I get a series of square binary images as in the picture below,
I want to find the red point, which is the point of intersection of four blocks (2 black and 2 white). For doing so, I use to get the sum of all pixel values along the diagonal directions of the square image, which is 45 deg and 135 deg respectively. The intersection of maximum pixel sum 45 deg line and minimum pixel sum 135 deg line is where my red point is.
Now that I get the co-ordinate of the red point in 45 deg-135 deg co-ordinate system, how to I transform them to earth co-ordinates?
In other words, say I have a point in 45deg-135deg co-ordinate system; How do I find the corresponding co-ordinate values in x-y co-ordinate system? What is the transformation matrix?
some more information that might help:
1) if the image is a 60x60 image, I get 120 values in 45deg-135deg system, since i scan each row followed by column to add the pixels.
I don't know much about matlab, but in general all you need to do is rotate your grid by 45 degrees.
Here's a helpful link; shows you the rotation matrix you need
wikipedia rotation matrix article
The new coordinates for a point after 2D rotation look like this:
x' = x \cos \theta - y \sin \theta.
y' = x \sin \theta + y \cos \theta.
replace theta with 45 (or maybe -45) and you should be all set.
If your red dot starts out at (x,y), then after the -45 degree rotation it will have the new coordinates (x',y'), which are defined as follows:
x' = x cos(-45) - y sin (-45)
y' = x sin (-45) + y cos (-45)
Sorry when I misunderstood your question but why do you rotate the image? The x-value of your red point is just the point where the derivative in x-direction has the maximum absolute value. And for the y-direction it is the same with the derivative in y-direction.
Assume you have the following image
If you take the first row of the image it has at the beginning all 1 and the for most of the width zeroes. The plot of the first column looks like this.
Now you convolve this line with the kernel {-1,1} which is only one nested loop over your line and you get
Going now through this result and extracting the position of the point with the highest value gets you 72. Therefore the x-position of the red point is 73 (since the kernel of the convolution finds the derivative one point too soon).
Therefore, if data is the image matrix of the above binary image then extracting your red point position is near to one line in Mathematica
Last[Transpose[Position[ListConvolve[{-1, 1}, #] & /#
{data[[1]],Transpose[data][[1]]}, 1 | -1]]] + 1
Here you get {73, 86} which is the correct position if y=0 is the top row. This method should be implemented in a few minutes in any language.
Remarks:
The approximated derivative which is the result of the convolution can either be negative or positive. This depends whether it is a change from 0 to 1 or vice versa. If you want to search for the highest value, you have to take the absolute value of the convolution result.
Remember that the first row in the image matrix is not always in top position of the displayed image. This depends on the software you are using. If you get wrong y values be aware of that.
Related
In a 3D space (x,y,z), you are given two points with no restrictions.
Let's say Point 1 = (15,10,-5), Point 2 = (-1, 0, 11)
An arbitrary point (denoted X in the image) is made by finding the mid-point between point 1 and point 2, in this case (7,5,6), and then y is incremented by 10 which creates a third point
Point 3 = (7,15,6)
Attached is an image to better portray these points
The problem is to find an equation that creates the orange line that links the points 1, 2 and 3. The line doesn't necessarily have to link on the bottom, but I assume it is easier to create an ellipse with these points than an inverse parabola.
It is rather simple to build a circle through these three points (note they must be non-collinear).
Make a plane containing given points, use arbitrary coordinate system in this plane. For example, point P1 is origin, vector P2-P1 defines OX axis, vector product of P2-P1 and P3-P1 defines normal N, and (P2-P1) x N defines OY axis
Solve "circle through three points" problem in this plane, find radius and center.
Transform center back into 3D.
Also note that there is infinite number of ellipses and parabolas through three points (until we define additional limitations),
I have asked a question similar to this before but have since got further and also didn't tag the question right and wanted to get a bit of help on the maths around the question if possible.
I have a 3D sphere with points evenly spaced on its surface of which I know the coordinates. From these coordinates I am trying to define the orientation of some spikes that are coming out of the surface of my sphere along the vector between the centre of the sphere and the point at which the coordinates lie.
The idea is these euler angles will be very helpful in later aligning the spikes so they are all in roughly the same orientation if I am am to box out all of the spikes from an image.
Since the coordinates on the sphere are evenly spaced i can just take the average x, y and z coordinates to give me the centre and I can then draw a vector from the centre to each coordinate in turn.
The euler angles I need to calculate in this case are initially around the z axis, then around the new y axis, and finally again around the new z axis.
My centre point is currently being defined as the average coordinate of all my coordinates. This works as the coordinates are evenly spaced around the sphere.
I then use the equation that states
cos(theta) = dot product of the two vectors / magnitude of each vector multiplied together
on the x and y axis. One of my vectors is the x and y of the vector i am interested in whilst the other is the y axis (0,1). This tells me the rotation around the z axis with the y axis being 0. I also calculate the gradient of the line on this 2D plane to calculate whether I am working between 0 and +180 or 0 and -180.
I then rotate the x axis about the angle just calculated to give me x' using a simple 2D rotation matrix.
I then calculate the angle in the same way above but this time around the y axis using x' and z' as my second vector (where z' = z).
Finally I repeat the same as stated above to calculate the new z'' and x'' and do my final calculation.
This gives me three angles but when I display in matlab using the quiver3 command I do not get the correct orientations using this method. I believe I just do not understand how to calculate euler angles correctly and am messing something up along the way.
