
DIM t(5) AS DOUBLE: DIM ra(4) AS DOUBLE: DIM decl(4) AS DOUBLE
DIM x(4) AS DOUBLE: DIM y(4) AS DOUBLE: DIM z(4) AS DOUBLE
DIM r(4) AS DOUBLE: DIM re(5) AS DOUBLE: DIM rr0(4) AS DOUBLE
DIM sne(5) AS DOUBLE: DIM cse(5) AS DOUBLE
DIM xe(5) AS DOUBLE: DIM ye(5) AS DOUBLE: DIM ze(5) AS DOUBLE
DIM rab(4) AS DOUBLE: DIM declb(4) AS DOUBLE
DIM raf(4) AS DOUBLE: DIM declf(4) AS DOUBLE
DIM x0(4) AS DOUBLE: DIM y0(4) AS DOUBLE: DIM z0(4) AS DOUBLE
DIM theta(4, 4) AS DOUBLE: DIM co(4, 5) AS DOUBLE

REM Get times.
GOSUB 1000
FOR i = 1 TO 5
REM Get sinusoids.
sne(i) = SIN(ww# * t(i)): cse(i) = COS(ww# * t(i))
NEXT
REM Get coordinates.
GOSUB 1100
REM Get barycentric Earth positions.
GOSUB 1200

PRINT "Done initializing; now searching..."
nn = 0: rel0# = 10 ^ 9
FOR ii = 0 TO nn
q# = .9791 + ii * .0001#: p# = 1 - q#

REM r final
FOR r0# = 213.87# TO 213.87# STEP .01#

REM dr/dtheta final
FOR rp# = 129.5# TO 129.5# STEP .1#

REM Get final dr/dt by successive approx.
REM d2r/dt2 = 0 @ latus rectum; tiny ave. value used below
rtt# = 3.05 / 10 ^ 5 / 365.25 ^ 2
rt# = rp# * ww#
rmid# = r0# - .5# * rt# * (t(4) - t(1)) + .125# * rtt# * (t(4) - t(1)) ^ 2
rt# = rt# * (rmid# / r0#) ^ 2

REM getting c.o.m. coords. by weighted ave. of Barbarossa & Frey coords.,
REM causes < 0.5" error
FOR j = 1 TO 4
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(4)) + .5# * rtt# * (t(j) - t(4)) ^ 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 Fit sinusoid + wavelet term to '54-'86-'87-'97.
GOSUB 1500
rel# = lx# ^ 2 + ly# ^ 2 + lz# ^ 2
IF rel# < rel0# THEN GOSUB 3000
NEXT rp#: NEXT r0#
IF ii = 1 THEN PRINT "compiled &"; ii; "/"; nn; "done...";
IF ii > 1 THEN PRINT ii; "/"; nn; "...";
NEXT ii

PRINT : PRINT "wavelet amplitude, AU:"; SQR(rel0#)
PRINT "best mass ratio, final radius, dr/dtheta & d2r/dt2:"
PRINT q0#, r00#, rp0#, rtt#
PRINT "t(1) radius, midrange of t(1) & t(4) radii: "
rmid# = rr0(4) - .5# * rt0# * (t(4) - t(1)) + .125# * rtt# * (t(4) - t(1)) ^ 2
PRINT rr0(1), rmid#
PRINT "circular period at this distance: ";
PRINT rmid# ^ 1.5# / SQR(kk# / (pi2# / 365.25#) ^ 2)

