Posted: Sun Jan 17, 1988 9:45 PM CST Msg: FGII-3316-8136 From: PKARN To: amsat, amsatdl Subj: P3C Burn Planning Bob McGwier and I have just completed a preliminary set of calculations for the Phase III-C motor burns. Our assumptions were as follows: 1. Initial Ariane orbit as per Jan King's elements, massaged by James Miller: 222 km x 36086 km, i=10 deg. 2. Desired final orbit is 1500 km x 36086 km, i = 57 deg. 3. All burns are done at apogee, which is also the ascending node (this is implied by the transfer orbit's argument of perigee being 180 degrees). The spacecraft thrust vector for all burns is perpendicular to the major axis of the orbit, i.e., it is parallel to the surface of the earth directly below it. 4. No attempt is made to change the argument of perigee. At an inclination of 57 degrees it will increase by 26 degrees/year; 3.5 years after launch it will be at 270 degrees and apogee will occur at 57 N latitude. In another 3.5 years it will be 0 degrees and apogee will again occur at the equator. This gives a useful service life of 7+ years to the Northern hemisphere. 5. At no point during the burn(s) should the instantaneous orbit perigee decrease below 500 km, except for the beginning of the first burn (since perigee starts at 222 km). This keeps the satellite from re-entering should the motor fail at an inopportune time (or provide significantly less thrust than expected -- the delta-vees are all estimated by dead reckoning that assumes a given fuel flow rate and specific impulse figure. It would be nice if we had on-board accelerometers to measure real-time motor performance...) The straightforward approach to planning the burn would be to take the initial and final orbits, compute the velocity vectors for each at apogee, and program the burn for the vector difference. This would require a delta-v of 1,322.3 m/sec with the spacecraft thrust vector oriented at an angle of 99.2 degrees with respect to the initial velocity vector. Because this angle is greater than 90 degrees, however, a thrust component opposes the spacecraft's initial velocity. If the motor were to fail in the middle of the burn, or if its performance was significantly less than expected, the spacecraft would re-enter and burn up on the next perigee. The conservative approach we took instead is as follows: 1. Execute a first burn with the spacecraft thrust vector exactly perpendicular to the initial orbit plane. This raises both perigee and inclination, though not to their final values. 2. Reorient and execute a second burn to reach the intended final orbit. The purpose of the first burn is to make the second one safe; that is, the first burn should be just long enough so that at no time during the second burn will the instantaneous perigee decrease below a safe value (we chose 500 km). When we worked out the trigonometry, we got the following values: Burn #1: 453.735 m/sec, giving an intermediate orbit of 856 km x 36086 km, inclination 26 degrees (gee, 26 degrees sounds familiar!) Burn #2: 903.899 m/sec at an angle of 101.8 degrees to the plane of the intermediate orbit, giving a final orbit of 1500 km x 36086 km, inclination 57 degrees. The first burn gives us an opportunity to calibrate the motor performance by analyzing the actual intermediate orbit. Quick and accurate ranging during this period will be essential. If the burn is a little short, we can just do another one with the same orientation to make up the difference. If it is a little long, we will be taking more of a "insurance dogleg" than is necessary but we'll probably still make it to the final orbit. Clearly it would be better to err on the short side. Even though the angle for the second burn partially opposes the initial velocity vector, the higher initial perigee keeps it safe. During the second