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.
Related
For a Bezier curve I know the P0,P1 and the
control point P2. (and I can calculate the Q0)
I want to find for a new curve, with Q1(t=0,25) == Q0(t=0,25),
P0==Q0(t=0.25) and P1==Q0(t=0.75),
the new control point P2.
I re-edit my initial post. I've added a picture.
In the picture the blue and red dots are my initial P0 and P1 points and the green dot is the initial control point P2.
Now, I want to find the new control point P2' if I now the purple points P1' and P2'.
In essence I want to draw 2 new curves,
a) from red point to first purple point and
b) from blue point to the second purple point
Any idea?
The question is a bit convoluted but I will give it a shot. To me it seems that OP wants the control points of the segment of the original curve from the parameter value t = 0.25 up to the parameter value t=0.75. This can be achieved using De Casteljau's algorithm twice. Since I haven't found a step-by-step version for quadratic curves on SO, I made one.
Step 0: Take your three control points (note that I name them in a more usual order but I use OP's colour scheme).
Step 1: Use De Casteljau's algorithm for t=0.25. This means that
P_01 = 0.25 * P_0 + (1 - 0.25) * P_1,
P_12 = 0.25 * P_1 + (1 - 0.25) * P_2 and
P_012 = 0.25 * P_01 + (1 - 0.25) * P_12.
Note the purple colour signifying that my P_012 corresponds to the left of OP's purple points.
Step 2: Rename things to simplify writing the second run of the algorithm.
Step 3: Use De Casteljau's algorithm for the second time. But be careful: the parameter value t=0.75 of the original curve corresponds to the parameter value t=2/3 of what is left from it. This means that
P'_01 = 2/3 * P'_0 + (1 - 2/3) * P'_1,
P'_12 = 2/3 * P'_1 + (1 - 2/3) * P'_2 and
P'_012 = 2/3 * P'_01 + (1 - 2/3) * P'_12.
Step 4: Rename things again and pick up the results:
P''_0 is the left of OP's purple points,
P''_1 is the green point OP is looking for and
P''_2 is the right of OP's purple points.
In section 25.8.1 Basic Polygon Rasterization of the Vulkan spec it says:
Barycentric coordinates are a set of three numbers, a, b, and c, each in the range [0,1], with a + b + c
= 1. These coordinates uniquely specify any point p within the triangle or on the triangle’s
boundary as
p = a * p_a + b * p_b + c * p_c
where p_a , p_b , and p_c are the vertices of the triangle. a, b, and c are determined by:
a = A(p, p_b, p_c) / A(p_a, p_b, p_c)
b = A(p, p_a, p_c) / A(p_a, p_b, p_c)
c = A(p, p_a, p_b) / A(p_a, p_b, p_c)
where A(l,m,n) denotes the area in framebuffer coordinates of the triangle with vertices l, m, and n.
Framebuffer coordinates technically have three components. This is specified in 24.5 Controlling the Viewport as:
The vertex’s framebuffer coordinates (x_f , y_f , z_f ) are given by [snip]
What precisely is the formula of the A function?
Is it either:
(a) the same as the formula given to calculate whether the triangle is back-facing or front-facing in 25.8.1, namely:
a = -0.5 * sum_i(x_f[i] * y_f[i+1] - x_f[i+1] * y_f[i])
That is, is it taken as read that the forumla of A does not use the z_f components of its arguments, and is purely a function of the (x_f, y_f) components? (ie It is calculating the area of the two dimensional projection of the triangle onto the x-y plane in framebuffer-space)
or (b), does A use all three framebuffer components? ie Does A return the area of the triangle in the full three-dimensional framebuffer-space (like shown here for example)
or (c) something else?
It actually doesn't matter, mathematically speaking. Whichever function you pick, you'll find that the particulars of the math divide out when computing the barycentric coordinate.
A barycentric coordinate is computed by taking the ratio of two areas. If you linearly project two co-planar triangles from 3D space to 2D space with the same projection, the ratio of their areas is unchanged (assuming that they have an area post-projection).
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 have a green vehicle which will shortly collide with a blue object (which is 200 away from the cube)
It has a Kinect depth camera D at [-100,0,200] which sees the corner of the cube (grey sphere)
The measured depth is 464 at 6.34° in the X plane and 12.53° in the Y plane.
I want to calculate the position of the corner as it would appear if there was a camera F at [150,0,0], which would see this:
in other words transform the red vector into the yellow vector. I know that this is achieved with a transformation matrix but I can't find out how to compute the matrix from the D-F vector [250,0,-200] or how to use it; my high-school maths dates back 40 years.
math.se has a similar question but it doesn't cover my problem and I can't find anything on robotices.se either.
