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*Subject*: R: [amsat-bb] OSCARLATOR charts on my webpage*From*: "i8cvs" <domenico.i8cvs@xxxxxx>*Date*: Sat, 30 Mar 2002 19:25:06 +0100

----- Original Message ----- From: Vince Fiscus, KB7ADL <vlfiscus@mcn.net> To: <amsat-bb@AMSAT.Org> Sent: Friday, March 29, 2002 9:19 PM Subject: Re: [amsat-bb] OSCARLATOR charts on my webpage > Will, > I found making the ground tracks and range circles the easy part. Back > when I used to fool around with the Oscarlocator, I wanted to write a > little computer utility that would use the epoch time, raan, mean motion, > and eccentricity from an element set and compute the time and longitude of > the first S-N EQX, nodal period & precession. I never did write it, but I > still think it would be a fun and very educational exercise to try. Need to > be able to computer the longitude angle of the vernal equinox, and the time > it takes the satellite at its current MA in the element set to reach the > equator. Inclination is needed to compute precession. > > 73 DE KB7ADL Hi Vince, Early 1994 i did an exercise on this matter to convert keplerians elements of RS 10/11 in to nodal ephemerides or time and longitude of equator crossing, EQX and i wrote a program for programmable calculators using magnetic cards like the Texas Instruments TI-59 and SR-52 This exercise was published in Radio Rivista 10/94 pages 40-44 and i hope it helps in developing a program for PC >From the Nautical Almanac i got the Sideral Time at Greenwich for the first of january 1994 at time 00:00:00 In the Almanac the Sideral Time Ts of the first point of Aries is given in degrees and it must be converted in fraction of a day of 24 h and thus. Ts = 99° 25' .424733 or 99°.42846481 ; 99.42846481 / 360 = 0,27619018 The Sideral Time Ts at Greenwich of coarse can be calculated for every year using a simple program instead of the Nautical Almanac. As you remember it was necessary to introduce Ts in to the old tracking programs and a list of Sideral Time at Greenwich was available for many years at that time and i am sure that routines for this calculations are available somewhere. In any case: The Longitude ( Le ) of the satellite ascending node in degrees W is given by Le= Ts - RAAN In this exercise for RS 10/11 we have the following keplerian elements: Epoch time: 94134.60454 Inclination: 82.93 deg RAAN : 353.67 deg Eccentricity: 0.0013 Mean Motion: 13.72337 rev/day Thus the satellite crosses the equator S-N EQX at Epoch 94134.60454 wich means the year 1994 on day 134 of the year or May 14 at time of fraction day 0.60454 Since the fraction of day is 0.60454 the equator crossing time EQX in hours and decimals is: 0.60454 x 24 = 14.50896 hours and decimals The Sideral Time in degrees at epoch day 134 or Ts(134) is computed with the following formula: Ts (134) = (((((134 - 1)/365.2422)+0.27619018)x 360)+(15 x 14.50896)= = 448.15396 deg In this formula the constants 365.2422 are the numbar of days of the average tropic-year and 15 are the degrees of the angular earth rotation in 1 hour Since the result 448.15396 deg is greater than 360 deg we must subtract 360 to it and so: Ts (134) at epoch = 448.15396 - 360 = 88.1539 deg Since Ts (134) is less than 360 deg than the longitude Le of the ascending node at epoch time must be calculated with: Le = (Ts + 360) - RAAN or ( 88.1539 + 360 ) - 353.67 = 94.4839 deg And hence RS 10/11 on day May 14 1994 crosses the equator ( EQX ) at longitude (W) 94.4839 deg (degrees west) at 14.50896 hour and decimals or 14:30:32.256 UTC To calculate the following EQX it is necessary to compute a) The nodal Period b) the increment of nodal longitude per orbit And hence for RS 10/11 first compute the orbit semimajor axis ( a ) with the following formula: 13 7.53766 x 10 1/3 a = ( ---------------------------- ) = 7369.505 Km 2 MM where: 7.53766 is a constant MM = satellite Mean Motion Now determine the parameter ( p ) of the ellipsis for the satellite orbit and for RS 10/11 we use the following formula and we get: 2 p = a ( 1 - e ) = 7369.492 Km where: a= semimajor axis in Km e = orbit eccentricity Now to compute the orbit perturbations of RS 10/11 we need to determine the constant ( A ) using the following formula: 7 2.377328 x 10 A= ---------------------- x MM = 6.0072371 2 p where: 2.377328 is a constant p= parameter of the ellipsis of the satellite orbit Now we can compute the nodal precession ( dRA )of the orbit in deg / day and for the RS 10/11 exercise we get dRA = A cos i = 0.739382 deg/day In this formula ( A ) is the above computed constant and ( i ) is the satellite orbit inclination i = 82.93 deg Since i < 90 deg dRA keep the signe minus because the orbital precession is from East to West in the opposite motion of the satellite and so the ascending node ( and RAAN as well ) approaches to the first point of Aries and thus: dRA = - 0.739382 deg/day Now for the RS 10/11 exercise we can compute the daily rate of change of rotation for the apsis line ( dW ) better know as the daily change of the argument of perigee. 