Dark energy      (appendice 3)
          The nex web address is for the first part of this paper:
          The second part (annex 1) is for show that énergy from the zone of contraction(clusters of galaxies) is enough for explain the Univers expansion accélération,the web page is:
            The third part(annex 2) of this paper ,show a total accélération expansion that include accélération from dark énergy ans accélération gravity,the web page is :
                      Before begin,i have found a french paper that show that the current accélération expansion begin there is about 6 billions years et before it
was a décélération expansion, i am going try to give it to you, first do the first next web page and after go on the forum message on the top ;
                 In this part ,we compare constant find ( in the preceding part) with the Hubble constant and we wil studied more ma version.
            Bégin with a équation that allow to get the Hubble constant :
Be Vmoy. the avérage speed, H the Hubble constant, R the lenght that go from the avérage contraction zone center,with this we can get
the équation that allow to get the Hubble constant:
      Vmoy.= HR    Vmoy. = R/T = HR     H = 1/T     T is the nécessary time for a avérage expansion speed  Vmoy.  for get a lenght R,
This time T  is a estimation of the Universe age. With use of  accélération from  dark énergy(dea),we get :
(1/2)(dea)T = HR = R/T = Vmoy.           dea = (constant)R 
(dea) = 2R//T^2 = (constant)R
2/T^2 = (constant)
2(H)^2 = (constant)    this 2 équations give us the 2 nexts équation:
[(4/3)(pi)G]d = (constant)
GM/(Rcritique)^3 = (constant)
d is the libération speed for a avérage contraction zone. For this 2 équations , i have consider the next 2 équations :
T = [{3/2(pi)G}^(1/2)](1/d)^(1/2)
T is the nécessary time for do R and if we consider the avérage speed (Vmoy.) is équal to (1/2)VL ou VL is the libération speed,then this time
come from the équarion : (1/2)VL = Vmoy. = R/T
T = R/(Vmoy.) = R/[(1/2)VL]
VL = [2GM/R]^(1/2)
We have find that: 2/(T)^2 = (constant) = 2(H)^2            this give :
[(4/3)(pi)G]d = (constant) = 2(H)^2      this give :
GM/(Rctitical)^3 = (constant)
Try a exemple for a estimatate M :
 Take a Hubble constant équal to (75 km/s) per millions parsec, as there is (3.086)(10)^19 km in a millions parsecs,then :
H = 75 km/s per (3.086)(10)^19 km , la corresponding value give :
(2.043)(10)^(-18) per second  ou T = (4.11)(10)^11 second (or 13 billions years).
With our 2 following équations, we are ready to estimate M and d :
[(4/3)(pi)G]d = (constant) = 2(H)^2 = 2(1/T)^2
GM/(Rcritical)^3 = (constant) = 2(H)^2 = 2(1/T)^2
If the Universe is 30 billions light years lenght for its Ray, with 100 billions of galaxy as Milky Way, with a mass (10)^13 solar mass ,and that
Rcritical  for a avérage contraction zone is estimated to (1/2) billions lignt years, this we give :
[30/(.5)]^3 = 216 000 (216 thousands contraction zone) and 100 billions divided by 216 thousands give about :
463 thousands galaxy (for a avérage contraction zone) and see if that is the same as M :
For M (included in a avérage contraction zone) : M = (1.64)(10)^49 KG = 824 535 galaxy mass as our, or (824.535)thousands mass as Milky Way,
this is (1.78) time the number that we have estimated witout our constant, this is a little leass that the first estimated value, and that give a good look
of the error of my estimations, this can not cancel my version of dark énergy and a wished estimated this version with the informations that we have.
     We get the densitée d = (4.23)(10)^(-26) kg/(m)^3
           I hope that i have not do too much errors, what i wish with this last calculus was to compare (constant) with Hubble constant and give
a estimation for M   , Rcritical    ,   and the dnsity d.