# How to remap a fisheye image?

I want to flatten a series of fisheye images by remapping them to a rectinlinear projection.

To achieve this, I need to be able to remap the pixels of the image using fisheye correction formulas for the x-, and y-coordinates. How can I achieve this? I have found this question and this question but I wonder how to use this for fisheye correction.

So far I've tried to use ImageTransformation for this, but I can't get the function to work properly.

  f[pt_] := With[{s = {.5, .5}},
Module[{rd, polarcoor, ru, newcoor},
rd = Norm[pt - s]^2/Norm[s];
polarcoor =
CoordinateTransform[{"Cartesian" -> "Polar", 2}, (pt - s)];
ru = 1*Tan[2*ArcSin[((polarcoor[[1]])/(2*1))]];
CoordinateTransform["Polar" -> "Cartesian", {ru, polarcoor[[2]]}]
]
]
ImageTransformation[image,f]


This should first translate the image coordinates to polar coordinates, then calculate the new r (ru=r undistorted), and than translate these back to cartesian coordinates. The ru = 1*Tan[2*ArcSin[((polarcoor[[1]])/(2*1))]] is based on the above links, with a random value chosen as f. It fi

I get an error message, saying the function doesn't map. Anyone have an idea how to fix this and any further suggestions on how to improve the code?

Update 2 22-10-2013 Changed the code to:

image = Import["https://i.stack.imgur.com/JDX9f.jpg"];
r[pt_] := Module[{rd, ru, polarcoor, a},
rd = Norm[pt];
ru = *Transformation formula*
a = ArcTan @@ (pt);
ru {Cos[a], Sin[a]}
]


Using ru = Sqrt[rd]; ImageTransformation[image, r, DataRange -> {{-1, 1}, {-1, 1}}] gives

Using ru = ArcTan[rd]; ImageTransformation[image, r, DataRange -> {{-2, 2}, {-2, 2}}] gives

Both look like a step in the right direction, with straightened lines, but I got them by trial-and-error so I don't know the correctness.

• your function must always return a position. Your CoordinateTransform[]s are failing for some values ("incompatible with the coordinate assumptions of the specified coordinate chart") Oct 21, 2013 at 15:39
• Seems to work if you evaluate f symbolically, e.g. g[{x_, y_}] = f[{x, y}]; ImageTransformation[image, g] Oct 21, 2013 at 17:32
• re your latest update - try restarting your kernel, it doesn't misbehave here. Oct 22, 2013 at 10:03
• @cormullion: Still not working here. Strange. Oct 22, 2013 at 10:12

It seems to me that your radial remapping is working okay, but the original image appears to have an asymmetric distortion, in that the vertical lines of the buildings converge towards a point which is slightly up from the centre of the image. I think it will be necessary to correct for that before doing the radial remapping. Here's a simple attempt to do the correction.

image = Import["https://i.stack.imgur.com/JDX9f.jpg"] ~ImagePad~ -7;

f[c_][{x_, y_}] := Module[{θ, ϕ},
ϕ = ArcTan[x, y];
θ = π/2 Sqrt[x^2 + y^2];
c (1 - Sin[θ]) + 2 θ/π {Cos[ϕ], Sin[ϕ]}]

ImageTransformation[image, f[{0, 0.2}],
DataRange -> {{-1, 1}, {-1, 1}}, PlotRange -> {{-1, 1}, {-1, 1}}]


I think this is an improvement, though far from perfect. Of course it all comes down to knowing the form of the distortion.

I misunderstood the aim of the question and provided a transformation which unrolls the fisheye image into a panorama. I will leave it here as it attracted a good number of upvotes so I assume it is of interest.

f[x0_, y0_][{th_, r_}] := (1 - 2 r/Pi) ({Sin[th], Cos[th]} - {x0, y0}) + {x0, y0}

ImageTransformation[image, f[0, 0.2],
DataRange -> {{-1, 1}, {-1, 1}}, PlotRange -> {{0, 2 Pi}, {0, Pi/2}}]


• Different way than I've been thinking (see update): For fish-eye undistortion, the radius from image centre to a pixel must be corrected. So I've tried to find that radius (by getting polar coordinates), apply a function to it and than translate the new polar coordinates back to Cartesian. Not getting it to work so far though. Oct 22, 2013 at 9:32
• I am not looking for a way to make panorama images, which is basically what you have achieved. Looks nice though! Oct 22, 2013 at 14:10
• @ErikP, sorry - I misunderstood the aim of the remapping. This must be the highest number of upvotes I've ever had for an answer which utterly fails to answer the question! Oct 22, 2013 at 19:31
• no problem, but do you have any ideas on how to solve the question, or give my current solution a nudge in the right direction? Oct 23, 2013 at 8:14
• @ErikP, I'll have another look. The asymmetry of the distortion makes it more complicated than it otherwise would be... Oct 23, 2013 at 9:06

"Fisheye projection" can mean multiple different things. The most common projection used by lenses is the equal-area one, for which you were using a formula:

$$r = 2f\sin(\theta/2)$$

Here $\theta$ is the angle under which a point is visible and $r$ is the distance of that point's projection from the optical axis on the image plane. Please see the link above for a sketch of the coordinate system. The parameter $f$ is the focal length of the lens, but for our purposes it just controls the radius of the 180-degree circle on the image plane.

The image you posted doesn't look like a true fisheye photo. It looks like it is a stitched panorama. You can see the stitching artefacts. Thus there is a chance that it doesn't use equal area projection. For simplicity I'm going to assume that it uses an equi-angular projection, i.e. $r = f\theta$.

To remap the image to a rectilinear projection, let us first choose a coordinate system as follows:

• put the origin in the middle
• let the half-width of the image be 1

We achieve this with DataRange -> {{-1, 1}, {-1, 1}} in ImageTransformation.

Choose a transformation function:

trafo = Function[{theta}, 2 Sin[theta/2]]; (* equal area *)

trafo = Function[{theta}, theta]; (* equi-angular *)


Let us set f now. If the 180-degree image circle filled out the frame, it would be

f = 1/trafo[Pi/2];


But it's clear from looking at the image that the field of view is larger than 180 degrees. We can see the ground in all directions. (BTW this is also a strong indicator that this is not a real fisheye image, as greater than 180-degree FoV lenses are rare.) So the radius of a 180-degree circle is less than 1. With some experimentation I found that 0.75 gives good results.

f = 0.75/trafo[Pi/2];


fishImg = Import["https://i.stack.imgur.com/JDX9f.jpg"]


f = 0.75/trafo[Pi/2];

Transform:

ImageTransformation[
fishImg,
Function[{px},
With[{r = Norm[px]},
trafo[ArcTan[r/f]] f/r px
]
],
DataRange -> {{-1, 1}, {-1, 1}},
PlotRange -> 2 {{-1, 1}, {-1, 1}}
]


The crop of the resulting image is set using PlotRange.

You can experiment with other projection functions too and see if you'll get better results. There won't be a large difference though.