Comet ISON: Hubble's Scientist's Claim?!?

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PostTue Sep 10, 2013 7:18 pm » by Chronicnerd


First off,
Love the news segments from the disclose folks...as well I know reporting news requires one to report what is said whether or not one might or might not believe it. As such... my assumption is that the author of the following article posted on Disclose was doing just this:

http://www.disclose.tv/news/Is_Comet_ISON_a_UFO_Hubbles_scientists_do_a_reality_check/93417

I would like to invite anyone interested in this to contemplate what I would consider the following "Reasonable Questions", but first I am going to point out why I have questions.


Image
"The comet itself does not have three pieces," White wrote. "They are an artifact from adding up the separate exposures. The comet does not look the same in each exposure because both the comet and the Hubble telescope are moving during the exposure. The comet is blurred, just as a picture taken out the window of a moving car will be blurred."


A few things to point out about the above statement by Richard White:
1.) The images he released with his explanation do not show the *full picture*
2.) Is it safe to assume *ALL BODIES* in this picture would have some form of blur?
3.) If the answer is "no only Ison would blur"... then I would have to say he is smokin' something strong.

So, let's take a look at the original image people were using to generate the "multi-prong" looking image everyone was thinking was "odd":

The *ORIGINAL* image can be manipulated and viewed here:
http://hla.stsci.edu/cgi-bin/display?image=hlsp_ison_hst_wfc3_130430_f606w_v1&autoscale=&title=Ison+130430+WFC3+F606W
(you need to move up and right just a bit when it first loads)


This is the *original* section of the very large snapshot taken that many people used as the starting point to get the "triangle" looking structure:
Image

When you hit "darken" a few times you get something like this:
Image

To be very clear, I went ahead and zoomed it a bit:
Image

Now, I invite you to look at the *original*, the *zoomed out darkened version*, and then the zoomed image and think about this for a second...

If the "visual artifacts" are from *both* Comet Ison moving and the telescope moving... then I would be willing to make a *wild* guess that *ALL OTHER BODIES SHOULD BE BLURRED*... even if they were far away...

Why?
Look at the *DISTANCE* between the 3 segments of the "visual artifact"...
Look at the *TAIL* of the Comet and how it actually *forms* in the shape of the 3 segments...
Look at all of the objects around the comet... and notice how the only "blurring" happening is due to *distance* and not *MOTION*...


Now... if I take a snapshot from a moving car ("just as a picture taken out the window of a moving car will be blurred") like such:
Image

Then I would expect *most things* to have some form of *blur* on them...

Do we see *ANY* form of blur on *ANY* of the celestial bodies around the Comet Ison area?

I find Richard White's explanation a bit ~confusing~ and ~perplexing~...

There is *SO* much motion that the comet's perspective blurs to the point where you have such great distances and *SKEWING* of the ~single solid object~?

So... if it is moving...and this is a "side view snapshot" (because the tail actually stretches out we know we are looking from some form of side view)... then I would expect it to blur in the vector of the motion... but not show *very clear* gaps between the various "exposures" they "combined"...

I know another thing that shows this same type of "visual" behavior:
Image

A fan... depending upon your exposure time, you can get things like:
Image


However, this shows a motion of "extensions" from a central point...
Hey... wait a minute...

Image

Yeah... well you guys decide... I just find it "odd" that they had such "blurring" that it is nothing more than the several exposures they combined... when *every other celestial body in the same image has no form of motion artifact...AT ALL*...

But... then again... I don't photograph comets for a living...so maybe there is some special case scenario for comets...

Or... it is a "simple of enough" explanation for most people to easily accept and forget about...

You be the judge...


Cheers,

-CN

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PostTue Sep 10, 2013 9:58 pm » by Nedroj


Good one (:
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PostTue Sep 10, 2013 10:03 pm » by Slith


Impressive post. Great thinking. Wow. :flop:
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PostTue Sep 10, 2013 11:02 pm » by Shaggietrip


Thank you for the post CN. I have explained this myself to people to help them understand. Verbally. I must say your post explains it very well. With the images provided.


