A friend gave me a logo which he created in Adobe Illustrator CS 5 with a rectangle in blue color (Cyan 100%, M 0%, Y 0%, B 0%) and saved it as PDF. I opened the PDF-file in Adobe Illustrator CS 6 17.0.0.
When I used the pipette to get the color information the result was: Cyan 74%, Magenta 24%, Y 0%, B 0%.
I tried to put two other rectangles over it. One with the original color: Cyan 100%, M 0%, Y 0%, B 0%
And one with the measured color: Cyan 74%, Magenta 24%, Y 0%, B 0%
The one with the original color looks 100% same like the logo. (different values...)
And the one with the measured color looks different. (same values...)
Any idea? Is it a bug?
Adobes pipette reads only the RGB values and calculates the CMYK values even if the document is in CMYK - Mode.
If you make a small calculation than you can see that CMYK can hold about 10^8 numbers (100 C; * 100 M; * 100 Y; * 100 K;) while RGB can only hold 16,7 * 10^6 (R 256; * G 256; * B 256) which gives a gab of 83*10^6 values. So there is more than one value in CMYK that gives the same RGB value as another CMYK value.
Related
I'm using Postgres 9.5 and I've just installed PostGIS for some extended functions. I have a table with (x,y) points and I want to find the rectangle that fits the maximum number of points. The constraint is that the rectangle side lenghts are fixed. So far I'm counting how many points are in the box without rotation. My points are centered around the origin, (0,0).
SELECT Sum(CASE
WHEN x > -5
AND x < 5
AND y > -10
AND y < 10 THEN 1
ELSE 0
END) AS inside_points,
Count(1) AS total_points
FROM track_t;
This query gives me the count of points inside a rectangle with origin (0,0) and lenghts x = 10 and y = 20.
From here I would create a helper table of rotated rectangle corner points (angle, x1, y1, x2, y2), then cross join to my data, and count over the points per angle, while GROUP BY angle. Then I can select which angle gives me the most points inside the rectangle.
But this seems a little old fashioned, and perhaps non-performant. Additionally, counting points inside a rotated rectangle is not a trivial calculation.
Are there more efficient and elegant ways, perhaps using Postgres Geometric Datatypes or PostGIS Box2D, to rotate a rectangle with fixed side lenghts, and then to count the number of points inside? The geometric functions look good, but they seem to provide minimum bounding boxes and not the other way around.
In addition to Postgresql, I'm using a Python framework that could be used in case SQL can't make this work.
Update: One thing I tried is to use Geometric Types, specifically BOX
SELECT deg, Box(Point(-5, -10), Point(5, 10)) * Point(1, Radians(deg))
FROM Generate_series(0, 360, 90) AS deg
Unforunately, the Rotate function by a Point doesn't work for Polygons.
I ended up by generating rectangle vertices, rotating those vertices, and then comparing the area of the rectangle (constant) with the area of the 4 triangles that are made by including the test point.
This technique is based on the parsimonious answer:
Make triangle. Suppose, abcd is the rectangle and x is the point then if area(abx)+area(bcx)+area(cdx)+area(dax) equals area(abcd) then the point is inside it.
The rectangles are defined by
A bottom left (-x/2,-y/2)
B top left (-x/2,+y/2)
C top right (+x/2,+y/2)
D bottom right (+x/2,-y/2)
This code then checks if point (qx,qy) is inside a rectangle of width x=10 and height y=20, which is rotated around the origin (0,0) by an angle with range of 0 to 180, by 10 degrees.
