Thanks for the link; I checked it. Their chart on p. 581, shows that the authors are confident that "Supernova 1006" was most likely >10deg South, of Barbarossa's orbit in that part of the sky (where Barbarossa's orbit is ~12deg S of the ecliptic). Also, Barbarossa would have been >90 away in ecliptic longitude, then.

But this is a fruitful area of investigation! Thanks for volunteering this information!

Didn't someone in this thread mention a little program that lets you plot the barycentre of a system? Could we try sticking in a Neptune mass planet and the brown dwarf planet barbi into that?

Thanks Jim! This seems to be the lightcurve that Brian Warner already has sent me (he sent it to me in numerical form, as I had requested), but it's helpful to have a graph of it.

Thanks! He was on my list but I didn't have this email address!

Jan. 15, 2010, 3PM CST To: Robert Stephens Re: lightcurves for 947 Monterosa and 1717 Arlon

Dear Mr. Stephens:

I am calculating the rotation axes of 947 Monterosa and 1717 Arlon. I want all existing lightcurves.

I see 1717 Arlon listed on your website. Would you send me the lightcurve in numerical form? Brian Warner recently sent me a lightcurve on Monterosa, in such form, and it was satisfactory.

Hi Joe, off the subject but I thought I'd best stick this down here before I forget it. From looking at the esu and its relationship to gravity, I get Q^2 = 4G h^2 b / r^2 k

Where b is the speed of gravity.

Rearranging for b I get a lightly lower speed of gravity than I think it is but I haven't worked it out to many decimal places.

With the numbers I've used (1.602E-19)^2 = 4 *6.678E-11 * (6.626E-34)^2 * b / (2.4E-12)^2 *9E 9 The speed of gravity comes out as about 1.1344E 25

Putting that speed of gravity into the equation x = c^2 / b^2 the answer differs from h by about point one. I think that's not too bad at all, as I think we could e talking about particles having a form factor, an rms value divided by the average value.

Sorry about the thread interruption.

A question on these asteroids. Asteroids do bang into each other but not like in the movies. Fairly gentle affairs. That has to effect the spins but I'm not too sure about the axis of spin. Not much mass, they should precess, yeah?

A little update on that. I thought I'd see what happens to the proton with this equation. Now I've taken the radius of the proton as being 1.321E-15 Something has to change here and k_e has to increase. Working it through, with that value of the speed of gravity, 1.1344E 25 I get k_e changing from 9E 9 to 2.9706E 16

Well we have c^2 = k_e / k_m so with these figures k_m changes to 0.3305 (That suggests that pi is involved someplace there) I took the speed of light as 2.998E 8

Now I think it makes sense to look at the equation, b^2 = k_e / k_m Where b is the speed of gravity. Then we'd get k_m very close to the numerical value of h

I think we have to think about, when a proton changes into a neutron and somehow loses most of its magnetic field. Does it spin a tiny it faster than light? Dumping its mag field into gravitational space, where of course it simply cannot be "seen"?

Review. These are relevant to Barbarossa, because at the end of the Mayan Long Count (Dec. 21, 2012AD) all four of the asteroids, with known rotation periods, whose rotation periods are just above the theoretical & empirical critical value 5.1287h (namely Davida, Laetitia, Monterosa, and Arlon) are within 3deg of conjunction or opposition to Barbarossa. That is, all four of the asteroids with the special rotation period, align with Barbarossa and the Sun at the end of the Mayan Long Count.

Furthermore, the two, of these four, whose rotation axes have been estimated in the journal literature, have the same rotation axis (according to my average of the determinations in the literature). Lightcurves for Monterosa and Arlon enable me to estimate their rotation axes too. If three, or all, of the four asteroids, have the same rotation axis, this might be enough to overcome the psychological "denial" of some people, and get them to realize that there is a phenomenon here worth investigating (by looking in the direction of Barbarossa, with a large telescope).

Progress on obtaining lightcurves of Monterosa and Arlon. Though a lightcurve or lightcurves for Arlon have been measured, I have no lightcurve data yet for Arlon, and no responses from any investigators.

