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There REALLY IS something strange in western sky around sunset...!!!

 
Dr. AstroModerator
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10/23/2013 02:00 PM

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Re: There REALLY IS something strange in western sky around sunset...!!!
...


Or maybe you're misreading your star chart. I provided a you with a photo of Ophiuchus, Venus, and all. I see nothing unidentified in it, do you? Venus just happens to be in the constellation you yourself mentioned, I do not believe that to be a coincidence.
 Quoting: Dr. Astro


It's not Venus, I've checked several time, with software, charts..etc. Totally different hours, movement, description.
Thanks anyway, but it ain't Venus. I just don't buy that. It doesn't matter what we say here, time will tell
 Quoting: pstrusi


Wrong. It's not "totally different hours" - it's in the same constellation, Ophiuchus. Would you like me to prove that to you mathematically?
 Quoting: Dr. Astro


I'm not doubting you, i would really like to see the calculations if possible and not too much trouble.
 Quoting: the gas man


Totally possible, no problem :). For the sake of simplicity and to make the post more manageable though I will stick to "back of the envelope" type calculations. That means simply using the mean orbital elements of the planets, ignoring perturbations, etc. I normally use a more complex series of formulas that include the major perturbations and other factors in my sun calculations, but since I intend to detail it in a post equation by equation rather than just dumping a excel sheet on you, it would take far too much time and forum space to do all that. Even the "back of the envelope" complexity is much higher than what you normally find in a GLP post and in any case it's good enough to get you an answer that is accurate to within a fraction of a degree or so. Finally, I will show you how you can take the coordinates from the answer and identify the constellation that contains those coordinates, not using "google sky" or some other program that "could be part of some conspiracy," but with a good old fashioned paper star atlas.

Last Edited by Astromut on 10/23/2013 02:01 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
bump
Be ye the master of all that surrounds thee.
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10/23/2013 04:59 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
Calling Dr. Astro...

Rumours of Ison outbursting !!!!

Can you confirm plz.
Dr. AstroModerator
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10/23/2013 05:01 PM

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Re: There REALLY IS something strange in western sky around sunset...!!!
Calling Dr. Astro...

Rumours of Ison outbursting !!!!

Can you confirm plz.
 Quoting: Anonymous Coward 43931421


I haven't seen any indication of that thus far. It appears to be following revised-downward projections of its brightness.

Last Edited by Astromut on 10/23/2013 05:02 PM
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Raymantheheretic

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10/23/2013 05:31 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
Look, I believe that Planet X exists, but all of you people saying you're seeing some big, red object need to post some proof. A picture, video, both. And not just one day, do it daily. You can't expect anyone to take you seriously without proof. And you shouldn't be posting on here if you're not going to post proof.

One guy even said it was as bright as, or brighter, than Venus. I mean really..come on now lol.
 Quoting: Anonymous Coward 1768581

My favorite. Too bad the original audio is gone though;

Same observer different day;

This is the one that kicked off the thread. Ho hum compared to the others imo but with the claim by the witness that it has been observed repeatedly in the same area low on the horizon and how tiny and distant it appears with out zoom it is intriguing;

Yet another one of a 'contrail' lacking the symmetry one would expect to see. Hmmm... ;


Photos from different days taken by SunnnyDaze;
OK here is what I got, it is a little disappointing, in print, but it was well worth the hassel to see this thing through the binoculars ... next week I will have the battery,for the attached camera ... may go buy it tomorrow, haha can't wait til you guys see this, it looked just like that vid (with the lady talking in the background who kept loosing focus) ... and exactly what JB was describing (blood chilling) sure glad he saw it first and warned us

here is when it first appeared to me (all of a sudden)
and what you would see "naked eye" from where I live WSW just to the right of Pike's Peak
you have to look really close (or magnify) to see it
[link to i272.photobucket.com]
this one is closer but still with landmarks
[link to i272.photobucket.com]
closer a little closer
[link to i272.photobucket.com]
a minute later
[link to i272.photobucket.com]
enhanced (cropped and enlarged only)
[link to i272.photobucket.com]
super enhanced every witch way but loose -trying to catch that V - like I said very disappoint not to be able to capture what I saw
[link to i272.photobucket.com]

so sorry but my camera can not zoom in enough... but this is good enough for you to know that it is out there ... and where it is, so someone with a real camera can get a GOOD picture ... but like I said, through the binoc s - it was OUTRAGEOUS!
 Quoting: SunnyDaze

sorry, forgot to change the date/time/stamp - still showing Aug instead of October - but the time and day is right

these are not to show you anything other than its location on my horizon ... as you can see, it appears very close to the same place - above the mt and in the sky

it is very faint in this shot - look to the rt of the tel pole
10/06/13
[link to i272.photobucket.com]

10/08/13
[link to i272.photobucket.com]

and in this one it is also faint, (don't be distracted by the chemtrails to the left ... it is almost in the middle of the shot (slightly to the rt) and appears to have moved left from the position in the previous shots - but so did the Sun - and it seems to be the same distance from the sun - but that is the best shot I got of it tonight - within two minutes it was cover by the chem-clouds

10/19/13
[link to i272.photobucket.com]

and Astro and Geo, just for you - fyi, this is what a contrail looks like with my camera

[link to i272.photobucket.com]

as you can see, there is a clear difference - and I know what a contrail looks like
 Quoting: SunnyDaze


Those videos are just the best examples of what I have gleaned so far from the first 60 pages sans Astro & the whole wrecking crew going to work on this thread full time.

Sure it's easy enough to cry contrail but imo that is as about as plausible as claiming a hard to see 'planet' is just bright enough to engulf a power line in front of it and not be considered merely lens flare.

Last Edited by Raymantheheretic on 10/23/2013 05:37 PM
#Geomagnetic_Storm#

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10/23/2013 05:48 PM

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Re: There REALLY IS something strange in western sky around sunset...!!!
Calling Dr. Astro...

Rumours of Ison outbursting !!!!

Can you confirm plz.
 Quoting: Anonymous Coward 43931421


I think you are talking about Comet Linear...


[link to www.universetoday.com]
Geoshill


Link to my Gaming Channel….
[link to m.youtube.com (secure)]
Karlos

User ID: 46194112
United Kingdom
10/23/2013 06:09 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
Calling Dr. Astro...

Rumours of Ison outbursting !!!!

Can you confirm plz.
 Quoting: Anonymous Coward 43931421


Not ISON, but one of the LINEARs is currently doing so.
C/2012 X1, to be exact.
It went from mag 14 to around 8.5 in only a day or two. Fingers crossed for another Holmes-like manifestation.
It's currently in Coma Berenices, so will be a viable target for me by about 5am. So long as the sky stays clear, I'm gonna try to bag it visually with a 8" SCT

It's on 'Spaceweather' today.


edited to add...
Oops, sorry Geo, didn't notice your post there...

Last Edited by Karlos on 10/23/2013 06:10 PM
Amityschild

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10/23/2013 06:45 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
New post on topic: Thread: INCOMING: From the Ends of the Heaven - What is in the Western Sky?

INCOMING: From the Ends of the Heaven - What is in the Western Sky?
Dr. AstroModerator
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10/23/2013 08:03 PM

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Re: There REALLY IS something strange in western sky around sunset...!!!
...


