Tim Nikias wrote:
> 
> I've already posted this under binaries.animations, but I'll ask here as
> well.
> 
> My problem is as follows:
> I have a particle with given speed and position. I want it to cross a
> certain position. If I have to consider gravity, I need means of finding
> if:
> -the given speed is sufficient at all

The force for flying is proportional to :
  - the profil exposed to the air speed (assume some constante, unless
    your object is rolling showing a different aspect...)
  - some exponential of the air speed (relative speed)
    For low velocity (a.k.a the usual one), you may approximate it with
    a simple square of the speed

Speed is sufficient when the force in the +y direction counter-balance the
 gravity.

    m.g/2  == a.v.v

    v = V-wind

> -which would be the minimum speed required

  Best is to compute that from what you expect.
  If you want to simulate a plane which can take off at 100 space unit/time unit
   (100 km/h, or 100 Miles/h or whatever), once you have fixed the mass of the
  object (as well as the gravity value), you automatically get the value 'a' in
the 
  equation. Then you fix it once and for all.

> -the direction the particle should head with the given/calculated speed

  ???? I do not understand.
  I assume particule are just perfect sphere, so heading do not have sens.
  If you mean the orientation of the trust for the engine, 
  it depend on the ratio you want or need to have between the horizontal
  acceleration (which is up to you) and the vertical acceleration
  (which is constrained by the previous formula, if you really want to
  supply some of the lifting power directly by the trust of the engine
  rather that by the suffisant velocity of the media around the object). 
  Keep in mind that the horizontal velocity is the integration of
  the horizontal acceleration minus the air resistance.
     ( i.e. :     dv = h_a - b.v.v )
  b should be also a constant (depend on the profil too).
  h_a is the horizontal acceleration
  v is the relative air speed

Typical WWI airplane heading match the horizontal velocity.
Some modern english airplane do not fly to take off, they are
lifted by the engine like rocket.

> 
> Hope this is not too much for someone to answer... ;)

Hope you can understand me... I do not know where you are going.

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