# Re: Spincronic shows his stupidity - again.

spintronic wrote:
On 2 Oct, 00:35, Earle Jones <earle.jo...@xxxxxxxxxxx> wrote:

In article <1191225335.924695.40...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,

spintronic <spintro...@xxxxxxxxxxx> wrote:

On 30 Sep, 22:09, Bob Casanova <nos...@xxxxxxxx> wrote:

On Sun, 30 Sep 2007 11:30:27 -0700, the following appeared
in talk.origins, posted by spintronic
<spintro...@xxxxxxxxxxx>:

On 30 Sep, 15:24, josephus <dogb...@xxxxxxxxxxxxx> wrote:

spintronic wrote:

On 29 Sep, 18:31, Ye Old One <use...@xxxxxxxxx> wrote:

Sure you did! O.k, we will walk you back to your room now, everything
will
be o.k. Watch the step!

sure sounds like a jealous drunk trollslob

Jealous of what? some lying old guys delusions? I dont think so!

Let him be, he will be dead soon. Then we can all have a rest!

You however need to rethink your GR.

You wouldn't know anything about GR. You have trouble with
freshman algebra, as shown by your "10 solutions to a
second-degree equation" howler.

8, solutions. Non of which you can prove wrong!

*
Spin: The fundamental theorem of algebra:

Every polynomial equation (rational coefficients) of degree 'n' has
exactly 'n' roots.

For more:

http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

Ya, you should try to read it.

You can have real & complex coefficients, with 2 solutions for each.

aX^2 - aX = 0, where a is the coefficient.

Exercise for the student:

What are the three roots of

x^3 -1 = 0

8 solutions.

spin does not know any real mathematics. 2 roots means only 2 answers for any quadratic not kinds of answers.

just to show how really ignorant spin is.

define m = gm/c^2 for any mass. this is the relativistic qualifier.

since v^2 = Gm(2/r-1/A) for any circular orbit A=r
and we get V2 = Gm/r

this is important relating to black hole event horizon.

GBH/m = c^2

when we look at time T' = T*(1- 1/n)

n = D/m

because at r = m
GBH/m == C^2
m = Gbh/C^2 see the above definition
then 1/m = C^2/GBH
then GBH/(m*C^2) == 1
then any multiplies of m show up as 1/(D/m) and because M and GBH/C^2*m cancel note that for values less than 1 for D D/M = 1+d/m because we measure from m not zero.
we get T1 = T*(1- 1/n) where n = D/M
since we have .004M when we divided by m we get 1.004

or we can have T1*(1 - 1/1.004) = 1 -.99960016
or T' = T*(.00039984)

0r T =(2501.0004) seconds per Tick of T'

none of these values resemble any numbers produced by spintropic.

GBH/c^2 = m
then m = the event horizon.

the diameter ( in our frame) is 2*m

for spins 1.5 solar mass he gets 2*10^30 *1.5

GM for our solar mas is 1.3251244E20 * 1.5 = 1.996966E20
divide by c^2 == 2215 km == m for our 1.5 solar BH.

for the calculation of spins bogus problems M and GBH do not appear in the calculation. for any distances above an event horizon.

and the matter circling a black hole aggregate and generate friction. they slow and all fall into the hole. because the Black Hole should be spinning, the excess atomic destruction exits the spinning pole and we get a plume of hot plasma. not matter but plasma.

there is a mathematical function that says a spinning mass will have a torque that throws matter upward and downward. except I dont know that kind of math. it has to do with gyroscopes.

josephus
--
I go sailing in the Summer and
look at STARS in the Winter.
"Everybody is igernant, jist on differt subjects"
Will Rogers Jr.
"it aint what you know that gets you in trouble
it is what you know that aint so"
Josh Billings.

.

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