Re: R: [amsat-bb] OSCARLATOR charts on my webpage

["Vince Fiscus, KB7ADL" <vlfiscus@xxxxxxx>]

I remember seeing this in various forms in astronomy books.  Once you have 
the equations, it's easy to write the code.  The fun part is learning the 
definitions and what it all means.


At 07:25 PM 3/30/02 +0100, "i8cvs" <domenico.i8cvs@tin.it> wrote:

>----- 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
>
>
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> >

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