Panoramic Pixel Calculator

[Joshua]

[I would like to open this stating that I've been writing this over the period of about a day, and my thoughts have been quite jumbled as I check, recheck, and find new information. Hopefully it will all make sense in the end.]

I'm using PTGui, which I've found really works well to produce equarectangular panoramas, but I'm having a personal issue with resolution. I took two panos, both with the same camera, lens, focal length, image resolution, ect, yet PTGui outputs different Max resolutions for both. One's about 218 megapixels and the other is 505. Both settings were produced when I clicked on Set Optimum Size>Maximum Size (No Loss of Detail), though i don't know how it came up with vastly different numbers for both; they were both 360*180 panos.

So, I figure there's some absolute value to the amount of pixels one could get under the same camera conditions, obviously. Now I just need help figuring it out. Problem is that I keep getting vastly different equations and values (118, 218, 233, 505), and don't know if any of them is right. I started to build a Pano Calculator in Excel based on Frank van der Pol's, reproduced here. I've got a Nikon D5000 with a Sigma 18-200 OS HSM, and I've figured out and plugged in all the equations into excel and got things working as they should so far.

Some of the equations so far are so:

1) Consider shooting portrait at 18mm. According to the pano calc, 18mm=47.39 degrees FOV for my camera. It takes 360 degrees to achieve a full pano, therefore it would take 360/47.39=7.59 images without any overlap whatsoever to produce a full pano. It then makes sense to multiply this to the corresponding height of the sensor, which has a resolution of 4288 x 2848, therefore 7.59*2848=21,616.32. This would be the width of the pano, and the hight would have to be 1/2 the width (for 180 degrees), therefore 21,616.32/2=10,808.16, leaving us with a total of 233 megapixels.

Though Photoshop struggles stitching what happened to be 15 images on my computer, I managed to have it work in small batches of 4 photographs using the Photomerge feature with a Spherical Projection and distortion correction on. Then I lined up each group manually (which was a piece of cake), and blended them all together. I then, using the offset filter, lined up both edges to produce a full 360 horizontal FOV and cropped it. Looking at the image size, it was 21,392 pixels wide, increadably close to the 21,616 estamate above (only 1.047% off). I figure this is accurate enough.

Here's the issue, before I did the photoshop test above I thought that perhaps this equation was wrong, as I was comparing it to Jeffrey Martin's 80 Gigapixel panorama of London. He states on his blog that he used a Canon 550D with a 80-400mm Sigma Lens at 400mm. Assuming he was shooting Portrait (which it doesn't matter, as the math works out for both the same) his lens would have a FOV of 2.134 degrees and a resolution of 3456. Therefore, (360/2.134)*3456=583,012 x 291,506 pixels which is 169.95 Gigapixels.

It has just occurred to me to check the sky, as it seems to be very consistent given the pano was shot over a three day period. Upon closer inspection, the sky seems to have somewhat noticeable posterization if zoomed into closely, and a passing bird seems to be uncannily out of focus and lacking in detail. I can only assume he photographed the sky at wide angle with much fewer shots considering how dynamic a sky can be, resulting in much fewer pixels. Also, of course, is the hole in the Nadir part of the image, subtracting even more.

Clearly, by now, I have gotten somewhat off track, and in fact feel like I have answered my own question simply out of doggedly obsessing about it for the last 24 hours. If anyone has any comments and such though, I would be glad to hear it. I may come back to add on conclusions about the London picture, as I was working on another test of it on another computer that should, very roughly, tell me how many degrees the different parts of it are.

[Dave Humphries]

Hi Joshua,

Welcome to the CiC forums from me

I don't do panoramas, but I think I followed your logic and maths (without checking the actual arithmetic ) and it all seemed to make sense to me.

Not sure it helps much though,

[inkista]

To me, 80 gigapixels and equirectangular would simply imply that the pano is around 12600x6300.

It does NOT imply to me that the final pano is at the full resolution at which it was shot. There are practical limits to what you can store and deliver on line. I know that I once had to scale a high-resolution (non-cubic) pano I stitched to 25% to get under Flickr's maximum 20MB file limit.

Also, about the bird and the sky: just some wild guessing, but remember that the zenith and nadir (the poles) of a sphere are going to be the most distorted when stored in equirectangular format, and that the corners of a cube are going to also show the most distortion. These things are rarely stored as spheres internally, but are warped on the fly, typically from six cube faces.

In addition, a 400mm lens gives you a thinner DoF than with a wider lens. I have a feeling that your pano photographer probably wasn't focusing on the bird when he took the member shot that included it. You can still get an appreciable amount of blur and misfocus with a 400mm lens, even if you're shooting at f/5.6 or f/8. I'm not disputing, though, that using a wider lens is a more likely explanation.

I do know what it's like to have panos eat your brain, though. I'm still puzzling over how the hell Eric Rougier did this one: http://www.fromparis.com/panoramas_q...nd_reflexions/

[Joshua]

Thanks Dave. By the time I actually posted the thread I kind of figured it all out myself, I just figured others could double check. I went through so much trouble different equations coming out to pretty close to other numbers, and I was just hesitant I got it wrong.

But yeah...I guess there was really no point in me posting it by the time I did...