IMFD IMFD- What is EIn?

Krishnan

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Hey y'all.
So as you may know already, I have been trying to get my feet wet with IMFD, so I have been trying to do a moon trip with it.
I decided to read the IMFD manual by Mark Lieberbaum, however, I saw lots of mentions of EIn, but I didn't quite understand what it is.
Dimitri from David Courtney's videos said that it was essentially relative inclination, but the manual said something different.
Also, does anyone know how to get the scenario files that are given in the Manual? I thought I found them after some searching online, but it's not a recording with annotations.
Thanks everyone!
 

jarmonik

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Yes, that wiki page is indeed confusing.

The image is correct and the manual says:

"EIn" is Escape/Ejection Inclination. An angle between escape vector and the orbital plane of the ship, it is defined as: π/2 − angle (n, E). In a transfer programs, It is also the angle between the orbital plane of the ship and the target position at the time of interception. In the equation "n" is a normal vector of ship's orbital plane and "E" is a position vector of a body or it can be an escape vector depending of program.

"EIn" is not a relative inclination "RIn" which is an angle between two orbital planes.
 

Krishnan

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Yes, that wiki page is indeed confusing.

The image is correct and the manual says:

"EIn" is Escape/Ejection Inclination. An angle between escape vector and the orbital plane of the ship, it is defined as: π/2 − angle (n, E). In a transfer programs, It is also the angle between the orbital plane of the ship and the target position at the time of interception. In the equation "n" is a normal vector of ship's orbital plane and "E" is a position vector of a body or it can be an escape vector depending of program.

"EIn" is not a relative inclination "RIn" which is an angle between two orbital planes.
Ohh, so it is the angle from the target position at intercept and the orbital plane of the ship, that's what I understand. Why do we have to keep it to a minimum though?
 

N_Molson

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I'd say because of fuel efficiency and duration / timing of the launch window, but that's a guess, I can't prove the maths behind it.
 

Krishnan

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A diagram or something would be useful. I don't quite understand it.
1624151690343.png
This I see as relative inclination
 

Arvil

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OK reading this thread I just had a sudden insight. Imagine your ship’s orbital inclination is x degrees to the ecliptic. Imagine you set up IMFD with off-plane trajectory, and if Mars’ position at interception is a few degrees in declination above your orbital plane. So, in order to intercept Mars correctly, you’ll have to rotate the plane of your orbit up (tilt the orbit) so that it intersects Mars orbit right where Mars will be when you get there. It’s like a plane alignment, takes lotsa gas. But, adjusting the eject date and interception date will move the transfer orbit and Mars’ location until they coincide. I believe it’s changing your argument of periapsis rather than the inclination. Much cheaper. Or something like that; sorta makes sense?
 
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