Longitude of ascending node

[downloaderfan]

Alright,i know this is the 4th thread in orbiter forum about this topic,but even after reading those and some websites i found from google,i'm having a hard time understanding what is the longitude of ascending node.

Perhaps the key lies in understanding this diagram:
[ame="http://en.wikipedia.org/wiki/Longitude_of_the_ascending_node"]Longitude of the ascending node - Wikipedia, the free encyclopedia[/ame]


If someone could explain me in terms of a total noob,i would appreciate it.
I'm finding it difficult to understand those angles in 3D.

Another reason i started this thread is because i wanted to know how to control the value of LAN during an ascent because when i launch a ship with a particular heading from a particular location,it has some value of LAN,and then if i launch another ship following the same heading after a while,i find that it has a different LAN,changing the position of nodes with my target as well.

Thank u

 

[Urwumpe]

Well, you have to remember that LAN in the Orbital elements is the LAN at the epoch J2000 (Julian Epoch, Year 2000) or: What the longitude of your ascending node would have been on January 1, 2000.

It changes over time in space, because your orbit is changed by Earths gravity (non-spherical gravity). And if you launch a few hours later on the same heading, it changes because Earth moves around the sun (and your orbit would be theoretically fixed in space) and because Earth rotates as well and with it the Greenwich meridian.

It is just a reference for describing the orientation of the orbit. In reality, nobody would calculate your orbit back in time. Instead, a coordinate system is used in the calculation, that is fixed for the situation at J2000, not the current rotating coordinate system of Earth.

You control the LAN just like you control the inclination. Generally speaking, you do the plane change burn at the northern most or southern most point of the orbit for that. (But better use align plane MFD anyway)

Understanding the angles is pretty easy: You start in an unrotated orbit around the equator. Your periapsis is at 0° longitude (in J2000). First you rotate by argument of periapsis. This angle moves the line connecting perigee and apogee inside your orbit plane.

Then you rotate by inclination, the axis is radial to Earth and passes through 0°N 0°E (at J2000).

Finally you rotate around the rotation axis of Earth (Again in J2000) by the longitude of the ascending node.

The J2000 coordinate system is fixed in space, it doesn't rotate with Earth: Earth rotates below your orbit while the angles remain constant.

 

[downloaderfan]

Well, you have to remember that LAN in the Orbital elements is the LAN at the epoch J2000 (Julian Epoch, Year 2000) or: What the longitude of your ascending node would have been on January 1, 2000.

It changes over time in space, because your orbit is changed by Earths gravity (non-spherical gravity). And if you launch a few hours later on the same heading, it changes because Earth moves around the sun (and your orbit would be theoretically fixed in space) and because Earth rotates as well and with it the Greenwich meridian.

It is just a reference for describing the orientation of the orbit. In reality, nobody would calculate your orbit back in time. Instead, a coordinate system is used in the calculation, that is fixed for the situation at J2000, not the current rotating coordinate system of Earth.

You control the LAN just like you control the inclination. Generally speaking, you do the plane change burn at the northern most or southern most point of the orbit for that. (But better use align plane MFD anyway)

Understanding the angles is pretty easy: You start in an unrotated orbit around the equator. Your periapsis is at 0° longitude (in J2000). First you rotate by argument of periapsis. This angle moves the line connecting perigee and apogee inside your orbit plane.

Then you rotate by inclination, the axis is radial to Earth and passes through 0°N 0°E (at J2000).

Finally you rotate around the rotation axis of Earth (Again in J2000) by the longitude of the ascending node.

The J2000 coordinate system is fixed in space, it doesn't rotate with Earth: Earth rotates below your orbit while the angles remain constant.

Thanks for the quick reply,
I will still have to think a little bit more about what u said,but i would like to ask something,cant i just adjust the orientation and take care of the LAN during the ascent itself(like the inc)?As plane change costs a lot of fuel.

 

[Urwumpe]

Thanks for the quick reply,
I will still have to think a little bit more about what u said,but i would like to ask something,cant i just adjust the orientation and take care of the LAN during the ascent itself(like the inc)?As plane change costs a lot of fuel.

Sure - you just need to launch at the right time. For the correct LAN, the launch window becomes important. Every sidereal day, you have at least one chance to fly into the desired LAN. Can also be two chances, if your spaceport and target inclination permits. (You can fly a northern trajectory or a southern trajectory to reach the target inclination, unless your latitude is equal to the equatorial inclination)

 

[downloaderfan]

Sure - you just need to launch at the right time. For the correct LAN, the launch window becomes important. Every sidereal day, you have at least one chance to fly into the desired LAN. Can also be two chances, if your spaceport and target inclination permits. (You can fly a northern trajectory or a southern trajectory to reach the target inclination, unless your latitude is equal to the equatorial inclination)

How can i come to know when it is the right time to launch for my desired value of LAN,perhaps some MFD you could point out...

 

[Urwumpe]

How can i come to know when it is the right time to launch for my desired value of LAN,perhaps some MFD you could point out...

