Yes, I might have misunderstood what keith... meant.  The way you 
understood it, the thing you are converting y p and r to is just (y,p) ?

But then it seems like a fairly crude approximation of overlap (because 
images aren't circles).  That could be useful, by being 10 times faster 
than a more accurate method, if the more accurate method were too slow.  
But all the effort is in coding and you likely need the more accurate 
method anyway to weed out the false positives, so why not write just the 
more accurate method.

I agree that using r when computing the direction vector doesn't seem to 
fit any meaning of "direction vector" I can think of.

The sin and cos of r are used with hfov/2 and vfov/2 to compute the 
relative positions of the corners given the center position.


On Thursday, January 20, 2022 at 11:57:25 AM UTC-5 chaosjug wrote:

> My interpretation of "direction vector" seems to be different. Not sure 
> what 
> aCuria meant, because using r for the vector does not make sense to me. I 
> thought I would just take y and p as vector in spherical coordinate and 
> then 
> compute the delta angle. For your example this would lead to a delta angle 
> of 
> 8 degrees.
>
> Am Donnerstag, 20. Januar 2022, 17:44:36 CET schrieb johnfi...@gmail.com:
> > My method does fail to cover the case that any image has y or p values
> > crossing 180 degrees (meaning directly behind the viewpoint of the
> > panorama). I didn't think that case needed to be covered.
> > 
> > Crossing the 360 point of direction relative to the origin, is an 
> ordinary
> > condition that would need to be covered. But that was not the problem I
> > was pointing out.
> > 
> > Consider an example:
> > Photo A has yaw minus 4 degrees, pitch 0 and hfov 10 degrees. Photo B has
> > yaw plus 4 degrees, pitch 0 and hfov 10 degrees. Clearly they overlap.
> > But A has direction vector at minus 90 degrees, while B has direction
> > vector at plus 90 degrees, so they are 180 degrees apart and only 10
> > degrees wide, so if you thought those angles were comparable measures, 
> you
> > would think the images did not overlap.
> > 
> > On Thursday, January 20, 2022 at 11:26:49 AM UTC-5 chaosjug wrote:
> > > Not sure if I get your point. The images are not circles so hfov is 
> only
> > > an
> > > estimate. The right value would be something between hfov and vfov I
> > > guess.
> > > But why is the origin special? Of course I would need to make sure to
> > > calculate the angle between 359° and 1° as 2°. But that is even more
> > > tricky
> > > when implementing your version with quadrangles. I'm pretty sure there 
> is
> > > some
> > > sort of transformation to avoid those problem...
> > > 
> > > Am Donnerstag, 20. Januar 2022, 17:15:47 CET schrieb 
> johnfi...@gmail.com:
> > > > The angles compared in your step 3 aren't actually in comparable 
> units.
> > > 
> > > So
> > > 
> > > > your method would miss overlaps near the origin (where two photos 
> each
> > > 
> > > have
> > > 
> > > > small y and p values but opposite signs on their y and p values).
> > > > 
> > > > On Thursday, January 20, 2022 at 11:04:47 AM UTC-5 
> keith...@gmail.com
> > > 
> > > wrote:
> > > > > Pseudocode:
> > > > > 1) convert y p r to a direction vector
> > > > > 2) compute angle between two direction vectors A and B
> > > > > 3) compare angle to hfovA/2 + hfovB/2
> > > > > 4) if the angle is smaller it’s possible it overlaps. Must check
> > > 
> > > rotation
> > > 
> > > > > of the two images to be sure
> > > > > 
> > > > > On Thu, 20 Jan 2022 at 10:37 PM, chaosjug <chao...@gmx.de> wrote:
> > > > >> Hi,
> > > > >> 
> > > > >> does someone have the formulas or a code example how to calculate 
> if
> > > > >> images
> > > > >> overlap? So starting with y,p,r,hfov and aspect ratio. I only 
> need an
> > > > >> estimate, so I hope I can neglect the lens type.
> > > > >> 
> > > > >> Regards
> > > > >> Stephan
>
>
>
>
>

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