"Matthew Pace" <mat### [at] lycoscom> wrote in message
news:mat### [at] netplexaussieorg...
> > right image_width/image_height*x
> >
> > This way you get the aspect ratio needed. You just had it opposite based
on
> > what you stated above. Maybe you missed this variation?
> >
> > If that wasn't the reason you found to do this correctly the other thing
> > that matters here is the image output resolution needs to match at least
the
> > aspect ratio. So... if image_height=480 and image_width=640, dividing
them
> > makes 1.333333 for the aspect ratio, then you could conceivably use any
> > other such resolution for the image output size as long as keeping the
same
> > ratio. Anyway, it's height divided by width for 'right'.
>
> Ok, thanks for the corrected version, I will try it out.  Can you
> clarify what the second part is?  Do you mean that if I start a scene at
> 640x480, I could render the image at 800x600 and have it look the same
> (except bigger), but if i were to render it the second time at 500x250 ,
> it would come out distorted?

Exactly. It's all about aspect ratio of the resulting resolution when you're
using image_width divided by image_height, since those two values are
creating a new different value to multiply by x (if x is your camera's left
to right direction). Should always be okay until you use non-default values
for 'up' or other camera changes.

Glad to help.

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