Understanding inclination and launch azimuth

[dimkinv]

Hello all,
I have a question, I'm flying orbiter on/off for a couple of years now, Lately I've tried to read about inclination and launch azimuth. Now the formula of launch azimuth is not so complex and placing values there is the easiest thing.
But I have a question I cannot understand. So for example, to launch from cape canaveral to the ISS you need to launch either about 42 or 135 azimuth. But I've thought that the difference between these two is pro-grade or retro-grade orbits. But from what I see you sometimes launch to 135 instead of 42 azimuth. whats the difference? (And if so then how do you launch to the same inclination but with retro-grade for example?)

With that in mind, I've searched youtube for graphical explanations and examples are pretty slim.
I'd be glad if someone could point me to a video with examples and even more glad if there is a simulator that can show what exactly happens from launch to orbit (orbiter doesn't show orbit lines in spectator view, like KSP for example)

Thanks a lot!

 

[jangofett287]

The image on this page https://en.wikipedia.org/wiki/Orbital_elements does a pretty good job of explaining Inclination and the other terms used to describe orbits.
As for launching, look at the ground track of an orbit:

It's fairly obvious there's 2 ways an orbiting vessel could pass over a launch site; going up or going down. I made a couple diagrams, where Theta (θ) is the angle between the ground track and North.

Obviously to get into a specific orbit you want to fly the heading made by that orbit's ground track, the small issue is that the launch site is not stationary. It's moving east as the earth rotates. This means any vessel launching from it has to aim slightly off the ground track in order to meet it once it's in orbit. This adjusted angle is the Launch Azimuth.
To launch retrograde just imagine the arrow on the diagrams is pointing the other way and then adjust for the earth's rotation.

(I'm not quite sure of the maths to calculate Launch Azimuth from the ground track angle. The correction required is going to depend on the Latitude of the Launch Site. I usually fly the XR2 which has enough DeltaV to sort things out in orbit.)

 

[boogabooga]

Hello all,
But I have a question I cannot understand. So for example, to launch from cape canaveral to the ISS you need to launch either about 42 or 135 azimuth. But I've thought that the difference between these two is pro-grade or retro-grade orbits. But from what I see you sometimes launch to 135 instead of 42 azimuth. whats the difference?

No, the resulting inclinations relative to the equator are the same. The difference between these two is the orbital element called the "longitude of the ascending node." Study the definition of this element and see if you can see why. Notice that both azimuths are generally eastward and therefore cannot be retrograde.

(And if so then how do you launch to the same inclination but with retro-grade for example?)

A retrograde orbit will have an equitorial inclination greater than 90 degrees and a prograde orbit will have one less than 90 degrees, so this is not possible. If you mean instead how to launch to the same orbital plane but in the retrograde direction instead of the prograde direction, you can approximate by:

retrograde launch azimuth = 360 degrees - prograde launch azimuth

but this neglects the rotation of the earth, etc.

But, this is about useless for getting to the ISS.

With that in mind, I've searched youtube for graphical explanations and examples are pretty slim. I'd be glad if someone could point me to a video with examples and even more glad if there is a simulator that can show what exactly happens from launch to orbit

Try this:

(orbiter doesn't show orbit lines in spectator view, like KSP for example)

You didn't look hard enough.
[ame="http://www.orbithangar.com/searchid.php?ID=4864"]Videnie orbit drawing 1.0[/ame]

 

[ADSWNJ]

Just a comment on the prograde vs retrograde orbits: if you are launching NE-E-SE directions then it's prograde (i.e. going with the spin of the Earth), and SW-W-NW would be retrograde. Whilst you could launch retrograde into the plane of the ISS, you would be travelling at orbital velocity in exactly the opposite direction to the ISS, so it would be impossible to rendezvous. (It would be a pretty spectacular collision though!) So when using a tool like Enjo's LaunchMFD, you will only ever see the launch azimuths in the same direction as the satellite is travelling.

Here's another little thought experiment for you: say you take off from Cape Canaveral, climb to 50km alt, and fly due West, rock solid on bearing 270, with unlimited fuel, and travelling say 400 m/s. You would essentially fly along the latitude line, parallel to the Equator, and end up back at the Cape, with you doing most of the work, and the Earth helping to spin the Cape back towards you as well. OK, so now do the same thing at successively faster airspeeds, up to 7500 m/s (orbital velocity) ... and despite setting your autopilot to fly a 270 heading, you find yourself sliding south to 28.5 degrees below the equator and back up to 28.5 degrees above again. This is the centripetal force acting on you to force the center of your orbit to go through the center of gravity of the earth (i.e. you can never be in a stable no-thrust orbit with the center of your orbital plane not going through the center of the earth). It's always fascinated me that at some small level of impact, airplanes have to correct for this effect in their flights any time they are not flying a Great Circle route (i.e. a 3D arc whose center is the center of gravity of the Earth).

 

[dimkinv]

Thank you all for the great answers. It is much clearer now!