Tuesday, 15th May, 2,012.
My Astronomy Course,Lesson 3.
I am very sorry. I missed out a vital step! Regarding my distance finder(Read ZZ37 again, please, and make this extra step!)(astronomical objects), you DIVIDE THE VELOCITIES BY FIVE,FIRST! So as to get apparent magnitudes and radial velocities ON A PAR!!
My platform was The Hubble Diagram. Where we compared magnitudes with velocities. This is where one of the 2-1’s showed up. Though Orthodoxy will deny it.
It is as though we had apples and oranges. Two quite different entities.
So we have to have two fruit that are digitally the same!
Now it should all work out!
So, the formula is: Apparent magnitudes(v) x radial velocities(rv)DIVIDED BY FIVE. Note ONLY the radial velocities get divided by five! DIGITALLY. Then you take THE SQUARE ROOT OF THE PRODUCT!!
In other words get the MEAN, The Geometrical Mean of Mags times Vels.
The VERY first step is to multiply the velocities(rv) by 1.25. To equalize vels and mags. Because mags go up in multiply increments of 2.5, while vels go up in multiply increments of 2. So by multiplying vels by 1.25, we bring them level on multiply increments of 2.5.
So altogether, we take the vels and multiply them by 1.25. Next we multiply the rv vels and v mags together, and get their square root. In other words we take The Geometric Mean of the two basic distance obtainers, rv velocities(divided by five) and v magnitudes.
Remove O.M. The Obscuring Matter(Extinction Effect) by basing to R, The Reddening, using COLOUR magnitudes.(It is VERY important NOT to use the orthodox magnitudes! Because they are wrong.)
Removing the obscuring matter OM RAISES the value of the result.
First reduce everything to the same units!(E.g. The O.M. equalized to a magnitude – for subtraction.)
Mass Effect also needs to come off. M.E. is Geometric Mean less Arithmetic Mean.
Take a batch of about 40 candidates. When you have all the distance figures for each candidate, plot maps in R.A. Right Ascension and Decl. Declination.
The ideal thing then(if possible) is to fit R.A. to Declination to get a 3d picture.
If you do EXACTLY as I have stated, the results should be STUNNING!!(Because the right figures in a patterned cosmos should picture out A DEFINITE PATTERN!!)
Don’t forget to divide those velocities by FIVE!!(Fortunately I remembered in time!)(To give you the CORRECT formula now!)