Regarding Monterosa, Brian Warner has sent me his 2007 lightcurve. The only other relevant lightcurve investigator (of ~20 I emailed) who has responded, is Z., who never responded directly, but did tell me through an intermediary, that he is refusing to release the 2003 lightcurve data on Monterosa. I've been told that Z's professional and semiprofessional colleagues agree, that Z has the right to refuse, with or without a reason. I don't want to mention this astronomer by name, because his refusal through an intermediary, despicable though it is, is less despicable than the others, who ignored my email altogether; so, he should be denounced less than they, not more. I've already visited my U.S. Representative's office personally to complain (to a staffer, though I met the Representative himself at the Republican caucus last week), and have complained to two of the tax-exempt foundations funding the IAU Minor Planet Center. I eventually was able to download a somewhat cryptic graph, of part or all of Z's lightcurve, which appears on Behrend's website, so I can get a somewhat inaccurate and ambiguous version of Z's data by measuring this graph with a ruler.

Determining Monterosa's axis from available data. Already (see above) I've estimated, from only Monterosa's stated amplitudes in its two extant lightcurves, and from the antisymmetry of Z's lightcurve graph, that Monterosa's axis differs only 9deg from the common axis of Laetitia and Davida.

Now, using Warner's lightcurve and taking measurements from the graph of Z's, I'll use my BASIC computer program (see next post for the program code) to estimate Monterosa's axis better.

Explanation of my computer program for finding asteroid axes from lightcurves.

I. Earlier methods. A. The oldest method ("amplitude-phase method")(simplest version). At one extreme: if the axis is perpendicular to the ecliptic, then the amplititude of the lightcurve is about constant. At the other extreme: if the axis lies in the ecliptic, then the amplitude becomes about zero twice per orbit, at opposite points depending on the direction of the tilt. Thus the tilt to the ecliptic can be estimated as an interpolation between these two extreme situations, and the direction of that tilt from the location of the extremes. This is basically the method by which I've estimated Monterosa's axis already. This method involves some ambiguity of solution, but published studies usually have shown good agreement.

B. Modern researchers like to model the asteroids as triaxial ellipsoids or even as convex polyhedra, trying to find not only the axis, but also the shape, from the lightcurve ("shape inversion method"). There is still much ambiguity of solution (see, inter alia, Drummond et al, Icarus, 2009), though the solutions generally agree with the earlier solutions made by the amplitude-phase method.

II. My method (the "Gauss map harmonic" method).

A. The spherical map. In differential geometry, the map from a surface to the unit sphere, taking unit normal vectors, to identical vectors as radii of the sphere, is called the "spherical map", a.k.a. "Gauss map". To sum over surface area, one sums over the unit sphere, weighted everywhere by "K inverse", i.e. the reciprocal of the Gaussian curvature, which is the Jacobian of the inverse Gauss map. I replace the rotating asteroid (assumed to be convex) with a rotating unit sphere, and integrate the brightness weighted by three factors: "K inverse", the albedo, and a photometric function. The product of these three factors, I call the "luminance function". The luminance function is found, by fitting the lightcurve data, as a rapidly convergent series of spherical harmonics.

B. Minkowski's problem. In 1903, Hermann Minkowski solved (lengthy article in German available from an online service) the problem: What restrictions are necessary and sufficient, for a function on the Gauss sphere (i.e., image of Gauss map), to be the Gaussian curvature of a real, closed, convex surface? Minkowski solved this only approximately, but later exact solutions came from other mathematicians. I follow Louis Nirenberg, Communications in Pure and Applied Mathematics, 6:337-394, 1953 (the part of Nirenberg's article, dealing with Minkowski's problem, starts on p. 377; at the very beginning of Nirenberg's article, is a misprint omitting the "-1" inverse sign from "K inverse"). The answer is (paraphrasing Nirenberg eqn. 12.1, p. 377, or equivalently eqn. 13.6, p. 381) that it is necessary and sufficient, for the three terms of order n=1 (the n=1, m=0 term and the two, sine & cosine, n=1, m=1 terms), in the spherical harmonic expansion (on the Gauss sphere) of "K inverse" (the reciprocal of the Gaussian curvature) to be zero.

