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Author Topic: Returning Apollo Question  (Read 2315 times)

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Offline Lightspeed

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24 March 2003, 01:09:50
Hello all, glad to see an operating forum.  I have held this question to see if I can work it out but I
can't, so I thought I would ask for help.  I am using the apollo saturn V add on and have made it to
the moon, landed, redocked and returned to earth.  All but the last went wonderful.  Returning to
earth, I had to make some wild orbit changes to get there, and then from atmospheric interface to
parachute opening was about 1.5 seconds...lol.  My speed was over 10K/sec and my angle of entry
was about straight down.  So.....  How can I angle this to a survivable angle and make my re-
entry's more realistic?  Kinda puts a downer on events, after all the hard work that had been done
up to that point.   And If you dont care to add, how do you know in advance where your going to
land?  My apollo 11 astronauts stepped out of thier capsule onto the dirt of the deserts of east
Africa.  lol.   Thanks in advance.


Lightspeed

Offline Lightspeed

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Reply #1 - 24 March 2003, 02:13:19
One last question guys off the subject of my last question, this will be about the ISS Inclination.  
Real life has it at 51.6  with the data I can find.  In orbiter it is 73 degrees.  Any ideas of why?


« Last Edit: 24 March 2003, 04:24:01 by Lightspeed »
Lightspeed

rjcroy

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Reply #2 - 24 March 2003, 04:18:09
I can answer your second question.   The numbers are different because they describe the same orbit in
different reference frames.   The OrbitMFD displays data in a geocentric-ecliptic reference frame,
whereas the inclination of earth orbiting satellites are usually quoted in the geocentric-equatorial
reference frame.  

The 'geocentric' part of the reference (the same for both frames) states that the centre of the universe
for this reference frame is the centre of the Earth.   So position = <0, 0, 0>  is the centre
of the earth for both reference frames.  

The other part (equatorial or ecliptic) defines what plane, passing through the centre of the earth, is
parallel with an orbit of inlination 0°.    If the Z axis is 'up' and 'down' then this plane defines all
positions where you have no z component in your position, i.e. .  (this would much easier to
explain with a diagram :) ).

In the geocentric-equatorial reference frame (ISS inclination 51.66°)  the plane is parallel to the
equator.  So an orbit running along the equator has an inlination of 0° in the equatorial frame.  

In the geocentric-ecliptic frame (used in Orbiter's Orbit MFD)  the plane is parallel to a line running
from the centre of the earth to the centre of the sun.  Because the Earth is tilted at an angle of 23.45°
in this reference frame values of inclination will not be the same.  

Unfortunately converting a ecliptic frame inclination to an equatorial frame is not as simple as
substracting 23.45° from the OrbitMFD value.

« Last Edit: 24 March 2003, 04:24:01 by rjcroy »

Offline Lightspeed

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Reply #3 - 24 March 2003, 04:18:44
Doing alot of testing tonight.  Please forgive my numerous questions.  I have been testing orbit
INC's and real time data.  I have always wondered if Orbiter was correct in time to the true
position of the ISS.  So I calculated my current launch time and took off within 2 mins of true time
to see where the ISS would be, comparing to where the ISS really was at that time.  It was
nowhere close.  But, the exact time that sunrise was happening on the other side of the world
was correct and I was happy to see that.  I have flown this STS-112 flight many times, you take off
with a heading of 42 degrees and the launch date is OCT 7th, 2002.  You will normally end up with
an INC of 73.11 or around that.  Now, launching at present time, March 24 (GMT)  you end up with
an INC of 59.55 with the same heading.  

Why am I doing this?  I would like to launch previous shuttle missions at the exact time and INC
and orbit parameters that the real shuttle did.  But I dont understand why the INC is changing
with date.  I first thought it was because of the earths rotation around the sun, but, this should
not change reference to the equator.  So....  Once again, if anyone can shed some light on this I
would appreciate it.


« Last Edit: 24 March 2003, 04:24:01 by Lightspeed »
Lightspeed

Offline Lightspeed

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Reply #4 - 24 March 2003, 04:24:01
Thanks Rjcroy for the info.   I will study your response and that might be why the INC values are
off in my 3rd question or post.  Thanks again.


« Last Edit: 24 March 2003, 04:24:01 by Lightspeed »
Lightspeed