Am 25.05.2014 09:19, schrieb Thomas de Groot:
> On 25-5-2014 7:58, skippy wrote:
>> Thinking the normals of the planes would be opposing each other (hence
>> y is 1
>> for bottom plane and -1 for top plane). Reference says the distance
>> value is how
>> far along the normal from the origin. So if the normal is going
>> y-negative, I'd
>> need a negative number to bring the plane back 'up' above the camera
>> and light
>> source.
>>
>> But I can't seem to make this idea work. Suggestions?
>>
>>
>
> Your normals are correct, however, you interpret wrongly the distance of
> the plane to the origin. -1 and -7 mean that the planes are positioned
> at -1 and -7 from the origin along the y-axis, irrespective of the
> normal /orientation/.

No Thomas, you're wrong there; the value does state a transition along 
the normal vector. The statement

     plane { <X,Y,Z>, D }

is equivalent to:

     plane { <X,Y,Z>, 0  translate <X,Y,Z>*D }


(This, by the way, is equivalent to saying that D specifies the 
(shortes) distance between the origin and the plane, with negative 
values indicating that the origin is on the "outside" of the plane, 
while positive values indicate that it is "inside".)


Thus, the plane

     plane { <0,1,0>, -1 }

passes through <0,-1,0>, while the plane

     plane { <0,-1,0>, -7 }

passes through <0,7,0>.

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