C. Fitting the luminance function to the axis and lightcurves. A priori, the luminance function (that is, the amount of light coming from the part of the asteroid surface nearly perpendicular to a given normal direction) can be anything, if the asteroid is white or black enough there, or the photometric formula is non-Lambertian enough, or its Gaussian curvature is small enough or large enough there. Really, asteroids' albedos usually are said to be practically constant, and photometric functions have been said to matter little; so, shape, i.e. Gaussian curvature, is the main factor. The large zero order and second order Fourier terms (three terms total) for n lightcurves, determine (via a linear system of 3*n equations) 3*n spherical harmonic terms (n=1, m=1, but not n>1, m=1, is prohibited because of Minkowski's problem, so it is difficult, though not impossible, for first order harmonics to arise from shape; rival causes of first order harmonics in the lightcurve, include non-principal axis rotation, moon asteroid(s), albedo variations, and experimental error), all with m=0 or m=2, of "K inverse", the reciprocal of the Gaussian curvature.

D. Choosing the most plausible axis. For many choices of asteroid rotation axis, the spherical harmonic terms required, will be such that "K inverse" is negative at some points on the unit sphere. Possible axes include only those for which "K inverse" is everywhere positive, which is about the same thing, as having all spherical harmonics small relative to the constant term P(0,0)(i.e. n=0, m=0). The likeliest asteroid shapes are almost round, so, if other things be equal, the likeliest axes are those which give the smallest sum of squares of magnitudes of the higher terms relative to P(0,0) ("small higher harmonics" criterion). Another way to find the true axis, is to eliminate all axes which imply negative luminance functions, then choose the remaining axis that lies nearest some principal axis of rotation (as estimated from "K inverse" by assuming constant density).

E. Test of program. By the "small higher harmonics" criterion alone, I find (from only two lightcurves, July 1986 & April 1990, in Dotto et al, A&A Supp, 1992) that 129 Antigone's axis is nearly perpendicular to the ecliptic (not an error; I used celestial coordinates). I find prograde rotation though there is almost a 180 degree ambiguity. I chose Antigone for the test, because Drummond (op. cit.) recently studied it with triaxial modeling and the Keck telescope, finding, though with considerable ambiguity, the likeliest axis as ecliptic coords. (200,55), which agrees with some of the earlier estimates.

F. Future improvements. At least three or four well-spaced lightcurves are recommended; this would remove the unfair advantage of the ecliptic poles. In my test (see "E") I considered third order harmonics also, to get 5*n = 10 spherical harmonic terms. Improvement can come from using an asteroid with more known lightcurves, or from insisting on principal axis rotation, as estimated assuming constant density.

G. Photometric functions. Those simple enough to be useful, include mainly Lambert's and Lommel-Seeliger's. Lambert's (i.e. cos(e) * cos(i) ) and Lommel-Seeliger's, both satisfy the Helmholtz law, which says essentially that light is reversible, so the photometric function, expressed as light seen per unit surface area, must be a symmetric function of cos(e) & cos(i), where e & i are the angles of reflection and incidence, resp. For an asteroid seen from Earth, the line of incidence about equals that of reflection, so Lommel-Seeliger's law reduces approximately to sqrt(cos(e) * cos(i)).

H. Experimental determination of true asteroid photometric function. From superficial glacial till rocks on my father's farm in central Iowa (mostly igneous Canadian shield rocks, with some Iowa limestone) I selected some of 1 to 10cm size that are mostly rather dark, none very white. They are unwashed. For more randomness, I cracked some. I spread them on a concrete floor and made camera lightmeter measurements, corrected for distance, at 30deg and 45 deg, with the line of illumination the same as the line of reflection. The best empirical fit, is (cos(e) * cos(i))^(5/6), that is, closer to Lambert's law than to Lommel-Seeliger's. This photometric law also minimized the sum of squared relative coefficients (i.e. coefficients divided by the P(0,0) coefficient) of the nonconstant spherical harmonic terms, for the best axis.

REM program "AstAx.bas" to find asteroid axes from lightcurves REM language: Basic REM author: Joseph C. Keller, M.D.; B.A., Harvard

REM program "AstAx.bas" to find asteroid axes from lightcurves REM language: Basic REM author: Joseph C. Keller, M.D.; B.A., Harvard

