Mapping points on non-linear scale - math

I have a time scale (seconds) that starts at 0 and ends at 360.
The width of the scale can vary—but let's say it's 240 pixels wide, making each each segment 40 pixels wide
<--40-->
|------|------|------|------|------|------|
0 1 3 10 30 120 360
How can I calculate which pixel I should use to plot a point, given an arbitrary time in seconds?

pixel_X = timeline_left_X + time * timeline_period / timeline_width ?

Related

variogram range is not unique within some specific degree

variogram of the dataset
I got this variogram from a set of spatial temporal data, but i am wondering why the 90 degree have such a 'small' range compare to other, is it my data problem or I can adjust some setting in the
v<-variogram(Thickness~Age+x+y,data,alpha=seq(0,135,45),beta=seq(0,45,45))
Any suggestion? Thank you
I will assume that you are referring to the range of the plotted graph and not the range parameter that is used in Geostatistics.
This is simply, because the longest distance is smaller at 90 degrees. For example, imagine that you have observations in a rectangle with width 100 and height 50. Then, when computing the variagrom, pair of points will be observed at the distance of at most 100 for 0 degrees, while, at most of 50 for 90 degrees.

Working with degrees of a circle in R

I have a bearing of X degrees and I wish to calculate other bearings either side of X. We all know a circle has 360 degrees but if I'm given a bearing of 0 degrees and I wish to calculate the bearings + and - 30 degrees of 0 how can I program this such that I am looking for 0 - 30 = 330 degs and 0 + 30 = 30 degrees?
The modulus operator (%%) can help you with that
x <- 0
(x+30) %% 360
# [1] 30
(x-30) %% 360
# [1] 330
In this case it will keep all values in the range [0, 360). See the ?"%%" help page for more information or even the wikipedia page for these types of operations

Scaling a window while keeping ratios the same

I have a resizable window, and a graph which consists of 11 lines of different values ranging from 0 to 1000. What is the math I would use to compute this?
I want to have the data ranging from 0 to 1000 be so that it equals 0-1000 pixels on screen. But if I resize my window to say 640 / 480, the graph will adjust only will be less detailed.
This is a simple proportion: if 640 pixels bar represents value of 1000, value of Y will represent 640 * Y / 1000 pixels bar.
You did not specify a programming language.
In HTML you can size anything as %, so you calculate size as % of maximum - i.e. 550 (out of 1000 max) = 55%
If you use % size it will automatically adjust with screen/window size.
If you specify your programming language of choice we may be able to help you more

Very Simple Math Question

Very simple math question.
Say I have an image with a point being tracked in it. Here are my variables:
Image Height
Image Width
Point (pixles from left) coordinate X
Point (pixles from top) coordinate Y
For example the width, I want it to return a value of -0.5, which represents the distance from the center, such that 1 would be the total right, and -1 would be the total left.
So, how would I calculate so that
The point was (width) a quarter way across the entire frame, or a half way across the left SIDE of the frame. The variables would equal:
Image width: 40
Point X: 10
I know this is basic, but I seriously am having a mind cramp right now O_o.
Thanks,
Christian
Xnew = 2*X/Width - 1
Ynew = 2*Y/Height - 1
Explanation:
X/Width gives you value from 0 (total left) to 1 (total right). 2*X/Width then gives a value from 0 (total left) to 2 (total right). Subtract 1 to get a value from -1 (total left) to 1 (total right).
The same for Y.
If image width is 40, and Point x is 10, then in "your" coordinates PointX will be 0.5 (assuming that coordinates are from -20 to 20). So:
PointX = 1 - 2 * (X / ImageWidth)
PointY = 1 - 2 * (Y / ImageHeight)
Checkout:
PointX = 1 - 2 * (10 / 40) = 0.5 (or 10 pixels to the right side)