Or maybe you're misreading your star chart. I provided a you with a photo of Ophiuchus, Venus, and all. I see nothing unidentified in it, do you? Venus just happens to be in the constellation you yourself mentioned, I do not believe that to be a coincidence.
 Quoting: Dr. Astro


It's not Venus, I've checked several time, with software, charts..etc. Totally different hours, movement, description.
Thanks anyway, but it ain't Venus. I just don't buy that. It doesn't matter what we say here, time will tell
 Quoting: pstrusi


Wrong. It's not "totally different hours" - it's in the same constellation, Ophiuchus. Would you like me to prove that to you mathematically?
 Quoting: Dr. Astro


I'm not doubting you, i would really like to see the calculations if possible and not too much trouble.
 Quoting: the gas man


Alright, so here we go. The first thing you need are the orbital elements of the planets, particularly earth and Venus. We shall simplify things down to "back of the envelope" calculations which will get us within a fraction of a degree of the right answer. We won't concern ourselves with all the planetary perturbations calculations, it would take me all night to type all that out, instead we'll focus on the mean orbital elements of the planets freed of perturbations for the epoch 1990.0. Here is a scan of a page containing those elements from the book "Practical Astronomy with Your Calculator" by Peter Duffett-Smith.
[link to dropcanvas.com]
The rest is just math. First we need to find the number of days since the epoch of those elements. In fact we can find the apparent position of Venus for a given day and time by converting to fractions of a day since epoch 1990.0. Let's use last night for this demonstration, say at 23:00 UT, which was just after sunset for my location. So October 22, 2013 23:00 UT. Let's convert this date and time to the Julian day number.
y = year
m = month
d = day (fraction of a day)
So here we have
y = 2013
m = 10
d = 22.95833
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
In this case JD = 2456588.45833333
We also want the JD for 1990.0, 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 Venus.
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 both Earth and Venus. In this case that equates to 8568.4260132099 degrees. Now obviously that's greater than 360, 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 "d' routine."
Mean anomaly M for earth (which I will call Me) then = 288.4260132098 degrees
Me = 288.4260132098
Next is the true anomaly, v. True anomaly is the angle between the real planet's position and its perihelion position. Normally this calls for an iterative process to solve Kepler's equation, but we will simplify things here by approximating the value with the equation of the center.
v = M + 360/pi * e * sin(M)
where
M = mean anomaly
e = eccentricity
We know the mean anomaly for earth above and eccentricity for earth = 0.016713. Therefore, earth's true anomaly (which I will call ve) equals 286.6090301972 degrees.
ve = 286.6090301972
Now we need to calculate heliocentric longitude, l. For earth I will call it Le.
l = (360/365.242191 * D/Tp) + 360/pi * e * sin(360/365.242191 * D/Tp + E - W)
where
D = days since 1990.0
Tp = Period
e = eccentricity
E = longitude at epoch 1990.0
W = longitude of the perihelion
Le = 8669.3774431972
Use the d' routine to get Le = 29.3774431972 degrees
Now we need the radius vector, that is to say, distance from the sun, r. For earth I will call this re.
r = (a*(1-e^2))/(1+e*cos(v))
where
a = semi-major axis of the orbit
e = eccentricity
v = true anomaly
re = 0.9949674832 AU
Now we repeat these calculations for Venus given the values from the table for Venus.
Mean anomaly for Venus = Mv = 210.6743831713 degrees
True anomaly for Venus = vv = 210.2781423101 degrees
heliocentric longitude for Venus = Lv = 341.7083783101 degrees
radius vector for Venus = rv = 0.7275574534 AU
Now we need the heliocentric latitude for Venus. This is given by
U = Arcsin(sin(Lv - Omega)*sin(i))
where
Lv = heliocentric longitude of Venus
omega = longitude of the ascending node of Venus
i = inclination of Venus
U = -3.3822083713 degrees
Now we need l' and r' which are the heliocentric longitude and radius vector of Venus respectively projected onto the plane of the ecliptic.
l' = arctan((sin(Lv-Omega)*cos(i))/cos(Lv-omega))+omega
Now 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. 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.
Following this routine, the result you should get for l' is 341.7048475079 degrees.
l' = 341.7048475079 degrees
r' is a simple one, it is simply rv * cos(U)
where
rv = radius vector of Venus
U = heliocentric latitude of Venus
r' = 0.7262901907 AU
Now we calculate the geocentric ecliptic longitude (lambda) and latitude (beta) of Venus. Since Venus is an inner planet the formula is this:
lambda = 180 + Le + arctan ((re*sin(l' - Le))/(re*sin(l'-Le)))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of venus projected onto ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
lambda = 256.0825668728 degrees
beta is calculated with the following formula:
beta = arctan((r'*tan(U)*sin(lambda-l'))/(re*sin(l'-Le))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of venus projected onto ecliptic
U = heliocentric latitude for Venus
r'= radius vector of Venus projected onto the plane of the ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
beta = -3.3298312039 degrees
For the final 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)
Obl = 23.43748
Now we just have to convert these coordinates from ecliptic to equatorial and we will have the geocentric equatorial coordinates of Venus at this timepoint.
Right ascension = arctan((sin(lambda)*cos(Obl)-tan(beta)*sin(Obl))/cos(lambda))​
where
lambda = 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. The result in this case should yield
right ascension = 253.8713410337 degrees. If you convert this to hours, minutes, seconds you get 16hr 55m 29.12s
declination = arcsin(sin(beta)*cos(obl)+cos(beta)*sin(obl)*sin(lambda))
where
lambda = geocentric ecliptic longitude
beta = geocentric ecliptic latitude
obl = obliquity of the ecliptic
declination = -26.0216957613 degrees. If you convert this to degrees, minutes, seconds you get -26d 1' 18.1"

So after all that, here are the calculated coordinates of Venus, which is accurate to within a few arcminutes in right ascension and about a quarter of a degree in declination (primarily because we didn't calculate for perturbations from the planets).
RA: 16hr 55m 29.12s
Dec: -26d 1' 18.1"
These are geocentric coordinates, but topocentric parallax will be insignificant at the distance of Venus, at least for naked eye observations like this (it only amounts to a dozen arcseconds or so, which you can't see by eye) so this is good enough as a general case for all observers. Now we just need to figure out which constellation contains these coordinates. I'm going to go to the university library and track down a star atlas. I'll use a 1991 copy of the Cambridge star atlas 2000.0 if it's not checked out (2000.0 is the epoch the atlas is in, not the date of publication). I'll shoot a video of it and upload it later tonight, but for now here are the calculations. Enjoy! :)
astrobanner2
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10/23/2013 08:12 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
Look, I believe that Planet X exists, but all of you people saying you're seeing some big, red object need to post some proof. A picture, video, both. And not just one day, do it daily. You can't expect anyone to take you seriously without proof. And you shouldn't be posting on here if you're not going to post proof.

One guy even said it was as bright as, or brighter, than Venus. I mean really..come on now lol.
 Quoting: Anonymous Coward 1768581

My favorite. Too bad the original audio is gone though;

Same observer different day;

This is the one that kicked off the thread. Ho hum compared to the others imo but with the claim by the witness that it has been observed repeatedly in the same area low on the horizon and how tiny and distant it appears with out zoom it is intriguing;

Yet another one of a 'contrail' lacking the symmetry one would expect to see. Hmmm... ;


Photos from different days taken by SunnnyDaze;
OK here is what I got, it is a little disappointing, in print, but it was well worth the hassel to see this thing through the binoculars ... next week I will have the battery,for the attached camera ... may go buy it tomorrow, haha can't wait til you guys see this, it looked just like that vid (with the lady talking in the background who kept loosing focus) ... and exactly what JB was describing (blood chilling) sure glad he saw it first and warned us

here is when it first appeared to me (all of a sudden)
and what you would see "naked eye" from where I live WSW just to the right of Pike's Peak
you have to look really close (or magnify) to see it
[link to i272.photobucket.com]
this one is closer but still with landmarks
[link to i272.photobucket.com]
closer a little closer
[link to i272.photobucket.com]
a minute later
[link to i272.photobucket.com]
enhanced (cropped and enlarged only)
[link to i272.photobucket.com]
super enhanced every witch way but loose -trying to catch that V - like I said very disappoint not to be able to capture what I saw
[link to i272.photobucket.com]

so sorry but my camera can not zoom in enough... but this is good enough for you to know that it is out there ... and where it is, so someone with a real camera can get a GOOD picture ... but like I said, through the binoc s - it was OUTRAGEOUS!
 Quoting: SunnyDaze

sorry, forgot to change the date/time/stamp - still showing Aug instead of October - but the time and day is right

these are not to show you anything other than its location on my horizon ... as you can see, it appears very close to the same place - above the mt and in the sky

it is very faint in this shot - look to the rt of the tel pole
10/06/13
[link to i272.photobucket.com]

10/08/13
[link to i272.photobucket.com]

and in this one it is also faint, (don't be distracted by the chemtrails to the left ... it is almost in the middle of the shot (slightly to the rt) and appears to have moved left from the position in the previous shots - but so did the Sun - and it seems to be the same distance from the sun - but that is the best shot I got of it tonight - within two minutes it was cover by the chem-clouds

10/19/13
[link to i272.photobucket.com]

and Astro and Geo, just for you - fyi, this is what a contrail looks like with my camera

[link to i272.photobucket.com]

as you can see, there is a clear difference - and I know what a contrail looks like
 Quoting: SunnyDaze


Those videos are just the best examples of what I have gleaned so far from the first 60 pages sans Astro & the whole wrecking crew going to work on this thread full time.

