Dmd panorama flatten image3/19/2023 IFloorY = (2 * l) || iCeilingY >= (2 * l)) continue ĬlrTopLeft = bm.GetPixel(iFloorX, iFloorY) ĬlrTopRight = bm.GetPixel(iCeilingX, iFloorY) ĬlrBottomLeft = bm.GetPixel(iFloorX, iCeilingY) ĬlrBottomRight = bm.GetPixel(iCeilingX, iCeilingY) calculated indices in Cartesian coordinates with trailing decimalsĬolor clrTopLeft, clrTopRight, clrBottomLeft, clrBottomRight ĭouble fBottomRed, fBottomGreen, fBottomBlue įor (i = 0 i = 0 & x = 0 & y = (2 * l) || iCeilingX >= (2 * l) || Int iFloorX, iCeilingX, iFloorY, iCeilingY for use in neighbouring indices in Cartesian coordinates assume the source image is square, and its width has even number of pixelsīitmap bm = new Bitmap("lillestromfisheye.jpg") īitmap bmDestination = new Bitmap(4 * l, l) īitmap bmBilinear = new Bitmap(4 * l, l) There are 2 resulting images, one with and one without bilinear interpolation. I’m plagiarising my own code from the image rotation with bilinear interpolation article for the bilinear interpolating parts. The centre of the image should be the centre of the “circle” (or that small planet, as it’s affectionately known). And I’m mapping the quadrants to the standard Cartesian quadrants because they make the math easier. They’re not necessary, but they make the programming easier. Here’s another diagram to show what happens when we iterate over the destination image:īefore we get to the code, here are 2 assumptions to simplify the process: Sines and cosines are periodic functions. It’s also equivalent to going from 0 radians to -2PI radians. Going from left to right is equivalent to going from 2PI radians and decreasing to 0 radians on the inscribed circle. Or the variable l slowly reduces to zero. So in the destination image, in raster coordinates, going from top to bottom is equivalent to going from outer inscribed circle of source image to centre of source image. What happens when circles are involved? Radius and angles, that’s what. Well, what do you get when you curl up a line? A circle. You may also notice that only the pixels within the inscribed circle of the source image is used. But that only mitigate the problem, not solve it. The simplest solution seems to be to get a higher resolution source image. But the source image has less pixel information. Because you’re trying to “guess” the pixels mapped in the destination image from the source image. The second problem… I don’t know if there’s a solution. I didn’t like that, so decided to give those up. But it doesn’t matter to me, and you will get a jagged top with unfilled pixels. The first problem can be solved by using a bigger destination image. You may also notice that the centre of the source image is very “squeezed”, and the pixels around there will be repeated in the resulting image. You may have noticed that the corners of the source image will not be in the resulting image. And the way we do that, is to “uncurl” the fisheye image. So given the fisheye image, we want to get that landscape image back. Your wrist forms the “right end” of the image. The tips of your fingers form the “left end” of the image. Hold your right hand, palm downwards, in front of you, thumb towards you, little finger away from you. Here’s how to visualise the fisheye image. Or so I understand.īasically, it’s like taking a normal (but panoramic is better) picture and “curling” it along its bottom or top (but much more complicated than that). It’s created by using fisheye lens in your camera, take a bunch of pictures using that, and use stereographic projection to get the final image. But first, what’s a 360 degree fisheye image? In this article, you will learn how to flatten a 360 degree fisheye image back to its landscape panoramic form.
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