XDV360 replacement to convert from fisheye photo to equirectangular

[jumpjack]

My fisheye camera came with XDV360 software, which converts in real time both images and videos from fisheye to "cropped equirectangular", i.e. shows only a small portion of the full dome (to be viewed with Head Mounted display).
But I think the SW is designed for fisheye cameras with 180° FOV or less, but mine has 235° FOV (actually seeing behind itself!), thus resulting in images squeezed in height.
I'd need a SW which allows customizing the FOV before reprojecting.
Any suggestion?

[jumpjack]

Example image, and drawing showing the problem.

235fisheye1.jpg

235fisheye-error1.png

[johnvanvliet]

like this


a very simple polar to cylindrical using the built in gimp tool

there are other tools that would do a better job , MATLAB comes to mind ( or the opensource math tool Octave -- runs matlab m files )

or even gmic ( there is a gimp plugin )

gmic 235.png -euclidean2polar 50%,50%,3,1 -o 123.png

[jumpjack]

This looks interesting, could you please provide some details about parameters used?
But there is some tuning needed yet: your first image appears mirrored and squeezed w.r.t. the one obtained with XDV360:

Yours:
sQ4EZPcU.jpg

Mine:
235fisheye_2.jpg

Additionally it looks like XDV360 crops the image when converting to equirectangular, I didn't notice it.

Instead, the vertical ratio looks the same:
235-comparison.png
(above XDV360 output, below your first image mirrored, enlarged 150% and cropped).

But the person looks "vertically compressed" in the head up display, so probably the problem is somewhere else...

About your second image, it looks not cropped, but it remembers me the "vertical squeezing" operated by XDV360.
I see the black area starts around pixel 600 of 800, which means 6/8=3/4=0.75 , while 235°/360°=0.65 , but I don't know if ratio must be the same, don't know if there is linearity or non-linearity involved in the reprojection... any help?!? I'm getting confused.

Are you able to obtain a circular image were the black area begins more far from the center? I'd like to try it in the head-up display. But I don't know where the black area should start from in the flat image to get such a circular image from it.

[johnvanvliet]

MMPS remapps to this
-- i RENAME the "project" program do to a naming collision with Proj4 and gmt

MMPSproject -i  orthographic -w 1600 -h 1600 -scale 1 -lat 90 -long 0 -f fisheye.ppm -out cor.ppm

after cropping the top half