Here's the code. It's taking 9 minutes to check 750k points, so there is definite room for improvement. Additionally, It can be parallelized once I upgrade to 9.6
with t as (select 10*0.5 as x, 20*0.5 as y, 17.0 as qx, -3.0 as qy)
select
z.angle
-- ABC area
--,abs(0.5*(z.ax*(z.by-z.cy)+z.bx*(z.cy-z.ay)+z.cx*(z.ay-z.by)))
-- CDA area
--,abs(0.5*(z.cx*(z.dy-z.ay)+z.dx*(z.ay-z.cy)+z.ax*(z.cy-z.dy)))
-- ABCD area
,abs(0.5*(z.ax*(z.by-z.cy)+z.bx*(z.cy-z.ay)+z.cx*(z.ay-z.by))) + abs(0.5*(z.cx*(z.dy-z.ay)+z.dx*(z.ay-z.cy)+z.ax*(z.cy-z.dy))) as abcd_area
-- ABQ area
--,abs(0.5*(z.ax*(z.by-z.qx)+z.bx*(z.qy-z.ay)+z.qx*(z.ay-z.by)))
-- BCQ area
--,abs(0.5*(z.bx*(z.cy-z.qx)+z.cx*(z.qy-z.by)+z.qx*(z.by-z.cy)))
-- CDQ area
--,abs(0.5*(z.cx*(z.dy-z.qx)+z.dx*(z.qy-z.cy)+z.qx*(z.cy-z.dy)))
-- DAQ area
--,abs(0.5*(z.dx*(z.ay-z.qx)+z.ax*(z.qy-z.dy)+z.qx*(z.dy-z.ay)))
-- total area of triangles with question point (ABQ + BCQ + CDQ + DAQ)
,abs(0.5*(z.ax*(z.by-z.qx)+z.bx*(z.qy-z.ay)+z.qx*(z.ay-z.by)))
+ abs(0.5*(z.bx*(z.cy-z.qx)+z.cx*(z.qy-z.by)+z.qx*(z.by-z.cy)))
+ abs(0.5*(z.cx*(z.dy-z.qx)+z.dx*(z.qy-z.cy)+z.qx*(z.cy-z.dy)))
+ abs(0.5*(z.dx*(z.ay-z.qx)+z.ax*(z.qy-z.dy)+z.qx*(z.dy-z.ay))) as point_area
from
(
SELECT
a.id as angle
-- bottom left (A)
,(-t.x) * cos(radians(a.id)) - (-t.y) * sin(radians(a.id)) as ax
,(-t.x) * sin(radians(a.id)) + (-t.y) * cos(radians(a.id)) as ay
--top left (B)
,(-t.x) * cos(radians(a.id)) - (t.y) * sin(radians(a.id)) as bx
,(-t.x) * sin(radians(a.id)) + (t.y) * cos(radians(a.id)) as by
--top right (C)
,(t.x) * cos(radians(a.id)) - (t.y) * sin(radians(a.id)) as cx
,(t.x) * sin(radians(a.id)) + (t.y) * cos(radians(a.id)) as cy
--bottom right (D)
,(t.x) * cos(radians(a.id)) - (-t.y) * sin(radians(a.id)) as dx
,(t.x) * sin(radians(a.id)) + (-t.y) * cos(radians(a.id)) as dy
-- point to check (Q)
,t.qx as qx
,t.qy as qy
FROM generate_series(0,180,10) AS a(id), t
) z
;
the results then are
angle;abcd_area;point_area
0;200;340
10;200;360.6646055963
20;200;373.409049054212
30;200;377.846096908265
40;200;373.84093170467
50;200;361.515248361426
60;200;341.243556529821
70;200;313.641801308188
80;200;279.548648061772
90;200;240
*100;200;200*
*110;200;200*
*120;200;200*
*130;200;200*
*140;200;200*
150;200;237.846096908265
160;200;277.643408923024
170;200;312.04311584956
180;200;340
Where the rotations of angles 100, 110, 120, 130 and 140 degrees then includes the test-point (indicated with *)
I use VB for my application and have a rounded Background Image including logo. Its also a bit anisotropic anisalized to prevent a sharp Logo. Because vb only allows color key-ing without any tolerance I have to replace the pixels with an Alpha value to an Pixel without one. In a nutshell I have to combine the Pixel of the screen with the Pixel of the background image. But the Problem is: I don't know the formula to get the result color of the Pixel.
This is called "Alpha Compositing" You can use the following equations, courtesy of the wikipedia page (click here), to get the resulting RGB values from two transparent RGB images. (only one has to be transparent, so the other one can be your screen/background with 100% alpha.)
These equations assume that you have one transparent object on top of another object, where R1 is the Red component of object #1, a1 is the alpha of object #1, and so on and so forth...
I'll show the highlights below
R = (R1 * a1 / 255) + (R2 * a2 * (255 - a1) / (255^2))
G = (G1 * a1 / 255) + (G2 * a2 * (255 - a1) / (255^2))
B = (B1 * a1 / 255) + (B2 * a2 * (255 - a1) / (255^2))
Alpha = a1 + (a2 * (255 - a1) / 255)
I also recommend skimming the wiki page because it's quite fascinating.
I need to find the average Edit: total 2D velocity given multiple 2D velocities (speed and direction). A few examples:
Example 1
Velocity 1 is 90° at a speed of 10 pixels or units per second.
Velocity 2 is 270° at a speed of 5 pixels or units per second.
The average velocity is 90° at 5 pixels or units per second.
Example 2
Velocity 1 is 0° at a speed of 10 pixels or units per second
Velocity 2 is 180° at a speed of 10 pixels or units per second
Velocity 3 is 90° at a speed of 8 pixels or units per second
The average velocity is 90° at 8 pixels or units per second
Example 3
Velocity 1 is 0° at 10 pixels or units per second
Velocity 2 is 90° at 10 pixels or units per second
The average velocity is 45° at 14.142 pixels or units per second
I am using JavaScript but it's mostly a language-independent question and I can convert it to JavaScript if necessary.