Transforming captured co-ordinates into screen co-ordinates

I think this is probably a simple maths question but I have no idea what's going on right now.
I'm capturing the positions of "markers" on a webcam and I have a list of markers and their co-ordinates. Four of the markers are the outer corners of a work surface, and the fifth (green) marker is a widget. Like this:
Here's some example data:
Top left marker (a=98, b=86)
Top right marker (c=119, d=416)
Bottom left marker (e=583, f=80)
Bottom right marker (g=569, h=409)
Widget marker (x=452, y=318)
I'd like to somehow transform the webcam's widget position into a co-ordinate to display on the screen, where top left is 0,0 not 98,86 and somehow take into account the warped angles from the webcam capture.
Where would I even begin? Any help appreciated
In order to compute the warping, you need to compute a homography between the four corners of your input rectangle and the screen.
Since your webcam polygon seems to have an arbitrary shape, a full perspective homography can be used to convert it to a rectangle. It's not that complicated, and you can solve it with a mathematical function (should be easily available) known as Singular Value Decomposition or SVD.
Background information:
For planar transformations like this, you can easily describe them with a homography, which is a 3x3 matrix H such that if any point on or in your webcam polygon, say x1 were multiplied by H, i.e. H*x1, we would get a point on the screen (rectangular), i.e. x2.
Now, note that these points are represented by their homogeneous coordinates which is nothing but adding a third coordinate (the reason for which is beyond the scope of this post). So, suppose your coordinates for X1 were, (100,100), then the homogeneous representation would be a column vector x1 = [100;100;1] (where ; represents a new row).
Ok, so now we have 8 homogeneous vectors representing 4 points on the webcam polygon and the 4 corners of your screen - this is all we need to compute a homography.
Computing the homography:
A little math:
I'm not going to get into the math, but briefly this is how we solve it:
We know that 3x3 matrix H,
H =
h11 h12 h13
h21 h22 h23
h31 h32 h33
where hij represents the element in H at the ith row and the jth column
can be used to get the new screen coordinates by x2 = H*x1. Also, the result will be something like x2 = [12;23;0.1] so to get it in the screen coordinates, we normalize it by the third element or X2 = (120,230) which is (12/0.1,23/0.1).
So this means each point in your webcam polygon (WP) can be multiplied by H (and then normalized) to get your screen coordinates (SC), i.e.
SC1 = H*WP1
SC2 = H*WP2
SC3 = H*WP3
SC4 = H*WP4
where SCi refers to the ith point in screen coordinates and
WPi means the same for the webcam polygon
Computing H: (the quick and painless explanation)
Pseudocode:
for n = 1 to 4
{
// WP_n refers to the 4th point in the webcam polygon
X = WP_n;
// SC_n refers to the nth point in the screen coordinates
// corresponding to the nth point in the webcam polygon
// For example, WP_1 and SC_1 is the top-left point for the webcam
// polygon and the screen coordinates respectively.
x = SC_n(1); y = SC_n(2);
// A is the matrix which we'll solve to get H
// A(i,:) is the ith row of A
// Here we're stacking 2 rows per point correspondence on A
// X(i) is the ith element of the vector X (the webcam polygon coordinates, e.g. (120,230)
A(2*n-1,:) = [0 0 0 -X(1) -X(2) -1 y*X(1) y*X(2) y];
A(2*n,:) = [X(1) X(2) 1 0 0 0 -x*X(1) -x*X(2) -x];
}
Once you have A, just compute svd(A) which will give decompose it into U,S,VT (such that A = USVT). The vector corresponding to the smallest singular value is H (once you reshape it into a 3x3 matrix).
With H, you can retrieve the "warped" coordinates of your widget marker location by multiplying it with H and normalizing.
Example:
In your particular example if we assume that your screen size is 800x600,
WP =
98 119 583 569
86 416 80 409
1 1 1 1
SC =
0 799 0 799
0 0 599 599
1 1 1 1
where each column corresponds to corresponding points.
Then we get:
H =
-0.0155 -1.2525 109.2306
-0.6854 0.0436 63.4222
0.0000 0.0001 -0.5692
Again, I'm not going into the math, but if we normalize H by h33, i.e. divide each element in H by -0.5692 in the example above,
H =
0.0272 2.2004 -191.9061
1.2042 -0.0766 -111.4258
-0.0000 -0.0002 1.0000
This gives us a lot of insight into the transformation.
