1. 1984 Dec, No.50 p.2-6

2. 1985 Feb, No.51 p.2-6

3. 1985 Apr, No.52 p.8-11

* James Miller G3RUH
*

Historical Note --------------- This was my first "serious" article about astro-numerical stuff. It was written at a time when few people had computers, and even fewer knew how to use them. So to some extent the material was aimed as being do-able on a hand calculator as well. Again, at the time there was a total dearth of technical matter as herein described, although I didn't know so. I was simply pouring it out as a kind of public service. On reflection it's a bit clumsy and today I'd do it differently, but I hope it's still useful. This was also the series that got me "noticed" by the then AO-10 command stations. I was invited to generate input to help plan attitude schedules for the satellite. This I did, and also wrote a lot of navigation software which is still in daily use (1994) with Oscar-13. JRM --------------------------------------------------------------------------- SUN'S UP! Part 1 by J.R. Miller G3RUH Introduction ------------ A grand numerical offering this month - but don't panic, self-training does not lead to blindness! In the last issue I said that I'd write about modelling the Sun's orbit, so here it is. This information can be used to predict eclipses/shadowing, to check spacecraft illumination, computerise an alamanac, to navigate or simply to calibrate your garden sun dial. I had hoped to include some applications this session, but the space-time warp got me so it's carried forward to the next issue. Then I'll include calculating the angle of the Sun above or below the Oscar-10 orbit plane, and the vital solar panel illumination coefficient. If there's still space I'll then show how shadowing (eclipse) is predicted. Most of these ideas are embodied in an updated version of the program OSCAR10.BAS which appeared in O.N. 45 (Dec 1983), renamed PLAN10 and coded now for the BBC computer. This is shortly to be on sale via AMSAT, as is a listing for conversion into other BASIC dialects. [Updated 1990 Oct]. Sun Model --------- The Sun revolves around the Earth (sorry Copernicus but it's all relative, mate) just like any other satellite. So finding its position is just as easy! Looking further down this spiel may look horrendous, but experience shows that full frontal brevity results in mass confusion and a flood of mail, so I've tried to include everything. The Sun has an almost circular orbit, e = 0.017, and the time between equator crossings, called a tropical year, is 365.2422 days. Perigee (perihelion) is slowly moving so the time between successve perigees or anomalistic period is 365.2596 days. The orbit plane (ecliptic) is inclined at 23.44 degrees to the equatorial plane. The Sun's ascending node DEFINES the celestial origin, (the First Point in Aries). Thus the Sun's RAAN is always zero. To complete the ephemeris we need a reference time (epoch), a position round the orbit (mean anomaly), and the perigee position (or equivalent) at that time. First I give the Sun's ephemeris for epoch 1990 Jan 0.0. To find the Sun's location at any other time first calculate the elapsed time since epoch (step 1). In step 2 mean anomaly etc are cranked on to the time of interest. What you do next depends on what answers you want. Step 3a gives basic rectangular coordinates, such as you might use for satellite illumination studies. It's also the stepping stone to the other coordinates. Step 3b converts to an Earth based frame which will yield lat/long if required. Step 4a produces the astronomic coordinates (declination and right ascension) which can also be reduced to lat/lon (step 4b). The equation of time is appended in step 4c. GHAA (Greenwich hour angle Aries) is the same quantity that also features in most satellite position calculations. IMPORTANT: Now Read This ------------------------ 1. All angles are in DEGREES, and if your calculator/computer works in radians then take appropriate measures. 2. INT(X) means take the integer part of X, and is defined as the integer just LESS than X. Thus INT(-1.6) is -2. Some machines will return -1, so take care. This definition is regular through zero. ATN means arctangent, ASN means arcsine. ASN(X) = ATN(X/(SQR(1-X*X)). SQR means square root. 3. In the following equations it is implicit that any angle A is in the range 0 <= A < 360 degrees. Before re-use any result should be so bounded if need be. Effect this by A = 360*(A-INT(A/360)). Failure to heed this suggestion may result in clumsy answers. 4. The coordinate transformations are obviously interlinked, and not unique. Rearrange them as you will.