REM initialize at double precision CLS : PRINT "initializing asteroid axis program" pi# = 4 * ATN(1): pi180# = pi# / 180 DIM p(5, 4) AS DOUBLE: DIM px(10, 17) AS DOUBLE DIM snsn(3, 24) AS DOUBLE: DIM cscs(3, 24) AS DOUBLE DIM s(7, 2) AS DOUBLE: DIM time(160, 2) AS DOUBLE: DIM mag(160, 2) AS DOUBLE DIM ax(17, 24) AS DOUBLE DIM ax2(17, 24) AS DOUBLE: DIM conv(4, 2) AS DOUBLE DIM x(3, 3) AS DOUBLE: DIM se(2, 2, 3) AS DOUBLE: DIM mu(2) AS DOUBLE DIM ss(10, 2, 2) AS DOUBLE: DIM s2(7) AS DOUBLE DIM sumsq(24) AS DOUBLE: DIM co(10) AS DOUBLE: DIM coprov(10) AS DOUBLE DIM c(10, 10) AS DOUBLE: DIM cc(10, 24) AS DOUBLE: DIM cc0(10, 24) AS DOUBLE DIM coord(3, 17, 24) AS DOUBLE: DIM coord2(3, 17, 24) AS DOUBLE

REM find ten tesseral harmonics REM (0,0),(2,0),(2,2(s,c)),(3,2(s,c)),(3,3(s,c)),(4,3(s,c)) GOSUB 1000 REM find areas of tesserae GOSUB 1200 REM find x,y,z coords of tesserae GOSUB 1400 REM Fourier analyze lightcurves GOSUB 4000 PRINT "done initializing"

REM randomly search for best fitting axis PRINT "randomly searching for best axis" PRINT "Have you entered the asteroid's helio- & geocentric" PRINT "normalized xyz coords in program lines 310-340?" PRINT "If not, end this run, and reprogram lines 310-340." REM x,y,z helio- & geocentric asteroid coords REM (celestial or ecliptic coords, but remember which) REM normalized celestial coords for se(Warner/Berger,helio/geo,x/y/z) REM Warner is JPL for 0h, Jan. 7, 2007, but lightcurve is only REM Jan. 7 data (Jan. 1 & 4 data had much bigger error) REM Berger is JPL for 0h Mar. 30, 2003 (data are Mar. 29.6) REM 310 se(1, 1, 1) = .3305#: se(1, 1, 2) = .8472#: se(1, 1, 3) = .4159# REM 320 se(1, 2, 1) = .6647#: se(1, 2, 2) = .657#: se(1, 2, 3) = .3557# REM 330 se(2, 1, 1) = -.796#: se(2, 1, 2) = .4988#: se(2, 1, 3) = .3428# REM 340 se(2, 2, 1) = -.5702#: se(2, 2, 2) = .6841: se(2, 2, 3) = .4548# REM for Antigone lightcurves use this subroutine for se(i,j,k) GOSUB 8000

REM find pole minimizing sum of (tesseral harmonic coeffs)^2 REM really, sum of (ci/c0)^2

ncounter = 300: sumsq0# = 10 ^ 12

PRINT "beginning trials of "; ncounter; " axes; completed:" FOR counter = 1 TO ncounter 700 GOSUB 2000 indicator = counter / 10 - INT(counter / 10) IF indicator < .05 THEN PRINT counter; " "; 800 NEXT counter REM print results GOSUB 5000

END

REM tesseral harmonics REM find sines, cosines 1000 FOR jj = 0 TO 23 th# = pi180# * 15 * jj FOR j = 1 TO 3 snsn(j, jj) = SIN(j * th#): cscs(j, jj) = COS(j * th#) NEXT j: NEXT jj FOR kk = 0 TO 16 x# = COS(pi180# * 10 * (17 - kk)) REM find associated Legendre functions GOSUB 1100 px(1, kk) = p(0, 0) px(2, kk) = p(2, 0) px(3, kk) = p(2, 2) px(4, kk) = p(2, 2) px(5, kk) = p(3, 2) px(6, kk) = p(3, 2) px(7, kk) = p(3, 3) px(8, kk) = p(3, 3) px(9, kk) = p(4, 3) px(10, kk) = p(4, 3) NEXT kk PRINT "finished finding 10 tesseral harmonics" RETURN

1400 FOR kk = 0 TO 16 lat# = pi180# * 10 * (kk - 8): sn# = SIN(lat#): cs# = COS(lat#) FOR jj = 0 TO 23 coord(3, kk, jj) = sn# lon# = pi180# * 15 * jj coord(1, kk, jj) = COS(lon#) * cs#: coord(2, kk, jj) = SIN(lon#) * cs# NEXT: NEXT RETURN

