## 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.
• 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 anglee 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)

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Created: 14 April 2021

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