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Anything that is NORTH of the equator is a possitive number +1 to +179 deg.
Anything that is SOUTH of the equator is a Negative number -1 to -179 deg.
*** UNCONFIRMED ***
First version....
The direction of the planet/moon/sun ROTATION determines which direction is 0 deg. and which is 180 deg.
ZERO DEG. is in the direction of the rotation, 180 deg. is in the other direction.
If you view the planet/moon/sun at the equator with north UP and south DOWN and the direction of the rotation is left to right like Kerbins then 0 deg. is to the EAST.
If the rotation is the opposite then 0 deg. is to the WEST.
If a planet/moon is tidal locked (Like the Mun, one face always pointing twords the parent object)
Then the 0 deg. is supposed to be taken from the parent object.
Second version...
0 deg. is taken from the ORBIT of the object (NOT the rotation) as viewed TWORDS the parent object.
View a planet/moon with the parent object behind it and 0 deg. will be twords the forward side of the orbit.
*** Problems with BOTH of those. ***
What if the planet/moon has a 90 deg. (Polar orbit)
In the first one if it's tidal locked you can't take the directions from the parent object because of the 90 deg. inclination of the orbit to the parent object.
What if the planet/moon has it's poles DIRECTLY on the orbit line/path.
In the second version with a 90 deg. incline of the equator it CAN'T be taken from the ORBITS direction.
If anyone has the CORRECT version of determining the direction of the 0 / 180 deg. of inclination I'd love to see a link.
But the launch site limits what inclinations you can reach -- to be exact, the longitude of the launch site defines the minimum inclination for any launch orginating at that site. This makes sense when you think about it -- if I launch from /here/, then obviously my orbit must pass over that location, so therefore it must reach at least this longitude.
KSC is located at 0 degrees, 5 minutes, 49 seconds (North) -- converted into fractional degrees, that is 0.097 degrees. Therefore, without heoric measures, you can't launch into an orbit with an inclanation of less that 0.097 from KSC: and if you have an appropiate mod installed, it will show your orbital inclination as 0.097 while sitting at the pad.
To launch into the minimum inclination for your launch site, launch due east [90 degrees on the Nav Ball] (or west, if you want a retrograde orbit). To launch into a polar orbit from any launch site, launch due north [0 degrees on the Nav Ball] (or south, if you want a retrograde orbit). To launch into intermediate inclinations requires math -- /lots/ of math. See http://www.orbiterwiki.org/wiki/Launch_Azimuth for the nasty details.
This is why KSC is located so close to the equator: as you've already learned, a launch azimouth of 45 degrees gives you an inclination of 45 degrees. However, be aware that this /won't/ be true when you start landing elsewhere (whether on Kerbin or other planatary bodies) and taking off again.
Now, in regards to your specific questions:
What is the inclination different when I launch at 225 vs 45 degrees?
Inclination tells you whether an orbit is prograde (0-90, 270-364.9999) or retograde (90.00001 - 269.99999. Picture a circle with 0 degrees marked as due east, with the number increasing as you move towards the top of the circle (north) and continues.
What is the dfference between AN / DN, is there any difference between the two?
When you are trying to match the orbit of another craft, each orbit is converted into a plane. The intersection of two planes is a line, but then we find where the two orbits intersect that line to produce four nodes -- two "Ascending", and two "Descending". However, only one of the two spacecraft involved is "active" at any given time (the one that you are controlling, of course), so only one pair matters, and these are marked on the orbital display as the "Ascending Node" [AN] and "Descending Node" [DN]. The designation as ascending or descending is really arbitrary (https://en.wikipedia.org/wiki/Orbital_node for the origin of the terms, if you are curious).
Plane change burns (made normal or anti-normal) are most effective at the AN or DN. If your orbit is highly eccentric (Pe and Ap are different), then the altitudes at the AN and DN will be different -- in which case, you can get the most bang for your buck by burning at whichever node has the /higher/ altitude.
It sounds like you are using MechJeb -- first, I'd recommend doing everything that MechJeb does for you by hand, especially rendezvous, a few dozen times before allowing MechJeb to take over. You'll learn alot more that way. :)
Setting that side, there are two points to be made for MechJeb:
1) I'm pretty confident that the "Inclination" field in the launch autopilot is /actually/ specifying the launch azimuth. From KSC this doesn't matter -- but from elsewhere...
