Paradoxes Resolved, Origins Illuminated - Requiem for Relativity
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Jim

1832 Posts

Posted - 14 Jan 2010 :  14:06:44  Show Profile  Reply with Quote
Dr Joe, There is a light curve posted at: www.minorplanetobserver.com/pdolc/947_IV. you might find useful.
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Joe Keller

USA
957 Posts

Posted - 14 Jan 2010 :  15:35:01  Show Profile  Reply with Quote
quote:
Originally posted by Stoat

Hi Joe, ... http://adsabs.harvard.edu/full/1977MNRAS.180..567S



Hi Bob!

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!

- Joe
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Joe Keller

USA
957 Posts

Posted - 14 Jan 2010 :  15:52:13  Show Profile  Reply with Quote
quote:
Originally posted by Jim

Dr Joe, There is a light curve posted at: www.minorplanetobserver.com/pdolc/947_IV. you might find useful.



Thanks Jim! The "947" suggests that it is indeed about asteroid 947 Monterosa. I tried the link a few times and it wouldn't work, though.

- Joe Keller
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Stoat

United Kingdom
964 Posts

Posted - 14 Jan 2010 :  17:12:16  Show Profile  Reply with Quote
This guy looks as though he's done or is about to do the lightcurve of 1717
http://members.dslextreme.com/users/rstephens/index_files/Page692.htm
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Jim

1832 Posts

Posted - 14 Jan 2010 :  22:43:54  Show Profile  Reply with Quote
Dr Joe, How about this: www.minorplanet observer.com/pdolc/947_Monterosa.htm
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Jim

1832 Posts

Posted - 14 Jan 2010 :  22:48:22  Show Profile  Reply with Quote
Dr Joe, Its: www.minorplanetobserver.com/pdolc/947_monterosa.htm
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Stoat

United Kingdom
964 Posts

Posted - 15 Jan 2010 :  06:24:22  Show Profile  Reply with Quote
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?
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Joe Keller

USA
957 Posts

Posted - 15 Jan 2010 :  15:25:53  Show Profile  Reply with Quote
quote:
Originally posted by Jim

Dr Joe, Its: www.minorplanetobserver.com/pdolc/947_monterosa.htm



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.

- Joe
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Joe Keller

USA
957 Posts

Posted - 15 Jan 2010 :  16:13:08  Show Profile  Reply with Quote
quote:
Originally posted by Stoat

...he's done or is about to do the lightcurve of 1717
http://members.dslextreme.com/users/rstephens/index_files/Page692.htm



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.

Sincerely,
Joseph C. Keller
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Stoat

United Kingdom
964 Posts

Posted - 17 Jan 2010 :  07:41:53  Show Profile  Reply with Quote
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?
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Stoat

United Kingdom
964 Posts

Posted - 18 Jan 2010 :  05:59:03  Show Profile  Reply with Quote
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"?
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Joe Keller

USA
957 Posts

Posted - 26 Jan 2010 :  21:22:37  Show Profile  Reply with Quote
Update on asteroid lightcurves


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.

Edited by - Joe Keller on 26 Jan 2010 23:41:44
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Joe Keller

USA
957 Posts

Posted - 26 Jan 2010 :  21:27:02  Show Profile  Reply with Quote
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

REM find assoc Legendre fns (Jahnke & Emde, p. 111)
1100 p(0, 0) = 1
REM p(1, 0) = x#
p(2, 0) = .5# * (3 * x# ^ 2 - 1)
REM p(1, 1) = SQR(1 - x# ^ 2)
REM p(2, 1) = -SQR(1 - x# ^ 2) * 3 * x#
REM p(3, 1) = -1.5# * SQR(1 - x# ^ 2) * (5 * x# ^ 2 - 1)
p(2, 2) = 3 * (1 - x# ^ 2)
p(3, 2) = 5 * x# * p(2, 2)
p(3, 3) = -15 * (1 - x# ^ 2) ^ 1.5#
p(4, 3) = 7 * x# * p(3, 3)
RETURN

REM find areas of parts of sphere
1200 th1# = 0: th2# = 15
FOR kk = 16 TO 8 STEP -1
GOSUB 1300
th1# = th2#: th2# = th2# + 10
NEXT kk
RETURN

1300 x# = (COS(th1# * pi180#) - COS(th2# * pi180#)) / 48
FOR jj = 0 TO 23
ax(kk, jj) = x#: ax(16 - kk, jj) = x#
NEXT jj
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

2710
c(1, j) = ss(j, 1, 1)
c(2, j) = ss(j, 1, 2)
RETURN

2712
c(3, j) = ss(j, 1, 1)
c(4, j) = ss(j, 2, 1)
c(5, j) = ss(j, 1, 2)
c(6, j) = ss(j, 2, 2)
RETURN