REM cross products for sector angles
FOR i = 2 TO 4
FOR j = 1 TO i - 1
IF i - j = 2 GOTO 400
cpz# = x0(i) * y0(j) - y0(i) * x0(j)
cpx# = y0(i) * z0(j) - z0(i) * y0(j)
cpy# = z0(i) * x0(j) - x0(i) * z0(j)
sn# = SQR(cpz# ^ 2 + cpx# ^ 2 + cpy# ^ 2) / rr0(i) / rr0(j)
theta(i, j) = ATN(sn# / SQR(1 - sn# ^ 2))
PRINT "sector "; j; i; ": "; theta(i, j); "rdn;";
PRINT theta(i, j) / (t(i) - t(j)); "rdn/d"
REM quadratic Euler-Maclaurin est. of areal speed
u2# = rr0(i) ^ 2 - rr0(i) * (rt0# + rtt# * (t(i) - t(4))) * (t(i) - t(j)) / 3
v2# = rr0(j) ^ 2 + rr0(j) * (rt0# + rtt# * (t(j) - t(4))) * (t(i) - t(j)) / 3
PRINT theta(i, j) * (u2# + v2#) * .25# / (t(i) - t(j)); "AU^2 / day"
400 NEXT j
NEXT i
PRINT "excess travel: ";
PRINT theta(2, 1) + theta(3, 2) + theta(4, 3) - theta(4, 1)
PRINT "apparent period: ";
PRINT (t(4) - t(1)) / 365.25# * pi2# / theta(4, 1)

REM PRINT "sinusoidal extrapolation: ";
xpred# = kx0# + sx0# * sne(5) + cx0# * cse(5)
ypred# = ky0# + sy0# * sne(5) + cy0# * cse(5)
zpred# = kz0# + sz0# * sne(5) + cz0# * cse(5)
REM PRINT "barycentric c.o.m. X, Y, Z for 5-th time: "
REM PRINT xpred#, ypred#, zpred#
REM PRINT "geocentric X, Y, Z: "
xpred# = xpred# - xe(5): ypred# = ypred# - ye(5): zpred# = zpred# - ze(5)
REM PRINT xpred#, ypred#, zpred#
decl# = ATN(zpred# / SQR(xpred# ^ 2 + ypred# ^ 2)) / pi180#
ra# = ATN(ypred# / xpred#) / pi180# + 180
PRINT "J2000 celestial coords., 5-th point:"; ra#, decl#
PRINT "RA"; INT(ra# / 15); "h";
a# = 4 * (ra# - INT(ra# / 15) * 15)
PRINT INT(a#); "m";
PRINT 60 * (a# - INT(a#)); "s"
PRINT "Decl"; -INT(-decl#); "deg";
a# = 60 * (-decl# - INT(-decl#))
PRINT INT(a#); "arcmin";
PRINT 60 * (a# - INT(a#)); "arcsec"

REM Find descending node: perp. to cross product & to normal to ecliptic.
REM Find inclination, argument of perihelion.
700 GOSUB 7000
PRINT "RA & Decl. of desc. node: ";
PRINT radnode# / pi180#, decldnode# / pi180#
PRINT "inclination: "; incl# / pi180#
PRINT "argument of perihelion: "; perih# / pi180# + 180


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
REM 0h Feb 23, 2009
t(5) = 54885.5
RETURN

