What reference direction determines the negative sign for inclination angles?

[Dunbaratu]

I'm trying to do some stuff with the KOS mod about orbital inclinations and I figured out how to calculate the inclination from the current velocity vector, but I have no idea what KSP uses to decide which tilt is as positive angle and which is a negative angle.

The inclination is simply this:

arccos( magnitude_of_y_component_of_velocity / magnitude_of_entire_velocity_vector ).

(since "Y" is always the axis of the SOI body's north pole direction.)

But the problem is I can't tell the difference between, say, inclination of +5 degrees or -5 degrees.

In other words, what does KSP use as the "reference direction" in this diagram from wikipedia:

And anything I can think of to use as a reference point is something that would keep changing:

It can't be based on which side of the orbit has the periopsis and which has the apoapsis because if it was then objects with matching inclinations but differently shaped orbits could have opposite inclination signs from each other if it worked like that.

It can't be based on the longitude lines of the body because the planet rotates once a day (it's day) underneath you and what had been an ascending node at a chosen longitude might eventually become a descending node at that longitude about half a day later as the planet turns on its axis.

It can't be based on which side of the planet its orbit around its parent body comes out, for the same reason - it rotates around once a year (it's year) as the reference flips around to the opposite half a year later.

For this reason in the real world astronomers use a reference point external to the solar system. The reference direction they use is to point at the constellation Ares. If your orbit goes above the equator when on the Ares-facing side of the planet, it's a positive inclination. If your orbit goes below the equator when on the Ares-facing side of the planet, it's a negative inclination.

But how does that work in KSP? There's nothing external to the solar system. What is the reference direction KSP uses to decid whether an inclination is positive or negative?

Edited October 16, 2013 by Steven Mading
setting prefix to solved.

[Meithan]

Quoting from maltesh's Kerbal Orbit Parameters:

"Reference Direction: In the universe of Kerbal Space Program, at UT=0.0, a ray drawn from Kerbin to Kerbol points about 0.09° east of the Reference Direction."

It's not a very intuitive reference, really.

However, orbital inclination doesn't really have a sign. I mean, there is no difference between +5° or -5°: in both cases, the angle between the planes is just 5°. You can see this from the diagram you posted: you can measure inclination by measuring the angle marked i, or by measuring the angle opposite to it to the left. It makes no difference.

Why do you need the sign?

[Dunbaratu]

Quoting from maltesh's Kerbal Orbit Parameters:

"Reference Direction: In the universe of Kerbal Space Program, at UT=0.0, a ray drawn from Kerbin to Kerbol points about 0.09° east of the Reference Direction."

It's not a very intuitive reference, really.

However, orbital inclination doesn't really have a sign. I mean, there is no difference between +5° or -5°: in both cases, the angle between the planes is just 5°. You can see this from the diagram you posted: you can measure inclination by measuring the angle marked i, or by measuring the angle opposite to it to the left. It makes no difference.

Why do you need the sign?

Here's why the sign matters: Two satellites which both have an inclination of +5 relative to the SOI body, and which cross the equator above the same point on the planet have zero inclination relative to each other. Two objects, one of which has an inclination of -5 and the other of which has an inclination of +5, which also cross the equator at the same point on the planet have an inclination difference of 10 relative to each other.

[Specialist290]

Assuming I'm interpreting the orbital parameter definitions correctly, it should be fairly easy to find the reference direction through one simple parameter: The Longitude of the Ascending Node.

The ascending node is the point where the orbit crosses over from "south" of the reference plane to "north." If you set the longitude of the ascending node to exactly 0 degrees, then that means that your orbit is going to cross the equator at a point where, if you drew a line straight up from the surface to your craft, that line would be pointing straight at the reference direction.

[Meithan]

Ah, then as Specialist290 says, what you're looking for is the longitude of the ascending node (LAN). If you want the two satellites to be on the same orbital plane, their inclinations and their LANs must be equal. In your example, your two satellites have LANs that differ in 180°, which is what you're interpreting as a minus sign in inclination. Also note that the actual value of the LAN is not important (so determining the reference direction is not necessary): as long as they're the same (and inclinations match), the orbits will be on the same plane.

[Dunbaratu]

Assuming I'm interpreting the orbital parameter definitions correctly, it should be fairly easy to find the reference direction through one simple parameter: The Longitude of the Ascending Node.