2714
c(7, j) = ss(j, 1, 1)
c(8, j) = ss(j, 2, 1)
c(9, j) = ss(j, 1, 2)
c(10, j) = ss(j, 2, 2)
RETURN

2750
FOR i = 1 TO 10
FOR j = 1 TO 24
cc(i, j) = cc0(i, j)
NEXT: NEXT
RETURN

REM solve matrix equation by row elimination
2800 i00 = 1: i0 = 2
GOSUB 2900
i00 = 3: i0 = 6
GOSUB 2900
i00 = 7: i0 = 10
GOSUB 2900
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

3200 ff# = .5# * a# * p# ^ 2 + b# * p# + c#
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

4300 t# = t# / 2000
GOTO 4210

4500
cc0(1, 1) = s(0, 1): cc0(2, 1) = s(0, 2)
cc0(3, 1) = s(3, 1): cc0(4, 1) = s(4, 1)
cc0(5, 1) = s(3, 2): cc0(6, 1) = s(4, 2)
cc0(7, 1) = s(5, 1): cc0(8, 1) = s(6, 1)
cc0(9, 1) = s(5, 2): cc0(10, 1) = s(6, 2)
FOR j = 2 TO 24
cc0(1, j) = cc0(1, 1): cc0(2, j) = cc0(2, 1)
cc0(3, j) = cc0(3, 1): cc0(4, j) = cc0(4, 1)
cc0(7, j) = cc0(7, 1): cc0(8, j) = cc0(8, 1)
NEXT j
FOR j = 2 TO 24
FOR i = 1 TO 2
a24# = snsn(i + 1, 1): b24# = cscs(i + 1, 1)
cc0(4 * i + 1, j) = cc0(4 * i + 1, j - 1) * b24# + cc0(4 * i + 2, j - 1) * a24
cc0(4 * i + 2, j) = cc0(4 * i + 2, j - 1) * b24# - cc0(4 * i + 1, j - 1) * a24
NEXT i: NEXT j
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 celestial coords 0h1Aug1986/0h2Apr1990, helio/geo, x/y/z
8000
i = 1: j = 1
GOSUB 8051
GOSUB 8100
j = 2
GOSUB 8052
GOSUB 8100
i = 2: j = 1
GOSUB 8053
GOSUB 8100
j = 2
GOSUB 8054
GOSUB 8100
RETURN

8051 a# = 20: b# = 7: c# = 50: d# = -16: e# = -34
RETURN
8052 a# = 19: b# = 42: c# = 58: d# = -15: e# = -17
RETURN
8053 a# = 11: b# = 34: c# = 16: d# = 10: e# = 12.5#
RETURN
8054 a# = 10: b# = 51: c# = 42: d# = 18: e# = 25
RETURN

REM convert celestial coords to xyz
8100
alpha# = pi180# * (a# * 15 + b# / 4 + c# / 240)
delta# = pi180# * (d# + e# / 60)
se(i, j, 3) = SIN(delta#): cs# = COS(delta#)
se(i, j, 1) = cs# * COS(alpha#)
se(i, j, 2) = cs# * SIN(alpha#)
RETURN

9900 END

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

Edited by - Joe Keller on 27 Jan 2010 16:42:43
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Stoat

United Kingdom
964 Posts

Posted - 27 Jan 2010 :  06:37:10  Show Profile  Reply with Quote
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.

Edited by - Stoat on 27 Jan 2010 06:39:26
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Joe Keller

USA
957 Posts

Posted - 27 Jan 2010 :  16:10:17  Show Profile  Reply with Quote
quote:
Originally posted by Stoat

...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.
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Stoat

United Kingdom
964 Posts

Posted - 28 Jan 2010 :  07:28:38  Show Profile  Reply with Quote
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.
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shando

Canada
208 Posts

Posted - 28 Jan 2010 :  08:30:48  Show Profile  Visit shando's Homepage  Reply with Quote
Have a look at "The Lost Book of Enki" ISBN: 1-59143-037-2
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nemesis

84 Posts

Posted - 28 Jan 2010 :  14:17:02  Show Profile  Reply with Quote
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.
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Joe Keller

USA
957 Posts

Posted - 28 Jan 2010 :  21:29:11  Show Profile  Reply with Quote
quote:
Originally posted by nemesis

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.




Great! Please post the results here!
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Joe Keller

USA
957 Posts

Posted - 28 Jan 2010 :  21:46:33  Show Profile  Reply with Quote
quote:
Originally posted by shando

Have a look at "The Lost Book of Enki" ISBN: 1-59143-037-2



Thanks for the reference!
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