REM light time correction
1050 FOR j = 1 TO 5
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 + 25.3# / 240)
declf(1) = 0 - pi180# * (5 + 56 / 60 + 12.7# / 3600)
rab(1) = pi180# * (11 * 15 + 3 / 4 + 12.28# / 240)
declb(1) = 0 - pi180# * (5 + 58 / 60 + 8.6# / 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 D4 & D5
rab(4) = pi180# * (11 * 15 + 21 / 4 + 39.99# / 240)
declb(4) = 0 - pi180# * (8 + 34 / 60 + 9.6# / 3600)
raf(4) = pi180# * (11 * 15 + 21 / 4 + 1.25# / 240)
declf(4) = 0 - pi180# * (8 + 50 / 60 + 20.4# / 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
a# = 0: i = 5
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#
rra# = .9899423000000001#: rrb# = .9901818999999999#
r# = rra# * b# + rrb# * a#: w# = msun#
GOSUB 1270
REM planets interpolated to nearest RA second, or Decl arc minute
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 4x4 linear systems to fit first-order sinusoid + wavelet term, to x,y,z
REM x:
1500 FOR i = 1 TO 4
co(i, 1) = 1: co(i, 2) = sne(i): co(i, 3) = cse(i): co(i, 5) = x(i)
i7 = 0
IF i = 2 THEN i7 = 1
IF i = 3 THEN i7 = -1
co(i, 4) = i7
NEXT i
FOR i = 1 TO 3
FOR j = i + 1 TO 4
qq# = co(j, i) / co(i, i)
FOR k = 1 TO 5
co(j, k) = co(j, k) - co(i, k) * qq#
NEXT: NEXT: NEXT
lx# = co(4, 5) / co(4, 4)
cx# = (co(3, 5) - co(3, 4) * lx#) / co(3, 3)
sx# = (co(2, 5) - co(2, 4) * lx# - co(2, 3) * cx#) / co(2, 2)
kx# = (co(1, 5) - co(1, 4) * lx# - co(1, 3) * cx# - co(1, 2) * sx#) / co(1, 1)
REM y:
FOR i = 1 TO 4
co(i, 1) = 1: co(i, 2) = sne(i): co(i, 3) = cse(i): co(i, 5) = y(i)
i7 = 0
IF i = 2 THEN i7 = 1
IF i = 3 THEN i7 = -1
co(i, 4) = i7
NEXT i
FOR i = 1 TO 3
FOR j = i + 1 TO 4
qq# = co(j, i) / co(i, i)
FOR k = 1 TO 5
co(j, k) = co(j, k) - co(i, k) * qq#
NEXT: NEXT: NEXT
ly# = co(4, 5) / co(4, 4)
cy# = (co(3, 5) - co(3, 4) * ly#) / co(3, 3)
sy# = (co(2, 5) - co(2, 4) * ly# - co(2, 3) * cy#) / co(2, 2)
ky# = (co(1, 5) - co(1, 4) * ly# - co(1, 3) * cy# - co(1, 2) * sy#) / co(1, 1)
REM z:
FOR i = 1 TO 4
co(i, 1) = 1: co(i, 2) = sne(i): co(i, 3) = cse(i): co(i, 5) = z(i)
i7 = 0
IF i = 2 THEN i7 = 1
IF i = 3 THEN i7 = -1
co(i, 4) = i7
NEXT i
FOR i = 1 TO 3
FOR j = i + 1 TO 4
qq# = co(j, i) / co(i, i)
FOR k = 1 TO 5
co(j, k) = co(j, k) - co(i, k) * qq#
NEXT: NEXT: NEXT
lz# = co(4, 5) / co(4, 4)
cz# = (co(3, 5) - co(3, 4) * lz#) / co(3, 3)
sz# = (co(2, 5) - co(2, 4) * lz# - co(2, 3) * cz#) / co(2, 2)
kz# = (co(1, 5) - co(1, 4) * lz# - co(1, 3) * cz# - co(1, 2) * sz#) / co(1, 1)
RETURN

REM update best
3000 rel0# = rel#: q0# = q#: r00# = r0#: rp0# = rp#: rt0# = rt#
FOR i7 = 1 TO 4
x0(i7) = x(i7): y0(i7) = y(i7): z0(i7) = z(i7): rr0(i7) = r(i7)
NEXT i7
kx0# = kx#: ky0# = ky#: kz0# = kz#
sx0# = sx#: sy0# = sy#: sz0# = sz#
cx0# = cx#: cy0# = cy#: cz0# = cz#
lx0# = lx#: ly0# = ly#: lz0# = lz#
RETURN

REM cross product of two orbital radius vectors
7000 a# = y0(1) * z0(4) - z0(1) * y0(4)
b# = z0(1) * x0(4) - x0(1) * z0(4)
c# = x0(1) * y0(4) - y0(1) * x0(4)
REM normal to ecliptic
aa# = 0: bb# = 0 - SIN(23.45 * pi180#): cc# = COS(23.45 * pi180#)
REM Find inclination.
cs# = (a# * aa# + b# * bb# + c# * cc#) / SQR(a# ^ 2 + b# ^ 2 + c# ^ 2)
incl# = ATN(SQR(1 - cs# ^ 2) / cs#)
REM Cross to get a vector in orbital & ecliptic planes.
a3# = b# * cc# - c# * bb#
b3# = c# * aa# - a# * cc#
c3# = a# * bb# - b# * aa#
REM Normalize.
d# = SQR(a3# ^ 2 + b3# ^ 2 + c3# ^ 2)
a3# = a3# / d#: b3# = b3# / d#: c3# = c3# / d#
REM Convert to celestial coords.
decldnode# = ATN(c3# / SQR(a3# ^ 2 + b3# ^ 2))
radnode# = ATN(b3# / a3#)
IF radnode# < 0 THEN radnode# = radnode# + pi#
cs# = (x0(4) * a3# + y0(4) * b3# + z0(4) * c3#) / rr0(4)
perih# = ATN(SQR(1 - cs# ^ 2) / cs#)
IF perih# < 0 THEN perih# = perih# + pi#
dthdt# = 2 * pi# / 2936.42# * (rmid# / rr0(4)) ^ 2
perih# = perih# + 6.71# * dthdt# - pi# / 2
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 0h Feb 23, 2009
DATA -.894536,0,.396574,0,.171904,0