2) MechJeb always makes plane changes at the /next/ AN or DN, even if your orbit is highly eccentric. On the other hand, MechJeb requires circular orbits to execute a Hohmann transfer, so in pratice it rarely matters.
3) MechJeb can and will perform a plane change burn as part of the "Fine tune closest approach" maneuver -- if the delta is fairly small (+/- 10 degrees seems reasonable). This will save you loads of fuel when trying to reach Minimus -- execute the Hohmann transfer /without/ matching planes (the orbit that results won't appear to intersect with Minimus -- that's normal) followed by a "Fine tune closest approach" (/now/ the orbit interesects with Minimus). This saves you fuel because the altiude (above Kerbin) is much higher when MechJeb creates the "Fine tune" manuver node than it would have been had you executed the plane change burn first. Also, note that this burn /won't/ occur at either the AN or DN -- you /can/ match planes with an arbitrary target at any point along your orbit, but you can't simply burn normal / anti-normal -- you have to burn a bit "off". Calculating how /much/ off you need to be is... Well, I can't even find the math to tell you how to calculate it. :) But MechJeb does the right math and it all works out in the end. :)
Does that mean that an inclination can be a negative value? Because I'm pretty sure that inclination only goes between 0 and +180 deg.
Longitude? Maybe you mean lattitude? Cause if i understand it correctly any orbit which doesn't cross the celectial body's surface goes through all the longitudes unless it's a synchronous orbit. It's actually an interesting definition but the problem is that it doesn't differentiate between a prograde and retrograde orbits.
At any rate here's a definition I came up with, after some additional reading, which (at least to me) seems to be comprehensive: inclination is an angle between a direction of ship's movement (prograde vector) at the ascending node and the reference direction.
My post was mainly thinking in terms of the direction you go from launch, so essentially, from the equator, which is also the orbital plane of Kerbin.
It was also only really aimed at resolving any confusion about why you could see the "same number" when clearly your orbit is anything but the same and why an orbit that "looks the same" but is travelling in opposite directions would have different numbers.
Edit:
That doesn't work either. If you launch from KSC toward the South East, your actual inclination will be -45 degrees. At the AN, it would measure +45.
The Mun has an inclination of zero. So, if you want to check your inclination in Kerbin orbit, set Mun as target, and hover the mouse over the AN/DN markers.
In other words, you can go either southeast or northeast at launch, inclination of both orbits will be roughly the same: +45 deg. Just the longitude of ascending node will be different.
I think the most practical solution is to use Kerbal engineer redux or the similair mod. The point of my question is not how to know my ship's inclination (I constantly have inclination in front of my eyes when flying) but to understand WHY my ship has this or that inclination.
Everything else is just measurement tradition.
1) My ageing brains memory of when I learned this stuff years ago.
2) The manner in which KSP and KER present this data.
Turns out there is no use of + or - in the "real world".
So, here's one line from the wiki for "orbital inclination"...
"The inclination of orbits of natural or artificial satellites is measured relative to the equatorial plane of the body they orbit if they do so close enough. The equatorial plane is the plane perpendicular to the axis of rotation of the central body."
Sorry, you are correct (and I checked it several times, because I knew I was going to get it wrong. Lattitude is correct.
However, with the word changed my statements is still true -- the maximum and minimum latitude a sattalite will visit is determined by its inclination. If your inclination is 45 degrees, then you'll never go further North than +45 degrees and never further South than -45 degrees latitude (as measured from the planet's surface, of course). This is why polar orbits are used so heavily for spy satallites (and, in KSP, required for resource scanners) -- they eventually /do/ pass over every point on the planet.
FYI: -20 degrees = 340 degrees = 700 degrees, in most contexts (including inclintaion). I assuemd, given the original question, that KSP (or, at least, the mod he is using to view it) uses only positive numbers for inclinations, thus "A simple example... bearing of 45 degrees ... inclination of 45 degrees ... bearing of 225 degrees ... inclination of 135 degrees ...". An inclination of 135 degrees could be represented just as well as -225 degrees.