REM choose random pole and check fit REM this can be speeded twofold by considering clockwise vs. counterclockwise REM (reversing sines but not cosines) REM and restricting to northward poles REM as written, rotation always counterclockwise viewed from pole REM because frame always right-handed 2000 x(1, 1) = RND(1) - .5#: x(1, 2) = RND(1) - .5#: x(1, 3) = RND(1) - .5# REM select region of interest x(1, 1) = -.1# + .2# * x(1, 1) x(1, 2) = -.4# + .2# * x(1, 2) x(1, 3) = .9# + .2# * x(1, 3) x(2, 1) = x(1, 2): x(2, 2) = -x(1, 1): x(2, 3) = 0 REM vector cross product x(3, 1) = -x(1, 3) * x(2, 2): x(3, 2) = x(1, 3) * x(2, 1) x(3, 3) = x(1, 1) * x(2, 2) - x(1, 2) * x(2, 1) REM normalize new frame vectors GOSUB 2100 REM find tesserae coords in new frame GOSUB 2150

FOR i2 = 1 TO 2 REM weight area of each tessera by Lommel-Seeliger law photometric function GOSUB 2200 REM find Fourier coefficient of lightcurve caused by each tesseral harmonic GOSUB 2400 NEXT i2 REM shift 2nd lightcurve repeatedly and solve for tesseral coeffs GOSUB 2600 REM find best fit for the two lightcurves REM for the asteroid in the two positions GOSUB 3000 RETURN

2100 FOR i = 1 TO 3 d# = 0 FOR j = 1 TO 3 d# = d# + x(i, j) ^ 2 NEXT j d# = 1 / SQR(d#) FOR j = 1 TO 3 x(i, j) = x(i, j) * d# NEXT j: NEXT i RETURN

2150 FOR kk = 0 TO 16 FOR jj = 0 TO 23 x# = coord(1, kk, jj): y# = coord(2, kk, jj): z# = coord(3, kk, jj) FOR i = 1 TO 3 coord2(i, kk, jj) = x# * x(1, i) + y# * x(2, i) + z# * x(3, i) NEXT i: NEXT jj: NEXT kk RETURN

2200 FOR kk = 0 TO 16 FOR jj = 0 TO 23 FOR i = 1 TO 2 mu(i) = 0 FOR j = 1 TO 3 mu(i) = mu(i) - se(i2, i, j) * coord2(j, kk, jj) NEXT j: NEXT i IF mu(1) < 0 OR mu(2) < 0 THEN GOTO 2300 REM Lommel-Seeliger law (my version of f(alpha) ) REM csalpha# = 0 REM FOR j = 1 TO 3 REM csalpha# = csalpha# + se(i2, 1, j) * se(i2, 2, j) REM NEXT j REM cspsi# = (csalpha# - mu(1) * mu(2)) / SQR((1 - mu(1) ^ 2) * (1 - mu(2) ^ 2 REM muu# = mu(1) + mu(2) REM f(alpha)=sin(psi/2) for small mu, 1 for large mu REM vs# = (1 - cspsi#) / 2 REM falpha# = SQR(vs#) * COS(muu# * pi# / 4) + SIN(muu# * pi# / 4) REM phot# = falpha# * mu(1) * mu(2) / muu# REM vs. Lambert's law: REM phot# = mu(1) * mu(2) REM simplified Lommel-Seeliger (alpha is small and mu(1)=mu(2) approx.) REM phot# = SQR(mu(1) * mu(2)) REM my empirical simplified Lommel-Seeliger phot# = (mu(1) * mu(2)) ^ (5 / 6) REM trial photometric law REM phot# = (mu(1) * mu(2)) ^ 2 2280 ax2(kk, jj) = ax(kk, jj) * phot# NEXT jj NEXT kk RETURN

2300 phot# = 0 GOTO 2280

REM zero the sums 2400 GOSUB 2450 REM use angle dependence of px(ii,kk) and convolute FOR kk = 0 TO 16 GOSUB 2460 FOR jj = 0 TO 23 conv(0, 2) = conv(0, 2) + ax2(kk, jj) FOR i = 1 TO 3 conv(i, 1) = conv(i, 1) + ax2(kk, jj) * snsn(i, jj) conv(i, 2) = conv(i, 2) + ax2(kk, jj) * cscs(i, jj) NEXT i: NEXT jj ss(1, 1, i2) = ss(1, 1, i2) + px(1, kk) * conv(0, 2) ss(2, 1, i2) = ss(2, 1, i2) + px(2, kk) * conv(0, 2) ii = 1 FOR i = 2 TO 3 ii = ii + 2 ss(ii, 1, i2) = ss(ii, 1, i2) - px(ii, kk) * conv(i, 2) ss(ii, 2, i2) = ss(ii, 2, i2) + px(ii, kk) * conv(i, 1) ii = ii + 2 ss(ii, 1, i2) = ss(ii, 1, i2) - px(ii, kk) * conv(i, 2) ss(ii, 2, i2) = ss(ii, 2, i2) + px(ii, kk) * conv(i, 1) NEXT i: NEXT kk FOR ii = 4 TO 10 STEP 2 ss(ii, 1, i2) = ss(ii - 1, 2, i2) ss(ii, 2, i2) = -ss(ii - 1, 1, i2) NEXT ii RETURN

