[Date Prev][Date Next][Thread Prev][Thread Next] - [Date Index][Thread Index][Author Index]

R: OSCARLATOR charts on my webpage




----- 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



AMSAT Top AMSAT Home