[Kryptid]

I feel at a bit of a loss, then.

[McQueen]

« Reply #18 on: 29/05/2021 11:15:51 » By crazy scientist.

I just wanted to applaud. Unfortunately that option is not available. Still. Well done.

[Zer0]

SideNote - How to Post or Upload an Image/Picture/Chart.


Step by Step...

1) First, look at the bottom of this box.
It has a long option mentioned
" Attachments and Other options ".
Click on the

• Sign and it will Expand & show more Options.

IMG_20210601_125257.jpg (266.84 kB . 1080x2156 - viewed 2083 times)

2) Click on the [ Choose File ] box...
And your System will automatically take you to your Images/Screenshots/Downloads Folder.
Choose Whichever File you wish to Upload.
(Allowed file types: gif, jpg, pdf, png, PNG, JPG, GIF, PDF, jpeg)
When you have finally made the Selection, you'd automatically return to the posting screen & now it should look somewhat like this.
(An indication that the File has been chosen & loaded Successfully)


IMG_20210601_130452.jpg (263.64 kB . 1080x2156 - viewed 2077 times)

3) Now note that You need to provide a Location inside the chat box window to specify where exactly you would wish the File to be Pasted.

The ' Cursor ' is the Target 💧Point where the ' File ' shall appear.
(blue arrow)
And Click on (Insert Attachment 0) option.
(red arrow)


IMG_20210601_131515.jpg (271.43 kB . 1080x2156 - viewed 2076 times)

4) Once the Above 3(three) steps are followed, this is how the End final result would look like...


[ Attachment = 0 ]
Indicates the File has been inserted Successfully.
(neon green arrow)


IMG_20210601_132155.jpg (274.42 kB . 1080x2156 - viewed 2077 times)

Now you may just Preview or Post your comment & see whether everything is in place.
(If Not, then you may easily select the Modify Post option and make necessary changes)

P.S. - The Arthur Schopenhauer portrait has a message hidden from sight, all one needs to do is simply drag the portrait towards the left of the screen, & the message to the right is revealed.


IMG_20210601_133119.jpg (248.99 kB . 1080x2156 - viewed 2072 times)

*Note* - I use a Cell/Mobil phone hence the Desktop version would slightly differ in visualization, but the Steps/Process would ideally remain the same.
👍
(Allowed file types: gif, jpg, pdf, png, PNG, JPG, GIF, PDF, jpeg
Restrictions: 12 per post, maximum total size 2500KB, maximum individual size 2500KB)

[captcass (OP)]

I feel at a bit of a loss, then.

I know. Very frustrating. I will be getting into this deeper over the next few days. I need to determine where the dilation gradient curve of the central mass has a dilation factor ~equal to the star masses at that radius. I think it is at the corotation circle, but I have not had time to get into that yet. Inside that limit, the stars should follow the planetary formula. Out side, the galactic formula. It won't be a gradual shift. Either the galactic gradient is predominant (deeper) or the star's. Whichever it is will determine the velocity.

A clue is where the the velocities fail to follow Kepler......

[captcass (OP)]

After reading just the abstract of this paper:  Andromeda chained to the box – dynamical models for M31: bulge and bar
Matias Blaña Díaz, Christopher Wegg, Ortwin Gerhard, Peter Erwin, Matthieu Portail, Michael Opitsch, Roberto Saglia, Ralf Bender at:
https://academic.oup.com/mnras/article/466/4/4279/2726589,
it seems apparent that the bulge dynamics, and mass estimates, are such that an accurate enough determination of the dilation gradient, along with the lack of individual star data, make it impossible to determine positively where the gradients equalize in M31, and I am thinking probably everywhere. What is needed is a very simple galaxy with a classic bulge, who's mass data is well verified. Any suggestions?

[captcass (OP)]

Thanks, ZerO, for the step by step.

The corotation circle of M31 is ~6 kpc. In the graphs, the anomalous behavior begins at ~8 kpc. EDIT: looking at the curve of the anomalies, is seems more likely they actually begin at the corotation circle, ~6 kpc. The data is probably not that accurate so as to show the initial deviations.