Figure 0.Showing astronomic and geocentric coordinate systems.Solar Ephemeris at 1990 Jan 0.0 ------------------------------- DAY0 726482 Day number of Epoch (origin approx. year BC 1) M0 356.6349 Mean anomaly at epoch L0 279.4033 Mean longitude along ecliptic EQC1 1.9151 Equation of centre harmonic coefficient EQC2 0.0200 " " " " OBLQ 23.4406 Obliquity of ecliptic (inclination) Step 1. Calculate number of days since epoch -------------------------------------------- Enter Year (e.g.1985), Day number (Jan 1 = 1), hours, minutes, seconds: YRDAY = INT( (YEAR-1)*365.25 ) - DAY0 Days from year difference FRDAY = ( HOUR + ( MIN + SEC/60 )/60 )/24 Fraction of day TIM = YRDAY + DAYNO + FRDAY Time in days since epoch Note: These formulae only valid 1900 Mar 01 - 2100 Feb 28. Step 2. Compute GHAA and True Longitude --------------------------------------- MA = M0 + 360 * TIM / 365.2596413 Sun's Mean Anomaly ML = L0 + 360 * TIM / 365.2421988 Mean Longitude GHAA = ML-180 + 360 * FRDAY Greenwich Hour Angle Aries TL = ML + EQC1*SIN(MA) + EQC2*SIN(2*MA) True Longitude Note: 1. 'Longitude' means angle from ascending node up the ecliptic. 2. The Sun's argument of perigee (AP) is given by AP = ML - MA Step 3a. Compute Unit Vector, Geocentric Rectangular Coordinates ---------------------------------------------------------------- Xsun = COS(TL) ( = COS(DEC) * COS(RA) ) Ysun = SIN(TL) * COS(OBLQ) ( = COS(DEC) * SIN(RA) ) Zsun = SIN(TL) * SIN(OBLQ) ( = SIN(DEC) ) Coordinate set is right-handed orthogonal located at the Earth's centre, X and Y axes in the equatorial plane; X points at the ascending node, or the first point in Aries (Vernal Equinox), and Y axis is at RA = 90 degrees; Z axis normal to the plane, pointing approximately at the Pole Star. Step 3b. Unit Vector, Geocentric, Greenwich Coordinates ------------------------------------------------------- C = COS(GHAA): S= SIN(GHAA) Xg = C * Xsun + S * Ysun ( = COS(LAT) * COS(-LON) ) Yg = -S * Xsun + C * Ysun ( = COS(LAT) * SIN(-LON) ) Zg = Zsun ( = SIN(LAT) ) Coordinate set is right-handed orthogonal located at the Earth's centre, X and Y axes in the equatorial plane. X points at intersection of Greenwich meridian and Equator, and Y axis points at longitude 90 degrees East. Z axis normal to the plane, pointing up through North Pole. This set is as the set above but rotated about the Z axis RAAN degrees East. Step 4a. Astronomic Coordinates -------------------------------- DEC = ASN (Zsun) Declination RA = ATN (Ysun/Xsun) Right Ascension if Xsun < 0 then RA = RA + 180 correct quadrant if RA < 0 then RA = RA + 360 ATN is assumed to yield a value in the range +- 90 degrees. RA is in the same quadrant as true longitude, TL. Step 4b. Geographic Coordinates ------------------------------- LAT = DEC Sun's Latitude, +North LON = GHAA - RA Sun's Longitude, +West Step 4c. Equation of Time ------------------------- EQOT = ( ML - RA ) * 4 Minutes Equation of time is the amount to be subtracted from 'sundial' time to give local time, maxima being -(14m 16s) around Feb 12th, and +(16m 24s) around Nov 3rd. The Sun is at maximum elevation at local noon minus EQOT. Test Data --------- The following example data are provided to help you to check your own calculations. EXAMPLE: Evaluate all data for, Aug 12th 1985 at 0145:00 GMT. All angles reduced to range 0 - 360 degrees. From diary: DAYNO = 224 YRDAY = -1826: FRDAY = 0.072917: TIM = -1601.927083 days MA 217.7751: ML 140.4681: TL 139.3144 GHAA 346.7181: RA 141.7354: DEC 15.0302 Xsun -0.758298: Xg -0.875425: LAT 15.0302 North Ysun 0.598108: Yg 0.407897: LON 204.9828 West Zsun 0.259328: Zg 0.259328: EQOT -(5m 4s) Sources and Accuracy -------------------- The solar data has been derived from: 1. The Astronomical Ephemeris 1984. ISBN:0-11-886919-1, HMSO & USGPO. 