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 Quick Question on Craft Exits 
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Joined: Tue Jun 12, 2007 11:52 pm
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Post Quick Question on Craft Exits
How do you change the angle of an exit on a craft? Thanks in advance.


Wed Jul 01, 2009 5:58 pm
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Post Re: Quick Question on Craft Exits
Code:
   AddExit = Exit
      Offset = Vector
         X = -13
         Y = 0
      Velocity = Vector //This
         X = -3
         Y = 0
      Radius = 10
      Range = 210


Wed Jul 01, 2009 6:07 pm
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Post Re: Quick Question on Craft Exits
I'm not sure you can.

Rockets and crates always have left/right facing exits and dropships have down facing exists.

Edit:

Roon3 wrote:
Code:
   AddExit = Exit
      Offset = Vector
         X = -13
         Y = 0
      Velocity = Vector //This
         X = -3
         Y = 0
      Radius = 10
      Range = 210


That's not it.
He's asking about changing the angle.

Like, making the exit at 45 degrees so actors can eject diagonally.


Wed Jul 01, 2009 6:08 pm
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Post Re: Quick Question on Craft Exits
Since vectors can be represented as angle and magnitude, couldn't you set a vector exclusively on those two?


Wed Jul 01, 2009 6:11 pm
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Post Re: Quick Question on Craft Exits
The velocity tells you where its facing and how fast the actor shot out will fly. (IE x = -15 y = - 15 will shoot the actor out diagonally with a velocity of 15)

Edit:
MaximDude, yes it is, try it.


Wed Jul 01, 2009 6:11 pm
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Post Re: Quick Question on Craft Exits
mail2345 wrote:
Since vectors can be represented as angle and magnitude, couldn't you set a vector exclusively on those two?


If it was that simple, wouldn't somebody do it by now?
I don't recall ever seeing a craft with angeled exits.

Roon3 wrote:
The velocity tells you where its facing and how fast the actor shot out will fly. (IE x = -15 y = - 15 will shoot the actor out diagonally with a velocity of 15)

Edit:
MaximDude, yes it is, try it.


Its something different.
The exit itself still stays the same.


Wed Jul 01, 2009 6:12 pm
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Post Re: Quick Question on Craft Exits
MaximDude wrote:
mail2345 wrote:
Since vectors can be represented as angle and magnitude, couldn't you set a vector exclusively on those two?


If it was that simple, wouldn't somebody do it by now?
I don't recall ever seeing a craft with angeled exits.

I've done it, and if I recall correctly, it was velocity.
viewtopic.php?f=61&t=12686


Wed Jul 01, 2009 6:16 pm
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Post Re: Quick Question on Craft Exits
Roon3 wrote:
Code:
   AddExit = Exit
      Offset = Vector
         X = -13
         Y = 0
      Velocity = Vector //This
         X = -3
         Y = 0
      Radius = 10
      Range = 210

Thanks Roon, that did it.

EDIT: I have another question. Is there a way to keep crafts from scuttling from being upside-down?


Wed Jul 01, 2009 6:22 pm
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REAL AMERICAN HERO
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Post Re: Quick Question on Craft Exits
Well....no. You could make the game flip the craft and change the sprite when it's are upside down (so it still looks upside down), but that won't change any of your offsets.


Wed Jul 01, 2009 7:25 pm
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Post Re: Quick Question on Craft Exits
Make the down thruster really freaking powerful and it could lift your craft up while it's upside down. I've found lighter craft, like small rockets (eg AAL Cannonball) can do this pretty easily.


Wed Jul 01, 2009 8:41 pm
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Post Re: Quick Question on Craft Exits
Yes, but it'll still auto scuttle.


Wed Jul 01, 2009 8:52 pm
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Post Re: Quick Question on Craft Exits
People are saying tilted exit angles are impossible, obviously nobody ever downloaded my Oil Faction, my dropship had tilted exits.
*Feels abandoned*

:dead:





Thu Jul 02, 2009 12:40 am
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Post Re: Quick Question on Craft Exits
Couldn't that be done by repeatedly forcing the ship out of AIMODE_SCUTTLE when upside-down? I recall seeing that somewhere in the forums...


Thu Jul 02, 2009 1:19 am
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Post Re: Quick Question on Craft Exits
That prevents manual scuttle.
You cant ajust status, btw.


Thu Jul 02, 2009 1:37 am
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