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For illustrating some properties of perspective I want to place the
viewing plane at a given distance from the camera. I think this has to
do with the direction vector, but also with the right vector.
If I put a perspective camera at a point <a,b,c> viewing in the
direction <0,0,1> how can I determine the combination of the right and
the direction vector to obtain the viewing plane, parallel to the
(X,Z)-plane at a distance d from the camera in the direction determined
by <0,0,1>?
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Hi hermans
Although i'm not sure i would guess that the distance of the viewing plane
is given by the length of the direction vector (compare to the illustration
at the docs 3.3.1.1). so to get what you want you should have to declare
direction as d*vnormalize(dicection_vector) where d is the distance of the
viewing plane to the cameras location.
AFAIK the right vector playes only a role with ortographic cameras where it
determines (with the up vector) the size of the image. it also has
consequences on the aspect ration of the image. it should have no effect on
the distance of the viewsing plane
Hope that helped
Regards Roman
hermans <sas### [at] telenet be> wrote:
> For illustrating some properties of perspective I want to place the
> viewing plane at a given distance from the camera. I think this has to
> do with the direction vector, but also with the right vector.
> If I put a perspective camera at a point <a,b,c> viewing in the
> direction <0,0,1> how can I determine the combination of the right and
> the direction vector to obtain the viewing plane, parallel to the
> (X,Z)-plane at a distance d from the camera in the direction determined
> by <0,0,1>?
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"hermans" <sas### [at] telenetbe> wrote in message
news:43ad6f61@news.povray.org...
> For illustrating some properties of perspective I want to place the
> viewing plane at a given distance from the camera. I think this has to do
> with the direction vector, but also with the right vector.
> If I put a perspective camera at a point <a,b,c> viewing in the direction
> <0,0,1> how can I determine the combination of the right and the direction
> vector to obtain the viewing plane, parallel to the (X,Z)-plane at a
> distance d from the camera in the direction determined by <0,0,1>?
Hi,
With the camera at <A,B,C> and a 'direction' vector of <0,0,1> then, if you
imagine a viewing plane 1 unit from the camera in the +z direction, the
image will fill a rectangle on the viewing plane that has a width as defined
by the 'x' component of the 'right' vector and a height as defined by the
'y' component of the 'up' vector (so long as the other components of the
'right' and 'up' vectors are 0).
The 'right' vector and the 'up' vector normally match the proportions of the
image, so for an 800 x 600 image you would use x*4/3 for the 'right' vector
and y for the 'up' vector (the default values). If these are not in
proportion the image gets distorted (as in the following example).
The following SDL illustrates that, so long as the RightVector is a multiple
of <1,0,0> and the UpVector is a multiple of <0,1,0>, then it should keep
the spheres in the corners of the image (despite potentially distorting the
image).
Regards, and Merry Xmas
Chris B.
#local A = 5;
#local B = 2;
#local C = 3;
#local RightVector = x*4/3;
#local UpVector = 2*y;
light_source { < -15, 10 ,-80> color rgb 1}
camera {location <A,B,C> direction <0,0,1>
right RightVector
up UpVector
}
sphere {<A-RightVector.x/2,B-UpVector.y/2,C+1>,0.05 pigment {color rgb
<50,0,0>}}
sphere {<A-RightVector.x/2,B+UpVector.y/2,C+1>,0.05 pigment {color rgb
<50,0,0>}}
sphere {<A+RightVector.x/2,B-UpVector.y/2,C+1>,0.05 pigment {color rgb
<50,0,0>}}
sphere {<A+RightVector.x/2,B+UpVector.y/2,C+1>,0.05 pigment {color rgb
<50,0,0>}}
box {-0.5,0.5 rotate <45,45,0> translate <A,B,C+2> pigment {color rgb
<1,0,1>}}
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