Update
This is now possibly natively as of 2.74. See my answer below.
Is it possible to render a panorama in cycles that is not a full 360x180 view?
E.g. 360° x 30°
?
How can I achieve this?
This is now possibly natively as of 2.74. See my answer below.
Is it possible to render a panorama in cycles that is not a full 360x180 view?
E.g. 360° x 30°
?
How can I achieve this?
This is not a very efficient way of doing this because it requires rendering the entire $360^\circ \times 180^\circ$ image and then cropping off a good chunk of it, but it seems to be the only option.
Here is a diagram:
$H$ is the height of the $360^\circ \times 180^\circ$ image, $h$ is the height of the cropped ($360°\times n^\circ$) image, and $n$ is the desired angle. When $n$ is $180^\circ$ we can see that $D$ would be $\frac{1}{2}$ the height of the image so $D=\frac{H}{2}$. From here it is just a little simple trigonometry and we can get:
$$ h = \sin(\frac{n}{2})\frac{H}{2}2 = \sin(\frac{n}{2})H $$
Using your example of $30^\circ$ we get $h = \sin(15) H ≈ 0.26 H$ A slight rearrangement gives $H ≈ \frac{h}{0.26}$ which you can use to plug in the desired height and get the height you must use to render.
This is now possible as of this commit (will be in 2.74).
Cycles: Adding field-of-view options to the equirectangular panorama camera
This patch adds the option to set minimum/maximum latitude/longitude values for the equirectangular panorama camera in Cycles, as discussed in
T34400
.
360°x180° puts out a 2d image that's distorted so that the angles are now x and y axis correct?
Just render the 360°x180° and cut the x120° (two thirds of the 180°) that you don't want off?