Hello,
ok seems like doing this:
iMovable movable = this.hulk.GetMovable();
csReversibleTransform transform = movable.GetTransform();
csVector3 newPos = transform.GetOrigin().add(
transform.GetFront().multiply(-elapsed_second *
speed));
csHitBeamResult hbr = meshTerrain.HitBeam(
new
csVector3(newPos.getX(),-100,newPos.getZ()),
new csVector3(newPos.getX(),100,newPos.getZ())
);
newPos = hbr.getIsect();
csHitBeamResult v1 = meshTerrain.HitBeam(
new
csVector3(newPos.getX(),-100,newPos.getZ()+0.1f),
new
csVector3(newPos.getX(),100,newPos.getZ()+0.1f));
csHitBeamResult v2 = meshTerrain.HitBeam(
new
csVector3(newPos.getX()-0.1f,-100,newPos.getZ()-0.1f),
new
csVector3(newPos.getX()-0.1f,100,newPos.getZ()-0.1f));
csHitBeamResult v3 = meshTerrain.HitBeam(
new
csVector3(newPos.getX()+0.1f,-100,newPos.getZ()-0.1f),
new
csVector3(newPos.getX()+0.1f,100,newPos.getZ()-0.1f));
csPlane3 plane = new csPlane3(
v1.getIsect(),
v2.getIsect(),
v3.getIsect()
);
csVector3 norm = plane.Normal();
norm.Normalize();
if (norm.getY() < 0.0f) {
norm = norm.multiply(-1.0f);
}
float ax = (float)-(Math.PI/2-Math.acos(norm.multiply(new
csVector3
(0,0,1))));
float az = (float)(Math.PI/2-Math.acos(norm.multiply(new
csVector3
(1,0,0))));
movable.SetTransform(new csOrthoTransform(
new csYRotMatrix3(this.hulkRotY)
.multiply(new csXRotMatrix3(ax))
.multiply(new csZRotMatrix3(az))
,newPos));
movable.UpdateMove();
yield the correct result...
http://www.allcolor.org/terrain-character-pos2.png
Is this correct ?
Thanks,
Quentin Anciaux
On Thursday 15 February 2007 10:54:44 Quentin Anciaux wrote:
> Hi, I tried...
>
> But I don't have the result I wish... see
> http://www.allcolor.org/terrain-character-pos.png
>
> To arrive at this, I first hitbeam the character position with the terrain
> to find the elevation.
> Then I do three hitbeams near the character to form a plane and get the
> normal, then I compute the angle with the X and Z axis... but this is not
> correct.... someone could help me ?
>
> I used the following code to position the character (please not this is in
> java).
>
> iMeshWrapper meshTerrain =
> this.app.getEngine().FindMeshObject("Terrain");
> CDebug.assertNotNull(meshTerrain);
> iMovable movable = this.hulk.GetMovable();
> csReversibleTransform transform = movable.GetTransform();
> csVector3 newPos = transform.GetOrigin().add(
> transform.GetFront().multiply(-elapsed_second *
> speed));
> csHitBeamResult hbr = meshTerrain.HitBeam(
> new csVector3(newPos.getX(),-100,newPos.getZ()),
> new csVector3(newPos.getX(),100,newPos.getZ())
> );
> newPos = hbr.getIsect();
> csHitBeamResult v1 = meshTerrain.HitBeam(
> new
> csVector3(newPos.getX()+0.1f,-100,newPos.getZ()),
> new
> csVector3(newPos.getX()+0.1f,100,newPos.getZ()));
> csHitBeamResult v2 = meshTerrain.HitBeam(
> new
> csVector3(newPos.getX()-0.1f,-100,newPos.getZ()),
> new
> csVector3(newPos.getX()-0.1f,100,newPos.getZ()));
> csHitBeamResult v3 = meshTerrain.HitBeam(
> new
> csVector3(newPos.getX()+0.1f,-100,newPos.getZ()+0.1f),
> new
> csVector3(newPos.getX()+0.1f,100,newPos.getZ()+0.1f));
> csPlane3 plane = new csPlane3(
> v1.getIsect(),
> v2.getIsect(),
> v3.getIsect()
> );
> csVector3 norm = plane.Normal();
> norm.Normalize();
> if (norm.getY() < 0.0f) {
> norm = norm.multiply(-1.0f);
> }
> movable.SetTransform(new csOrthoTransform(
> new csYRotMatrix3(this.hulkRotY).multiply(
> new csZRotMatrix3( (float)
> (Math.PI/2-Math.acos(norm.multiply(new
> csVector3(0,0,1)))) ).multiply(
> new
> csXRotMatrix3( (float) (Math.PI/2-Math.acos(norm.multiply(new
> csVector3(1,0,0)))) )
> )
> )
> ,newPos));
> movable.UpdateMove();
>
> and I used the following code to position the camera:
>
> csReversibleTransform spriteTransform = this.hulk.GetMovable()
> .GetTransform();
> csReversibleTransform cameraTransform = camera.GetTransform();
> csVector3 cameraOrigin = spriteTransform.GetOrigin().add(new
> csVector3
> (0,2,-8));
> cameraTransform.SetOrigin(cameraOrigin);
> csVector3 lookAt =
> cameraOrigin.subtract(spriteTransform.GetOrigin()).multiply(-1.0f);
> cameraTransform.LookAt(lookAt,this.cameraOrientation);
>
>
> Quentin Anciaux
>
> On Thursday 15 February 2007 00:49:18 Andrew Robberts wrote:
> > csReversibleTransform has a LookAt function that will compute a look-at
> > transform.
> >
> > So, you can use this to compute a transform that will "look" towards the
> > normal of the terrain. You can do this by doing the following:
> >
> > csReversibleTransform rt;
> > rt.LookAt(terrainNormal, csVector3(0,1,0));
> > csMatrix3 mat = rt.GetT2O();
> > meshWrapper->GetMovable()->SetTransform(mat);
> >
> > This should orient your mesh so that some part of it is facing down the
> > terrain, but it probably won't be the correct part. I can't remember off
> > the top of my head which way the lookat will orient, but you can fix this
> > with a bit of trial and error. What you'll have to do is rotate your
> > character and then apply the lookat transform.
> >
> > For example, if your character's back faces the direction you want its
> > front to face then you can change line 3 to:
> >
> > csMatrix3 mat = rt.GetT2O() * csYRotMatrix3(3.14f);
> >
> > 3.14f is a half-turn rotation.
> >
> > If your character isn't even standing upright, then you'll probably have
> > to apply a rotation around the X axis:
> >
> > csMatrix3 mat = rt.GetT2O() * csXRotMatrix3(3.14f * 0.5f); // half of
> > 3.14fis a quarter turn
> >
> > If neither is sufficient then you might have to combine the two in some
> > way:
> >
> > csMatrix3 mat = rt.GetT2O() * csYRotMatrix3(3.14f) * csXRotMatrix3(3.14f
> > * 0.5f);
> >
> > or something similar. If you play around with it then you should get a
> > feel for transforming your character and manage to get him facing the
> > right way.
> >
> > --
> > Andrew Robberts
> >
> > On 2/14/07, Quentin Anciaux <[EMAIL PROTECTED]> wrote:
> > > Hi,
> > >
> > > I've another question now that I know how to find the elevation. I
> > > would like
> > > the character displayed on the surface not only to be shown at the
> > > correct elevation but also to "rotate" according to the vector
> > > perpendicular (normal ?) from the terrain... How can I do ?
> > >
> > > Thank you very much, any help is appreciated.
> > >
> > > Quentin Anciaux
> > >
> > > -----------------------------------------------------------------------
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