[Zer0]

You are most Welcome Sir!
🙂

Feel sorry the OP got off on a wrong footing...Glad that things have finally settled down...& Happy to watch this go towards a fruitful direction.
✌️

(To Everyone participating, & Others who might join in the future, let's strive towards a conclusion in a civilized manner)
🙏
Edit - I forgot to use the golden word...Please!
🙃




P.S. - Happy sailing in Friendly waters Good ol' Captain!
⛵

[captcass (OP)]

I like people's pertinent questions because it helps me think and move froward. I did a brief email to my list last night clarifying the radius of the Sun I used and the transition between the gradients and 19 more downloaded the paper. I use Avstats, as they show when the whole paper has been viewed. There were also 38 partial downloads. Including the partials, both Avstats and Webylizer agree that 93% of my list has downloaded the paper at least in part. That could mean several things, including that some didn't get past the abstract  and that some went right to the velocities section and skipped the rest.

[captcass (OP)]

Playing with galactic gradients I found out this:
As per:
https://academic.oup.com/pasj/article/69/1/R1/2632658 :
"An approximate estimation of the mass inside the solar circle can be obtained for a set of parameters of R₀ = 8 kpc and V₀ = 200 to 238 km s−1, assuming spherical distribution of mass, as
M₀ = R₀V²/G = (7.44 to1.05)×10¹⁰M_⊙∼10¹¹M_⊙
with G being the gravitational constant, and the solar rotation velocity V₀ being related to R₀ as V₀ = (A − B)R₀, where A and B are the Oort's constants (Oort 1965), which are determined by measuring the radial velocity and proper motion of a nearby star. See Kerr and Lynden-Bell (1986) for a review regarding the Oort constants, and tables 1 and 2 for recent values."

So....
Using 231 km/s for the Sun’s V_o,
1 kpc = 3.08567758128*10¹⁹ m
At 8 kpc in the MW, for the Sun:
M_o = (R*V²)/G =
((2.468542065024*10^20)*(231000^2))/(6.67430*10^-11) =
1.97359832689189374166579266739583177262034*10⁴¹ kg =
9.9225657460*10¹⁰ M_☉
Vs
M_o = R(c² – T₀c²)/c² =
(2.468542065024*10^20)*(((299792458^2)-(299792458^2*0.9999994062788036))/(6.67430*10^-11))) =
1.97359832685608858237940840531023542229746*10⁴¹ kg =
9.9225657458*10¹⁰ M_☉
Where:
V_o = √c² - T₀c²
T₀ = (c² – V²)/c²

M31 at corotation = 6 kpc:
M_o = (R*V²)/G =
((1.851406548768*10^20)*(239000^2))/(6.67430*10^-11) =
1.58449865112711337518541270245568823696867*10⁴¹ kg =
7.9663079493*10¹⁰ M_☉
VS
M_o = R(c² – T₀c²)/c² =
 (1.851406548768*10^20)*((299792458^2-(299792458^2*0.9999993644431615))/(6.67430*10^-11) =
1.58449865108127202096258645581425443195541*10⁴¹ kg =
7.9667084875*10¹⁰ M_☉
 

[Bored chemist]

That could mean several things, including that some didn't get past the abstract   and that some went right to the velocities section and skipped the rest.

It could mean they laughed at it and then used it...

" (I am sitting in the smallest room of my house. I have your review before me. In a moment it will be behind me.)"

I'm aware of the irony that I take the role of a critic but,...

[captcass (OP)]

It could mean they laughed at it and then used it...

I have absolutely no idea what that means.

The point is, all my equations derive physical phenomena through c and the time dilation factor, alone. No mass is required, except to slow time. Even the amount of mass can be derived from just c and T₀.

[Bored chemist]

I have absolutely no idea what that means.

The point is that, these days, the fact that someone looked at your paper shows that
(1) the title seemed interesting
(2) their antivirus software didn't flag it as a threat.

It does not say anything about the quality of the content of the file.

It could not do so; the decision to download it is made before you know what is in the file.

The fact that you keep telling us how many people have looked at a file tells us that you do not recognise how irrelevant it is.

[captcass (OP)]

The point is that,

...the downloads are the direct result of me emailing formulas, not a title. I did another mailing last night with what I just posted yesterday and 12 more downloaded the full paper. That brings me to 82% complete downloads and 75 partial.

[Bored chemist]

The point is...

... just slightly beyond your grasp.
They can't judge the downloads until they download them.

[captcass (OP)]

They can't judge the downloads

...unless they download. You don't seem to realize how difficult it is to get people to download....a download rate of 100%, which I now have if the partials are considered, is totally unheard of.... The normal successful response rate is 1.5%.

[jeffreyH]

The point is that,

...the downloads are the direct result of me emailing formulas, not a title. I did another mailing last night with what I just posted yesterday and 12 more downloaded the full paper. That brings me to 82% complete downloads and 75 partial.

You appear to be branching out from grifting into email spam. Bravo you!