2450 FOR ii = 1 TO 10 FOR j = 1 TO 2 ss(ii, j, i2) = 0 NEXT: NEXT RETURN

2460 FOR i = 0 TO 3 FOR j = 1 TO 2 conv(i, j) = 0 NEXT: NEXT RETURN

REM solve 10x10 linear system to find coeffs of tesserals REM 10x24 matrix on Right, so best phase shift of 2nd lightcurve is found REM Left matrix has 2x2, 4x4 & 4x4 blocks REM make matrices REM fill Left matrix 2600 GOSUB 2700 REM fill Right matrix GOSUB 2750 REM solve matrix equation GOSUB 2800 RETURN

2700 FOR j = 1 TO 2 GOSUB 2710 NEXT j FOR j = 3 TO 6 GOSUB 2712 NEXT j FOR j = 7 TO 10 GOSUB 2714 NEXT j RETURN

2900 FOR i = i00 TO i0 FOR ii = i00 TO i0 IF ii = i THEN GOTO 2970 q# = c(ii, i) / c(i, i) FOR j = i00 TO i0 c(ii, j) = c(ii, j) - c(i, j) * q# NEXT j FOR j = 1 TO 24 cc(ii, j) = cc(ii, j) - cc(i, j) * q# NEXT j 2970 NEXT ii NEXT i FOR i = i00 TO i0 d# = 1 / c(i, i) FOR j = 1 TO 24 cc(i, j) = cc(i, j) * d# NEXT j NEXT i RETURN

REM find minimum sum of criterion REM and compare to master minimum, sumsq0# 3000 sumsqbest# = 10 ^ 12 FOR j = 1 TO 24 s# = 0 FOR i = 2 TO 10 s# = s# + (cc(i, j) / cc(1, j)) ^ 2 NEXT i IF s# < sumsqbest# THEN GOSUB 3090 sumsq(j) = s# NEXT j REM interpolate to find smallest possible and replace master best if needed GOSUB 3100 RETURN

3090 sumsqbest# = s#: j0 = j RETURN

3100 i0 = j0 - 1: k0 = j0 + 1 IF i0 = 0 THEN LET i0 = 24 IF k0 = 25 THEN LET k0 = 1 c# = sumsqbest#: b# = .5# * (sumsq(k0) - sumsq(i0)) a# = sumsq(k0) + sumsq(i0) - 2 * c# p# = -b# / a# REM find minimum of parabola GOSUB 3200 sumsqbest# = ff# IF ff# < sumsq0# THEN GOSUB 3300 RETURN

REM set new best tesseral coeffs, axis 3300 FOR i = 1 TO 10 c# = cc(i, j0): b# = .5# * (cc(i, k0) - cc(i, i0)) a# = cc(i, i0) + cc(i, k0) - 2 * c# GOSUB 3200 co(i) = ff# NEXT i FOR j = 1 TO 3 x0(j) = x(1, j) NEXT j sumsq0# = sumsqbest# 3390 RETURN

REM find first 4 (0th through 3rd order) REM Fourier coeffs. for each of two lightcurves REM omega for Monterosa & for Antigone, resp. REM om# = 2 * pi# * 24 / 5.1655# 4000 om# = 2 * pi# * 24 / 4.9572# REM could make Doppler correction to om#, using JPL ephemeris radial velocity i1 = 1: nn = 40 REM zero the sums GOSUB 4020 GOSUB 4200 REM (for Antigone only) undo Dotto's Bowell & Zappala corrections REM find alpha#, the angle between Earth & Sun REM & undo Zappala, Bowell corrections (p. 196, Dotto etal, A&ASupp, 1992) GOSUB 7000 i1 = 2: nn = 28 GOSUB 4020 GOSUB 4200 REM for Antigone only GOSUB 7000 REM make permanent Left matrix GOSUB 4500 RETURN

