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I am developing an application for the iPhone using opencv. I have to use the method solvePnPRansac:

http://opencv.willowgarage.com/documentation/cpp/camera_calibration_and_3d_reconstruction.html

For this method I need to provide a camera matrix:
__ __
| fx 0 cx |
| 0 fy cy |
|_0 0 1 _|

where cx and cy represent the center pixel positions of the image and fx and fy represent focal lengths, but that is all the documentation says. I am unsure what to provide for these focal lengths. The iPhone 5 has a focal length of 4.1 mm, but I do not think that this value is usable as is.

I checked another website:

http://docs.opencv.org/modules/calib3d/doc/camera_calibration_and_3d_reconstruction.html

which shows how opencv creates camera matrices. Here it states that focal lengths are measured in pixel units.

I checked another website:

http://www.velocityreviews.com/forums/t500283-focal-length-in-pixels.html

(about half way down) it says that focal length can be converted from units of millimeters to pixels using the equation: fx = fy = focalMM * pixelDensity / 25.4;

Another Link I found states that fx = focalMM * width / (sensorSizeMM); fy = focalMM * length / (sensorSizeMM);

I am unsure about these equations and how to properly create this matrix.

Any help, advice, or links on how to create an accurate camera matrix (especially for the iPhone 5) would be greatly appreciated,

Isaac

p.s. I think that (fx/fy) or (fy/fx) might be equal to the aspect ratio of the camera, but that might be completely wrong.

UPDATE:

Pixel coordinates to 3D line (opencv)

using this link, I can figure out how they want fx and fy to be formatted because they use it to scale angles relative to their distance from the center. therefore, fx and fy are likely in pixels/(unit length) but im still not sure what this unit length needs to be, can it be arbitrary as long as x and y are scaled to each other?

Community
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Isaac
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5 Answers5

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You can get an initial (rough) estimate of the focal length in pixel dividing the focal length in mm by the width of a pixel of the camera' sensor (CCD, CMOS, whatever).

You get the former from the camera manual, or read it from the EXIF header of an image taken at full resolution. Finding out the latter is a little more complicated: you may look up on the interwebs the sensor's spec sheet, if you know its manufacturer and model number, or you may just divide the overall width of its sensitive area by the number of pixels on the side.

Absent other information, it's usually safe to assume that the pixels are square (i.e. fx == fy), and that the sensor is orthogonal to the lens's focal axis (i.e. that the term in the first row and second column of the camera matrix is zero). Also, the pixel coordinates of the principal point (cx, cy) are usually hard to estimate accurately without a carefully designed calibration rig, and an as-carefully executed calibration procedure (that's because they are intrinsically confused with the camera translation parallel to the image plane). So it's best to just set them equal to the geometrical geometrical center of the image, unless you know that the image has been cropped asymmetrically.

Therefore, your simplest camera model has only one unknown parameter, the focal length f = fx = fy.

Word of advice: in your application is usually more convenient to carry around the horizontal (or vertical) field-of-view angle, rather than the focal length in pixels. This is because the FOV is invariant to image scaling.

Francesco Callari
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  • Thank you for your response, I found it informative and very helpful. Just one more quick question. Given the field of view angle, I have found online that fy = fx = cx/(tan((horizontalAngle)/2)) is this the formula you recommend? – Isaac Feb 06 '13 at 17:54
  • also, can I still get a vector relative to my camera in this way if I construct my fx in this way: http://stackoverflow.com/questions/3107771/pixel-coordinates-to-3d-line-opencv?rq=1 – Isaac Feb 06 '13 at 18:03
  • upvote for pointing out the FoV is advantageous as a parameter since it is scale-independent – hAcKnRoCk Mar 09 '18 at 12:29
6

The "focal length" you are dealing with here is simply a scaling factor from objects in the world to camera pixels, used in the pinhole camera model (Wikipedia link). That's why its units are pixels/unit length. For a given f, an object of size L at a distance (perpendicular to the camera) z, would be f*L/z pixels.

So, you could estimate the focal length by placing an object of known size at a known distance of your camera and measuring its size in the image. You could aso assume the central point is the center of the image. You should definitely not ignore the lens distortion (dist_coef parameter in solvePnPRansac).

In practice, the best way to obtain the camera matrix and distortion coefficients is to use a camera calibration tool. You can download and use the MRPT camera_calib software from this link, there's also a video tutorial here. If you use matlab, go for the Camera Calibration Toolbox.

Milo
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  • Thank you for your answer, I think I will simply video tape a chessboard with the iphone 5 camera and use the second link you sent. Do you think there might be a premade camera matrix and distortion coefficient for the iphone 5? – Isaac Feb 11 '13 at 17:35
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    @Isaac: I don't know of any premade camera matrix for Iphone 5, but you could post the calibration result for yours ;-) – Milo Feb 11 '13 at 19:19
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Here you have a table with the spec of the cameras for iPhone 4 and 5. The calculation is:

double f = 4.1;
double resX = (double)(sourceImage.cols);
double resY = (double)(sourceImage.rows);
double sensorSizeX = 4.89;
double sensorSizeY = 3.67;
double fx = f * resX / sensorSizeX;
double fy = f * resY / sensorSizeY;
double cx = resX/2.;
double cy = resY/2.;
user2743760
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1

Try this:

func getCamMatrix()->(Float, Float, Float, Float)
{

    let format:AVCaptureDeviceFormat? = deviceInput?.device.activeFormat
    let fDesc:CMFormatDescriptionRef = format!.formatDescription

    let dim:CGSize = CMVideoFormatDescriptionGetPresentationDimensions(fDesc, true, true)

    // dim = dimensioni immagine finale
    let cx:Float = Float(dim.width) / 2.0;
    let cy:Float = Float(dim.height) / 2.0;

    let HFOV : Float = format!.videoFieldOfView
    let VFOV : Float = ((HFOV)/cx)*cy

    let fx:Float = abs(Float(dim.width) / (2 * tan(HFOV / 180 * Float(M_PI) / 2)));
    let fy:Float = abs(Float(dim.height) / (2 * tan(VFOV / 180 * Float(M_PI) / 2)));

    return (fx, fy, cx, cy)
}
Gigi
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1

Old thread, present problem.

As Milo and Isaac mentioned after Milo's answer, there seems to be no "common" params available for, say, the iPhone 5.

For what it is worth, here is the result of a run with the MRPT calibration tool, with a good old iPhone 5:

[CAMERA_PARAMS]
resolution=[3264 2448]
cx=1668.87585
cy=1226.19712
fx=3288.47697
fy=3078.59787
dist=[-7.416752e-02 1.562157e+00 1.236471e-03 1.237955e-03 -5.378571e+00]

Average err. of reprojection: 1.06726 pixels (OpenCV error=1.06726)

Note that dist means distortion here.

I am conducting experiments on a toy project, with these parameters---kind of ok. If you do use them on your own project, please keep in mind that they may be hardly good enough to get started. The best will be to follow Milo's recommendation with your own data. The MRPT tool is quite easy to use, with the checkerboard they provide. Hope this does help getting started !

Eric Platon
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