Flight Question Having Trouble Calculating Launch Azimuth

[Ross]

Well, I'm trying to use the ascent autopilot in the DGIV-2. I want to fly to the moon. (Launching from Cape Canaveral)

I am having trouble figuring out what angle to launch at though. I wait until the moons orbital path is somewhat over mine, then try using the arcsin(cos(i)/cos(l)), where i,l, are inclination and latitude. The angle it gives me is 0.8 or something like that, and I'm pretty sure that's in degrees (according to the DG-IV calculator). So what gives? Am I missing something?

 

[Wishbone]

That's right, take a look at Moon's orbit in Map MFD. So you've got either to change planes or perform an out-of-plane TLI (trans-lunar injection). Would heartily suggest using LunarTransfer MFD available at http://koti.mbnet.fi/jarmonik/Orbiter.html .

 

[Tommy]

The Moon's inclination is lower than your launch latitude. This means you will launch at a heading of 90 degrees when the Moon's groundtrack passes as close as it will get. THen you will either need to align planes with the Moon, or perform an off-plane transfer. LTMFD is the easiest way to do that, but IMFD and Trans-X will also work.

This will be the case any time the target's inclination is less than your launch latitude.

 

[statickid]

your missing the fact that a heading of .8 degrees is pretty much going straight north! :thumbup:

---------- Post added at 08:27 AM ---------- Previous post was at 07:13 AM ----------

http://www.orbiterwiki.org/wiki/Launch_Azimuth

This should help you out. Also I don't know what you are calculating, or where you are getting your numbers from, but you should be getting a domain error for doing arcsine of n>1!!!

 

[Ross]

Thanks

Thanks for the replies everybody :hello:

I understand now, I guess I just don't really understand why the launch latitude has to be below the inclination of the target object. I guess I would have to go through the proof of the equation?

 

[Wishbone]

Please bear in mind that the best way to get to orbit is going straight (that is in one plane, without any crosstrack excursions). The plane is fixed in space, it is only the Earth that rotates. If so, then the plane should at launch cross the parallel where the launch site is. Take an apple and try to make a full-diameter cut through a "Cape Canaveral point". You can't do it with equatorial cuts. Try to make a cut through the "pole" of the apple - there would be only one option to do that (inclination of 90 degrees).

 

[Tommy]

I guess I just don't really understand why the launch latitude has to be below the inclination of the target object

You cannot launch directly into an inclination that is less than your starting latitude. Look at it this way - launching at 90 degrees heading results in the lowest possible inclination, ANY other heading and your inclination will be higher. If I launch from a latitude of 30 degrees, then that becomes the northermost part of my orbit. From there, my vessel will drift down to 30 degrees south latitude, then back up to 30 degrees north. My inclination will be 30 degrees.

If, while in Orbit, you watch both your current latitude and your inclination you will find that they match twice per Orbit (if we don't care about north or south). In other words, if my inclination is 30 degrees, at one point in my orbit I will be at my most northernmost point, which will be at a latitude of 30 degrees north, and it happens again at the southernmost point, at 30 degrees south latitude. At those two points, (the peaks of the sine wave of our groundtrack) where current latitude equals the inclination, the vessel is traveling due East.

You can verify this easily. Launch a DG at three different headings - 80, 90, ad 100. You will find that 90 degrees results in the lowest inclination, and the 80 degree and 110 degree headings result in basically identical inclinations, which will be higher than your starting latitude.

The are two ways to achieve an inclination of less than the launch latitude. You can either change the plane one you are in orbit, or you can use a "dog leg" during the launch (which has the effect of changing your launch latitude to that of the "dog leg" rather than the actual lift off.)

 

[statickid]

Thanks for the replies everybody :hello:

I understand now, I guess I just don't really understand why the launch latitude has to be below the inclination of the target object. I guess I would have to go through the proof of the equation?

naw don't go through the proof its just that the cosine of the top number will be bigger than the bottom one, meaning the bottom number will go into it at least once. since no angles have a sine >1, then there are no solutions to the equation.

 

[Ross]

Wow I still can't believe how helpful you guys are. Thanks.

Made it to the moon in 8.5 days

 

[statickid]

comegrabyourlations!!