Launch MFD should be perfectly capable of that.

 

[downloaderfan]

Launch MFD should be perfectly capable of that.

Thanks for all ur info,still little bit confused,if im orbiting earth,then the reference direction(from that diagram) refers to some point in the earth's equator,right?What point would it be?The first point of areis?

After a closer look at launch MFD,when i click the option of inclination,it asks me to input INC and LAN with respect to the ecliptic frame,not the earth's equator...with respect to the earth's equator,it only asks for inc.(Unless I'm interpreting it wrong)

 

[martins]

I find it easiest to describe the elements like this:

The 6 scalar elements describe the orbit of an object, where two of them (semi-major axis and eccentricity) describe the shape of the orbit, three (inclination, argument of periapsis and longitude of ascending node) describe the orientation of the orbit in space, and the last (true anomaly, or mean anomaly at epoch or at a reference date) describes the position of the orbiter along the orbit.

The three elements describing the orientation are essentially the Euler angles of the orbit's rotation, where the longitude of ascending node describes a rotation around the axis perpendicular to the reference plane.

Edit: Of course, the three rotational elements depend on the choice of reference frame (e.g. ecliptic/equatorial)

 

[downloaderfan]

Thanks a lot martin,i got a basic general idea of what the different terms of an orbit describe,but for a better understanding,let me ask the question in this way,

Lets say im in an orbit of inclination 47 degrees and a LAN of 50 degrees with respect to the earth's equator...what does that mean?

BTW please accept my personal gratitude for making orbiter

 

[Urwumpe]

Lets say im in an orbit of inclination 47 degrees and a LAN of 50 degrees with respect to the earth's equator...what does that mean?

That your orbit ground track passes between 47° North and South latitude and that your relative inclination to an Orbit with 0° LAN and 47° Equatorial is 50 Degress.

 

[martins]

:hesaid:
Essentially, the inclination describes the amplitude of the ground track, while the LAN describes the horizontal shift (notwithstanding the Earth's rotation, which also shifts the ground track).

In linear algebra terms, (and forgetting about argument of periapsis) starting from an orbit in the equatorial plane, with periapsis aligned with vernal equinox, first rotate around an axis in the equatorial plane perpendicular to vernal equinox by 47 degrees, then rotate around normal to equatorial plane by 50 degrees.

 

[downloaderfan]

That your orbit ground track passes between 47° North and South latitude and that your relative inclination to an Orbit with 0° LAN and 47° Equatorial is 50 Degress.

Oh i get it now,50 degrees counterclockwise rotation wrt to the orbit having LAN 0 degrees.

The only thing i would like to know now is how to use launch MFD to get my desired LAN,should i take off a few seconds before time to intersection in launch MFD?

 

[Cras]

Oh i get it now,50 degrees counterclockwise rotation wrt to the orbit having LAN 0 degrees.

The only thing i would like to know now is how to use launch MFD to get my desired LAN,should i take off a few seconds before time to intersection in launch MFD?

Launch when the time to intersection equals the time to reach half of orbital velocity.

If you have launched that vehicle with Launch MFD before, that value will be displayed, so launch when the intersection countdown reaches that value.

 

[Enjo]

The best time to launch is also given more intuitively by offplane correction being at 0* (having the time described by Cras)

---------- Post added at 08:10 PM ---------- Previous post was at 08:07 PM ----------

After a closer look at launch MFD, when i click the option of inclination,it asks me to input INC and LAN with respect to the ecliptic frame,not the earth's equator...with respect to the earth's equator,it only asks for inc.(Unless I'm interpreting it wrong)

Have an even closer look. Suffix the inc lan input with C for eCliptic and Q for eQuatorial

 

[Miner1]

Actually, the relative inclination between an orbit with [math](\Omega_1,i_1) = (50,47)[/math] and an orbit with [math](\Omega_2,i_2) = (0,47)[/math] is approximately 36 degrees.

[math]cos(i_{rel})=cos(i_1)cos(i_2)cos(\Omega_1-\Omega_2)+sin(i_1)sin(i_2)[/math]

 

[downloaderfan]

Alright,i took off with the delta glider as per Cras's instructions in the default 'Launch and dock to the ISS' challenge that comes with orbiter 2010,and i had a relative inclination less than 0.1 degrees when i got into orbit.Thanks for all ur help guys. :thankyou:

 

[Enjo]

Another Soul saved

 

[downloaderfan]

Another question,lets say i want to arrive earth with inclination 11 degrees and LAN 48.42 degrees with respect to the ecliptic from the moon.Whats the inclination and LAN i should have in a lunar orbit so as to arrive earth without(or with least)plane change?

 

[Enjo]

I believe that you should simply depart with ecliptic inclination, and after leaving lunar SOI, use IMFD to put you into that orbit.

 

[Cras]

I believe that you should simply depart with ecliptic inclination, and after leaving lunar SOI, use IMFD to put you into that orbit.

That is how I have done it.