Thanks again and stay well. :cheers:
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PostTue Sep 10, 2013 11:17 pm » by Doogle


Looks like another winged planet. Maybe this is the fucker they meant all along. :D

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PostTue Sep 10, 2013 11:23 pm » by Fatdogmendoza


Doogle wrote:Looks like another winged planet. Maybe this is the fucker they meant all along. :D


Image

Doogle. you Peaky Blinder you... :mrgreen:
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all these Social media C@*ts can sick my fkn dick, although it feels good


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PostTue Sep 10, 2013 11:31 pm » by Doogle


Fatdogmendoza wrote:
Doogle wrote:Looks like another winged planet. Maybe this is the fucker they meant all along. :D


Image

Doogle. you Peaky Blinder you... :mrgreen:


Feeky chucker, no razors in my cap, I don't even wear one. I was being more than half serious. I think CNerd has made some pretty good observations.
I did laugh when I heard the geezer explaining the picture the other day, but what do I know(?) I'm still using polaroid.

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PostSat Nov 02, 2013 8:29 pm » by sammygarfield


Chronicnerd wrote:But... then again... I don't photograph comets for a living...so maybe there is some special case scenario for comets...

Or... it is a "simple of enough" explanation for most people to easily accept and forget about...

You be the judge...

Cheers,

-CN


Here is is a video from someone that seems to photograph comets for a living that includes the raw images from the hubble telescope and explanations to how the image came to be: http://www.youtube.com/watch?v=jbLe6kEd6mE

Another example of the same effect is here: http://imageshack.us/a/img845/4871/hu4j.jpg

Does this help clarify anything?

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PostSat Nov 02, 2013 8:43 pm » by The57ironman


sammygarfield wrote:
Chronicnerd wrote:But... then again... I don't photograph comets for a living...so maybe there is some special case scenario for comets...

Or... it is a "simple of enough" explanation for most people to easily accept and forget about...

You be the judge...

Cheers,

-CN


Here is is a video from someone that seems to photograph comets for a living that includes the raw images from the hubble telescope and explanations to how the image came to be: http://www.youtube.com/watch?v=jbLe6kEd6mE

Another example of the same effect is here: http://imageshack.us/a/img845/4871/hu4j.jpg

Does this help clarify anything?

.


:clapper:

.....thank you and welcome to the forum

:cheers: :hiho: Sammy

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PostSun Feb 01, 2015 3:01 pm » by Astronut


Chronicnerd wrote:First off,
If the "visual artifacts" are from *both* Comet Ison moving and the telescope moving... then I would be willing to make a *wild* guess that *ALL OTHER BODIES SHOULD BE BLURRED*... even if they were far away...

Nope. It's due to parallax; ISON is much, much closer to Hubble than the distant stars. And yes, it's actually close enough that you get a differential motion between ISON and the background stars on the celestial sphere. You can regard the stars as fixed points on the celestial sphere and the equatorial coordinate grid, but you cannot do the same for ISON. You must calculate the apparent position of ISON over the time the images were taken, taking into account the parallax induced by Hubble's orbit around earth.
Chronicnerd wrote:Why?
Look at the *DISTANCE* between the 3 segments of the "visual artifact"...
Look at the *TAIL* of the Comet and how it actually *forms* in the shape of the 3 segments...

The distance between the 3 segments is simply due to the time between the exposures. The tail will continue to point in the same direction because it is a translation of ISON's apparent position due to parallax, not a rotation. You can calculate for yourself exactly how much ISON's position should be displaced over time due to its motion and due to parallax from Hubble's orbit. By calculating the apparent position of ISON over time as seen from Hubble you can even recreate the 3 segment image by plugging in the times the images were taken and graphing it out. Here are the equations needed to do the calculation:
The first thing you need are the orbital elements of earth, ISON, and Hubble. Here is a scan of a page containing the orbital elements of earth from the book "Practical Astronomy with Your Calculator" by Peter Duffett-Smith.

