See site in english Voir le site en francais
Website skin:
home  download  forum  link  contact

Welcome, Guest. Please login or register.
Did you miss your activation email?

Login with username, password and session length

Author Topic: calculating launch headings  (Read 7849 times)

0 Members and 1 Guest are viewing this topic.

Offline Stuart

  • Newbie
  • *
  • Posts: 14
  • Karma: 0
27 July 2005, 17:00:42
Hi everyone.

Anyone know of a quick way of getting a launch azimuth from the MFD's? Eg, I'm trying to launch to Vespucci which is on an
orbital inclination of 21.56 degrees. Is there a quick way for finding out what heading to launch on to minimise the need for
correction later? (this may be a stupid question... but then when it comes to maths I'm pretty stupid...)


Offline Simonpro

  • Hero Member
  • *****
  • Posts: 546
  • Karma: 0
Reply #1 - 27 July 2005, 18:24:40
azimuth = arcsin ( cos (inclination) / cos (lattitude) )


-------------------------------

Offline Stuart

  • Newbie
  • *
  • Posts: 14
  • Karma: 0
Reply #2 - 28 July 2005, 13:25:26
Thanks Simonpro - yeah I'd seen that. I was just wondering whether there was a way of working it out quickly from the
instruments which would be more in keeping with my slapdash, seat-of-pants approach to things!

Is arcsin just the inverse of sin? So in WinXP's calculator you'd do (cos(21.56) / cos(28.52)) where 21.56 is the inclination
I want to be on & 28.52 is my current latitude. Well this gives you 1.058. If you check inv & hit sin my calculator tells me
it's an Invalid input for function. Does this mean I should use a real scientific calculator? Boy... I think it's in the loft
somewhere...


Offline C3PO

  • Full Member
  • ***
  • Posts: 159
  • Karma: 0
Reply #3 - 28 July 2005, 13:56:40
Quote
Stuart wrote:
Does this mean I should use a real scientific calculator? Boy... I think it's in the loft
somewhere...

WHAT???

I've got my HP48GX programmed to calculate the Azimuth just by entering the lattitude, inclination, selecting North or
South (ascending or decending) direction, and viola!:)

I always have the calculator within arms length when Orbiter is running, and if I ever get into space, it's going with
me!!!:lol:

C3PO


Offline Stuart

  • Newbie
  • *
  • Posts: 14
  • Karma: 0
Reply #4 - 28 July 2005, 15:12:07
C3PO - it is as I thought. My slapdash, haphazard attitude lets me down again! Those teachers were right. Well if I ever get
into space I think I might have to tag along with you! :)



fort

  • Guest
Reply #5 - 28 July 2005, 17:58:41
I do not understand ... This left now a month ago or two and it is completely unperceived past ... However, for somebody who is
interested in the formulas, with mathematics, of which me...when i understand ( not so often...)

Made by Washington Kuhlmann , alias BEEP:

http://www.orbithangar.com/searchauth.cfm?search=washington%20kuhlmann

http://orbit.m6.net/v2/read.asp?id=24048&highl=equation




A azimuth of launching ? North, South? Two operations and you´ll have all. Not a calculation to be made.


And thank you CP3O for what it is of this calculation during launchings to the lower part of the equatorial plan :

http://orbit.m6.net/v2/read.asp?id=9667&highl=azimuth


I pained weeks since to be understood before reading that...


Offline Stuart

  • Newbie
  • *
  • Posts: 14
  • Karma: 0
Reply #6 - 29 July 2005, 12:46:00
Ah! Thanks Fort. That'll help.
I did try using the formula -
azimuth = arcsin ( cos (inclination) / cos (latitude) )
as Simonpro suggested. It's just that, as I said earlier, before I get to the arcsin part I have a value of >1 (for Cape
Canaveral's lattitue and a desired inclination of 21 degrees), which you can't arcsin.

But I did notice that if I reversed the equation (don't forget I have a very slippery grasp of maths indeed) -
azimuth = arcsin ( cos (lattitude) / cos (inclination)) then I get 71 degrees which seemed to put me on the correct
heading... eh? Surely a happy coincidence? Boy I wish I'd paid more attention at school.


Offline Simonpro

  • Hero Member
  • *****
  • Posts: 546
  • Karma: 0
Reply #7 - 29 July 2005, 15:01:38
It is definately inclination over lattitude :)

Example:
Mir/ISS/Etc orbit at around 52 degrees inclination, and the cape is at about 28:

Az=arcsin(cos(52)/cos(28))
Az=arcsin(0.6594627)
Az=41.3degrees


-------------------------------

Offline Simonpro

  • Hero Member
  • *****
  • Posts: 546
  • Karma: 0
Reply #8 - 29 July 2005, 15:02:22
Damn, just noticed - you're trying to get into an orbit that is of lower inclination than your launch base. This is
impossible. The minimum inclination orbit you can reach has equivalent inclination to the lattitude of your laucnh
site. :)


-------------------------------

Offline Stuart

  • Newbie
  • *
  • Posts: 14
  • Karma: 0
Reply #9 - 29 July 2005, 17:32:53
Simonpro - ahhhhhhhhhhhhhhh! [clink of penny dropping]
So anything over 28 degrees would have given me the right answer. Wow.
Phew - well, now I've docked with Vespucci & now just have to plot a course somewhere easy!
Thanks guys!


Offline Simonpro

  • Hero Member
  • *****
  • Posts: 546
  • Karma: 0
Reply #10 - 30 July 2005, 10:01:28
No probs, enjoy! :)


-------------------------------

Offline StarLost

  • Hero Member
  • *****
  • Posts: 683
  • Country: Canada ca
  • Karma: 17
Reply #11 - 30 July 2005, 22:02:21
Hmmm... (sound of head pounding at my temples).

A clarification here please. You are referring to launch only when saying that latitude gives the minimum orbit inclination?
Then change the inclination after orbit injection?

At first blush, it almost sounds like you're saying that GEO is impossible unless you launch from the equator, and I know
that is not reality.

Maybe I need to indulge in ethanol again.


Offline Simonpro

  • Hero Member
  • *****
  • Posts: 546
  • Karma: 0
Reply #12 - 31 July 2005, 00:55:41
Okay, without any in-orbit plane changes you can't reach an orbit of lower inclination than your launch lattitude.

Most GEOSAT launches perform a plane change as part of their insertion into orbit, before their speed is increased
too much.
The formula, and min inc rule, only apply if you hold constant azimuth throughout orbit insertion :)


-------------------------------

fort

  • Guest
Reply #13 - 31 July 2005, 09:51:33


    '...the minimum inclination orbit you can reach has equivalent inclination to the latitude of your launch site.'


   Thank you for the last comments . I had so often read this, but just this, here or there , without explanations ...   Now,
it is clear.

    Good day.

« Last Edit: 31 July 2005, 09:51:33 by fort »