Sure it's easy enough to cry contrail but imo that is as about as plausible as claiming a hard to see 'planet' is just bright enough to engulf a power line in front of it and not be considered merely lens flare.
 Quoting: Raymantheheretic



Thanks, Rayman!!
Anonymous Coward
User ID: 48778624
Brazil
10/23/2013 08:16 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
LINEAR?!! LOL its magnitude changed just 1 to 2 days ago!
this thread is about witnessings from weeks, even 4 years ago
Eleven-15

User ID: 41838234
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10/23/2013 08:26 PM

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Re: There REALLY IS something strange in western sky around sunset...!!!
astro, you may believe anything that you want ... and I may as well - but there is no way, that anyone can tell what that object is - using MY pictures ...

I looked at every dot that appeared in the sky for over a half an hour ... some were clouds, and most were con-trail - which were easily distinguishable but what I saw - was neither ... neither

you did not see what I saw - that's as plain and simple as I can get - the picture I took, just to verify its location could not possibly be used to determine what it is - and that sort of judgment call on your part, is what is causing you to loose all your credibility on GLP

believe it or not - it is not a contrail, because you say so

IMO the only one who has failed to grasp the significance of a post as unique as this one has been is you - because you have completly closed your mind to the possibility that there is something out there - that isn't on the sky-map - THAT ISN'T a CONTRAIL ... while the rest of us are obviously OPEN to that possibility and being rewarded for our efforts - no matter what you choose to believe - or not
 Quoting: SunnyDaze


He has always had credibility. That is just you maybe...
 Quoting: #Geomagnetic_Storm#


kissup
Anyone who has read GLP for any period of time has acquired good knowledge mixed in with the 99% noise.

The trick is learning to filter out the noise.

Quote from Anonymous Coward (AC)
Anonymous Coward
User ID: 1702426
United States
10/23/2013 08:33 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
wtf is Astronut still doing posting wall of disinfo gibberish in this thread? Holy shit that is some straight up psycho shit yo.

Astronut is like the Harry Potter guy in this clip, with transhuman metrosexual nerd glasses.


[link to www.youtube.com (secure)]
Anonymous Coward
User ID: 39528572
Colombia
10/23/2013 08:37 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
wtf is Astronut still doing posting wall of disinfo gibberish in this thread? Holy shit that is some straight up psycho shit yo.

Astronut is like the Harry Potter guy in this clip, with transhuman metrosexual nerd glasses.


[link to www.youtube.com (secure)]
 Quoting: Anonymous Coward 1702426


rofl

Awesome
Anonymous Coward
User ID: 48829352
Brazil
10/23/2013 08:47 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
Look, I believe that Planet X exists, but all of you people saying you're seeing some big, red object need to post some proof. A picture, video, both. And not just one day, do it daily. You can't expect anyone to take you seriously without proof. And you shouldn't be posting on here if you're not going to post proof.

One guy even said it was as bright as, or brighter, than Venus. I mean really..come on now lol.
 Quoting: Anonymous Coward 1768581

My favorite. Too bad the original audio is gone though;

Same observer different day;

This is the one that kicked off the thread. Ho hum compared to the others imo but with the claim by the witness that it has been observed repeatedly in the same area low on the horizon and how tiny and distant it appears with out zoom it is intriguing;

Yet another one of a 'contrail' lacking the symmetry one would expect to see. Hmmm... ;


Photos from different days taken by SunnnyDaze;
OK here is what I got, it is a little disappointing, in print, but it was well worth the hassel to see this thing through the binoculars ... next week I will have the battery,for the attached camera ... may go buy it tomorrow, haha can't wait til you guys see this, it looked just like that vid (with the lady talking in the background who kept loosing focus) ... and exactly what JB was describing (blood chilling) sure glad he saw it first and warned us

here is when it first appeared to me (all of a sudden)
and what you would see "naked eye" from where I live WSW just to the right of Pike's Peak
you have to look really close (or magnify) to see it
[link to i272.photobucket.com]
this one is closer but still with landmarks
[link to i272.photobucket.com]
closer a little closer
[link to i272.photobucket.com]
a minute later
[link to i272.photobucket.com]
enhanced (cropped and enlarged only)
[link to i272.photobucket.com]
super enhanced every witch way but loose -trying to catch that V - like I said very disappoint not to be able to capture what I saw
[link to i272.photobucket.com]

so sorry but my camera can not zoom in enough... but this is good enough for you to know that it is out there ... and where it is, so someone with a real camera can get a GOOD picture ... but like I said, through the binoc s - it was OUTRAGEOUS!
 Quoting: SunnyDaze

sorry, forgot to change the date/time/stamp - still showing Aug instead of October - but the time and day is right

these are not to show you anything other than its location on my horizon ... as you can see, it appears very close to the same place - above the mt and in the sky

it is very faint in this shot - look to the rt of the tel pole
10/06/13
[link to i272.photobucket.com]

10/08/13
[link to i272.photobucket.com]

and in this one it is also faint, (don't be distracted by the chemtrails to the left ... it is almost in the middle of the shot (slightly to the rt) and appears to have moved left from the position in the previous shots - but so did the Sun - and it seems to be the same distance from the sun - but that is the best shot I got of it tonight - within two minutes it was cover by the chem-clouds

10/19/13
[link to i272.photobucket.com]

and Astro and Geo, just for you - fyi, this is what a contrail looks like with my camera

[link to i272.photobucket.com]

as you can see, there is a clear difference - and I know what a contrail looks like
 Quoting: SunnyDaze


Those videos are just the best examples of what I have gleaned so far from the first 60 pages sans Astro & the whole wrecking crew going to work on this thread full time.

Sure it's easy enough to cry contrail but imo that is as about as plausible as claiming a hard to see 'planet' is just bright enough to engulf a power line in front of it and not be considered merely lens flare.
 Quoting: Raymantheheretic


hesright

If you believe the first video is a contrail, you must be out of your mind. Every kid recognize a contrail, come on!

This is something that nobody knows what it is.
Dr. AstroModerator
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10/23/2013 09:06 PM

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Re: There REALLY IS something strange in western sky around sunset...!!!
wtf is Astronut still doing posting wall of disinfo gibberish in this thread? Holy shit that is some straight up psycho shit yo.

Astronut is like the Harry Potter guy in this clip, with transhuman metrosexual nerd glasses.


[link to www.youtube.com (secure)]
 Quoting: Anonymous Coward 1702426


Just because you're not intelligent enough to understand simple math does not make it "psycho shit" homie. Ya feel me yo?
astrobanner2
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10/23/2013 09:19 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
...