I realise that I should show some code that I've tried, but I don't know where to start. I would be very grateful if somebody could help me to solve this.
ROS provides the tf library which allows you to transform between frames. You can simply set a static transform between the pose of your camera and the pose of your desired location. Then, you can get the pose of any point detected by your camera in the reference frame of your desired point on your robot. ROS tf will do everything you need and everything I explain below.
The longer answer is that you need to construct a transformation tree. First, compute the static transformation between your two poses. A pose is a 7-dimensional transformation including a translation and orientation. This is best represented as a quaternion and a 3D vector.
Now, for all poses in the reference frame of your kinect, you must transform them to your desired reference frame. Let's call this frame base_link and your camera frame camera_link.
I'm going to go ahead and decide that base_link is the parent of camera_link. Technically these transformations are bidirectional, but because you may need a transformation tree, and because ROS cares about this, you'll want to decide who is the parent.
To convert rotation from camera_link to base_link, you need to compute the rotational difference. This can be done by multiplying the quaternion of base_link's orientation by the conjugate of camera_link's orientation. Here's a super quick Python example:
def rotDiff(self,q1: Quaternion,q2: Quaternion) -> Quaternion:
"""Finds the quaternion that, when applied to q1, will rotate an element to q2"""
conjugate = Quaternion(q2.qx*-1,q2.qy*-1,q2.qz*-1,q2.qw)
return self.rotAdd(q1,conjugate)
def rotAdd(self, q1: Quaternion, q2: Quaternion) -> Quaternion:
"""Finds the quaternion that is the equivalent to the rotation caused by both input quaternions applied sequentially."""
w1 = q1.qw
w2 = q2.qw
x1 = q1.qx
x2 = q2.qx
y1 = q1.qy
y2 = q2.qy
z1 = q1.qz
z2 = q2.qz
w = w1 * w2 - x1 * x2 - y1 * y2 - z1 * z2
x = w1 * x2 + x1 * w2 + y1 * z2 - z1 * y2
y = w1 * y2 + y1 * w2 + z1 * x2 - x1 * z2
z = w1 * z2 + z1 * w2 + x1 * y2 - y1 * x2
return Quaternion(x,y,z,w)
Next, you need to add the vectors. The naive approach is to simply add the vectors, but you need to account for rotation when calculating these. What you really need is a coordinate transformation. The position of camera_link relative to base_link is some 3D vector. Based on your drawing, this is [-250, 0, 200]. Next, we need to reproject the vectors to your points of interest into the rotational frame of base_link. I.e., all the points your camera sees at 12.53 degrees that appear at the z = 0 plane to your camera are actually on a 12.53 degree plane relative to base_link and you need to find out what their coordinates are relative to your camera as if your camera was in the same orientation as base_link.
For details on the ensuing math, read this PDF (particularly starting at page 9).
To accomplish this, we need to find your vector's components in base_link's reference frame. I find that it's easiest to read if you convert the quaternion to a rotation matrix, but there is an equivalent direct approach.
To convert a quaternion to a rotation matrix:
def Quat2Mat(self, q: Quaternion) -> rotMat:
m00 = 1 - 2 * q.qy**2 - 2 * q.qz**2
m01 = 2 * q.qx * q.qy - 2 * q.qz * q.qw
m02 = 2 * q.qx * q.qz + 2 * q.qy * q.qw
m10 = 2 * q.qx * q.qy + 2 * q.qz * q.qw
m11 = 1 - 2 * q.qx**2 - 2 * q.qz**2
m12 = 2 * q.qy * q.qz - 2 * q.qx * q.qw
m20 = 2 * q.qx * q.qz - 2 * q.qy * q.qw
m21 = 2 * q.qy * q.qz + 2 * q.qx * q.qw
m22 = 1 - 2 * q.qx**2 - 2 * q.qy**2
result = [[m00,m01,m02],[m10,m11,m12],[m20,m21,m22]]
return result
Now that your rotation is represented as a rotation matrix, it's time to do the final calculation.
Following the MIT lecture notes from my link above, I'll arbitrarily name the vector to your point of interest from the camera A.
Find the rotation matrix that corresponds with the quaternion that represents the rotation between base_link and camera_link and simply perform a matrix multiplication. If you're in Python, you can use numpy to do this, but in the interest of being explicit, here is the long form of the multiplication:
def coordTransform(self, M: RotMat, A: Vector) -> Vector:
"""
M is my rotation matrix that represents the rotation between my frames
A is the vector of interest in the frame I'm rotating from
APrime is A, but in the frame I'm rotating to.