http://dropcanvas.com/6eyoa

Hubble Space Telescope Orbit
Orbit Epoch (Julain Day) 2456412.251
Orbital period (years) Tp 0.0001821
Longitude at Epoch 358.4212
Longitude of Perigee 59.6612
eccentricity 0.0002971
semi-major axis (km) 6934.189
inclination 28.4694
Longitude of ascending node 230.3343
Argument of perigee 189.3269

C/2012 S1 (ISON) Orbit
Perihelion Date (Julian Day) 2456625.264
Argument of Perihelion w 345.54102
Longitude of Perihelion w' 641.2278526
Longitude of Ascending Node O 295.6868325
Inclination i 62.16095792
Eccentricity e 1
Perihelion Distance (AU) q 0.012466817

The rest is just math. First we need to find the number of days since the epoch of those elements.

The first step is to convert the date and time of when we want to know ISON's apparent position as seen from Hubble to a julian day number.
y = year
m = month
d = day (fraction of a day)
The procedure to convert this to Julian day number format is as follows:
If m = 1 or 2, subtract 1 from y and add 12 to m, otherwise y' = y and m' = m
If the date is > or = 1582 October 15 calculate:
A = the integer part of y'/100
B = 2-A+integer part of A/4
otherwise B = 0
if y' is negative calculate C = integer of ((365.25 x y')-.75)
otherwise C = integer part of (365.25 x y')
D = integer of (30.6001 x (m' + 1))
JD = B + C + D + d + 1720994.5

For calculating the position of earth relative to ISON we also want the JD for 1990.0 (which is the epoch of earth's orbital elements), which equals 2447891.5. Therefore, the days since epoch 1990.0

which we will call D, equals 8696.9583333335.
D = 8696.9583333335
We will first calculate the position of earth at this timepoint before we do the calculations for ISON.
Mean anomaly = 360/365.242191 * D/Tp + E - W
where
D = days since epoch 1990.0
Tp = Period
E = Longitude at epoch (1990.0)
W = Longitude of perihelion
Refer to the above scanned page for all of these values for Earth. In most cases that will give you a number greater than 360 for any date past 1991 because earth will have gone around the sun more than once, so you need to reduce it down to a value within 360 degrees using the following formula:
d = starting value (8568.4260132099 in this case)
(d/360 - integer of (d/360))*360 = d'
if d' < 0, then 360 + ((d/360 - integer of (d/360))*360) = d'
d' is the new value. This formula will be reused later on, I will simply refer to it as the "360 degree routine."

Next is the true anomaly, v. True anomaly is the angle between the real planet's position and its perihelion position. This calls for an iterative process to solve Kepler's equation (E - e*sin(E)

= M).
We start by making an approximation of E = E0 = M
Find the value of
sigma = E-e*sin(E-M)
where
M = mean anomaly
e = eccentricity
if sigma < 10^-6 radians then take the present value of E as the correct solution, otherwise continue for another iteration
For the next iteration find delta E = sigma/(1-e*cos(E0))
E1 = E0 - delta E
Substitute E1 back in the equation above to find a new value for sigma and repeat this routine as necessary until convering on a solution where sigma < 10^-6 radians.

tan (v/2) = Sqrt((1+e)/(1-e))*tan(E)
where
v = true anomaly
e = eccentricity
E = solution to kepler's equation from the iterative process above

find tan(v/2), take the arctangent and multiply the result by 2. You need to evaluate the numerator and denominator within the arctan function separately in order to do a quadrant disambiguation routine. If the numerator is positive and the denominator negative, add 180 to the result of the arctan function (after converting the result to degrees of course). If the numerator is negative and the denominator positive, then add 360 to the arctan result unless that makes it greater than 360 degrees, in which case add 180 degrees. Finally, if both the numerator and denominator are negative, then add 180 degrees to the result of the arctan function. Otherwise if you have made no other additions and the result is negative, add 180 degrees, or else keep the result of the arctan function without adding or subtracting anything.

Now you have v, the true anomaly.