It's not Venus, I've checked several time, with software, charts..etc. Totally different hours, movement, description.
Thanks anyway, but it ain't Venus. I just don't buy that. It doesn't matter what we say here, time will tell
 Quoting: pstrusi


Wrong. It's not "totally different hours" - it's in the same constellation, Ophiuchus. Would you like me to prove that to you mathematically?
 Quoting: Dr. Astro


I'm not doubting you, i would really like to see the calculations if possible and not too much trouble.
 Quoting: the gas man


Alright, so here we go. The first thing you need are the orbital elements of the planets, particularly earth and Venus. We shall simplify things down to "back of the envelope" calculations which will get us within a fraction of a degree of the right answer. We won't concern ourselves with all the planetary perturbations calculations, it would take me all night to type all that out, instead we'll focus on the mean orbital elements of the planets freed of perturbations for the epoch 1990.0. Here is a scan of a page containing those elements from the book "Practical Astronomy with Your Calculator" by Peter Duffett-Smith.
[link to dropcanvas.com]
The rest is just math. First we need to find the number of days since the epoch of those elements. In fact we can find the apparent position of Venus for a given day and time by converting to fractions of a day since epoch 1990.0. Let's use last night for this demonstration, say at 23:00 UT, which was just after sunset for my location. So October 22, 2013 23:00 UT. Let's convert this date and time to the Julian day number.
y = year
m = month
d = day (fraction of a day)
So here we have
y = 2013
m = 10
d = 22.95833
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
In this case JD = 2456588.45833333
We also want the JD for 1990.0, 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 Venus.
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 both Earth and Venus. In this case that equates to 8568.4260132099 degrees. Now obviously that's greater than 360, 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 "d' routine."
Mean anomaly M for earth (which I will call Me) then = 288.4260132098 degrees
Me = 288.4260132098
Next is the true anomaly, v. True anomaly is the angle between the real planet's position and its perihelion position. Normally this calls for an iterative process to solve Kepler's equation, but we will simplify things here by approximating the value with the equation of the center.
v = M + 360/pi * e * sin(M)
where
M = mean anomaly
e = eccentricity
We know the mean anomaly for earth above and eccentricity for earth = 0.016713. Therefore, earth's true anomaly (which I will call ve) equals 286.6090301972 degrees.
ve = 286.6090301972
Now we need to calculate heliocentric longitude, l. For earth I will call it Le.
l = (360/365.242191 * D/Tp) + 360/pi * e * sin(360/365.242191 * D/Tp + E - W)
where
D = days since 1990.0
Tp = Period
e = eccentricity
E = longitude at epoch 1990.0
W = longitude of the perihelion
Le = 8669.3774431972
Use the d' routine to get Le = 29.3774431972 degrees
Now we need the radius vector, that is to say, distance from the sun, r. For earth I will call this re.
r = (a*(1-e^2))/(1+e*cos(v))
where
a = semi-major axis of the orbit
e = eccentricity
v = true anomaly
re = 0.9949674832 AU
Now we repeat these calculations for Venus given the values from the table for Venus.
Mean anomaly for Venus = Mv = 210.6743831713 degrees
True anomaly for Venus = vv = 210.2781423101 degrees
heliocentric longitude for Venus = Lv = 341.7083783101 degrees
radius vector for Venus = rv = 0.7275574534 AU
Now we need the heliocentric latitude for Venus. This is given by
U = Arcsin(sin(Lv - Omega)*sin(i))
where
Lv = heliocentric longitude of Venus
omega = longitude of the ascending node of Venus
i = inclination of Venus
U = -3.3822083713 degrees
Now we need l' and r' which are the heliocentric longitude and radius vector of Venus respectively projected onto the plane of the ecliptic.
l' = arctan((sin(Lv-Omega)*cos(i))/cos(Lv-omega))+omega
Now 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. 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.
Following this routine, the result you should get for l' is 341.7048475079 degrees.
l' = 341.7048475079 degrees
r' is a simple one, it is simply rv * cos(U)
where
rv = radius vector of Venus
U = heliocentric latitude of Venus
r' = 0.7262901907 AU
Now we calculate the geocentric ecliptic longitude (lambda) and latitude (beta) of Venus. Since Venus is an inner planet the formula is this:
lambda = 180 + Le + arctan ((re*sin(l' - Le))/(re*sin(l'-Le)))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of venus projected onto ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
lambda = 256.0825668728 degrees
beta is calculated with the following formula:
beta = arctan((r'*tan(U)*sin(lambda-l'))/(re*sin(l'-Le))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of venus projected onto ecliptic
U = heliocentric latitude for Venus
r'= radius vector of Venus projected onto the plane of the ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
beta = -3.3298312039 degrees
For the final 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)
Obl = 23.43748
Now we just have to convert these coordinates from ecliptic to equatorial and we will have the geocentric equatorial coordinates of Venus at this timepoint.
Right ascension = arctan((sin(lambda)*cos(Obl)-tan(beta)*sin(Obl))/cos(lambda))​
where
lambda = 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. The result in this case should yield
right ascension = 253.8713410337 degrees. If you convert this to hours, minutes, seconds you get 16hr 55m 29.12s
declination = arcsin(sin(beta)*cos(obl)+cos(beta)*sin(obl)*sin(lambda))
where
lambda = geocentric ecliptic longitude
beta = geocentric ecliptic latitude
obl = obliquity of the ecliptic
declination = -26.0216957613 degrees. If you convert this to degrees, minutes, seconds you get -26d 1' 18.1"

So after all that, here are the calculated coordinates of Venus, which is accurate to within a few arcminutes in right ascension and about a quarter of a degree in declination (primarily because we didn't calculate for perturbations from the planets).
RA: 16hr 55m 29.12s
Dec: -26d 1' 18.1"
These are geocentric coordinates, but topocentric parallax will be insignificant at the distance of Venus, at least for naked eye observations like this (it only amounts to a dozen arcseconds or so, which you can't see by eye) so this is good enough as a general case for all observers. Now we just need to figure out which constellation contains these coordinates. I'm going to go to the university library and track down a star atlas. I'll use a 1991 copy of the Cambridge star atlas 2000.0 if it's not checked out (2000.0 is the epoch the atlas is in, not the date of publication). I'll shoot a video of it and upload it later tonight, but for now here are the calculations. Enjoy! :)
 Quoting: Dr. Astro

So, after this humongous amount of mathematical jargon effort, your precision is within a few arc-minutes and a quarter of grade (i believe this is 15 arc minutes?) for 20+ years interval,after, if, you find some old astronomy book reference?
how exactly this calculus is more reliable than the method used by, for example stellarium, VSOP87, wich has an accuracy of 1 ARC-SECOND between 2000 BC and 6000 AD for Venus, this is, at least, 60 times more accurate than what you are doing for a far greater interval of time and is constantly updated by and for mathematicians, astronomers and planetariums all around the world since is open source?
Inquiring minds want to know...
the gas man

User ID: 42208693
United States
10/23/2013 09:21 PM

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Re: There REALLY IS something strange in western sky around sunset...!!!
...


It's not Venus, I've checked several time, with software, charts..etc. Totally different hours, movement, description.
Thanks anyway, but it ain't Venus. I just don't buy that. It doesn't matter what we say here, time will tell
 Quoting: pstrusi


Wrong. It's not "totally different hours" - it's in the same constellation, Ophiuchus. Would you like me to prove that to you mathematically?
 Quoting: Dr. Astro


I'm not doubting you, i would really like to see the calculations if possible and not too much trouble.
 Quoting: the gas man