2. Explanatory Supplement to the Astronomical Ephemeris. 1961 Rev. 1974. ISBN: 0-11-880578-9, HMSO (presently out of print, new edn. in prep.) Angular errors are unlikely to exceed 0.2' of arc. The data may be used for a decade forward or backwards from 1990 with graceful degradation. These equations describe a 'mean' Sun's motion. Discrepancies compared with an almanac, are created partly by planetary purturbations to the Earth's orbit, partly by equating Universal time with Ephemeris time and partly to numerical adjustments deliberately embodied in almanac tables. It's the Earth/Moon centre of gravity and not the Earth's centre which orbits the Sun. There are other sources too - consult the references. SUN'S UP! Part 2 by J.R. Miller G3RUH This article is a continuation of the first part (O.N 50, December 1984) on the general theme of the relationship between the Sun and an Earth satellite. I'm using Oscar-10 in particular, but the methods are in general applicable to any of the others. First a few notes about vectors and then how to calculate solar elevation above the orbit plane, and the solar panel illumination coefficient. Eclipses will be dealt with in part 3. Vectors ------- Before proceeding I think it would be useful to introduce a few things about vectors. A vector is a neat way of describing a point or a line in space, and vectors can be added, subtracted and multiplied, as can ordinary numbers which are merely one-dimensional vectors. In our context a vector consists of three numbers, representing coordinates along the three axes. For example a point 1 unit along the X-axis, and zero units along the Y and Z axes can be compactly written < 1, 0, 0>. A line at 45 degrees to the X and Y axes, and in the same plane is < 0.71, 0.71, 0> A line equidistant from all three axes is < 0.58, 0.58, 0.58>, and so on. (Sketch them). The length of a vector < X, Y, Z > is given by L = SQR( X*X + Y*Y + Z*Z ). So you will see that all the above vectors are of length 1.0, and are not suprisingly called Unit Vectors. The angle between two unit vectors is found by what is called their scalar (dot) product. Denoting unit vectors A and B by < XA, YA, ZA > and < XB, YB, ZB> then their included angle ANG is found from: Cos(ANG) = XA*XB + YA*YB + ZA*ZB = < A > dot < B > = A.B The length of vector A projected onto unit vector B (or vice versa) is also given by their scalar product, though in this case A need not necessarily be of unit length. This is called 'resolving A onto B'. If the scalar product of two unit vectors is unity, then the vectors are coincident. If the scalar product of two vectors is zero then they are at right angles (90 deg) to each other. Vectors also very obligingly obey some familiar rules A+B = B+A, and A.B = B.A PROVIDED they are expressed in the same coordinate system. Verify these easy concepts for yourself using the three unit vectors above, and inventing some of your own. For further reading try "Advanced Engineering Mathematics", E. Kreyszig, John Wiley and Sons, ISBN 0-471-88941-5 or almost any similar title. Solar Elevation above Oscar-10's Orbit -------------------------------------- As a first example of using the sun's position let us work out the elevation of the Sun above (or below) Oscar-10's orbit plane. To do this we need a vector to describe the Sun's position, and another vector to represent the satellite's orbital plane; a perpendicular to the plane provides just this. Then the scalar (dot) product of these two gives the cosine of the angle 'twixt them. We want the angle above the plane which is 90 minus this (sketch it), i.e. the arcsine of their scalar product. Denoting the orbit normal components by < Xnorm, Ynorm, Znorm >, the Sun's vector as before by < Xsun, Ysun, Zsun >, and the solar elevation by SEL we have: SEL = ASN( Xsun*Xnorm + Ysun*Ynorm + Zsun*Znorm ) where for a plane, given its RAAN and inclination INCL the normal in celestial coordinates is: Xnorm = SIN(INCL) * SIN(RAAN) Ynorm = - SIN(INCL) * COS(RAAN) Znorm = COS(INCL) (Verify this for yourself with some sketches, trying various values for INCL and RAAN. For example if INCL = 0 then the normal vector becomes < 0,0,1 >, ie straight up the Z-axis, and so on). Continuing the example (Pt.1) for 1985 Aug 12th 0145:00 GMT, DATIM = 224.072917, Oscar-10 INCL = 25.6 deg and RAAN = 159.8 - 0.1724 * DATIM = 121.2 deg, we find: Xnorm = 0.3696 Xsun = -0.7583 norm . Sun Ynorm = 0.2238 Ysun = 0.5981 = 0.0874 Znorm = 0.9018 Zsun = 0.2593 and hence SEL = ASN(0.0874) = +5 deg. (Fig. 1) When solar elevation approaches zero, as it does twice a year, the sun lies in the satellite orbital plane, and hence Oscar-10 passes through the earth's shadow on each orbit. Also the smallest angular diameter of the Earth, seen from maximum range is +-ATN(6378/41000) = +-9 deg, so whenever the solar elevation is less than this (as in the above example) an eclipse is inevitable at some time round the orbit. You can also see that in 1987 the value hovers around zero for most of the year, so we can expect some frequent changes to the operating schedule in consequence.

Figure 1.OSCAR-10 SOLAR ELEVATION 1984-7. The maximum possible value is the sum of Oscar-10 and the Sun's inclinations, i.e. 25.6 + 23.4 = 49 deg.Illumination Coefficient ------------------------ From the satellite's point of view, perhaps the single most important parameter is the solar panel illumination coefficient, for without sunlight a non-nuclear powered spacecraft dies. You will know from the pictures how Oscar-10's panels are arranged, and as the satellite spins at 30 r.p.m. or so about the axis of symmetry each panel is illuminated in turn. Obviously for maximum effect the Sun needs to be perpendicular to the spin axis, whilst if it becomes co-axial then there will be no illumination. So to analyse this problem we need a vector representing the satellite's attitude. If we assume that Oscar-10 is nominally oriented, then the spin axis is exactly parallel to the orbit semi major axis (pointing in the apogee to perigee direction), so this gives us our attitude vector. And, as before, the cosine of the angle between Sun and spin axis is given by the scalar product of their two unit vectors. If we call this the Sun Angle, then 0 deg = no illumination, 90 deg = 100% illumination. We already have the Sun's vector, in celestial coordinates, so it remains to translate a unit length along the spin axis into the same coordinate set by rotating it through the various angles that define the orbit attitude; argument of perigee (AP), inclination (INCL) and right ascension of ascending node (RAAN). All elliptic orbit calculation programs do this, though probably a bit obscured. The answer fully worked out is: Xa = Cos(AP) * Cos(RAAN) - Sin(AP) * Cos(INCL) * Sin(RAAN) Ya = Cos(AP) * Sin(RAAN) + Sin(AP) * Cos(INCL) * Cos(RAAN) Za = Sin(AP) * Sin(INCL) The sun angle (SA) and illumination coefficient (ILL) are thus: SA = ACS( Xa*Xsun + Ya*Ysun + Za*Zsun) ILL = 100 * SIN( SA ) (percent) Worked example: Continuing as before for 1985 August 12 @ 0145 GMT. First work out argument of perigee for DATIM = 224.072917: AP = 336.6 + 0.2834*DATIM = 400.1 = 40.1 deg (see O.N. 50), and we already have RAAN = 121.2 and INCL = 25.6 from above. Evaluating we find Xa = -0.8931 Xsun = -0.7583 a . sun = 0.9608 Ya = 0.3534 Ysun = 0.5981 SA = ACS(0.9608) = 16 deg Za = 0.2783 Zsun = 0.2593 ILL = 28% A value of SA in the range 0 to 90 means that the Sun is illuminating somwhere between the spacecraft's top and midriff, while a value 90-180 degrees, the lower half. Exactly 90 degrees is the ideal condition, side-on.