4020 FOR i = 0 TO 6 s(i, i1) = 0 NEXT i denom# = 0 RETURN

4200 FOR n1 = 1 TO nn READ t#, m# REM change the next block of lines, if different lightcurves are used

REM this line is for MonterosaWarner REM time(n1, i1) = t# * .00001# * om#: mag(n1, i1) = 10 ^ (m# * .0004#)

REM this block is for Antigone (measured from journal graph) IF t# = 0 OR t# > 99 THEN GOTO 4300 GOTO 4350 4210 m# = (m# - 50) * .006186# time(n1, i1) = t# * 2 * pi# mag(n1, i1) = 10 ^ (m# * .4#) NEXT n1 time(0, i1) = 2 * time(1, i1) - time(2, i1) time(nn + 1, i1) = 2 * time(nn, i1) - time(nn - 1, i1) a# = (time(nn + 1, i1) + time(nn, i1)) * .5# b# = (time(0, i1) + time(1, i1)) * .5#: leng# = a# - b# - 2 * pi# denom# = 0 FOR ii = 1 TO nn delt# = .5# * (time(ii + 1, i1) - time(ii - 1, i1)): t# = time(ii, i1) REM deal equitably with excess length of time interval bb# = t# - b#: aa# = a# - t# wt# = leng# IF aa# < wt# THEN LET wt# = aa# IF bb# < wt# THEN LET wt# = bb# wt# = wt# * delt#: denom# = denom# + wt# s(0, i1) = s(0, i1) + mag(ii, i1) * wt# FOR i = 1 TO 3 s(2 * i - 1, i1) = s(2 * i - 1, i1) + mag(ii, i1) * 2 * SIN(i * t#) * wt# s(2 * i, i1) = s(2 * i, i1) + mag(ii, i1) * 2 * COS(i * t#) * wt# NEXT i: NEXT ii denom# = 1 / denom# FOR n1 = 0 TO 6 s(n1, i1) = s(n1, i1) * denom# NEXT n1 RETURN

5000 PRINT "results of trials of "; ncounter; " random axes:" PRINT "best axis, xyz coordinates: "; PRINT x0(1), x0(2), x0(3) REM PRINT "best (i.e., those with minimum sum of negative albedos," PRINT "giving observed lightcurve Fourier coefficients)" REM PRINT "sum of negative albedos: "; sumsq0# PRINT "min. sum of squares of relative coefficients above zeroth: "; sumsq0# PRINT "tesseral harmonic coefficients for asteroid shape/albedo:" FOR ii = 1 TO 10 PRINT co(ii) NEXT ii RETURN

7000 REM cs# = 0 REM FOR i7 = 1 TO 3 REM cs# = cs# + se(j7, 1, i7) * se(j7, 2, i7) REM NEXT i7 REM alpha# = ATN(SQR(1 - cs# ^ 2) / cs#) / pi180# IF i1 = 1 THEN alpha# = 6.3# IF i1 = 2 THEN alpha# = 13.2# zappala# = 1 + .013# * alpha# REM I'm correcting intensity instead of log(intensity); rough compensation: zappala# = zappala# * (1 + .2# ^ 2 / 2 / 3) REM the factor 0.013 is for Type M asteroids; this applies to Dotto's paper REM which includes Antigone REM Lambert-based phase correction for homogeneous 2-sphere REM bowell# = ((pi# - alpha#) * cs# - SIN(alpha#) * COS(2 * alpha#)) / pi# REM I keep the distance correction & remove only the phase correction REM Gehrels' "Asteroids II" (& "Asteroids (I)") checked out at ISU REM so I'll use Gehrels & Tedesco, AJ 84:1079, 1979, Table II, p. 1080 REM empirical formula for 6.3deg & 13.2deg, resp. (alpha as given by Dotto) IF i1 = 1 THEN bowell# = 10 ^ (-.4# * .23125#) IF i1 = 2 THEN bowell# = 10 ^ (-.4# * .5148#) s(0, i1) = s(0, i1) * bowell# FOR i = 1 TO 6 s(i, i1) = s(i, i1) * bowell# * zappala# NEXT RETURN