Alright, so here we go. The first thing you need are the orbital elements of the planets, particularly earth and Venus. We shall simplify things down to "back of the envelope" calculations which will get us within a fraction of a degree of the right answer. We won't concern ourselves with all the planetary perturbations calculations, it would take me all night to type all that out, instead we'll focus on the mean orbital elements of the planets freed of perturbations for the epoch 1990.0. Here is a scan of a page containing those elements from the book "Practical Astronomy with Your Calculator" by Peter Duffett-Smith.
[link to dropcanvas.com]
The rest is just math. First we need to find the number of days since the epoch of those elements. In fact we can find the apparent position of Venus for a given day and time by converting to fractions of a day since epoch 1990.0. Let's use last night for this demonstration, say at 23:00 UT, which was just after sunset for my location. So October 22, 2013 23:00 UT. Let's convert this date and time to the Julian day number.
y = year
m = month
d = day (fraction of a day)
So here we have
y = 2013
m = 10
d = 22.95833
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
In this case JD = 2456588.45833333
We also want the JD for 1990.0, 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 Venus.
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 both Earth and Venus. In this case that equates to 8568.4260132099 degrees. Now obviously that's greater than 360, 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 "d' routine."
Mean anomaly M for earth (which I will call Me) then = 288.4260132098 degrees
Me = 288.4260132098
Next is the true anomaly, v. True anomaly is the angle between the real planet's position and its perihelion position. Normally this calls for an iterative process to solve Kepler's equation, but we will simplify things here by approximating the value with the equation of the center.
v = M + 360/pi * e * sin(M)
where
M = mean anomaly
e = eccentricity
We know the mean anomaly for earth above and eccentricity for earth = 0.016713. Therefore, earth's true anomaly (which I will call ve) equals 286.6090301972 degrees.
ve = 286.6090301972
Now we need to calculate heliocentric longitude, l. For earth I will call it Le.
l = (360/365.242191 * D/Tp) + 360/pi * e * sin(360/365.242191 * D/Tp + E - W)
where
D = days since 1990.0
Tp = Period
e = eccentricity
E = longitude at epoch 1990.0
W = longitude of the perihelion
Le = 8669.3774431972
Use the d' routine to get Le = 29.3774431972 degrees
Now we need the radius vector, that is to say, distance from the sun, r. For earth I will call this re.
r = (a*(1-e^2))/(1+e*cos(v))
where
a = semi-major axis of the orbit
e = eccentricity
v = true anomaly
re = 0.9949674832 AU
Now we repeat these calculations for Venus given the values from the table for Venus.
Mean anomaly for Venus = Mv = 210.6743831713 degrees
True anomaly for Venus = vv = 210.2781423101 degrees
heliocentric longitude for Venus = Lv = 341.7083783101 degrees
radius vector for Venus = rv = 0.7275574534 AU
Now we need the heliocentric latitude for Venus. This is given by
U = Arcsin(sin(Lv - Omega)*sin(i))
where
Lv = heliocentric longitude of Venus
omega = longitude of the ascending node of Venus
i = inclination of Venus
U = -3.3822083713 degrees
Now we need l' and r' which are the heliocentric longitude and radius vector of Venus respectively projected onto the plane of the ecliptic.
l' = arctan((sin(Lv-Omega)*cos(i))/cos(Lv-omega))+omega
Now 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. 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.
Following this routine, the result you should get for l' is 341.7048475079 degrees.
l' = 341.7048475079 degrees
r' is a simple one, it is simply rv * cos(U)
where
rv = radius vector of Venus
U = heliocentric latitude of Venus
r' = 0.7262901907 AU
Now we calculate the geocentric ecliptic longitude (lambda) and latitude (beta) of Venus. Since Venus is an inner planet the formula is this:
lambda = 180 + Le + arctan ((re*sin(l' - Le))/(re*sin(l'-Le)))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of venus projected onto ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
lambda = 256.0825668728 degrees
beta is calculated with the following formula:
beta = arctan((r'*tan(U)*sin(lambda-l'))/(re*sin(l'-Le))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of venus projected onto ecliptic
U = heliocentric latitude for Venus
r'= radius vector of Venus projected onto the plane of the ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
beta = -3.3298312039 degrees
For the final 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)
Obl = 23.43748
Now we just have to convert these coordinates from ecliptic to equatorial and we will have the geocentric equatorial coordinates of Venus at this timepoint.
Right ascension = arctan((sin(lambda)*cos(Obl)-tan(beta)*sin(Obl))/cos(lambda))​
where
lambda = 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. The result in this case should yield
right ascension = 253.8713410337 degrees. If you convert this to hours, minutes, seconds you get 16hr 55m 29.12s
declination = arcsin(sin(beta)*cos(obl)+cos(beta)*sin(obl)*sin(lambda))
where
lambda = geocentric ecliptic longitude
beta = geocentric ecliptic latitude
obl = obliquity of the ecliptic
declination = -26.0216957613 degrees. If you convert this to degrees, minutes, seconds you get -26d 1' 18.1"

So after all that, here are the calculated coordinates of Venus, which is accurate to within a few arcminutes in right ascension and about a quarter of a degree in declination (primarily because we didn't calculate for perturbations from the planets).
RA: 16hr 55m 29.12s
Dec: -26d 1' 18.1"
These are geocentric coordinates, but topocentric parallax will be insignificant at the distance of Venus, at least for naked eye observations like this (it only amounts to a dozen arcseconds or so, which you can't see by eye) so this is good enough as a general case for all observers. Now we just need to figure out which constellation contains these coordinates. I'm going to go to the university library and track down a star atlas. I'll use a 1991 copy of the Cambridge star atlas 2000.0 if it's not checked out (2000.0 is the epoch the atlas is in, not the date of publication). I'll shoot a viyOudeo of it and upload it later tonight, but for now here are the calculations. Enjoy! :)
 Quoting: Dr. Astro




Thank you, Dr!dasbier

Last Edited by Snot Bubble on 10/23/2013 09:24 PM
"Stupid Internet. I don't know why everyone is so impressed with it." - Pamela Anderson
Anonymous Coward
User ID: 4117918
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10/23/2013 09:27 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
This is just funny... is the object still there lately next to the sun as it sets?

It's been cloudy where I live for the past month or so..
..can't believe that anyone who just thought this was venus and didn't believe there any possible evidence to back up such claims, would not just post a comment and move along.

Sometimes, if you watch carefully, people reveal their motives through their actions.
Dr. AstroModerator
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User ID: 48376296
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10/23/2013 09:40 PM

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Re: There REALLY IS something strange in western sky around sunset...!!!
So, after this humongous amount of mathematical jargon effort, your precision is within a few arc-minutes and a quarter of grade (i believe this is 15 arc minutes?) for 20+ years interval,after, if, you find some old astronomy book reference?
how exactly this calculus is more reliable than the method used by, for example stellarium, VSOP87, wich has an accuracy of 1 ARC-SECOND between 2000 BC and 6000 AD for Venus, this is, at least, 60 times more accurate than what you are doing for a far greater interval of time and is constantly updated by and for mathematicians, astronomers and planetariums all around the world since is open source?
Inquiring minds want to know...
 Quoting: Anonymous Coward 25603040


Quarter of a degree in declination, a few arcminutes in right ascension. Not bad for calculations you can do by hand with minimal effort. It's not more accurate than VSOP87, but VSOP87 isn't something you can practically do by hand with just a pocket calculator. You seem to be missing the point here. The point is that I can show mathematically that Venus should be in Ophiuchus, you'd don't even have to rely on Stellarium which is frequently accused of being a "conspiracy." You're right; Stellarium is open source, so is VSOP87, there is no way in hell they can be "conspiracies" to hide the truth, but you can't reasonably lay out the entirety of calculations from VSOP87 line by line in a brief forum post and for the purposes of this post (naked eye observations), the accuracy achievable with a pocket calculator is good enough.

Last Edited by Astromut on 10/23/2013 09:41 PM
astrobanner2
Anonymous Coward
User ID: 1702426
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10/23/2013 09:45 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
...


Wrong. It's not "totally different hours" - it's in the same constellation, Ophiuchus. Would you like me to prove that to you mathematically?
 Quoting: Dr. Astro


I'm not doubting you, i would really like to see the calculations if possible and not too much trouble.
 Quoting: the gas man