I was hoping someone more knowledgeable than me could take a glance over my planned method of euler angle calculation and spot any flaws.
Thanks.
Imagine you have a grid of sample points of a function z = f(x, y) where 1 < x < N and 1 < y < N. The formula is not given, but just the raw data, that could be for example the grey level of an image.
I would like to find, given a point A, whose x and y coordinates are given (and z is known from the data, so A is a vertex of the surface) a number M of points that lie on the circumference of the circle with center in A and radius R that are a good approximation of a circular "cloth" draped on the imaginary surface described by the data points. Imagine also that the edges of the surface are a triangle mesh.
The biggest constraint in the approximation is that the sum of the length of the edges of the resulting polygon is constantly R * 2 * PI, so that moving the A point across the surface would just change the M points but never the sum of their reciprocal distances. The draping doesn't need to be perfect, it would be nice though to be as close as possible to the surface., or always on one side of the surface, above or below.
Could anybody give me a pointer to something to read about this? Is this a known problem?
I feel that the problem is not completely formulated, I'd already like some help to give a complete description of it.
I have 2 circles that collide in a certain collision point and under a certain collision angle which I calculate using this formula :
C1(x1,y1) C2(x2,y2)
and the angle between the line uniting their centre and the x axis is
X = arctg (|y2 - y1| / |x2 - x1|)
and what I want is to translate the circle on top under the same angle that collided with the other circle. I mean with the angle X and I don't know what translation coordinates should I give for a proper and a straight translation!
For what I think you mean, here's how to do it cleanly.
Think in vectors.
Suppose the centre of the bottom circle has coordinates (x1,y1), and the centre of the top circle has coordinates (x2,y2). Then define two vectors
support = (x1,y1)
direction = (x2,y2) - (x1,y1)
now, the line between the two centres is fully described by the parametric representation
line = support + k*direction
with k any value in (-inf,+inf). At the initial time, substituting k=1 in the equation above indeed give the coordinates of the top circle. On some later time t, the value of k will have increased, and substituting that new value of k in the equation will give the new coordinates of the centre of the top circle.
How much k increases at value t is equal to the speed of the circle, and I leave that entirely up to you :)
Doing it this way, you never need to mess around with any angles and/or coordinate transformations etc. It even works in 3D (provided you add in z-coordinates everywhere).
I want to calculate the angle between two vectors a and b. Lets assume these are at the origin. This can be done with
theta = arccos(a . b / |a| * |b|)
However arccos gives you the angle in [0, pi], i.e. it will never give you an angle greater than 180 degrees, which is what I want. So how do you find out when the vectors have gone past the 180 degree mark? In 2D I would simply let the sign of the y-component on one of the vectors determine what quadrant the vector is in. But what is the easiest way to do it in 3D?
EDIT: I wanted to keep the question general but here we go. I'm programming this in c and the code I use to get the angle is theta = acos(dot(a, b)/mag(a)*mag(b)) so how would you programmatically determine the orientation?
This works in 2D because you have a plane defined in which you define the rotation.
If you want to do this in 3D, there is no such implicit 2D plane. You could transform your 3D coordinates to a 2D plane going through all three points, and do your calculation inside this plane.
But, there are of course two possible orientations for the plane, and that will affect which angles will be > 180 or smaller.
I came up with the following solution that takes advantage of the direction change of the cross product of the two vectors:
Make a vector n = a X b and normalize it. This vector is normal to the plane spanned by a and b.
Whenever a new angle is calculated compare it with the old normal. In the comparison, treat the old and the current normals as points and compute the distance between them. If this distance is 2 the normal (i.e. the cross product a X b has flipped).
You might want to have a threshold for the distance as the distance after a flip might be shorter than 2, depending on how the vectors a and b are oriented and how often you update the angle.
One solution that you could use:
What you effectively need to do is create a plane that one of the vectors is coplanar to.
Getting the cross product of both vectors will create a plane, then is you get the normal of this plane, you can get the angle between this and the vector you need to get the signed angle for, and you can use the angle to determine the sign.
If the angle is greater than 90 degrees, then it is below the created plane; less than 90 degrees, and it is above.
Depending on cost of calculations, the dot product can be used at this stage instead of the angle.
Just make sure that you always calculate the normals by the same order of vectors.
This is useable more easily if you're using the XYZ axes, and that's what you're comparing against, since you already have the vectors needed for the plane.
There are possbly more efficient solutions, but this is one I came up with.
Edit: clarification of created vectors
a X b = p. This is perpendicular to both a and b.
Then, do either:
a X p or b X p to create another vector that is the normal to the plane created by the 2 vectors. Choice of vector depends on which you're trying to find the angle for.
Strictly speaking, two 3D vectors always have two angles between them - one below or equal to 180, the other over or equal to 180. Arccos gives you one of them, you can get the other by subtracting from 360. Think of it that way: imagine two lines intersect. You have 4 angles there - 2 of one value, 2 of another. What's the angle between the lines? No single answer. Same here. Without some kind of extra criteria, you can not, in theory, tell which of the two angle values should be taken into account.
EDIT: So what you really need is an arbitrary example of fixing an orientation. Here's one: we look from the positive Z direction. If the plane between the two vectors contains the Z axis, we look from the positive Y direction. If the plane is YZ, we look from the positive X direction. I'll think how to express this in coordinate form, then edit again.