As already said, for the hfov the following possibilities exist: 1. The hfov is calculated from the mid of the left pixel to the mid of the right pixel 2. The hfov is measured from the left of the left pixel area to the right of the right pixel area
Further more, for the reference point of the pixels is possible: A The reference point is in the mid of the pixels B The reference point is at the top, left point of the pixels Combining this, 4 possibilities would exist ( 1A, 1B, 2A, 2B ). In my example formulas I assumed (but didn't know surely) that only the combinations 1A and 2B are possibly. Of course, I didn't derive these formulas and it may be difficult to see what I wanted to say. But you can simply insert the values x = 0 and x = W-1 (first formula) in order to get alpha = - hfov and alpha = + hfov . Or x = 0 and x = W (second formula) in order to also get alpha = - hfov and alpha = + hfov . The difference is that for alpha = + hfov you must insert x = W and not x = W-1 in the second formula. This corresponds to the combination 2 B : The x coordinate of the right point in the rightmost pixel area is W and not W-1 in this case. If I find the formulas of Helmut Dersch too complicated, I could try to find out by numerical experiments which case (1A, 2B) is used there. Florian Königstein schrieb am Sonntag, 15. Mai 2022 um 21:05:02 UTC+2: > Ok, another correction: > The second formula would indicate that the hfov is measured from the left > of the left pixel area to the right of the right pixel area. The first > formula would indicate that the hfov is calculated from the mid of the left > pixel to the mid of the right pixel. > > Florian Königstein schrieb am Sonntag, 15. Mai 2022 um 21:01:43 UTC+2: > >> ... Sorry for repeated deletion and re-posting, but there seems to be no >> possibility the edit the message ... >> >> One question is e.g.: What is the exact meaning of the hfov (horizontal >> field of view): Is it the angle measured from the left of the leftmost >> pixel area to the right of the rightmost pixel or from the mid of the left >> to the mid of the right pixel ? >> >> The formulas in the optimizer are faily complicated. But assume that for >> a pixel the angle between the optical axis and the ray going through the >> "reference point" of the pixel area was calcualted, call it alpha. Assume >> the y coordinate is so that the y position is in the middle [[ y = >> (H-1)/2 if the reference point of each pixel is in the middle of the pixel >> area ]]. >> >> W and H are the width and height of the image in pixels. >> >> Assume in the optimizer the angle alpha is calculated as: >> alpha = arctan( (x- (W-1)/2) / ((W-1)/2) * tan(hfov / 2) ) >> This would indicate that the reference point is in the middle of the >> pixels. >> >> If the calculation would be: >> alpha = arctan( (x - W/2) / (W/2) * tan(hfov / 2) ) >> This would indicate that the reference point is in the left (and upper) >> pixel area. >> >> The actual formulas are more complicated, but it should be possible to >> get this information out of them. >> >> [email protected] schrieb am Sonntag, 15. Mai 2022 um 20:41:33 >> UTC+2: >> >>> If control points would link a pixel in one image to a pixel in another >>> they could have integer coordinates. >>> But if they link a 4.5 pixel wide and 2.5 pixel high leave of a tree to >>> the same leaf in a different photo the leaf might be 300.45 pixels more to >>> the left and 2.4 pixels higher in one picture than in the other. >>> >>> If you drop the digits after the comma the world might get a >>> playstation-1-look. >>> >> -- A list of frequently asked questions is available at: http://wiki.panotools.org/Hugin_FAQ --- You received this message because you are subscribed to the Google Groups "hugin and other free panoramic software" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/hugin-ptx/9e04dfad-778f-48dd-8bda-54a51eeb8ff0n%40googlegroups.com.