2 dW = A ( 2 - 2.5 sin i ) = - 2.776107 deg/day In this formula ( A ) is the above computed constant and ( i ) is the satellite orbit inclination. Since the inclination i > 63.43 deg the value of dW must be negative because the apsis line thant join the apogee with perigee rotate in the opposite direction of the satellite motion. Of coarse if i < 63.43 deg the signe of dW must be changed positive because the apsis line rotate in the same direction of the satellite motion. If i = 63.43 deg the above formula shows that dW is about 0.001037 deg/day and so the orbit is very stable because the perturbation due of the oblateness of the earth is almost completely cancelled out. With the above computed dW we can now calculate the satellite nodal period wich is the time needed by the satellite to orbit the earth starting from one equator crossing time to the next one. For the RS 10/11 exercise we first compute the anomalistic period wich is the time needed by the satellite to orbit the earth from one perigee to the next one and thus: Anomalistic period = 1440 / MM = 104.9304945 minutes In this formula 1440 = 24 x 60 is the numbar of minutes in one day of 24 hours MM = RS 10/11 Mean Motion or 13.72337 rev/day The nodal period can be computed using the following formula: dW Nodal period = anomal. period - ( --------------- x anomal. period ) MM x 360 Replacing in it the already computed figures for the RS 10/11 exercise we get Nodal period = 104.9894568 minutes or 1.74982424267 hours and decimals It is evident that in the case of this exercise for RS 10/11 the nodal period of 104.9894568 minutes is greater than the anomalistic period of 104.9304945 minutes because the whole orbital plane rotates over itself in the opposite direction of the satellite motion and thus the satellite take more time to go from one equator crossing to the next one than the time it take to go from one perigee to the next one. The time of nodal period must be added to the previous time of the EQX to get the next EQX and in this exercise for RS 10/11 on day May 14 1994 we get: Time of equator crossing (hour and decimals) 14.50896000000 + Nodal period ( hour and decimals ) 1.74982424267= Next time of EQX ( hour and decimals ) 16.25878424267 Next EQX = 16.25878424267 = 16:15:31.6233 UTC Now we must compute the increment of nodal longitude ( dL ) in deg from one EQX to the next one using the following formula: dL = ( nodal period x 0.2506847407 ) - ( dRA / MM ) in this formula 0.2506847407 deg/min is the angular speed of rotation of the earth and came from the fact that the earth rotate by 360 deg in respect to a star or a sideral day in a time of 23 hour, 56 minutes, 04 sec or 23.93444444 hour and decimals and thus 23.93444444 x 60 = 1436.066667 minutes The angular speed of rotation of the earth is : 360 / 1436.066667 = 0.2506847407 deg/min Replacing figures in the above formula for the RS 10/11 exercice the increment of longitude of the ascending node is : dL = 26.373132 deg/orbit This longitude increment must be added to the longitude of the ascending node of the previous orbit and in this exercise for RS 10/11 Longitude of Equator crossing ( deg W ) 94.483900 + Increment of nodal longitud ( deg ) 26.373132 = Longitude of the next EQX ( deg W ) 120.857032 Using satellite nodal ephemerides or EQX the time of equator crossing is intended as UTC time and the longitude of the equator crossing is expressed in deg West that is beginning from Greenwich at 0 deg of reference longitude and going to the west direction by 360 degrees I hope that this exercise will be usefull to write a little program in BASIC to be used in DOS for quick EQX calculation and educational exercise to try. Also i am very interested to receive a copy of it Have a good job 73" de i8CVS Domenico > Via the amsat-bb mailing list at AMSAT.ORG courtesy of AMSAT-NA. > To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org > ---- Via the amsat-bb mailing list at AMSAT.ORG courtesy of AMSAT-NA. To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org

**References**:**Re: OSCARLATOR charts on my webpage***From:*Vince Fiscus, KB7ADL

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