Visual Basic program "VB 2019 Sagittarius"

VB2019 Simultion of our Galagy, including Sagittarius 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 simulation is performed in cycles.
At each cycle the following 3 parameters are calculated:

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
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.
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.

The first three lines show these results of the three cycles (indicated as "t-2","t-1 and "t")
Line #4 shows the resuls when the star reaches perihelium. The three parameters used are: tt103(i), angle03(i) and sqrt(a3)/pc.
The parameter tt103(i) indicates the time. The parameter angle03(i) indicates the angle W (captial w) and the parameter sqrt(a3)/pc the distance.
In line #5 the three parameters are:anglep03, anglep13 and anglep23.
anglep03= angle03(i) - angle03old(i). anglep13 = anglep03*century/(tt103(i)-tt103old(i)). anglep23= anglep13 *3600
In line #6 the three parameters are:anglei(i), angle03(i)-anglei(i) and tt103(i)-tt103i(i)
angle03(i) is the longitude of the pericenter W. anglei(i) is the initial value of angle03(i).
In line #7 the three parameters are:angle03(i), angle13(i) and angle23(i).
angle03(i) is the longitude of pericenter W. angle13(i) = angle03(i)*century/(tt103(i)-tt103i(i)) .
angle23(i)= angle13(i) *3600

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

Line #1 shows: anglep03,anglep13 and anglep23
anglep03= angle03(i) - angle03old(i). anglep13 = anglep03*century/(tt103(i)-tt103old(i)). anglep23= anglep13 *3600
Line #2 shows sfx, anglei(i), angle03(i)-anglei(i), and tt103(i)-tt103i(i)
GETCOEF1 is a subroutine, which calculates the average of all the simulated results of angle03(i)-anglei(i). The result is sfx.
angle03(i) is the longitude of the pericenter W. anglei(i) is the initial value of angle03(i).
tt103(i) is the time when the star reaches the pericenter W. tt103i(i) is the time first event.
Line #3 shows: angle03(i),angle13(i) and angle23(i)
angle03(i) is the longitude of pericenter W. angle13(i) = angle03(i)*century/(tt103(i)-tt103i(i)) .
angle23(i)= angle13(i) *3600
Line #4 shows: tt103i(i),tt103(i),angle13c and angle23c
tt103(i) indicates the time when star is at Pericenter. tt103i(i) indicates the first event. angle13c = sfx*century/tt1. angle23c= angle13c *3600

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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