Need help with Orbit Inclination

[Cattasraafe]

So I've been trying to figure out how this works. I've been using mechjeb for launches and executing maneuver nodes and I know on orbit inclination 0 is equatorial and 90 is polar, but where can I find specifics on angles in between?

[KerikBalm]

I don't think I really understand the question.

0 is prograde equatorial, 90 is polar, between 0 and 90 is an inclined orbit.

Between 90 and 180 is an inclined retrograde orbit.

An inclination of 180 is a retorgrade orbit.

[Cattasraafe]

yeah that's what I was trying to figure out. I was just getting confused since equatorial orbit on the navball is marked as 90. I remembered telling me another easy way to find my inclination is to see where the target is on the navball and Subtract 90 hince if your target is lining up on 270 subtracting 90 would leave you with the 180 inclination and so forth. 

[bewing]

Hmmm. You've been confused by an oversimplification. The numbers on the navball are the degrees of a compass. 0 is north, 90 is east, 180 is south, 270 is west. All celestial bodies rotate east, by definition. So if you go east, you are going prograde with the planet. But going east is only "equatorial" if you happen to launch from the equator -- which we do on Kerbin, but nowhere else.

If you launch north or south, you will always put yourself in a polar orbit.

"Inclination" is measured by the value of the Ascending Node, when compared to a pure equatorial orbit -- for example, compared to the orbit of Mun around Kerbin.

The way you have been given is just a quick and dirty calculation that gives the right answer -- but do not look deeper for any meaning in it, because there is none to be found.

 

[KerikBalm]

yea, the navbal is the compass direction, The inclination is your angle to the equator. So you're right about the 90 degree difference.

TO have an inclination of 180, you launch heading 270.

To have an inclination of 0, you launch heading 90.

To have an inclination of 90, you launch heading 0 or 180 (north or south), they are both polar orbits.

[Nich]

 Are you attempting to launch directly into minimus is orbit? You can only do that twice a day and it's either 90+6° or 90-6° 

[OhioBob]

On ‎2‎/‎21‎/‎2016 at 0:33 AM, Nich said:

 Are you attempting to launch directly into minimus is orbit? You can only do that twice a day and it's either 90+6° or 90-6° 

 

That will get you close, but computing the required flight azimuth is actually a bit more complicated than that.  The equations are,

β = β_I – γ

where,

sin β_I  = cos i / cos L

tan γ = (V_eq cos L cos β_I) / (V_o – V_eq cos i)

β is the launch azimuth, i is the orbit inclination, L is the launch site latitude, V_eq is the velocity of the planet’s rotation at the equator, V_o is the velocity of the space vehicle immediately after launch, β_I is the inertial launch azimuth, and γ is a small correction to account for the velocity contribution due to the rotation of the planet.  In KSP, L = -0.097^o and V_eq = 174.94 m/s.

For your example, i = 6^o, and let’s say V_o = 2280 m/s.  Therefore, we have

β_I  = arcsin[ cos(6) / cos(-0.097) ] = 84.0008^o

γ = atan[ (174.94*cos(-0.097)*cos(84.0008)) / (2280–174.94*cos(6)) ] = 0.4973^o

β = 84.0008 – 0.4973 = 83.5035^o

[Nich]

My heading indicator only gives me whole degrees. But yes I do target minimus and make adjustments to keep an or dn less then .2

[Chaos_Klaus]

@OhioBob 

Well ... If you neglect those 0.1° and set L=0, the equations simplify like that::

tan γ = 0

γ = 0° 

and

sin β_I  = cos i

β_I  = arcsin(cos (i)) = arcsin(sin(i+90°)) = i+90°

 

So i+90° actually is a pretty solid way to go here.

[Tourist]

See, this is why I like this forum. What other gaming community can you think of where the answer to a game-play question leads to maths being deployed. And not just any maths either, but the kind that involves symbols and stuff. Don't get get me wrong, I don't really understand the maths that gets used here.... I'm a trial and error mission planner, but I like that this is the kind of place where this happens... hanging with the smart kids. Now I'm just going to go over there and bounce a ball, see how many times I can bounce it in an hour and then try to break that score.  

[AlextheBodacious]

The sad thing is it's either going to be locked or moved to somewhere else--or both.

[OhioBob]

4 hours ago, Chaos_Klaus said:

@OhioBob 

Well ... If you neglect those 0.1° and set L=0, the equations simplify like that::

tan γ = 0

γ = 0° 

and

sin β_I  = cos i

β_I  = arcsin(cos (i)) = arcsin(sin(i+90°)) = i+90°

 

So i+90° actually is a pretty solid way to go here.

If L=0, then

sin β_I  = cos i

tan γ = (V_eq cos β_I) / (V_o – V_eq cos i)

γ is not zero.

[Chaos_Klaus]

Ok. Sorry. You are right. I was seeing what I wanted to see.