How to calculate a specific distance inside of a picture?

Question

sorry for my bad english. I have the following problem:

Lets say the camera of my mobile device is showing this picture.

In the picture you can see 4 different positions. Every position is known to me (longitude, latitude).

Now i want to know, where in the picture a specific position is. For example, i want to have a rectangle 20 meters in front and 5 meters to the left of me. I just know the latitude/longitude of this point, but i don't know, where i have to place it inside of the picture (x,y). For example, POS3 is at (0,400) in my view. POS4 is at (600,400) and so on.

Where do i have to put the new point, which is 20 meters in front and 5 meters to the left of me? (So my Input is: (LatXY,LonXY) and my result should be (x,y) on the screen)

I also got the height of the camera and the angles of x,y and z - axis from the camera.

Can i use simple mathematic operations to solve this problem?

Thank you very much!

Answer 1

The answer you want will depend on the accuracy of the result you need. As danaid pointed out, nonlinearity in the image sensor and other factors, such as atmospheric distortion, may induce errors, but would be difficult problems to solve with different cameras, etc., on different devices. So let's start by getting a reasonable approximation which can be tweaked as more accuracy is needed.

First, you may be able to ignore the directional information from the device, if you choose. If you have the five locations, (POS1 - POS4 and camera, in a consistent basis set of coordinates, you have all you need. In fact, you don't even need all those points.

A note on consistent coordinates. At his scale, once you use the convert the lat and long to meters, using cos(lat) for your scaling factor, you should be able to treat everyone from a "flat earth" perspective. You then just need to remember that the camera's x-y plane is roughly the global x-z plane.

Conceptual Background The diagram below lays out the projection of the points onto the image plane. The dz used for perspective can be derived directly using the proportion of the distance in view between far points and near points, vs. their physical distance. In the simple case where the line POS1 to POS2 is parallel to the line POS3 to POS4, the perspective factor is just the ratio of the scaling of the two lines:

Scale (POS1, POS2) = pixel distance (pos1, pos2) / Physical distance (POS1, POS2)
Scale (POS3, POS4) = pixel distance (pos3, pos4) / Physical distance (POS3, POS4)
Perspective factor = Scale (POS3, POS4) / Scale (POS1, POS2)

So the perspective factor to apply to a vertex of your rect would be the proportion of the distance to the vertex between the lines. Simplifying:

Factor(rect) ~= [(Rect.z - (POS3, POS4).z / ((POS1, POS2).z - (POS3, POS4).z)] * Perspective factor.

Answer

A perspective transformation is linear with respect to the distance from the focal point in the direction of view. The diagram below is drawn with the X axis parallel to the image plane, and the Y axis pointing in the direction of view. In this coordinate system, for any point P and an image plane any distance from the origin, the projected point p has an X coordinate p.x which is proportional to P.x/P.y. These values can be linearly interpolated.

In the diagram, tp is the desired projection of the target point. to get tp.x, interpolate between, for example, pos1.x and pos3.x using adjustments for the distance, as follows:

tp.x = pos1.x + ((pos3.x-pos1.x)*((TP.x/TP.y)-(POS1.x/POS1.y))/((POS3.x/POS3.y)-(POS1.x/POS1.y))

The advantage of this approach is that it does not require any prior knowledge of the angle viewed by each pixel, and it will be relatively robust against reasonable errors in the location and orientation of the camera.

Further refinement

Using more data means being able to compensate for more errors. With multiple points in view, the camera location and orientation can be calibrated using the Tienstra method. A concise proof of this approach, (using barycentric coordinates), can be found here.

Since the transformation required are all linear based on homogeneous coordinates, you could apply barycentric coordinates to interpolate based on any three or more points, given their X,Y,Z,W coordinates in homogeneous 3-space and their (x,y) coordinates in image space. The closer the points are to the destination point, the less significant the nonlinearities are likely to be, so in your example, you would use POS 1 and POS3, since the rect is on the left, and POS2 or POS4 depending on the relative distance.

(Barycentric coordinates are likely most familiar as the method used to interpolate colors on a triangle (fragment) in 3D graphics.)

Edit: Barycentric coordinates still require the W homogeneous coordinate factor, which is another way of expressing the perspective correction for the distance from the focal point. See this article on GameDev for more details.

Two related SO questions: perspective correction of texture coordinates in 3d and Barycentric coordinates texture mapping.

Answer 2

I see a couple of problems.

The only real mistake is you're scaling your projection up by _canvasWidth/2 etc instead of translating that far from the principal point - add those value to the projected result, multiplication is like "zooming" that far into the projection.

Second, dealing in a global cartesian coordinate space is a bad idea. With the formulae you're using, the difference between (60.1234, 20.122) and (60.1235, 20.122) (i.e. a small, latitude difference) causes changes of similar magnitude in all 3 axes which doesn't feel right.

It's more straightforward to take the same approach as computer graphics: set your camera as the origin of your "camera space", and convert between world objects and camera space by getting the haversine distance (or similar) between your camera location and the location of the object. See here: http://www.movable-type.co.uk/scripts/latlong.html

Third your perspective projection calculations are for an ideal pinhole camera, which you probably do not have. It will only be a small correction, but to be accurate you need to figure out how to additionally apply the projection that corresponds to the intrinsic camera parameters of your camera. There are two ways to accomplish this: you can do it as a post multiplication to the scheme you already have, or you can change from multiplying by a 3x3 matrix to using a full 4x4 camera matrix:http://en.wikipedia.org/wiki/Camera_matrix with the parameters in there.

Using this approach the perspective projection is symmetric about the origin - if you don't check for z depth you'll project points behind you onto you screen as if they were the same z distance in front of you.

Then lastly I'm not sure about android APIs but make sure you're getting true north bearing and not magnetic north bearing. Some platform return either depending on an argument or configuration. (And your degrees are in radians if that's what the APIs want etc - silly things, but I've lost hours debugging less :) ).

Answer 3

If you know the points in the camera frame and the real world coordinates, some simple linear algebra will suffice. A package like OpenCV will have this type of functionality, or alternatively you can create the projection matrices yourself:

http://en.wikipedia.org/wiki/3D_projection

Once you have a set of points it is as simple as filling in a few vectors to solve the system of equations. This will give you a projection matrix. Once you have a projection matrix, you can assume the 4 points are planar. Multiply any 3D coordinate to find the corresponding 2D image plane coordinate.