[-191.9061;-111.4258] defines the translation of your points (in pixels)
[0.0272 2.2004;1.2042 -0.0766] defines the affine transformation (which is essentially scaling and rotation).
The last 1.0000 is so because we scaled H by it and
[-0.0000 -0.0002] denotes the projective transformation of your webcam polygon.
Also, you can check if H is accurate my multiplying SC = H*WP and normalizing each column with its last element:
SC = H*WP
0.0000 -413.6395 0 -411.8448
-0.0000 0.0000 -332.7016 -308.7547
-0.5580 -0.5177 -0.5554 -0.5155
Dividing each column, by it's last element (e.g. in column 2, -413.6395/-0.5177 and 0/-0.5177):
SC
-0.0000 799.0000 0 799.0000
0.0000 -0.0000 599.0000 599.0000
1.0000 1.0000 1.0000 1.0000
Which is the desired result.
Widget Coordinates:
Now, your widget coordinates can be transformed as well H*[452;318;1], which (after normalizing is (561.4161,440.9433).
So, this is what it would look like after warping:
As you can see, the green + represents the widget point after warping.
Notes:
There are some nice pictures in this article explaining homographies.
You can play with transformation matrices here
MATLAB Code:
WP =[
98 119 583 569
86 416 80 409
1 1 1 1
];
SC =[
0 799 0 799
0 0 599 599
1 1 1 1
];
A = zeros(8,9);
for i = 1 : 4
X = WP(:,i);
x = SC(1,i); y = SC(2,i);
A(2*i-1,:) = [0 0 0 -X(1) -X(2) -1 y*X(1) y*X(2) y];
A(2*i,:) = [X(1) X(2) 1 0 0 0 -x*X(1) -x*X(2) -x];
end
[U S V] = svd(A);
H = transpose(reshape(V(:,end),[3 3]));
H = H/H(3,3);
A
0 0 0 -98 -86 -1 0 0 0
98 86 1 0 0 0 0 0 0
0 0 0 -119 -416 -1 0 0 0
119 416 1 0 0 0 -95081 -332384 -799
0 0 0 -583 -80 -1 349217 47920 599
583 80 1 0 0 0 0 0 0
0 0 0 -569 -409 -1 340831 244991 599
569 409 1 0 0 0 -454631 -326791 -799
Due to perspective effects linear or even bilinear transformations may not be accurate enough.
Look at correct perspective mapping and more from google on this phrase, may be this is what you need...
Since your input area isn't a rectangle of the same aspect-ratio as the screen, you'll have to apply some sort of transformation to do the mapping.
What I would do is take the proportions of where the inner point is with respect to the outer sides and map that to the same proportions of the screen.
To do this, calculate the amount of the free space above, below, to the left, and to the right of the inner point and use the ratio to find out where in the screen the point should be.
alt text http://img230.imageshack.us/img230/5301/mapkg.png
Once you have the measurements, place the inner point at:
x = left / (left + right)
y = above / (above + below)
This way, no matter how skewed the webcam frame is, you can still map to the full regular rectangle on the screen.
Try the following: split the original rectangle and this figure with 2 diagonals. Their crossing is (k, l). You have 4 distorted triangles (ab-cd-kl, cd-ef-kl, ef-gh-kl, gh-ab-kl) and the point xy is in one of them.
(4 triangles are better than 2, since the distortion doesn't depend on the diagonal chosen)
You need to find in which triangle point XY is. To do that you need only 2 checks:
Check if it's in ab-cd-ef. If true, go on with ab-cd-ef, (in your case it's not, so we proceed with cd-ef-gh).
We don't check cd-ef-gh, but already check a half of it: cd-gh-kl. The point is there. (Otherwise it would have been ef-gh-kl)
Here's an excellent algorythm to check if a point is in a polygon, using only it's points.
Now you need only to map the point to the original triangle cd-gh-kl. The point xy is a linear combination of the 3 points:
x = c * a1 + g * a2 + k * (1 - a1 - a2)
y = d * a1 + h * a2 + l * (1 - a1 - a2)
a1 + a2 <= 1
2 variables (a1, a2) with 2 equations. I guess you can derive the solution formulae on your own.
Then you just make a linear combinations of a1&a2 with the corresponding points' co-ordinates in the original rectangle. In this case with W (width) and H (height) it's
X = width * a1 + width * a2 + width / 2 * (1 - a1 - a2)
Y = 0 * a1 + height * a2 + height / 2 * (1 - a1 - a2)
More of how to do this in objective-c in xcode, related to jacobs post, you can find here: calculate the V from A = USVt in objective-C with SVD from LAPACK in xcode
The "Kabcsh Algorithm" does exactly this: it creates a rotation matrix between two spaces given N matched pairs of positions.
http://en.wikipedia.org/wiki/Kabsch_algorithm

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