If you're going to be using a bunch of angles, I would just calculate each speed,
vx = v * cos(theta),
vy = v * sin(theta)
then sum the x velocities and the y velocities separately as vector components and divide by the total number of velocities,
sum(vx) / total v, sum(vy) / total v
and then finally calculate the final speed and direction with your final vx and vy. The magnitude of the speed can be found by a simple application of pythagorean theorem, and then final angle should just be tan-1(y/x).
Per example #3
vx = 10 * cos(90) + 10 * cos(0) = 10,
vy = 10 * sin(90) + 10 * sin(0) = 10
so, tan-1(10/10) = tan-1(1) = 45
then a final magnitude of sqrt(10^2 + 10^2) = 14.142
These are vectors, and you should use vector addition to add them. So right and up are positive, while left and down are negative.
Add your left-to-right vectors (x axis).
Example 1 = -10+5 = -5
Example 2 = -8 = -8
Example 3 = 10 = 10. (90 degrees is generally 90 degrees to the right)
Add you ups and downs similarly and you get these velocities, your left-to-right on the left in the brackets, and your up-to-down on the right.
(-5, 0)
(-8,0)
(10, 10)
These vectors contain all the information you need to plot the motion of an object, you do not need to calculate angles to plot the motion of the object. If for some reason you would rather use speeds (similar to velocity, but different) and angles, then you must first calculate the vectors as above and then use the Pythagorean theorem to find the speed and simple trigonometry to get the angle. Something like this:
var speed = Math.sqrt(x * x + y * y);
var tangeant = y / x;
var angleRadians = Math.atan(tangeant);
var angleDegrees = angleRadians * (180 / Math.PI);
I'll warn you that you should probably talk to someone who know trigonometry and test this well. There is potential for misleading bugs in work like this.
From your examples it sounds like you want addition of 2-dimensional vectors, not averages.
E.g. example 2 can be represented as
(0,10) + (0,-10) + (-8, 0) = (-8,0)
The speed is then equal to the length of the vector:
sqrt(x^2+y^2)
To get average:
add each speed, and then divide by the number of speeds.
10mph + 20mph / 2 = 15
12mph + 14mph + 13mph + 16mph / 4 = 14 (13,75)
This is not so much average as it is just basic vector addition. You're finding multiple "pixel vectors" and adding them together. If you have a velocity vector of 2 pixels to the right, and 1 up, and you add it to a velocity vector of 3 pixels to the left and 2 down, you will get a velocity vector of 1 pixel left, and 1 down.
So the speed is defined by
pij = pixels going up or (-)down
pii = pixels going right or (-)left
speedi = pii1 + pii2 = 2-3 = -1 (1 pixel left)
speedj = pij1 + pij2 = 1-2 = -1 (1 pixel down)
From there, you need to decide which directions are positive, and which are negative. I recommend that left is negative, and down is negative (like a mathematical graph).
The angle of the vector, would be the arctan(speedj/speedi)
arctan(-1/-1) = 45 degrees
I want to calculate the coverage of cmyk color + the coverage of white.
I only have no clue of how to calculate the white.
if c, m, y, and k go from 0 to 1 is then white:
w = 1 - (c+m+y+k/4)
or is it:
w = 1- c-m-y-k
(and the clamped to be 0 or above so it's not minus something)
Or is it more complex then that?
In one "standard definition" of CMYK, converting from RGB goes like this:
K=min(R,G,B)
C=(1-R-K)/(1-K)
M=(1-G-K)/(1-K)
Y=(1-B-K)/(1-K)
I think the level of white in a color would just be W=1-K, but I'm not entirely sure if that's what you're asking...
In ColorDialog why are the max values for saturation and hue are 240 and 239 respectively? To what do they correspond?
"In Windows, the HSL and HSV spaces are usually remapped to a scale between 0 to 240 so that colors can be represented with a 32-bit value."
See http://msdn.microsoft.com/en-us/library/aa511283.aspx
Hue is expressed as an angle around the color wheel, in this case multiplied by 2/3 to stay under 240 degrees. The highest value is 239 because 240 = 0 just as 360 degrees = 0 degrees on a compass.
The max values are slightly crunched to accommodate HDTV blacker blacks (etc).
The correspond to the maximum values in the scale of HSV :). 240 is divisible by 2,3,4,5,6,8,10,12,15,16,20,24,30,40,60,80,120 etc (did I miss any?). Much more versatile for doing colour palettes than 256.
239 will be because 240 is exactly the same as 0 (as it's a circle).