4313dim (01/18/83)
I have been trying to listen to UOSAT for about 3 months now
with the frustration of trying to determine when the next pass will
be. Many a time I have missed it due to miscalculations or having
to wait up to 1 hour for its crossing. Having had enough of this, I
have devised a simple program for the HP33 series calculator that,
given UOSAT data from this net, calculates the passes of UOSAT quite
accurately -- even two weeks ahead is within a few minutes! This
program could be adapted to any calculator; the program is self
explanatory.
Note that this program does not factor in the drag coefficient given
in the UOSAT data. This program was not meant to be that
sophisticated. Also, for best results, use the latest info for the
initial conditons of the satellite and orbital data.
The following registers are used:
0 number of iterations. Once the calculator has stopped, the
display will read the number of passes past what was
computed, ie, it will have a negative number for each time
the program is run past the last iteration.
2 This contains the present longitude for each calculation.
Initialize it with the information from the UOSAT bulletin.
3 Longitude increment in decimal form from the UOSAT bulletin.
4 the constant 360. Used for modulo 360 calculation.
5 time of equinox. Initialized from the the UOSAT bulletin in
decimal form (hours local or GMT)
6 time per revolution (hours) in decimal form from UOSAT
bulletin.
7 constant 24 for the modulo 24 calculation.
This is the program for a HP 33C calculator:
01 RCL 2 ;CALCULATE FOR NEW LONGITUDE
02 RCL 3 ;ADD THE INCREMENT
03 +
04 STO 2 ;ADD BACK TO LONGITUDE
05 RCL 4 ;LONGITUDE MUST BE MODULO 360 DEGREES
06 -
07 (g)X>0 ;CORRECT IF > 360 DEGREES (MODULO 360)
08 STO 2
09 RCL 5 ;NOW CALCULATE TIME OF EQUATOR CROSSING
10 RCL 6 ;ADD INCREMENT
11 +
12 STO 5
13 RCL 7 ;CORRECT IF >24 HOURS (MODULO 24)
14 -
15 (g)X>0
16 STO 5 ;IF >= 24 HOURS, SUBTRACT 24
17 0 ;CHECK FOR NUMBER OF ITERATIONS
18 RCL 0 ;FETCH NUMBER OF ITERATIONS TO GO
19 1
20 - ;DECREMENT SAME
21 STO 0 ;AND STORE IT
22 (f)X<=Y ;IF ZERO OR LESS, STOP
23 R/S ;STOP HERE. IF WANT ANOTHER PASS, HIT R/S
24 GTO 01 ;FOR EACH PASS, CALCULATOR WILL STOP WITH
;NUMBER OF PASSES DONE
An example of how this works is given:
On Jan. 7, the UOSAT data was given as:
Equatorial crossing @ 15:25:31 GMT and the Long. was 4.5 deg.
west. Therefore, the time was 10:25:31 EST which is about
10.425 hours decimal. This was entered into register 5.
4.5 was entered into register 2.
The period for the revolution of the craft was given as
94.787 minutes/pass which is 1.58 hours per pass. This is
entered into register 6. The longitudinal increment is
entered into register 3. This was given as 23.696
deg./pass.
Now, to find out how many times (iterations) are needed,
realizing that there are about 15 orbits per day, figure the number
of iterations as days times 15. I wanted the passes for the afternoon
of Jan. 17, which is 10 days after Jan. 7, or 150 iterations.
This was entered into register 0. Now start the program at
01 and wait. When it stops, register 2 will have 7.425 EST
and register 5 will be 318.9 deg. The angle here is too low
for any use; use anything between +- 15 deg. from your longitude
is sure to work. By hitting the R/S button once, the
time and long. registers are incremented once. Again look
at the values to see if they are within range. This single
iteration may be repeated. After four single iterations
(ie, pressing R/S four times) -4 will be displayed. The
time in register 5 will be 13.745 hours (13:45 EST) and the
angle will be 53.7 deg. One more iteration yields 15.325 hours
(15:20 EST) and 77.4 deg. Being in NJ (~74 deg W) I found that
I could hear (with only a 5/8 wave and a mobile rig which
proves that you don't need anything special) the 13:45 EST
pass (54 deg W) and the 15:20 EST pass (77 deg W).
Note that the result in register 5 is in hours decimal form.
Convert this into the normal hours:mins for convenience.
Also note that this is the equator crossing time. It takes
about 3.8 deg./min. for the craft to make its way up from the
equator, so for NJ (~41 deg N) add about 11 min. to the time
for the overhead crossing. This yielded a 15:31 pass that
WB2HJX will attest to.
Please direct any comments, etc. to this newsgroup or to me,
houxf!4313dim. Enjoy.
Scott McLellan KA2JIVkarn (01/18/83)
For those with personal computers, I can recommend several good programs to give azimuth/elevation data as functions of time for satellites such as Uosat; if there is interest, I'll post further details. For those who would like to use Scott's HP-33 program, I would be happy to post equator crossings derived from the NASA prediction bulletins I receive for the following satellites, if there is sufficient interest: Oscar 8,9 RS 3 thru 8 Noaa 6,7 Phil Karn, KA9Q/2