VB2019 Simultion of our Galagy, including sagitarius and 10 mini Black holes

Technical Data

Technical Data - Display 2 subtest 1

The right side of the bottom section contains the following data:
time/sec angle r min
"t" tt1 = 159698136800 angle00 = 162.37429431609 sqrt(a0)/pc = 0.000592310119256820
"t-1" tt1 - tdelta(i) = 159698136600 angle01 = 162.37901776951 sqrt(a1)/pc = 0.000592310118518200
"t-2" tt1 - 2*tdelta(i) = 159698136400 angle02 = 162.38374122282 sqrt(a2)/pc = 0.000592310119663260
"t min" tt103(i) = 159698136621 angle03(i) = 162.37850817662 sqrt(a3)/pc = 0.000592310118507230
"p03" anglep03 = 0.01998264328 anglep13 = 0.12275979307 anglep23 = 441.93525508186
"tmin" anglei(i) = 161.81503098032 angle03(i)-anglei(i) = 0.56347719629 tt103(i)-tt103i(i) = 158717156291
"tmin" angle03(i) = 162.378508117662 angle13(i) = 0.01120357016 angle23(i) = 40.33285259600
In order to understand the meaning of the above it is important to know that the the simulation is performed in cycles.
At each cycle the following 3 parameters are calculated:
  1. The time of the calculation. This is the cycle number, multiplied with "delta time". In this case "delta time" = 200.
    For line #1 this is: 798490684 cycles * 200 seconds = tcnt*tdelta = 159698136800 seconds
  2. The longitude of the pericenter W (capital W), which is equal to Omega + w
    w (small w) is the argument of the pericenter. See also: Carpe Diem and Sagittarius.htm#par4
    For line #1 this is: 162.37429431609 degrees.
  3. The square root (of the distance squared) divided by pc. That means the distance in par seconds.
    For line #1 this is 0.000592310119256820 par sec.
In many cases the parameters are indicated as par(i). The (i) identifies the object indicted and runs from 0 to 10. i=0 is Sagittarius A. i=1 is BH#1. i=2 is BH#2 etc. The actual numbers of the stars used are stored in the array source(i).
In a small number of cases the parameters are indicated as pari(i). The i in this case means initial value. In the first cycle that par(i) is calculated pari(i) is set equal to par(i). The parameter par(i)-pari(i) then indicates the increase (or decrease) of par(i) after the start of the simulation.

Technical Data - Curve Display 6b subtest 7

The right side of the bottom section contains the following data:
time/sec angle r min
"p03" anglep03 = 0.01998264328 anglep13 = 0.12275979307 anglep23 = 441.93525508186
sfx = 0.549881267824498 anglei(i) = 161.81503098032 angle03(i)-anglei(i) = 0.56347719629 tt103(i)-tt103i(i) = 158717156291
tcnt = 798526807 angle03(i) = 162.378508117662 angle13(i) = 0.01120357016 angle23(i) = 40.33285259600
tt103i(i) = 980980330 tt103(i) = 159698136621 angle13c = 0.01086608362 angle23c = 39.11790106185
id1 	a 	e 	i () 	omega  	w () 	Tp (yr) 	P (yr) 	Kmag 	q (AU) 	v (%c) 	dv      m0
S1 	0.5950 	0.5560 	119.14 	342.04 	122.30 	2001,800 	166.0 	14.70 	2160.7 	0.55 	0.03    12.40
S2 	0.1251 	0.8843 	133.91 	228.07 	66.25 	2018,379 	16.1 	13.95 	118.4 	2.56 	0.00	13.6
S4 	0.3570 	0.3905 	80.33 	258.84 	290.80 	1957,400 	77.0 	14.40 	1779.7 	0.57 	0.01    12.2
S6 	0.6574 	0.8400 	87.24 	85.07 	116.23 	2108,610 	192.0 	15.40 	860.3 	0.94 	0.00     9.2
S8 	0.4047 	0.8031 	74.37 	315.43 	346.70 	1983,640 	92.9 	14.50 	651.7 	1.07 	0.0     13.2
S9 	0.2724 	0.6440 	82.41 	156.60 	150.60 	1976,710 	51.3 	15.10 	793.2 	0.93 	0.02     8.2
S12 	0.2987 	0.8883 	33.56 	230.10 	317.90 	1995,590 	58.9 	15.50 	272.9 	1.69 	0.01     7.6
S13 	0.2641 	0.4250 	24.70 	74.50 	245.20 	2004,860 	49.0 	15.80 	1242.0 	0.69 	0.01    10.
S14 	0.2863 	0.9761 	100.59 	226.38 	334.59 	2000,120 	55.3 	15.70 	56.0 	3.83 	0.06    10.
S62     0.0905 	0.9760 	 72.76 	122.61 	 42.62 	2003,330 	 9.9 	16.10 	16.4 	7.03 	0.04    10.
S66 	1.5020 	0.1280  128.50 	 92.30 	134.00  1771,000       664.0 	14.80 	10712.4 0.21 	0.02     1 