The ascending node is the point where the orbit crosses over from "south" of the reference plane to "north." If you set the longitude of the ascending node to exactly 0 degrees, then that means that your orbit is going to cross the equator at a point where, if you drew a line straight up from the surface to your craft, that line would be pointing straight at the reference direction.

That is not true, because the reference direction does not rotate with the planet. If the reference direction were, hypothetically at 0 degrees longitude at some moment in time, it would be 180 degrees longitude exactly half a day later as the planet rotates and the reference direction does not.

[Dunbaratu]

Ah, then as Specialist290 says, what you're looking for is the longitude of the ascending node (LAN). If you want the two satellites to be on the same orbital plane, their inclinations and their LANs must be equal. In your example, your two satellites have LANs that differ in 180°, which is what you're interpreting as a minus sign in inclination. Also note that the actual value of the LAN is not important (so determining the reference direction is not necessary): as long as they're the same (and inclinations match), the orbits will be on the same plane.

I'm not the only one interpreting it as a minus sign. the game ITSELF is doing that. When you select a target it chooses to arbitrarily call the inclination either positive or negative and I am trying to determine what it's using to decide that. Even if my LAN and my target's LAN are at the same spot, if our inclinations differ they will sometimes be called positive and sometimes negative. If inclinations are not signed, then what is the game saying when it says my inclination is "-2.1 degrees"?

[rhoark]

arccos( magnitude_of_y_component_of_velocity / magnitude_of_entire_velocity_vector )

That only holds at the equator. As you approach the orbit's extremes of latitude, the y component approaches 0. The magnitude of the velocity vector can be constant (circular orbit). Therefore the formula as given can report anything in the range of the inclination, up to 90.

[Superfluous J]

All I know is that if its negative, I need to push up on my maneuver node. If it's positive, I need to push down. I've found that this is the case always, whether it be ascending node or descending node, or if the node is ahead or behind my orbit.

I assumed that the sign didn't actually matter and was just there to tell the player how to modify the maneuver node. Seems far to useful to be just a lucky chance.

Now that I think about it, it seems that it'd be negative if you are heading (celesital) SOUTH when you cross it, and positive if you're heading (celesital) NORTH. I think this because no matter what you do, map mode always has Kerbol's (and Kerbal's) North Pole facing up.

[Meithan]

I'm not the only one interpreting it as a minus sign. the game ITSELF is doing that. When you select a target it chooses to arbitrarily call the inclination either positive or negative and I am trying to determine what it's using to decide that. Even if my LAN and my target's LAN are at the same spot, if our inclinations differ they will sometimes be called positive and sometimes negative. If inclinations are not signed, then what is the game saying when it says my inclination is "-2.1 degrees"?

I see. Could it be that the game measures inclination in the range [-90°, 90°] instead of [0°,180°]? Usually, 0° means an equatorial prograde orbit, 90° is a polar orbit, and 180° is an equatorial retrograde orbit.

[NefariousParable]

You are confusing two separate parameters in the determination of the orientation of the orbit. There exists three elements that determine the orientation of the orbital plane - The first is inclination (i), the second is the Right Ascension of the Ascending Node [RAAN] [in your diagram it is referred to as longitude of ascending node] (capital Omega), and the third is the argument of the periapsis (lower Omega or w). These parameters are all independent from one another and are used to essentially constrain each position in the orbit to a fixed position in space - alleviating the conundrum you were alluding to by stating that Kerbin spins.

Now, the only element that requires the Reference Direction as a parameter in its calculation is the RAAN (longitude of ascending node). That's it. The back story of the Reference Direction is that it is an arbitrarily chosen convention vector in space used as an external positional reference - like the 12 on a standard clock - Why start at 12? Why not zero? Convention. (Well there's more to it than that but we'll leave it there for another time.)

The calculation of the inclination of the orbit is just as you had defined in that equation, which clearly you already know. What seems to be confusing you is the determination of the sign of the resultant of that equation - due to the mathematical behavior of the arccos function it will never be a negative value - which begs your question: "Where do the (-) signs come from?"

The answer: Convention and relative positioning.

Let us say an object is traveling in an orbit around Kerbin. Since the object is in orbit around Kerbin we define our reference plane to be the equator of Kerbin. Any point above this plane is in the positive domain, and any point below this plane is in the negative domain. It is to this plane we will measure our inclination angles. Now let us observe the object again - this time we notice that it is traveling with some inclination.