(main installment of another email debate, sent tonight, March 9, 2009)

Dear Dr. *******,

I've found that for this, what you called "leading with the pedigree" indeed does have the highest batting average (getting to first base, at least), though some call it obnoxious! With something like this, the pedigree is a way to slow down their finger on the way to the delete key.

Are you willing to debate me about this on Coast to Coast? Or anywhere? Is anyone?

You cited Standish's article. One way to rebut Standish is by this order-of-magnitude calculation: Standish claims that 0.5% correction to Neptune's mass, removes Uranus' residuals (though the dying Harrington disagreed). If Neptune has 1/20,000 solar mass, the correction is 1/4,000,000 solar mass. Typically the Neptune-Uranus distance is roughly comparable to the Uranus-Sun distance. In a long segment of Uranus' orbit, the Sun's gravitational acceleration alters Uranus' position by the order of one Uranus radius, seen as one radian. In that same segment, the 0.5% change in Neptune's mass might alter Uranus' position 1/4,000,000 radian = 1/20 arcsec. The residuals were thought to be much more than 1/10 arcsec; Standish considered 1/10 arcsec to be about the same as zero.

One of my Harvard classmates told me a Harvard grad student privately admitted faking data. If that happens even at Harvard, then it happens almost everywhere. Have you seen Standish's computer program? (My latest is posted on Dr. Van Flandern's website, self-contained, self-explanatory and only six screens long.) Standish used USNO data: isn't that Harrington's data, the same Harrington who disagreed with Standish? Did Standish make the grad student do it over and over again until it came out zero? There are many possible things that might have caused Standish to be wrong.

Regarding your suggestion to send it to journals, I've submitted my findings to a dozen journals without luck. None sent it for peer review. This also, is an indictment of the system.

I haven't emailed everyone, only a few hundred authorities, including the top two guys at that comet bureau. I've talked to about a score of astronomy & physics professors in their offices; they're almost always friendly but aren't very knowledgeable about this particular specialty, (with one possible exception I'm working on now) don't seem to have access to adequate telescope time, and my impression is that although they aren't sure I'm wrong, they don't feel powerful enough to do anything about it if I'm right. There is monumental over-centralization and groupthink.

I have Planet X on four DSS sky surveys now (all three red, and the optical IR plate; none of the four relevant blue filter plates - too faint in blue?), perfectly fitting an orbit. But here is one especially quick thing that one would have time to check out, that might persuade one not to dismiss this out of hand:

According to Paul Wesson's p=J/M^2 constant, the known solar system's angular momentum is 30x too small.

Sincerely,
Joe Keller

(email sent March 3, 2009; no response of any kind)

To: R. O. Harmon, Chair, Physics & Astronomy, Ohio Wesleyan Univ.
R. F. "Rock" Jones, President, OWU
E. H. Burtt, Honors Advisor

Dear sirs:

A college ultimately relies not on tuition, alumni or corporate donations, or tax funding. A college ultimately relies on veritas, truth. A college which becomes famous for discovering truth, will succeed. A college which does little to promote truth, will fail fiscally. This is especially important as we enter this Great economic Depression (the difference between this Depression and the last one is that this time the government is already bankrupt).

An amateur astronomer friend who lives in Columbus, asked me to ask OWU to look, with its 30 inch telescope, for an object which I have, now, identified with an accurate orbit on four out of four relevant sky surveys, and on a 2007 photo (by Steve Riley) with an 11 inch telescope, and also on three Feb. 2009 photos (by an anonymous philanthropist, by Joan Genebriera, and by the U. of Iowa) with telescopes ranging from 15 to >16 inches. The response then, was that after an OWU underling scoffed at this as pseudoscience, an OWU associate professor told me that he had a rule against talking to laymen.