Now we need to calculate heliocentric longitude, l.
l = (360/365.242191 * D/Tp) + 360/pi * e * sin(360/365.242191 * D/Tp + E - W)
where
D = days since epoch
Tp = Period
e = eccentricity
E = longitude at epoch
W = longitude of the perihelion

Use the 360 degrees routine to get l within 360 degrees.
Now we need the radius vector, that is to say, distance from the sun, r.
r = (a*(1-e^2))/(1+e*cos(v))
where
a = semi-major axis of the orbit
e = eccentricity
v = true anomaly

Now we repeat these calculations for Hubble given the values for Hubble's orbit above.
For Hubble we're dealing with an inclined orbit, so we also need to find the latitude. This is given by
U = Arcsin(sin(L - Omega)*sin(i))
where
L = true longitude of Hubble (same formula as heliocentric longitude for earth)
omega = longitude of the ascending node
i = inclination
Now we need to find the geocentric longitude; true longitude consists partly of values which are inclined to the equator by Hubble's inclination.

L' = v+w'-O
where
v = true anomaly
w' = longitude of perigee
O = argument of perigee
Perform the 360 degree routine on L'
Now we re-orient L' back into the plane of the equator so that we can find the geocentric longitude with respect to the vernal equinox, effectively geocentric right ascension.

geocentric right ascension = arctan((sin(L')*cos(e))/cos(L'))+O
where L' equals the result from above
e = eccentricity
O = argument of perigee
Remember to evaluate the arctan function using the quadrant disambiguation routine described above

Now we need to find the geocentric longitude. We now know the angle of Hubble east of the vernal equinox, but we need to know the number of degrees the vernal equinox is of the 0 line of longitude, the prime meridian, in other words, we need to know Greenwhich mean sidereal time.

GMST = 18.697374558+24.0657098244191*(JD-2451545)
GMST*15 = GMST in degrees
Perform the 360 degree routine on GMST in degrees to get the value within 360 degrees.
Now take the geocentric right ascension and subtract GMST in degrees. This is the eastern

longitude of Hubble. If the value is negative, it's west longitude, if the value is positive it is east longitude. Save this value and the latitude calculated above for later.

Now we need to calculate the position of ISON relative to earth. The orbital elements are listed above. We will approximate ISON's orbit as a parabolic orbit to aid in the ease of calculation.

First find W
W = (0.0364911624/(q*sqrt(q))) * d
where
q = perihelion distance
d = days since perihelion (perihelion date in Julian Days - JD)

Now we need to solve s^3 + 3s - W = 0

First approximate s = s0 = W/3
calculate sigma
sigma = s0^3 +3*s0 - W
if sigma < 10^-6 degrees then take s as the correct value otherwise continue with the iterations and proceed below
s1= (2*s0^3 + W)/(3(s0^2 + 1))
substitute s1 back into the formula above to find sigma and repeat until the value is within the accuracy of sigma <10^-6 degrees. Then take that value of s and proceed
Find the true anomaly
v = 2*arctan (s)
Find the distance from the sun
r = q*(1+s^2)
where
q = perihelion distance in AU
find heliocentric ecliptic longitude
l = v+w'
where
v = true anomaly
w'= longitude of perihelion
find the heliocentric ecliptic latitude
U = Arcsin(sin(l - O)*sin(i))
where
l = heliocentric ecliptic longitude
O = argument of perihelion
i = inclination

Now we need l' which is the heliocentric longitude projected onto the plane of the ecliptic.
l' = arctan((sin(l-O)*cos(i))/cos(l-O))+O
where
l = heliocentric ecliptic longitude
O = argument of perihelion
i = inclination
remember to perform the quadrant disambiguation routine on the arctan function

Now we calculate the geocentric ecliptic longitude (lam) and latitude (beta) of comet ISON. If ISON's radius vector from the sun is less than earth's at the time you are evaluating you can regard it as an inner planet in which case the formula is this:
lam = 180 + Le + arctan ((ri*sin(Le-l'))/(re-ri*sin(Le-l')))
where
Le = heliocentric longitude of earth
re = radius vector of earth
rI = radius vector of ISON
l' = heliocentric longitude of ISON projected onto ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula.