The value of this inclination has a magnitude found from that equation (0^o < i < 180^o), but ends up having TWO signs - both positive and negative (Take a close look on either side of the orbit, one is positive and one is negative of the same number). The orbit is divided into two halves, the ascending plane, and the descending plane separated by two nodes, the Ascending Node and Descending Node - which can be seen from the diagram in your original post. One half lies above and one half dips below the reference plane (equator) into the southern hemisphere - it is on this lower side, after the Descending Node, that you receive the negative sign. All this sign means is that your orbit has descended into the southern hemisphere or below the reference plane.

Now, you might think that adding a second object would bungle all of this math and such. This is not so; the angle of inclination between two space craft is just the difference between each of their angles of inclinations with respect to the equatorial reference plane.

i_ab = i_a - i_b = arccos(V_aZ/mag(V_a)) - arccos(V_bZ/mag(V_b))

Written this way it is clear to see how this number can easily be negative now, but note - the same number appears at the opposite side of the orbit only as a positive number. Where each appears depends on the relative positioning of the two orbits.

In summation - the polarity of the inclination angle, when written, is typically referred only as positive (0^o < i < 180^o) - because mathematically speaking it can only be positive (unless you want an imaginary number). Convention and usefulness, however, dictate that there be a distinction between whether the object is in the ascending plane or descending plane, i.e. above or below the plane of reference. This distinction becomes readily apparent in what 5thHorseman stated - it is extremely useful for inclination changes, such as the question: Thrust in which direction is going to be needed to match another object's orbital plane.

If you see a negative number for an inclination - that means you are looking at the Descending Node (DN in game) and if your ship is near that node it means you are about to pass below the plane of reference. In the case of a targeted second object you are about to pass below the intersection of the two object's orbital planes. It is all about defining what position you are currently at in the orbit, using relative positioning.

I hope this answers most of your question because this subject is notorious for causing confusion; It all ends up being relatively simple trig and vector algebra - mixed with some strange conventional usage.

- Nef

[Specialist290]

That is not true, because the reference direction does not rotate with the planet. If the reference direction were, hypothetically at 0 degrees longitude at some moment in time, it would be 180 degrees longitude exactly half a day later as the planet rotates and the reference direction does not.

Take another look at that diagram you posted. The longitude of the ascending node isn't being measured from the planet's surface; it's being measured from the reference direction. The planet's rotation has nothing to do with it.

[Dunbaratu]

All I know is that if its negative, I need to push up on my maneuver node. If it's positive, I need to push down. I've found that this is the case always, whether it be ascending node or descending node, or if the node is ahead or behind my orbit.

I assumed that the sign didn't actually matter and was just there to tell the player how to modify the maneuver node. Seems far to useful to be just a lucky chance.

Now that I think about it, it seems that it'd be negative if you are heading (celesital) SOUTH when you cross it, and positive if you're heading (celesital) NORTH. I think this because no matter what you do, map mode always has Kerbol's (and Kerbal's) North Pole facing up.

Any orbit will do both of those things. It will contain one spot where it goes south across the equator and on the opposite side a spot where it goes north across it.

[Dunbaratu]

You are confusing two separate parameters in the determination of the orientation of the orbit. There exists three elements that determine the orientation of the orbital plane - The first is inclination (i), the second is the Right Ascension of the Ascending Node [RAAN] [in your diagram it is referred to as longitude of ascending node] (capital Omega), and the third is the argument of the periapsis (lower Omega or w). These parameters are all independent from one another and are used to essentially constrain each position in the orbit to a fixed position in space - alleviating the conundrum you were alluding to by stating that Kerbin spins.

Now, the only element that requires the Reference Direction as a parameter in its calculation is the RAAN (longitude of ascending node). That's it. The back story of the Reference Direction is that it is an arbitrarily chosen convention vector in space used as an external positional reference - like the 12 on a standard clock - Why start at 12? Why not zero? Convention. (Well there's more to it than that but we'll leave it there for another time.)

The calculation of the inclination of the orbit is just as you had defined in that equation, which clearly you already know. What seems to be confusing you is the determination of the sign of the resultant of that equation - due to the mathematical behavior of the arccos function it will never be a negative value - which begs your question: "Where do the (-) signs come from?"

The answer: Convention and relative positioning.

Let us say an object is traveling in an orbit around Kerbin. Since the object is in orbit around Kerbin we define our reference plane to be the equator of Kerbin. Any point above this plane is in the positive domain, and any point below this plane is in the negative domain. It is to this plane we will measure our inclination angles. Now let us observe the object again - this time we notice that it is traveling with some inclination.