Of course neither of them have bothered to check my data or calculations. This is a far cry from LeVerrier who jumped on a train to check Lescarbault's. I am a cumlaude graduate of Harvard college in (pure) Mathematics. Recently I worked nominally half time, for a year under an NSF grant, doing the math and Fortran programming for an engineering group at Iowa State Univ., whose faculty leader received an award for the project. This doesn't prove that my (elementary though somewhat laborious) calculations are accurate, but only that they are likely to be. I've been working steadily on these calculations for two years and gradually have improved their accuracy.

This is not about whether you might think your fellows are "great guys" or whether I have a "union card". The people shovel money to colleges that make great discoveries. Ultimately, that's the only reason the people shovel money to colleges. The only reason people want to go to OWU, or Harvard, or any other college, is that it's known for something.

Consider the risk/benefit structure: by imaging the body now, with an hours-long time exposure near opposition, while nights are still long and the object is on the meridian at midnight, using a Wratten 29 deep red filter to remove mercury & sodium vapor lamp light pollution, you could produce an irrefutable image of this known reddish, sunlit object, complete with a track consistent with its 200 AU distance. Now, this discovery is languishing, because the willfully ignorant are not persuaded by heliocentric great circles through "disappearing dots" on sky surveys, nor by elementary statistical analyses of pixels on photos taken by small telescopes.

Essentially all the information I have, is in the updates I post to Dr. Tom Van Flandern's messageboard at www.metaresearch.org under my name, Joe Keller. The ALPO and "Bad Astronomy" messageboards censored me, but Dr. Van Flandern gave me academic freedom. Why not look? Maybe your college could be the first to tell the public about literally the biggest discovery in the Solar System, since the Sun.

Sincerely,
Joseph C. Keller, M. D.

Joe, Have you tried the HST? They have a proposal page or did have one and have looked at everything else, lots of it much less interesting.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />Joe, Have you tried the HST? They have a proposal page or did have one and have looked at everything else, lots of it much less interesting.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Thanks for the heads-up! I had searched for Hubble Space Telescope photos of the region, and found none at the appropriate times. This was expected, because the HST is not a wide field (Schmidt) camera; it has performed much less than a complete sky survey.

As soon as I post this, I'll send as an HST proposal, my very post to this messageboard which contains my latest computer program. It's the most convincing short letter I could write.

Thanks again! Maybe others who read this web site, could write to the HST bureau and ask them to look for Barbarossa.

Another person to whom to write, is Dr. Neil deGrasse Tyson. Tyson made one of the most civilized responses I've ever gotten to an unsolicited email. Most astronomy professors to whom I send unsolicited email (indeed I never saw the phrase "unsolicited communication" until I saw astronomy professors primly using it) act like Victorian virgins accosted by a sailor.

Tyson took a mature approach. He doesn't control a big telescope himself, but talks to many professional astronomers, to the public, and to statesmen and journalists.

Gregorian Calendar Start, and Julian Date 0: Barbarossa Perihelia

Sources:

John Herschel, "Outlines of Astronomy", 1849, as cited in Wikipedia article "Julian Day";

AF Aveni, "Empires of Time", 1989, p. 129.

In my post above, with my latest program, I mention that the four sky survey positions, let me solve Barbarossa's orbit. Using equal areas in equal times, I find that Barbarossa's latest perihelion was: 1997.2 (observation) + 6.8 (time to cover remaining 0.78deg to latus rectum, including correction for decreasing angular speed) - 434.3 (time from perihelion to latus rectum, by exact integral calculus, assuming P=6339.93, e=0.610596) = 1569.7 AD. Pope Gregory's new calendar started Oct. 15, 1582 (Gregorian date) i.e. 1582.8.

Julian dates begin 4713.0 BC. If Barbarossa's perihelion were 1583.0 AD, then the Julian date system (established 1583) would begin at the previous Barbarossa perihelion, if Barbarossa's period were 6295 yr (my estimate: 6340 yr).