Figure 2.OSCAR-10 NOMINAL ILLUMINATION computed as above for 1984 to 1987.An illumination factor of 28% is unacceptably low, and the managers will have had to take corrective action long before this happens, by reorienting the spacecraft. Now it is clearly desirable to keep the antennae pointing at the Earth as far as possible, so the favoured option is to 'twist' the satellite in the plane of the orbit only. A twist of about +30 degrees has been in use now since Autumn 1984, and has meant best access considerably before apogee. In terms of the mathematics shown above, an in-plane twist of TW is modelled quite simply by replacing argument of perigee AP in the equations by AP-TW ! Sketch it, and verify that an in terms of attitude an in-plane twist is appears identical to an AP change.

Figure3.OSCAR-10 ILLUMINATION. The effect of a +30 degree twist and the nominal curve. In particular note the improvement in illumination obtained around the beginning of 1985. However you can also see that the +30 degree twist will be no good for the period of our example, Autumn 1985.

Figure 4.OSCAR-10 ILLUMINATION showing the effect of a -30 degree adjustment again superimposed on the nominal. Now the null in August 1985 is raised to 50%, still not too good, but better than the original of less than 20%. A reorientation like this will have to be implemented around June 1st.This would also mean peak activity now concentrated after apogee, in much the same way as is presently done before it. In addition, as I remarked in O.N.49 page 3, eclipses are due in August 1985 around MA 85-130/256 when the transponder will have to be switched off. All in all an interesting time ahead. You might like to anticipate events of 1986-7, using the graphs. Visualising these effects is made immeasurably easier with my simple 3-D model (O.N.44 et seq) which comprises a hardboard ellipse and small globe. It can be made in a couple of hours. A table lamp a few metres away can be used to represent the Sun, and casts a shadow nicely for eclipse studies. In part 3: ECLIPSES. SUN'S UP! Part 3 by J.R. Miller G3RUH This is the final part of a series in which I've looked at the relationship between the Sun and a satellite. The first (O.N.50, December 1984) described a proper Sun model, whilst in the second (O.N.51, February 1985) vectors were explained and the results used to investigate sun angles and their effect on a satellite's solar illumination. In this part I deal with eclipses. Though I'm using Oscar-10 for my examples the methods apply to any Earth satellite, low orbiters like UOSAT, RS and NOAA through to OSCAR-10 and the geostationary comsats. Satellite Eclipses ------------------ A satellite is said to be 'eclipsed' when it is within the Earth's shadow, and the term 'shadowed' means the same thing. You can visualise what is happening in figure 5, which shows the shadow zone and a satellite within it.