REM Antigone Jul-Aug 1986 DATA 0.1,68,1,61,2,66,4,45,4.5,41.5,8.5,41,9,36,14,28.5,14.5,17,14.6,23,15,26 DATA 21.5,36,22,37,27,40,27.5,40.5,32.5,38.5,33,39,37.5,40.5,38,32.5 DATA 41.9,49.5,42.1,55,42.4,54,42.6,61,47.4,66.5,47.5,68,47.6,68.5 DATA 700,72,800,79,900,74,1000,62,1100,47,1200,21,1300,20,1400,26.5 DATA 1500,36,1600,48,1700,66.5,1800,83,1900,81.5,2000,71.5

REM Antigone Apr 1990 DATA 0,55.5,100,36.5,200,24,300,22.5,400,30.5,500,39,600,53.5,700,68 DATA 49.5,70,51.5,74,52,77.5,56,75.5,56.5,76,59.5,76.5 DATA 60,77,64,68,67,65,68.5,53,85.5,33,86,34 DATA 1300,37,1400,44,1500,52,1600,58.5,1700,63,1800,73,1900,68,2000,57

Hi Joe, if we go back six thousand years, we're talking about the birth of "modern" civilization. The Ubaid period in Mesopotamia (Catal Huyuk, in Anatolia as well?).

Straight away we are into conspiracy theory. The best "contact" myth we have is from Mesopotamia. Oanes, a being who comes out of the sea in the Persian Gulf, is a man who has a human head inside of a fish's head! He instructs the people in maths, art and culture. Later another comes out of the sea to give advice on the flood (explosion of Santorini) Now that's a gift to conspiracists, not for oil; smart money was always on water; but to grab ancient documents, that's why Iraq had to be invaded.

Now scientists in general, are not going to want to be seen as giving any credence to controversial ideas. If someone asks them, "what if Joe Keller is right?" they'll simply say, "wait two years and see."

Whatever happens, they will queue up to stick the knife in you. About the only person I can think of, who has the money to point a telescope where you want, is that millionaire guy Bigelow (sp?) he's supposed to be rather cantankerous but needs as needs must, as they say.

...Oannes, a being who comes out of the sea in the Persian Gulf, is a man who has a human head inside of a fish's head! ...but to grab ancient documents, that's why Iraq had to be invaded. ...

I didn't know that about Oannes. Checking, I find, that some paraphrase Berossus, as saying Oannes was a man "underneath", which sounds like a space suit.

Do you have more evidence, for this rumor I've seen elsewhere too, that the Iraq war is for seizing Sumerian/Babylonian records? I realize that such evidence would be difficult to get, because the operation would be secret.

Hi Joe, I don't like conspiracy theories at all, I think people fall in love with their own and before you know it the tail wags the dog. I did once get involved in the twin towers debate and it was a nightmare. There was a design fault in the towers. The viscoelastic dampers on each floor truss should have been destructively tested, as they could become load bearing. They exploded when the fires cooled down. Well, there's a FEBA alarm system on the sea bed, just of the American coast. That will have recorded the twin towers coming down. the signatures of dampers going off will have been recorded. The building's owners, and the insurers wanted that design fault kept quiet.

On the stealing of artifacts, there was general looting but there was also a steal to order thing going on. Truck loads of stuff left the country. Almost certainly, senior government officials were helping their collector friends get away with loot. Though the reason for the war is to do with clean water supplies for the whole region. Water is the most valuable commodity after all.

On Oanes, yeah it does sound like a space suit, and there's no god and magic ornament to the story.

On page 162 of contactee Stan Romanek's book "Messages", there is a chart showing a planetary alignment which is supposed to signify a date. This chart shows a tenth planet. An astronomer interpreted this as the plutino Eris (aka Xenia). Using a computer program called TheSky he found the alignment corresponds to the date Sept. 21, 2012. But this assumes that "planet 10" is in fact Eris and not an unknown body like Barbarossa. I am going to try to contact the publisher and see if the heliocentric longitudes of the planets in the chart at that date are known. It might be interesting to compare them to the estimated position of Barbarossa in 2012.

On page 162 of contactee Stan Romanek's book "Messages", ...the plutino Eris (aka Xenia). ...Sept. 21, 2012. ...I am going to try to contact the publisher and see if the heliocentric longitudes of the planets in the chart at that date are known. It might be interesting to compare them to the estimated position of Barbarossa in 2012.