I have data like this
x-axis data values
-20.49, -12.23, -9.99, -1.00 0 , 1.12, 2.23, 3.45, 4.56, 8.99, 20.99, 30.23
y-axis data values
10,20,20,40,50,60........
I would like to transform above given data into xy coordinate system.
Please have look at the image.
For eg:
along x-axis (min, max ) data value (-20.49, 30.23),
along y-axis (min, max ) data value (10,60)
now if I want plot data(-20.49, 10) in image,
the X coordinate is going to be =200,
and Y-coordinate going to be = 220.
Like this I want plot all data fits within the range of rectangle.
Hope this gives all details
Thanks
This is more of the math question, not related to any programming language. And speaking about Actionscript 3, it has Y axis going from top to bottom, not from bottom to top. Anyway: If you have two points on an axis that you want to map to screen coordinates of your choice, record xmin as lesser native value, xmax as greater native value, and coordinates as xleft and xright. Then, when you need to receive a screen coordinate for your given x, you calculate the xcoord value as:
xcoord = xleft + (x - xmin)*(xright - xleft)/(xmax - xmin);
Similar approach will net you correct values for the Y axis.
Related
what's the algorithm to be able locate and display places around me within a particular distance such as 100m,using easting and northing and name of the place where I'm based .
To be more clear, lets suppose I'm based in charing cross and I want to find all places within 100m using easting and northing data for example, easting =10000m and easting=20000m.
Thank you
Pythagoras is the relevant maths.
If your position is (x,y) then you can calc a distance to any other point (x2,y2) with:
distance = sqrt((x2-x)^2 + (y2-y)^2)
So you could just loop over all points, calc their distance and order the results by nearest first.
For large data sets this may become impractical, in which case you'll want to partition the points into large rectangles. The first stage then is to identify which rectangle your (x,y) is within and the adjacent rectangles, then loop through all points in those rectangles. You need the adjacent rectangles because your (x,y) might be right on the boundary of its rectangle.
More generally this partitioning approach comes under the general heading of spatial hashing. For very large areas you want a tree structure known as a quadtree, that breaks large areas down into smaller and smaller regions, but that might be overkill for what you want.
I am assuming by Cartesian coordinates you also mean linear. If you are trying to do this using actual earth coordinates this answer gets more complicated (as we aren't on a flat earth). For simple linear coordinates you could do something like:
bool contains( x, y)
{
return (x >= minx) && (x <= maxx) && (y >= miny) && (y <= maxy);
}
The min, max coordinates would be your current position + how far out you wanted to go. I think this is what you wanted to know. If you need accurate earth coordinates you might look into some geospatial libraries. If you need and estimate you can still use the algorithm above but I would use something like Rhumb lines to calculate the min, max coordinates.
I am using the Octave plot function to plat a 2D graph. I want to scale the x axis by 2x (no change to y axis). i.e. the distance between each unit in x axis is doubled.
How do I do that in Octave?
You have to use the parameter "dataaspectratio" of the function "set" to modify the aspect ratio of the plot (as the self explaining parameter name indicates). dataaspectratio accepts a 3-element vector, with the scales for X,Y and Z, so if you want X to be double as the Y axis scale you should use:
set(gca,'dataaspectratio',[2 1 1])
In Google maps, the closer one gets to the pole, the more strechted out the map gets and sp each pixel of map represents less movment (asymtotically to 0 at the north pole)
I'm looking for a formula to connect the width of a pixel in degrees to the latitute (i.e. the real world distance represented by a pixel on the map). I have some data points here for zoom level 12 (IIRC)
Lat Width
0 0.703107352
4.214943141 0.701522096
11.86735091 0.688949038
21.28937436 0.656590105
30.14512718 0.60989762
35.46066995 0.574739011
39.90973623 0.541457085
41.5085773 0.528679228
44.08758503 0.507194173
47.04018214 0.481321842
48.45835188 0.468430215
51.17934298 0.442887842
63.23362741 0.318394373
72.81607372 0.208953319
80.05804956 0.122131316
90 0
The reason for doing this is I want to input lat/lng pairs and sort out exactly what pixel they would be located with respect to 0,0
I might be wrong but are you sure thos points are the pixel height? They seem to be a cosine which would be the pixel width not the height.