Figure 5.Illustrating conditions of a satellite's solar eclipse.It should be obvious that the only condition necessary for an eclipse is that the distance marked UD (Umbral Distance) should be less than the radius of the Earth, since this is the size of the shadow. From the triangle, distance UD = Rsat * Sin(UA), so we need to work out Rsat and the Umbral Angle UA. Remembering the notes about vectors (pt. 2), the cosine of UA is simply the scalar (dot) product of a unit vector AWAY from the Sun, and a unit vector towards the satellite, i.e., denoting Cos(UA) by CUA: CUA = - (Xsun*Xsat + Ysun*Ysat + Zsun*Zsat) Since it's possible to get UD less than one Earth radius but on the SUNNY side of Earth, we must also check that the satellite is on the dark side, i.e. UA less than 90 degrees which means that CUA must be positive. Sun and Satellite's Unit Position Vector ---------------------------------------- The Sun unit vector calculations were dealt with in part 1 of this series. For the satellite we have to do the full orbital calculation, i.e. find mean anomaly (MA), thence eccentric anomaly (EA) and the in-plane coordinates (X and Y). These are then rotated through argument of perigee (AP), inclination (INCL) and then right ascension of the ascending node (RAAN) until we arrive at the satellite's position vector in celestial coordinates, the same as we used for the Sun. First solve Kepler's equation (in radians) MA = EA - EC*Sin(EA) for EA given MA where EC is eccentricity. This equation is 'transcendental' which means it's a sod, and has to be resolved iteratively (see appendix). Then for compactness in what follows denote: CEA = Cos(EA): SEA = Sin(EA) CAP = Cos(AP): SAP = Sin(AP) CNC = Cos(INCL): SNC = Sin(INCL): where INCL is Inclination. CRA = Cos(RAAN): SRA = Sin(RAAN) Rsat = SMA*(1-EC*CEA): where SMA = Semi Major Axis. X = (CEA-EC)/(1-EC*CEA): Unit satellite vector in Y = SQR(1-EC*EC)*SEA/(1-EC*CEA): orbit plane. Xsat = X*(CAP*CRA - SAP*CNC*SRA) + Y*( -SAP*CRA - CAP*CNC*SRA) Ysat = X*(CAP*SRA + SAP*CNC*CRA) + Y*( -SAP*SRA + CAP*CNC*CRA) Zsat = X*(SAP*SNC) + Y*( CAP*SNC) These calculations are embodied in all elliptic orbit programs (they use longitude of ascending node rather than RAAN because they work in geocentric coordinates). Equations above, though correct are inefficient and in the appendix they are presented more nicely. Eclipse - Worked Example ------------------------ Continuing the example started in parts 1&2, for Oscar-10 on 1985 August 12 at 0145 GMT. If you flog through the arithmetic you will find: MA = 129.3, AP = 40.1, RAAN = 121.2 (all in degrees - see pt1. page 2). Since EC = 0.61 we find EA = 147.9 degrees, and using a semi major axis of SMA = 26100 km you get, in addition to a headache: Xsat = 0.7864 Xsun = -0.7583 (as before. part 1 test data) Ysat = -0.5923 Ysun = 0.5981 Zsat = -0.1755 Zsun = 0.2593 Rsat = 39582 km CUA = 0.9961 UA = 5.09 deg UD = 3510 km The Earth's (equatorial) radius is 6378 km and, since the umbral distance is only 3510 km, at 0145 GMT August 12 1985 Oscar-10 will be eclipsed. Observability ------------- An MA of 129 degrees, (91/256 of the orbit) is 100 minutes before apogee, and the event will be observable for an hour or so (on the telemetry) by stations in Europe, Africa, the middle East and South America. The transponder will not, of course, be switched on! This 'eclipse season' lasts from 10th August to 3rd September 1985,and is illustrated on page 3 of O.N.49, Oct 1984 with an eclipse diagram. OSCAR-10 Planning Software -------------------------- These ideas and others are embodied in an updated version of the program which appeared as a pull-out centrefold in O.N. 45 (December 1983). Written in BBC BASIC and called PLAN10 it's available from AMSAT-UK. [ Updated 1990 Oct ]. It is well commented