CODEWITHSUNDEEP
opencvgeometryhough-transformpolar-coordinatescartesian-coordinates
onaclov2000
onaclov2000

Reputation: 5851

How to find a line from polar coordinates (Hough Transform Confusion)

I recently started a CV course and am going through old homeworks (the current ones aren't released). I've implemented a Hough Lines function, I loop through each point, if it's an edge, then I loop through 0-180 (or -90 to 90) theta values, and calculate rho, and finally store in an array.

When I tried to convert back from Polar Coordinates, I can find an X,Y pair (using rho * sin(theta), and rho * cos(theta)), however I don't understand how to convert that to a line in Cartesian space. To have a line you need either 2 points or a point and a direction (assuming ray then of course)

I just understand where the point is.

I've done some searching but can't seem to quite find the answer, folks tend to say, polar tells you x, then bam you have a line in cartesian, but I seem to be missing that connection where the "bam" was.

What I mean is described here; Explain Hough Transformation

Also Vector/line from polar coordinates Where it's asked how do I draw a line from polar coords, which the response was well here's x and y. but to me never mentions rest of that solution.

Is the line somehow related to y = mx+b where m is theta and b is rho?

If not how do I convert back to a line in cartesian space.

EDIT: After reviewing Sunreef's answer, and trying to convert so y was on it's own side, I discovered this answer as well: How to convert coordinates back to image (x,y) from hough transformation (rho, theta)?

It appears what I think I'm looking for is this

m = -cotθ

c = p*cosecθ

EDIT#2 I found some other examples on the net. Basically yes I'll need rho * sin(theta) and rho*cos(theta)

The other part that was messing me up was that I needed to convert to radians, once i did that, I started getting good results.

Upvotes: 3

Views: 5302

Answers (2)

MBo
MBo

Reputation: 80287

You are right that you can get some base point at the line as

 (X0, Y0) = (rho * cos(theta), rho * sin(theta))

and you can find (unit) direction vector of this line as perpendicular to normal:

(dx, dy) = ( -sin(theta), cos(theta))

Upvotes: 4

Sunreef
Sunreef

Reputation: 4552

Taken from Wikipedia:

The non-radial line that crosses the radial line ϕ = ɣ perpendicularly at the point (r0, ɣ) has the equation: r(ϕ) = r0 * sec(ϕ - ɣ).

If I suppose that the coordinates you have for your line are ɣ and r0, then you can rewrite this equation like this:

r(ϕ) * cos(ϕ) * cos(ɣ) + r(ϕ) * sin(ϕ) * sin(ɣ) - r0 = 0

And we know that when translating polar to cartesian coordinates, if we have a point P(r, ϕ) in the polar plane, then its coordinates in the cartesian plane will be:

x = r * cos(ϕ)

y = r * sin(ϕ)

So the equation above becomes a line equation as follows:

x * cos(ɣ) + y * sin(ɣ) - r0 = 0

This is the equation of your line in cartesian coordinates.

(Tell me if you see some mistakes, I did that quickly)

Upvotes: 1

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