MontBlanc2012
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Here is a straightforward challenge designed to test skills in orbital mechanics:
Ordinarily, this (somewhat arbitrary) orbit transfer could be constructed from a plane change manoeuvre; a retrograde burn to lower periapsis; and then another retrograde burn to lower apoapsis - so, three sequential burns in all.
However, the Delta Glider in question is fuel-constrained with just 2,030 m/s of dV for the main engines and 10 m/s in the RCS tanks. The goal of this challenge is, then, to find a more efficient way of making the orbital transfer - one that fits with this rather tight fuel budget.
The scenario for this challenge is given below. Just cut & paste into a txt file and save in the Orbiter Scenarios file as a .scn file.
---------- Post added 02-18-18 at 12:58 AM ---------- Previous post was 02-17-18 at 06:29 AM ----------
SPOILER ALERT:
For those interested: one solution to the challenge is a two-burn solution which, in TransX coordinates, is given below:
The first burn is 1422 m/s and the second burn is 604 m/s making a total of around 2028 m/s. You can try this sequence out and you should end up with an insertion into the target orbit to within about 0.01 degrees. If anyone can do better than this, please report.
A stock Delta Glider is in a circular orbit around Mars with an orbital radius of 9,000 km and coplanar with the plane of the ecliptic. For obscure operational reasons, the Delta Glider needs to transfer to a an orbit as follows:
Periapsis radius - 3,960 km
Apoapsis radius - 4,500 km
Inclination - 45 degrees
Longitude of ascending node - 135 degrees
Argument of periapsis - 65 degree
Ordinarily, this (somewhat arbitrary) orbit transfer could be constructed from a plane change manoeuvre; a retrograde burn to lower periapsis; and then another retrograde burn to lower apoapsis - so, three sequential burns in all.
However, the Delta Glider in question is fuel-constrained with just 2,030 m/s of dV for the main engines and 10 m/s in the RCS tanks. The goal of this challenge is, then, to find a more efficient way of making the orbital transfer - one that fits with this rather tight fuel budget.
The scenario for this challenge is given below. Just cut & paste into a txt file and save in the Orbiter Scenarios file as a .scn file.
Code:
BEGIN_DESC
END_DESC
BEGIN_ENVIRONMENT
System Sol
Date MJD 51982.0320109622
Help CurrentState_img
END_ENVIRONMENT
BEGIN_FOCUS
Ship GL-02
END_FOCUS
BEGIN_CAMERA
TARGET GL-02
MODE Extern
POS 4.000000 -84.078820 -30.845422
TRACKMODE TargetRelative
FOV 50.00
END_CAMERA
BEGIN_SHIPS
GL-02:DeltaGlider
STATUS Orbiting Mars
RPOS -6436164.352 0.020 6290929.391
RVEL -1524.8126 0.0000 -1560.0150
AROT -11.011 -49.870 -19.751
AFCMODE 7
PRPLEVEL 0:0.044404 1:0.005
NAVFREQ 586 466 0 0
XPDR 0
HOVERHOLD 0 1 0.0000e+000 0.0000e+000
AAP 0:0 0:0 0:0
SKIN BLUE
END
END_SHIPS
---------- Post added 02-18-18 at 12:58 AM ---------- Previous post was 02-17-18 at 06:29 AM ----------
SPOILER ALERT:
For those interested: one solution to the challenge is a two-burn solution which, in TransX coordinates, is given below:
Code:
Burn 1:
MJD: 51982.18271
Prograde: -890.905 m/s
Outward: 74.329 m/s
Plane: 1105.666 m/s
Burn 2:
MJD: 51982.27468
Prograde: -542.827 m/s
Outward: -32.921 m/s
Plane: -261.894 m/s
The first burn is 1422 m/s and the second burn is 604 m/s making a total of around 2028 m/s. You can try this sequence out and you should end up with an insertion into the target orbit to within about 0.01 degrees. If anyone can do better than this, please report.