burn, the instantaneous perigee will decrease from 856 km through a minimum of 500 km at 338 m/sec into the burn before climbing back up to 1500 km at 903.899 m/sec. The total delta-v for these two maneuvers is 1,357.6 m/sec, only 35 m/sec greater than the "straight through the gauntlet" path. We consider this cheap insurance. Depending on how far the tanks can be emptied beyond that, we are left with perhaps 100 m/sec or so of propellant margin. One thing we might do with this extra margin (assuming it is real!) is to wait until the argument of perigee increases to 240 degrees in the fall of 1990 and then burn it at an ascending node. This would increase the inclination by a degree or so, slowing the argument of perigee increase slightly. KA9Q & N4HY Posted: Thu Jan 21, 1988 1:05 AM CST Msg: CGII-3322-7757 From: BMCGWIER To: amsat Subj: James Miller's program Most of you have found the easy "fix" that makes James very handy program work with your IBM or clone BASIC. I leave it here for completeness. I will be done with the simulation of the burns for PIII-C by tommorrow and will post the details of that study here and then post them to Germany along with the details and drawings that Phil and I did for the calculations. To James: Thanks this was helpful. 10 DEF FND(X)=180!*X/3.1415927# 20 DEF FNR(X)=3.1415927#*X/180! 100 T$="B.ESA_KEP": REM Convert ESA Launch Data to Keplerian Elements 110 REM 120 REM Version 1.1 Last modified 1987 Dec 29 130 REM 140 REM (C)1988 J.R.Miller G3RUH 150 REM 160 REM Free use encouraged provided credit given to author 170 REM 180 REM Note: RAD(X) is equivalent to PI/180*X, where PI=3.141592654 190 REM DEG(X) is equivalent to 180/PI*X 200 : 210 REM ESA orbital data for P3c (de Jan King, 87/Dec/25) 220 HP = 222.504 :REM Perigee altitude, km 230 HA = 36076.636# :REM Apogee altitude, km 240 IN = 9.997 :REM Inclination, deg 250 WP = 178.148 :REM Argument of perigee, deg 260 LAN= -135.541 :REM Longitude of ascending node relative to Kourou 270 TLAN= -9 :REM .. refered to T_liftoff + TLAN. (deg East, sec) 280 TA = 127.554 :REM True anomaly, deg 290 TE = 4797.1 :REM Epoch time relative to T_liftoff, sec 300 : 310 REM Additional data 320 RV = 0 :REM Orbit Number at TE 330 LK = -52.7016 :REM Longitude of Kourou, deg E 340 REM Constants 350 GM = 398600! :REM Earth gravitational constant km^3/s^2 360 RE = 6378.14 :REM Earth equatorial radius, km 370 GHA0 = 98.8897 :REM Greenwich hour angle at 1988 Jan 0.0, deg 380 WE = 360.9856473# :REM Earth rotation rate, deg/day 390 : 400 PRINT "Please enter time of LIFT-OFF; day, hr, min, sec" 410 PRINT "e.g. 1988 Apr 01 @ 1200 UTC, enter; 92, 12, 0, 0" 420 INPUT DL,HL,ML,SL 430 TL = DL+(HL+(ML+SL/60)/60)/24 :REM Lift_off time, days 440 : 450 A = (HA+HP)/2+RE :REM Semi major axis 460 E = (HA-HP)/2/A :REM Eccentricity 470 N0 = SQR(GM/A/A/A) :REM Nominal mean motion 480 EA = 2*ATN(SQR((1-E)/(1+E))*TAN(FNR(TA/2))) :REM Eccentric anomaly 490 MA = EA-E*SIN(EA) :REM Mean anomaly 500 LAN =LAN + LK :REM Longitude ascending node deg East .. 510 TLAN=TL + TLAN/86400! :REM .. at time TLAN, days 520 GHAA=GHA0 + WE*TLAN :REM GHA (Longitude) Aries at TLAN, deg W 530 RA = GHAA + LAN :REM Right Ascension of Ascending Node .. 540 RA = RA-360*INT(RA/360) :REM .. reduced to range 0 - 360 deg 550 TE = TL + TE/86400! :REM Epoch time, days 560 PRINT 570 PRINT"Nominal Keplerian Elements for P3c" 580 PRINT"----------------------------------" 590 PRINT"Epoch year ",1988 600 PRINT"Epoch time ",TE," days" 610 PRINT"Inclination ",IN," deg" 620 PRINT"R.A.A.N ",RA," deg" 630 PRINT"Eccentricity ",E 640 PRINT"Arg perigee ",WP," deg" 650 PRINT"Mean Anomaly ",FND(MA)," deg" 660 PRINT"Nom. Mean motion ",FND(N0)*240," rev/day" 670 PRINT"Decay ",0," rev/d/d" 680 PRINT"Revolution No. ",RV," - " 690 PRINT"Semi major axis ",A," km" 700 STOP