Maximum/minimum distance calulation

In order to calculate the maximum (or minimum) distance a function ft = a + bt +ct^2 is used
The maximum is reached when dft/dt = 0 or when b+2ct=0 or when t=-b/2c
To calculate the three parameters a,b and c we use three measurements f0,f1 and f2 at t=0,t=1 and t=2
To calculate those three parameters we calulate the sum of minimum square distance between (ft - (a+bt+ct^2))^2 for t=0,1 and 2.
For t=0 we get (f0 - a)^2. For t=1 we get (f1 - (a+b+c))^2. For t=2 we get (f2 - (a+2b+4c))^2
The sum = (f0 - a)^2 + (f1 - (a+b+c))^2 + (f2 - (a+2b+4c))^2
Sum = (f0^2 - 2f0*a + a^2) + (f1^2 - 2*f1*(a+b+c) + (a+b+c)^2) + (f2^2 -2*f2*(a+2b+4c) +(a+2b+4c)^2
Sum = (f0^2 - 2f0*a + a^2) + (f1^2 - 2*f1*(a+b+c) + a^2 + 2ab + 2ac+ b^2 +2bc + c^2 + (f2^2 -2*f2*(a+2b+4c) + a^2 + 4ab + 8ac+ 4b^2 +16bc + 16c^2

dSum
-- = 0   -2f0 + 2a    -2f1  +2a+2b+2c    -2f2 + 2a+4b+8c = 0  or    f0 +f1 +f2 = a + a+b+c  + a+2b+4c = 3a+3b+5c   (1)  
da

 
 

dSum 
-- = 0                -2f1 + 2a+2b+2c    -4f2 + 4a+8b+16c = 0 or      f1 +2f2 =    a+b+c  + 2a+4b+8c = 3a+5b+9c  (2)     
db 

         


dSum
-- = 0                -2f1 + 2a+2b+2c   -8f2 + 8a+16b+32c = 0 or     f1 +4f2 =    a+b+c   + 4a+8b+16c = 5a+9b+17c (3)     
dc 



   (1) -  (2)  = f0+f1+f2-f1-2f2        = 3a+3b+5c -3a-5b-9c        or   f0-f2       = -2b-4c           (4)  
  5(1) - 3(3)  = 5f0+5f1+5f2-3f1-12f2   =  15a+15b+25c-15a-27b-51c  or   5f0+2f1-7f2  =-12b-26c         (5)
 -6(4) +  (5)  = -6f0+6f2 +5f0+2f1-7f2  = 12b+24c -12b -26c         or    -f0+2f1-f2  = -2c     or  c = (f0-2f1+f2)/2   (6)      
     (4)   -2b = f0-f2+ 4c       or -2b = f0-f2+2f0-4f1+2f2    or   -2b = 3f0 -4f1 + f2          or b = (3f0-4f1+f2)/-2

     (1)    3a = f0+f1+f2-3b-5c =  f0+f1+f2 -3((3f0-4f1+f2)/-2) - 5*((f0-2f1+f2)/2)
      *2    6a = 2f0+2f1+2f2 + 9f0-12f1+3f2  -5f0+10f1-5f2  = 6f0                                or a = fo
The minimum is at: t = -b/2c = ((3f0-4f1+f2)/-2) / 2*(f0-2f1+f2)/2 = -(3f0-4f1+f2)/ (f0-2f1+f2)


Reflection 1


Reflection 2


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Created: 14 April 2021

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