After a little trigonometry the pixel height adjusts to the formula:
where R is the earth radius, phi is the latitude and h is the height of a pixel in the equator.
This formula does not adjust to your points, that's why I asked if it was the width instead.
Anyway if you want so much precision that you cannot use the approximation in the previous answer you should also consider the R variable with the latitude and even with that I don't think you'll get the exact result.
Update:
Then the formula would be a cosine. If you want to take the variable radius of the earth the formula would be:
where R is the radius of the earth and d(0) is your pixel width at the equator. You may use this formula for R assuming the eearth to be an ellipsoid:
with a = 6378.1 (equator) and b = 6356.8 (poles)
While I am not sure what "height of a pixel" means, the plot of data (shown below) seems to fit the equation
y = a + bx + cx^2 + dx^3 where y = height, x = latitude
with coefficients
a = 7.0240278979641990E-01
b = 3.7784208874521786E-04
c = -1.2602864112736206E-04
d = 3.8304225582846095E-07
The general approach to find the equation is to first plot the data, then hypothesize the type of function, and then do a regression to find the coefficients.
With this toolbox I was performing calibration of my camera.
However the toolbox outputs results in matrix form, and being a noob I don't really understand mathy stuff.
The matrix is in the following form.
Where R is a rotation matrix, T is a translation vector.
And these are the results I got from the toolbox. It outputs values in pixels.
-0.980755 -0.136184 -0.139905 217.653207
0.148552 -0.055504 -0.987346 995.948880
0.126695 -0.989128 0.074666 371.963957
0.000000 0.000000 0.000000 1.000000
Using this data can I know how much my camera is rotated and distance of it from the calibration object?
The distance part is easy. The translation from the origin is given by the first three numbers in the rightmost column. This represents the translation in the x, y, and z directions respectively. In your example, the camera's position p = (px, py, pz) = (217.653207, 995.948880, 371.963957). You can take the Euclidean distance between the camera's location and the location of the calibration object (cx, cy, cz). That is it would just be sqrt( (px-cx)2 + (py-cy)2 + (pz-cz)2 )
The more difficult part regards the rotation which is captured in the upper left 3x3 elements of the matrix. Without knowing exactly how they arrived at this, you're somewhat out of luck. That is, it's not easy to convert that back to Euler Angles, if that's what you want. However, you can transform those elements into a Quaternion Rotation which will give you the unique unit vector and angle to rotate the camera to that orientation. The specifics of the computation are provided here. Once you have the Quaternion rotation, you can easily apply it to the vectors n = (0, 0, 1), up = (0, 1, 0) and right = (1, 0, 0) to get the normal (direction the camera is pointed), up and right vectors. The right vector is only useful if you are interested in slewing the camera left or right from its current position.
I'm guessing the code uses the 'standard' formation - then you will find more details in the opencv library docs or their book.
I have chart that plots values of Y-axis less than 1 (0.1, 0.24, etc).
When the chart is built, Y-axis label just shows 0 at the origin and no other values along the axis.
My solution was a little ugly. Data series labels show up with the correct % signs, but the % signs do not show up on the Y axis.
Y axis: set Y axis range from 1-100 (or 1-120 so there is room to display the data labels that say 100%) Make sure you set your tick marks to Auto or a number between 1 and 100, otherwise you'll get one long bar of tick marks running all the way up the y axis.
Data series: on the series tab, change your Value Field formula to multiply your data series by 100.
Data series labels: on the series data label properties window > general tab, copy the formula from your data series Value Field into the Label Data value field. round it to 0 decimal points and add &"%" to the end to concatenate a percent to the end of it.
That way the right numbers display for my data series and my Y axis, but the % just doesn't show up on the Y axis. not perfect yet. and too much work.
I just tested this and it seems this happens in SSRS 2005 and not 2008. You can fix this by right clicking on your chart and going to properties. It should bring up the full Chart Properties, go to the Y Axis tab and set Major Gridlines at a value like .25/.5 or whatever you want.