Alright, so here we go. The first thing you need are the orbital elements of the planets, particularly earth and Venus. We shall simplify things down to "back of the envelope" calculations which will get us within a fraction of a degree of the right answer. We won't concern ourselves with all the planetary perturbations calculations, it would take me all night to type all that out, instead we'll focus on the mean orbital elements of the planets freed of perturbations for the epoch 1990.0. Here is a scan of a page containing those elements from the book "Practical Astronomy with Your Calculator" by Peter Duffett-Smith.
[link to dropcanvas.com]
The rest is just math. First we need to find the number of days since the epoch of those elements. In fact we can find the apparent position of Venus for a given day and time by converting to fractions of a day since epoch 1990.0. Let's use last night for this demonstration, say at 23:00 UT, which was just after sunset for my location. So October 22, 2013 23:00 UT. Let's convert this date and time to the Julian day number.
y = year
m = month
d = day (fraction of a day)
So here we have
y = 2013
m = 10
d = 22.95833
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
In this case JD = 2456588.45833333
We also want the JD for 1990.0, 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 Venus.
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 both Earth and Venus. In this case that equates to 8568.4260132099 degrees. Now obviously that's greater than 360, 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 "d' routine."
Mean anomaly M for earth (which I will call Me) then = 288.4260132098 degrees
Me = 288.4260132098
Next is the true anomaly, v. True anomaly is the angle between the real planet's position and its perihelion position. Normally this calls for an iterative process to solve Kepler's equation, but we will simplify things here by approximating the value with the equation of the center.
v = M + 360/pi * e * sin(M)
where
M = mean anomaly
e = eccentricity
We know the mean anomaly for earth above and eccentricity for earth = 0.016713. Therefore, earth's true anomaly (which I will call ve) equals 286.6090301972 degrees.
ve = 286.6090301972
Now we need to calculate heliocentric longitude, l. For earth I will call it Le.
l = (360/365.242191 * D/Tp) + 360/pi * e * sin(360/365.242191 * D/Tp + E - W)
where
D = days since 1990.0
Tp = Period
e = eccentricity
E = longitude at epoch 1990.0
W = longitude of the perihelion
Le = 8669.3774431972
Use the d' routine to get Le = 29.3774431972 degrees
Now we need the radius vector, that is to say, distance from the sun, r. For earth I will call this re.
r = (a*(1-e^2))/(1+e*cos(v))
where
a = semi-major axis of the orbit
e = eccentricity
v = true anomaly
re = 0.9949674832 AU
Now we repeat these calculations for Venus given the values from the table for Venus.
Mean anomaly for Venus = Mv = 210.6743831713 degrees
True anomaly for Venus = vv = 210.2781423101 degrees
heliocentric longitude for Venus = Lv = 341.7083783101 degrees
radius vector for Venus = rv = 0.7275574534 AU
Now we need the heliocentric latitude for Venus. This is given by
U = Arcsin(sin(Lv - Omega)*sin(i))
where
Lv = heliocentric longitude of Venus
omega = longitude of the ascending node of Venus
i = inclination of Venus
U = -3.3822083713 degrees
Now we need l' and r' which are the heliocentric longitude and radius vector of Venus respectively projected onto the plane of the ecliptic.
l' = arctan((sin(Lv-Omega)*cos(i))/cos(Lv-omega))+omega
Now 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. 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.
Following this routine, the result you should get for l' is 341.7048475079 degrees.
l' = 341.7048475079 degrees
r' is a simple one, it is simply rv * cos(U)
where
rv = radius vector of Venus
U = heliocentric latitude of Venus
r' = 0.7262901907 AU
Now we calculate the geocentric ecliptic longitude (lambda) and latitude (beta) of Venus. Since Venus is an inner planet the formula is this:
lambda = 180 + Le + arctan ((re*sin(l' - Le))/(re*sin(l'-Le)))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of venus projected onto ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
lambda = 256.0825668728 degrees
beta is calculated with the following formula:
beta = arctan((r'*tan(U)*sin(lambda-l'))/(re*sin(l'-Le))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of venus projected onto ecliptic
U = heliocentric latitude for Venus
r'= radius vector of Venus projected onto the plane of the ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
beta = -3.3298312039 degrees
For the final 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)
Obl = 23.43748
Now we just have to convert these coordinates from ecliptic to equatorial and we will have the geocentric equatorial coordinates of Venus at this timepoint.
Right ascension = arctan((sin(lambda)*cos(Obl)-tan(beta)*sin(Obl))/cos(lambda))​
where
lambda = 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. The result in this case should yield
right ascension = 253.8713410337 degrees. If you convert this to hours, minutes, seconds you get 16hr 55m 29.12s
declination = arcsin(sin(beta)*cos(obl)+cos(beta)*sin(obl)*sin(lambda))
where
lambda = geocentric ecliptic longitude
beta = geocentric ecliptic latitude
obl = obliquity of the ecliptic
declination = -26.0216957613 degrees. If you convert this to degrees, minutes, seconds you get -26d 1' 18.1"

So after all that, here are the calculated coordinates of Venus, which is accurate to within a few arcminutes in right ascension and about a quarter of a degree in declination (primarily because we didn't calculate for perturbations from the planets).
RA: 16hr 55m 29.12s
Dec: -26d 1' 18.1"
These are geocentric coordinates, but topocentric parallax will be insignificant at the distance of Venus, at least for naked eye observations like this (it only amounts to a dozen arcseconds or so, which you can't see by eye) so this is good enough as a general case for all observers. Now we just need to figure out which constellation contains these coordinates. I'm going to go to the university library and track down a star atlas. I'll use a 1991 copy of the Cambridge star atlas 2000.0 if it's not checked out (2000.0 is the epoch the atlas is in, not the date of publication). I'll shoot a video of it and upload it later tonight, but for now here are the calculations. Enjoy! :)
 Quoting: Dr. Astro

So, after this humongous amount of mathematical jargon effort, your precision is within a few arc-minutes and a quarter of grade (i believe this is 15 arc minutes?) for 20+ years interval,after, if, you find some old astronomy book reference?
how exactly this calculus is more reliable than the method used by, for example stellarium, VSOP87, wich has an accuracy of 1 ARC-SECOND between 2000 BC and 6000 AD for Venus, this is, at least, 60 times more accurate than what you are doing for a far greater interval of time and is constantly updated by and for mathematicians, astronomers and planetariums all around the world since is open source?
Inquiring minds want to know...
 Quoting: Anonymous Coward 25603040


hesright
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10/23/2013 09:51 PM

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Re: There REALLY IS something strange in western sky around sunset...!!!
...


It's not Venus, I've checked several time, with software, charts..etc. Totally different hours, movement, description.
Thanks anyway, but it ain't Venus. I just don't buy that. It doesn't matter what we say here, time will tell
 Quoting: pstrusi


Wrong. It's not "totally different hours" - it's in the same constellation, Ophiuchus. Would you like me to prove that to you mathematically?
 Quoting: Dr. Astro


I'm not doubting you, i would really like to see the calculations if possible and not too much trouble.
 Quoting: the gas man