If ISON's radius vector from the sun is greater than earth's at the time you are evaluating you can regard it as an outer planet in which case the formula is this:
lam = arctan((re*sin(l'-Le)/(rI-re*cos(l'-Le)))+l'
where
Le = heliocentric longitude of earth
re = radius vector of earth
rI = radius vector of ISON
l' = heliocentric longitude of ISON projected onto ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula.

beta is calculated with the following formula:
beta = arctan((r'*tan(U)*sin(lam-l'))/(re*sin(l'-Le))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of ISON projected onto ecliptic
U = heliocentric latitude for ISON
r'= radius vector of ISON

Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula.

Distance of ISON from earth
reI=sqrt(rI^2+re^2-2*re*ri*cos(l'-Le)*cos(U))
where
Le = heliocentric longitude of earth
re = radius vector of earth
rI = radius vector of ISON
l' = heliocentric longitude of ISON projected onto ecliptic
U = heliocentric latitude for ISON

For the next step we will need the obliquity of the ecliptic. This can be calculated with the following formula, which will need time "t" centuries from 2000.0. We can calculate t thusly:
t = ((JD-2451545)/36525)
where JD = julian day number
t = 0.1380823637
Obliquity of the ecliptic (Obl) can then be calculated with this formula:
Obl = 23.43928-0.013*t+0.555*(10^-6)(t^3)-0.0141(10^-8)*(t^4)

Now we just have to convert these coordinates from ecliptic to equatorial and we will have the geocentric equatorial coordinates of ISON at this timepoint.
Right ascension = arctan((sin(lam)*cos(Obl)-tan(beta)*sin(Obl))/cos(lam))
where
lam = geocentric ecliptic longitude
beta = geocentric ecliptic latitude
obl = obliquity of the ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the

arctan function in this formula.

declination = arcsin(sin(beta)*cos(obl)+cos(beta)*sin(obl)*sin(lam))
where
lam = geocentric ecliptic longitude
beta = geocentric ecliptic latitude
obl = obliquity of the ecliptic

Now we have the geocentric coordinates for ISON, but we need to account for the parallax induced by

Hubble's orbit.

First find u
u = atan(0.996647*tan(U))
where U = geocentric latitude of Hubble

Next find p*sin(theta')
p*sin(theta') = 0.996647*sin(u)+(r/6378140)*sin(theta)
where
u = u above
r = altitude of hubble (convert to meters)
theta = geocentric lattitude of hubble

find p*cos(theta')
p*cos(theta') = cos(u)+r/6378140*cos(theta)
where
u = u above
r = altitude of hubble (convert to meters)
theta = geocentric lattitude of hubble

pi = (atan((6378140+r)/149597870700)*3600)/reI)/3600
where
r = altitude of Hubble (in meters)
reI = distance of ISON from earth

LST = (GMST*15)+Lh
where
GMST = Greenwhich mean sidereal time
Lh = geocentric longitude of Hubble
Perform the 360 degree routine to get LST in degrees for the equation below

delta RA = atan((p*cos(theta')*sin(pi)*sin(LST-RA))/(cos(declination)-p*sin(theta')*sin(pi)*cos(LST-RA))
where
RA = geocentric right ascension of ISON declination = geocentric declination of ISON.
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula.

subtract delta RA from the geocentric right ascension of ISON to get the parallax corrected right ascension of ISON which we will call the hubble centric RA

gamma = atan(tan(p*sin(theta')/p*cos(theta'))cos(.5*delta RA)(1/cos(LST-.5*(geocentric RA + hubble centric RA)))

delta declination = (p*sin(theta')*sin(pi)*sin(gamma-geocentric declination))/(sin(gamma)-p*sin(theta')*sin(pi)*cos(gamma-geocentric declination))

subtract delta declination from the geocentric declination to get the parallax corrected declination of ISON which we will call the hubble centric declination. Now we have both the hubble centric RA and hubble centric declination for this point in time. If you repeat these equations over multiple points in time covering the imaging times for the Hubble images you can generate a list of coordinates for ISON over the imaging session and graph the predicted trail of ISON as seen from Hubble over that period of time. Doing so results in the following graph which matches with the images taken by Hubble:
Image


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