The value of this inclination has a magnitude found from that equation (0^o < i < 180^o), but ends up having TWO signs - both positive and negative (Take a close look on either side of the orbit, one is positive and one is negative of the same number). The orbit is divided into two halves, the ascending plane, and the descending plane separated by two nodes, the Ascending Node and Descending Node - which can be seen from the diagram in your original post. One half lies above and one half dips below the reference plane (equator) into the southern hemisphere - it is on this lower side, after the Descending Node, that you receive the negative sign. All this sign means is that your orbit has descended into the southern hemisphere or below the reference plane.

Now, you might think that adding a second object would bungle all of this math and such. This is not so; the angle of inclination between two space craft is just the difference between each of their angles of inclinations with respect to the equatorial reference plane.

i_ab = i_a - i_b = arccos(V_aZ/mag(V_a)) - arccos(V_bZ/mag(V_b))

Written this way it is clear to see how this number can easily be negative now, but note - the same number appears at the opposite side of the orbit only as a positive number. Where each appears depends on the relative positioning of the two orbits.

In summation - the polarity of the inclination angle, when written, is typically referred only as positive (0^o < i < 180^o) - because mathematically speaking it can only be positive (unless you want an imaginary number). Convention and usefulness, however, dictate that there be a distinction between whether the object is in the ascending plane or descending plane, i.e. above or below the plane of reference. This distinction becomes readily apparent in what 5thHorseman stated - it is extremely useful for inclination changes, such as the question: Thrust in which direction is going to be needed to match another object's orbital plane.

If you see a negative number for an inclination - that means you are looking at the Descending Node (DN in game) and if your ship is near that node it means you are about to pass below the plane of reference. In the case of a targeted second object you are about to pass below the intersection of the two object's orbital planes. It is all about defining what position you are currently at in the orbit, using relative positioning.

I hope this answers most of your question because this subject is notorious for causing confusion; It all ends up being relatively simple trig and vector algebra - mixed with some strange conventional usage.

- Nef

Are you claiming that the orbital inclination number KSP reports isn't globally the same sign for the entire orbit? That you get a different number if you look at the ascending node versus the descending node? If that's all you meant it could have been explained a lot faster that way instead of telling me what I already knew.

I was operating on the belief that KSP applies the sign universally to your entire orbit, which is why it didn't make any sense to me. If it gives you a different sign depending on which node you're highlighting then my confusion entirely goes away based on that one fact alone. If that's what it does then it makes a lot more sense. I'll have a chance to try and see if this is true later tonight (not near a computer with KSP on it at the moment).

I was already aware that actual astronomers only use degrees from 0 to 180 for this. But KSP used negative numbers and so knowing what actual astronomers do didn't help figure out what the negative numbers meant. If the sign flips depending on which node you're looking at at the time, then it would all make sense. All orbits with a nonzero inclination will BOTH cross northward of the reference plane at some point and cross southward of it at some point. If all the negative sign that KSP gives means is that it's an alternate way of saying "this is the descending node" then it's simple. If it was giving a negative sign on both the ascending and descending node, which is what I thought it would do, that was what I didn't understand.

Edited October 15, 2013 by Steven Mading

[Dunbaratu]

Take another look at that diagram you posted. The longitude of the ascending node isn't being measured from the planet's surface; it's being measured from the reference direction. The planet's rotation has nothing to do with it.

Uhm. I know. That's why I posted what you quoted where I pointed out that the claim that it depends on where on the planet you cross the equator going northward cannot be true.

[Superfluous J]

Any orbit will do both of those things. It will contain one spot where it goes south across the equator and on the opposite side a spot where it goes north across it.

Right. And one will be positive and one will be negative. And using what I said I could predict which one is which each time.

I'm no orbital mechanic. I just do this stuff by dragging maneuver nodes around. I don't even pay attention to which one is the ascending and and which is the descending node. I just find the most convenient/farthest away and plan my burn there.

[Dunbaratu]

Right. And one will be positive and one will be negative.

I hadn't noticed that KSP puts a different sign on the reported inclination depending on which node you're highlighting. Normally, an orbit's inclination is a number that is reported as being constant for the entire orbital ellipse, so when KSP showed a negative sign for the inclination, I thought it was reporting the inclination of the orbit as a whole. That's why I was trying to figure out how it decides an orbit (the entire orbit) has a negative inclination. It turns out that's not what it's doing anyway so that renders the question moot.