Alright, so here we go. The first thing you need are the orbital elements of the planets, particularly earth and Venus. We shall simplify things down to "back of the envelope" calculations which will get us within a fraction of a degree of the right answer. We won't concern ourselves with all the planetary perturbations calculations, it would take me all night to type all that out, instead we'll focus on the mean orbital elements of the planets freed of perturbations for the epoch 1990.0. Here is a scan of a page containing those elements from the book "Practical Astronomy with Your Calculator" by Peter Duffett-Smith.
[link to dropcanvas.com]
The rest is just math. First we need to find the number of days since the epoch of those elements. In fact we can find the apparent position of Venus for a given day and time by converting to fractions of a day since epoch 1990.0. Let's use last night for this demonstration, say at 23:00 UT, which was just after sunset for my location. So October 22, 2013 23:00 UT. Let's convert this date and time to the Julian day number.
y = year
m = month
d = day (fraction of a day)
So here we have
y = 2013
m = 10
d = 22.95833
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
In this case JD = 2456588.45833333
We also want the JD for 1990.0, 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 Venus.
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 both Earth and Venus. In this case that equates to 8568.4260132099 degrees. Now obviously that's greater than 360, 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 "d' routine."
Mean anomaly M for earth (which I will call Me) then = 288.4260132098 degrees
Me = 288.4260132098
Next is the true anomaly, v. True anomaly is the angle between the real planet's position and its perihelion position. Normally this calls for an iterative process to solve Kepler's equation, but we will simplify things here by approximating the value with the equation of the center.
v = M + 360/pi * e * sin(M)
where
M = mean anomaly
e = eccentricity
We know the mean anomaly for earth above and eccentricity for earth = 0.016713. Therefore, earth's true anomaly (which I will call ve) equals 286.6090301972 degrees.
ve = 286.6090301972
Now we need to calculate heliocentric longitude, l. For earth I will call it Le.
l = (360/365.242191 * D/Tp) + 360/pi * e * sin(360/365.242191 * D/Tp + E - W)
where
D = days since 1990.0
Tp = Period
e = eccentricity
E = longitude at epoch 1990.0
W = longitude of the perihelion
Le = 8669.3774431972
Use the d' routine to get Le = 29.3774431972 degrees
Now we need the radius vector, that is to say, distance from the sun, r. For earth I will call this re.
r = (a*(1-e^2))/(1+e*cos(v))
where
a = semi-major axis of the orbit
e = eccentricity
v = true anomaly
re = 0.9949674832 AU
Now we repeat these calculations for Venus given the values from the table for Venus.
Mean anomaly for Venus = Mv = 210.6743831713 degrees
True anomaly for Venus = vv = 210.2781423101 degrees
heliocentric longitude for Venus = Lv = 341.7083783101 degrees
radius vector for Venus = rv = 0.7275574534 AU
Now we need the heliocentric latitude for Venus. This is given by
U = Arcsin(sin(Lv - Omega)*sin(i))
where
Lv = heliocentric longitude of Venus
omega = longitude of the ascending node of Venus
i = inclination of Venus
U = -3.3822083713 degrees
Now we need l' and r' which are the heliocentric longitude and radius vector of Venus respectively projected onto the plane of the ecliptic.
l' = arctan((sin(Lv-Omega)*cos(i))/cos(Lv-omega))+omega
Now 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. 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.
Following this routine, the result you should get for l' is 341.7048475079 degrees.
l' = 341.7048475079 degrees
r' is a simple one, it is simply rv * cos(U)
where
rv = radius vector of Venus
U = heliocentric latitude of Venus
r' = 0.7262901907 AU
Now we calculate the geocentric ecliptic longitude (lambda) and latitude (beta) of Venus. Since Venus is an inner planet the formula is this:
lambda = 180 + Le + arctan ((re*sin(l' - Le))/(re*sin(l'-Le)))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of venus projected onto ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
lambda = 256.0825668728 degrees
beta is calculated with the following formula:
beta = arctan((r'*tan(U)*sin(lambda-l'))/(re*sin(l'-Le))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of venus projected onto ecliptic
U = heliocentric latitude for Venus
r'= radius vector of Venus projected onto the plane of the ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
beta = -3.3298312039 degrees
For the final 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)
Obl = 23.43748
Now we just have to convert these coordinates from ecliptic to equatorial and we will have the geocentric equatorial coordinates of Venus at this timepoint.
Right ascension = arctan((sin(lambda)*cos(Obl)-tan(beta)*sin(Obl))/cos(lambda))​
where
lambda = 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. The result in this case should yield
right ascension = 253.8713410337 degrees. If you convert this to hours, minutes, seconds you get 16hr 55m 29.12s
declination = arcsin(sin(beta)*cos(obl)+cos(beta)*sin(obl)*sin(lambda))
where
lambda = geocentric ecliptic longitude
beta = geocentric ecliptic latitude
obl = obliquity of the ecliptic
declination = -26.0216957613 degrees. If you convert this to degrees, minutes, seconds you get -26d 1' 18.1"

So after all that, here are the calculated coordinates of Venus, which is accurate to within a few arcminutes in right ascension and about a quarter of a degree in declination (primarily because we didn't calculate for perturbations from the planets).
RA: 16hr 55m 29.12s
Dec: -26d 1' 18.1"
These are geocentric coordinates, but topocentric parallax will be insignificant at the distance of Venus, at least for naked eye observations like this (it only amounts to a dozen arcseconds or so, which you can't see by eye) so this is good enough as a general case for all observers. Now we just need to figure out which constellation contains these coordinates. I'm going to go to the university library and track down a star atlas. I'll use a 1991 copy of the Cambridge star atlas 2000.0 if it's not checked out (2000.0 is the epoch the atlas is in, not the date of publication). I'll shoot a video of it and upload it later tonight, but for now here are the calculations. Enjoy! :)
 Quoting: Dr. Astro



astrobanner2
Anonymous Coward
User ID: 4117918
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10/23/2013 09:58 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
If it's venus, why have I seen both Venus and this object simultaneously?

Why does google sky have no trouble identifying venus and every other known star and planet but does not identify this one?

It's about as bright as Venus so it should be no problem unless it's a new "star" or a UFO evidently being viewed all over the world..
AzaMorey_HollyWood

User ID: 47646043
Australia
10/23/2013 10:09 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
If it's venus, why have I seen both Venus and this object simultaneously?

Why does google sky have no trouble identifying venus and every other known star and planet but does not identify this one?

It's about as bright as Venus so it should be no problem unless it's a new "star" or a UFO evidently being viewed all over the world..
 Quoting: Anonymous Coward 4117918


this question...i like t.
SO MANY ASSHOLES....SO FEW LIGHT SABERS!

"AZAMOREY"

:SNAPPER:
Dr. AstroModerator
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10/23/2013 10:13 PM

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Re: There REALLY IS something strange in western sky around sunset...!!!
If it's venus, why have I seen both Venus and this object simultaneously?
 Quoting: Anonymous Coward 4117918

Some people are seeing contrails, some people are describing Venus, some people are describing Antares. Depends on the specific case. I had no trouble seeing Venus and contrails simultaneously the last few weeks as well.

The whole Venus calculations started because someone said they saw the "object" in Ophiuchus and that Venus was at a totally different location. Venus IS in Ophiuchus, that is the point. If your software shows otherwise, chances are you aren't using the software right.
astrobanner2
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User ID: 48136264
Spain
10/23/2013 10:14 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
I am pissing myself laughing.

You can discount every single Astro's post as total bullshit if you want to.

What amazes me is the lack of pure common sense.

If there was indeed a REAL and VISIBLE object in the western sky, do you really think that thousands of amateurs who are interested in astronomy would not take to their respective forums and report it? Sorry but no censorship would be able to contain the amount of posts if there was something out there.

I mean come on. Why is this bullshit threat even going? Has common sense ceased to exist?

I am sad for GLP and people who keeping this shite alive. PLease divert your attention to REAL issues.
Anonymous Coward
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10/23/2013 10:19 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
...


Wrong. It's not "totally different hours" - it's in the same constellation, Ophiuchus. Would you like me to prove that to you mathematically?
 Quoting: Dr. Astro


I'm not doubting you, i would really like to see the calculations if possible and not too much trouble.
 Quoting: the gas man