and serves as a useful comprehensive and comprehensible set of reference equations for solving the elliptic orbit problem, WITH DRAG. It can be dissected and refashioned to individual needs. It can be translated into other BASIC dialects. As coded, a sample PLAN10 output looks typically like the following, though as is usual with BASIC programs, the time increment and most else can be changed at will. You will notice the data is the same as for the examples in this series. SATELLITE PREDICTIONS --------------------- Location: G3RUH Lat 52.208 Lon 0.059 Ephemeris Dated: 1985 Day 224 OSCAR-10 YEAR: 1985 ORBIT: 1627 ILL: 28% SOLAR EL: 5 Deg DAY:HHMM MA/256 MODE RANGE EL AZ SQ UMD ECL? -------------------------------------------------------------- 224:0100 75 B 34114 19 203 25 1.07 + 224:0115 80 B 35320 18 204 22 0.85 ECL 224:0130 86 B 36396 17 205 20 0.66 ECL 224:0145 91 B 37348 16 206 17 0.55 ECL 224:0200 97 B 38178 14 207 15 0.58 ECL 224:0215 102 B 38891 13 208 13 0.73 ECL 224:0230 108 B 39488 12 209 12 0.94 ECL 224:0245 113 B 39971 10 210 10 1.18 + etc Wither? ------- If you have found this little series useful - or found it induced terminal boredom, I should like to know. I am quite happy to scribble more on the numerical aspects of satellites - but you may not be interested! Soon I propose to write about drag, and maybe about the non-spherical Earth. Perhaps you could suggest topics. Please don't forget the SAE though. [ I got 2 replies. One said that the article was "mind-blowing", and the other that I couldn't spell whither. JRM ] APPENDIX - Satellite Position Calculation ----------------------------------------- Equations are given in BASIC, as they've been tested in that form. By separating out the coordinate transformations the number of multiplies and additions are much reduced. The test at line 2140 is not set to umpteen decimal places. Newton's method is a 'second order' algorithm, so as coded will converge to within 0.002*0.002/2 radians, (for Oscar-10, this is equivalent to about 50 metres worst case position error, vastly better than the ephemeris accuracy). For definitions see part 1. 2000 REM Calculate Satellite Position 2010 REM ---------------------------- 2020 REM Assumes mean anomaly (MA), Argument of perigee (AP) and Right 2030 REM Ascension of Ascending Node (RAAN) to have been calculated 2040 REM as well as inclination (INCL), eccentricity (EC), semi-major axis 2050 REM (SMA) and semi-minor axis (SMB). All angles in RADIANS. 2060 2070 REM Solve MA = EA - EC*SIN(EA) for EA given MA. (Newton's Method) 2080 EA = MA: REM Initial solution 2090 REPEAT 2100 C = COS(EA): S = SIN(EA) 2110 DNOM = 1-EC*C 2120 DE = (EA - EC*S - MA)/DNOM: REM Change to EA for better 2130 EA = EA - DE: REM solution by this amount 2140 UNTIL ABS(DE) < 0.002: REM until converged. Coding as 2150 REM fast as possible. 2160 Rsat = SMA*DNOM: REM Geocentric distance 2170 2180 REM Now compute satellite coordinates in plane of ellipse 2190 C = COS(EA): S = SIN(EA) 2200 X1 = SMA*(C - EC) 2210 Y1 = SMB*S: REM Note: SMB=SMA*SQR(1-EC*EC) 2220 Z1 = 0 2230 2240 REM Rotate coordinates about Z through Argument of perigee 2250 C= COS(AP): S= SIN(AP) 2260 X2 = X1*C - Y1*S 2270 Y2 = X1*S + Y1*C 2280 Z2 = Z1 2290 2300 REM Rotate coordinates about X through Inclination 2310 C=COS(INCL):S=SIN(INCL) 2320 X3 = X2 2330 Y3 = Y2*C 2340 Z3 = Y2*S 2350 2360 REM Rotate about Z through RA of Ascending Node, and normalise 2370 C = COS(RAAN): S = SIN(RAAN): REM with respect to Rsat 2380 Xsat = (X3*C - Y3*S)/Rsat 2390 Ysat = (X3*S + Y3*C)/Rsat 2400 Zsat = (Z3 )/Rsat 2410 2420 REM Satellite's position unit vector in celestial coordinates 2430 REM now calculated. For geocentric coordinates use RAAN-GHAA 2440 REM above, instead of RAAN.

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