[Meithan]

I hadn't noticed that KSP puts a different sign on the reported inclination depending on which node you're highlighting. Normally, an orbit's inclination is a number that is reported as being constant for the entire orbital ellipse, so when KSP showed a negative sign for the inclination, I thought it was reporting the inclination of the orbit as a whole. That's why I was trying to figure out how it decides an orbit (the entire orbit) has a negative inclination. It turns out that's not what it's doing anyway so that renders the question moot.

Ah, that indeed clarifies everything. Have you confirmed that the indicated inclination has a different sign at the ascending and descending nodes?

By the way, are you sure that's the correct expression for inclination? In principle you need to do arccos(a_z/|a|), where a is any vector perpendicular to the orbital plane. But if the Y axis always points north (that it, parallel to the planet's rotation axis), then you can't use the Y-component to obtain inclination; it would only work for equatorial orbits.

In principle, one way to obtain inclination, if you know the full position and velocity vectors, is to use the (specific) angular momentum vector:

Then, inclination is just:

[purpletarget]

I see. Could it be that the game measures inclination in the range [-90°, 90°] instead of [0°,180°]? Usually, 0° means an equatorial prograde orbit, 90° is a polar orbit, and 180° is an equatorial retrograde orbit.

The game measures the An from 0-180, same as IRL systems. However it just displays the Decending node using the negative angle, but it's always just the negative counterpart to the An.

Uhm. I know. That's why I posted what you quoted where I pointed out that the claim that it depends on where on the planet you cross the equator going northward cannot be true.

Don't get too hung up on the Longtitude term with regards to the RAAN/LAN. Specialist290 is right on the money, as is NefariousParable. The Longtitude of the Ascending Node is not based on the longtitude of the planet, so the rotation of the planet doesn't matter. Indeed the planets position in the Orbit doesn't matter either. It's a longtitude as measured on the celestial sphere, and 0 degrees is the reference direction.

When discussion this measurement on Earth, or indeed anywhere in our solarsystem, that reference direction is based on the First Point of Aries, a nice bright easily identifiable star, which by virtue of being several dozen light years away, means that the parallax effects in measuring angles from one side of the solar system to the other are effectively negligible. So the RAN is measured as a counterclockwise angle from that line between the CoM of the body being orbited and the Reference direction, along the equatorial plane, to the point where the Subject orbit crosses the plane from South, to North. That's the Longtitude of the Acending Node, and that doesn't change based on the rotation of the planet.

For a little more granularity of some of these terms, the OP's diagram, and how they all work, you might want to check out the following:

http://forum.kerbalspaceprogram.com/threads/24623-Tutorial-Video-Krash-Test-Kerbals-How-to-science%21?p=303586&viewfull=1#post303586

[Dunbaratu]

The game measures the An from 0-180, same as IRL systems. However it just displays the Decending node using the negative angle, but it's always just the negative counterpart to the An.

Don't get too hung up on the Longtitude term with regards to the RAAN/LAN. Specialist290 is right on the money, as is NefariousParable. The Longtitude of the Ascending Node is not based on the longtitude of the planet, so the rotation of the planet doesn't matter. Indeed the planets position in the Orbit doesn't matter either. It's a longtitude as measured on the celestial sphere, and 0 degrees is the reference direction.

When discussion this measurement on Earth, or indeed anywhere in our solarsystem, that reference direction is based on the First Point of Aries, a nice bright easily identifiable star,

Already known. Already mentioned in my original post.

[Dunbaratu]

By the way, are you sure that's the correct expression for inclination? In principle you need to do arccos(a_z/|a|), where a is any vector perpendicular to the orbital plane. But if the Y axis always points north (that it, parallel to the planet's rotation axis), then you can't use the Y-component to obtain inclination; it would only work for equatorial orbits.

It's correct at the equator. I wasn't intending to claim it was a general case formula for all points of the orbit, but I admit that wasn't clear.

[Meithan]

Thanks purpletarget, I think you're right on the money. I'll see for myself later tonight. I guess putting a "free" minus sign to the inclination at the descending node is not such a bad idea.

Just a precision on the First Point of Aries: it used to be (two thousand years ago) in the constellation Aries, more or less aligned with the star Ras Hammel (Alpha Arietis), but since then, the precession of the equinoxes have displaced it and it is currently in the constellation Pisces. The name still sticks in astronomy lingo to refer to the direction of the vernal equinox in the current epoch.

[Dunbaratu]

Oh, with the fact that the sign changes depending on which node you're looking at, I changed the thread to solved.