Alright, so here we go. The first thing you need are the orbital elements of the planets, particularly earth and Venus. We shall simplify things down to "back of the envelope" calculations which will get us within a fraction of a degree of the right answer. We won't concern ourselves with all the planetary perturbations calculations, it would take me all night to type all that out, instead we'll focus on the mean orbital elements of the planets freed of perturbations for the epoch 1990.0. Here is a scan of a page containing those elements from the book "Practical Astronomy with Your Calculator" by Peter Duffett-Smith.
[link to dropcanvas.com]
The rest is just math. First we need to find the number of days since the epoch of those elements. In fact we can find the apparent position of Venus for a given day and time by converting to fractions of a day since epoch 1990.0. Let's use last night for this demonstration, say at 23:00 UT, which was just after sunset for my location. So October 22, 2013 23:00 UT. Let's convert this date and time to the Julian day number.
y = year
m = month
d = day (fraction of a day)
So here we have
y = 2013
m = 10
d = 22.95833
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
In this case JD = 2456588.45833333
We also want the JD for 1990.0, 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 Venus.
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 both Earth and Venus. In this case that equates to 8568.4260132099 degrees. Now obviously that's greater than 360, 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 "d' routine."
Mean anomaly M for earth (which I will call Me) then = 288.4260132098 degrees
Me = 288.4260132098
Next is the true anomaly, v. True anomaly is the angle between the real planet's position and its perihelion position. Normally this calls for an iterative process to solve Kepler's equation, but we will simplify things here by approximating the value with the equation of the center.
v = M + 360/pi * e * sin(M)
where
M = mean anomaly
e = eccentricity
We know the mean anomaly for earth above and eccentricity for earth = 0.016713. Therefore, earth's true anomaly (which I will call ve) equals 286.6090301972 degrees.
ve = 286.6090301972
Now we need to calculate heliocentric longitude, l. For earth I will call it Le.
l = (360/365.242191 * D/Tp) + 360/pi * e * sin(360/365.242191 * D/Tp + E - W)
where
D = days since 1990.0
Tp = Period
e = eccentricity
E = longitude at epoch 1990.0
W = longitude of the perihelion
Le = 8669.3774431972
Use the d' routine to get Le = 29.3774431972 degrees
Now we need the radius vector, that is to say, distance from the sun, r. For earth I will call this re.
r = (a*(1-e^2))/(1+e*cos(v))
where
a = semi-major axis of the orbit
e = eccentricity
v = true anomaly
re = 0.9949674832 AU
Now we repeat these calculations for Venus given the values from the table for Venus.
Mean anomaly for Venus = Mv = 210.6743831713 degrees
True anomaly for Venus = vv = 210.2781423101 degrees
heliocentric longitude for Venus = Lv = 341.7083783101 degrees
radius vector for Venus = rv = 0.7275574534 AU
Now we need the heliocentric latitude for Venus. This is given by
U = Arcsin(sin(Lv - Omega)*sin(i))
where
Lv = heliocentric longitude of Venus
omega = longitude of the ascending node of Venus
i = inclination of Venus
U = -3.3822083713 degrees
Now we need l' and r' which are the heliocentric longitude and radius vector of Venus respectively projected onto the plane of the ecliptic.
l' = arctan((sin(Lv-Omega)*cos(i))/cos(Lv-omega))+omega
Now 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. 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.
Following this routine, the result you should get for l' is 341.7048475079 degrees.
l' = 341.7048475079 degrees
r' is a simple one, it is simply rv * cos(U)
where
rv = radius vector of Venus
U = heliocentric latitude of Venus
r' = 0.7262901907 AU
Now we calculate the geocentric ecliptic longitude (lambda) and latitude (beta) of Venus. Since Venus is an inner planet the formula is this:
lambda = 180 + Le + arctan ((re*sin(l' - Le))/(re*sin(l'-Le)))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of venus projected onto ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
lambda = 256.0825668728 degrees
beta is calculated with the following formula:
beta = arctan((r'*tan(U)*sin(lambda-l'))/(re*sin(l'-Le))
where
Le = heliocentric longitude of earth
re = radius vector of earth
l' = heliocentric longitude of venus projected onto ecliptic
U = heliocentric latitude for Venus
r'= radius vector of Venus projected onto the plane of the ecliptic
Note that you need to follow the same quadrant disambiguation routine as listed above for the arctan function in this formula. The result in this case should yield
beta = -3.3298312039 degrees
For the final 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)
Obl = 23.43748
Now we just have to convert these coordinates from ecliptic to equatorial and we will have the geocentric equatorial coordinates of Venus at this timepoint.
Right ascension = arctan((sin(lambda)*cos(Obl)-tan(beta)*sin(Obl))/cos(lambda))​
where
lambda = 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. The result in this case should yield
right ascension = 253.8713410337 degrees. If you convert this to hours, minutes, seconds you get 16hr 55m 29.12s
declination = arcsin(sin(beta)*cos(obl)+cos(beta)*sin(obl)*sin(lambda))
where
lambda = geocentric ecliptic longitude
beta = geocentric ecliptic latitude
obl = obliquity of the ecliptic
declination = -26.0216957613 degrees. If you convert this to degrees, minutes, seconds you get -26d 1' 18.1"

So after all that, here are the calculated coordinates of Venus, which is accurate to within a few arcminutes in right ascension and about a quarter of a degree in declination (primarily because we didn't calculate for perturbations from the planets).
RA: 16hr 55m 29.12s
Dec: -26d 1' 18.1"
These are geocentric coordinates, but topocentric parallax will be insignificant at the distance of Venus, at least for naked eye observations like this (it only amounts to a dozen arcseconds or so, which you can't see by eye) so this is good enough as a general case for all observers. Now we just need to figure out which constellation contains these coordinates. I'm going to go to the university library and track down a star atlas. I'll use a 1991 copy of the Cambridge star atlas 2000.0 if it's not checked out (2000.0 is the epoch the atlas is in, not the date of publication). I'll shoot a video of it and upload it later tonight, but for now here are the calculations. Enjoy! :)
 Quoting: Dr. Astro

So, after this humongous amount of mathematical jargon effort, your precision is within a few arc-minutes and a quarter of grade (i believe this is 15 arc minutes?) for 20+ years interval,after, if, you find some old astronomy book reference?
how exactly this calculus is more reliable than the method used by, for example stellarium, VSOP87, wich has an accuracy of 1 ARC-SECOND between 2000 BC and 6000 AD for Venus, this is, at least, 60 times more accurate than what you are doing for a far greater interval of time and is constantly updated by and for mathematicians, astronomers and planetariums all around the world since is open source?
Inquiring minds want to know...
 Quoting: Anonymous Coward 25603040


Or you know ...that other method, which I like to call "look up in the sky and spot something and say WTF!!!"
Stewtown

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10/23/2013 10:28 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
wtf is Astronut still doing posting wall of disinfo gibberish in this thread? Holy shit that is some straight up psycho shit yo.

Astronut is like the Harry Potter guy in this clip, with transhuman metrosexual nerd glasses.


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 Quoting: Anonymous Coward 1702426


rofl

Awesome
 Quoting: Anonymous Coward 39528572


Ugh, I can't take it anymore, I had to put Astro on my "ignore" list, blatant dis info propaganda.
Dr. AstroModerator
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10/23/2013 10:44 PM

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Re: There REALLY IS something strange in western sky around sunset...!!!
wtf is Astronut still doing posting wall of disinfo gibberish in this thread? Holy shit that is some straight up psycho shit yo.

Astronut is like the Harry Potter guy in this clip, with transhuman metrosexual nerd glasses.


[link to www.youtube.com (secure)]
 Quoting: Anonymous Coward 1702426


rofl

Awesome
 Quoting: Anonymous Coward 39528572


Ugh, I can't take it anymore, I had to put Astro on my "ignore" list, blatant dis info propaganda.
 Quoting: Stewtown


Yeah that's it, ignore the guy who actually puts his money where his mouth is.
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10/23/2013 11:10 PM
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Re: There REALLY IS something strange in western sky around sunset...!!!
So, after this humongous amount of mathematical jargon effort, your precision is within a few arc-minutes and a quarter of grade (i believe this is 15 arc minutes?) for 20+ years interval,after, if, you find some old astronomy book reference?
how exactly this calculus is more reliable than the method used by, for example stellarium, VSOP87, wich has an accuracy of 1 ARC-SECOND between 2000 BC and 6000 AD for Venus, this is, at least, 60 times more accurate than what you are doing for a far greater interval of time and is constantly updated by and for mathematicians, astronomers and planetariums all around the world since is open source?
Inquiring minds want to know...
 Quoting: Anonymous Coward 25603040


Quarter of a degree in declination, a few arcminutes in right ascension. Not bad for calculations you can do by hand with minimal effort. It's not more accurate than VSOP87, but VSOP87 isn't something you can practically do by hand with just a pocket calculator. You seem to be missing the point here. The point is that I can show mathematically that Venus should be in Ophiuchus, you'd don't even have to rely on Stellarium which is frequently accused of being a "conspiracy." You're right; Stellarium is open source, so is VSOP87, there is no way in hell they can be "conspiracies" to hide the truth, but you can't reasonably lay out the entirety of calculations from VSOP87 line by line in a brief forum post and for the purposes of this post (naked eye observations), the accuracy achievable with a pocket calculator is good enough.
 Quoting: Dr. Astro

No, apparently you are missing the point wich is, i would have to trust your (or others using the same method) simplified hand writing calculations with a precision of arc minutes (wich is by no means doubting on the integrity of the person doing it)saying, in this case venus is in ophicious...(or whatever the name of the constelation actually is) or a program, for example stellarium, wich say venus is in scorpio now (albeit on one extreme) with a precision of arc seconds? wich version is more reliable in your opinion in order to contrast the actual view of the sky with an estimated trjectory to estblish if there´s some anomaly o discard it, wich is the point in this discussion, have a reliable contrast between what is expected and what is actually seen?
By the way, interesting debate.





GLP