"""
APrime = []
i = 0
for component in A:
APrime.append(component * M[i][0] + component * M[i][1] + component * m[i][2])
i += 1
return APrime
Now, the vectors from camera_link are represented as if camera_link and base_link share an orientation.
Now you may simply add the static translation between camera_link and base_link (or subtract base_link -> camera_link) and the resulting vector will be your point's new translation.
Putting it all together, you can now gather the translation and orientation of every point your camera detects relative to any arbitrary reference frame to gather pose data relevant to your application.
You can put all of this together into a function simply called tf() and stack these transformations up and down a complex transformation tree. Simply add all the transformations up to a common ancestor and subtract all the transformations down to your target node in order to find the transformation of your data between any two arbitrary related frames.
Edit: Hendy pointed out that it's unclear what Quaternion() class I refer to here.
For the purposes of this answer, this is all that's necessary:
class Quaternion():
def __init__(self, qx: float, qy: float, qz: float, qw: float):
self.qx = qx
self.qy = qy
self.xz = qz
self.qw = qw
But if you want to make this class super handy, you can define __mul__(self, other: Quaternion and __rmul__(self, other: Quaternion) to perform quaternion multiplication (order matters, so make sure to do both!). conjugate(self), toEuler(self), toRotMat(self), normalize(self) may also be handy additions.
Note that due to quirks in Python's typing, the above other: Quaternion is only for clarity. You'll need a longer-form if type(other) != Quaternion: raise TypeError('You can only multiply quaternions with other quaternions) error handling block to make that into valid python :)
The following definitions are not necessary for this answer, but they may prove useful to the reader.
import numpy as np
def __mul__(self, other):
if type(other) != Quaternion:
print("Quaternion multiplication only works with other quats")
raise TypeError
r1 = self.qw
r2 = other.qw
v1 = [self.qx,self.qy,self.qz]
v2 = [other.qx,other.qy,other.qz]
rPrime = r1*r2 - np.dot(v1,v2)
vPrimeA = np.multiply(r1,v2)
vPrimeB = np.multiply(r2,v1)
vPrimeC = np.cross(v1,v2)
vPrimeD = np.add(vPrimeA, vPrimeB)
vPrime = np.add(vPrimeD,vPrimeC)
x = vPrime[0]
y = vPrime[1]
z = vPrime[2]
w = rPrime
return Quaternion(x,y,z,w)
def __rmul__(self, other):
if type(other) != Quaternion:
print("Quaternion multiplication only works with other quats")
raise TypeError
r1 = other.qw
r2 = self.qw
v1 = [other.qx,other.qy,other.qz]
v2 = [self.qx,self.qy,self.qz]
rPrime = r1*r2 - np.dot(v1,v2)
vPrimeA = np.multiply(r1,v2)
vPrimeB = np.multiply(r2,v1)
vPrimeC = np.cross(v1,v2)
vPrimeD = np.add(vPrimeA, vPrimeB)
vPrime = np.add(vPrimeD,vPrimeC)
x = vPrime[0]
y = vPrime[1]
z = vPrime[2]
w = rPrime
return Quaternion(x,y,z,w)
def conjugate(self):
return Quaternion(self.qx*-1,self.qy*-1,self.qz*-1,self.qw)
I am just starting to learn jython, and just have a question which I cannot seem to get right.
From my text, I am to create a picture that is 640 x 480 pixels, and then, using a loop, pixel by pixel set the color to a calculation for r, g, b which we have already been given.
I can create a picture, I can set variables, however I cannot seem to go any further in creating a loop to set each pixel colour.
I know its only simple, but just wandering if anyone can help me out here.
xrange() will create a generator which yields integers in a range. for will loop once per element of an iterable.
for row in xrange(480):
for col in xrange(640):
...
This may help you to iterate through the pixels.
picture = makeEmptyPicture(400,200)
pixels = getPixels(picture)
#make an empty picture and get the pixels
for px in getPixels(picture):
x=getX(px)
y=getY(px)
r = (sin(x * radian * id[1]) * cos(y * radian * id[4]) + 1) * ord(StringID[0]) * 2.5
g = (sin(x * radian * id[2]) * cos(y * radian * id[5]) + 1) * ord(StringID[0]) * 2.5
b = (sin(x * radian * id[3]) * cos(y * radian * id[6]) + 1) * ord(StringID[0]) * 2.5
newColor=makeColor(255 - r, 255 - g, 255 - b)
setColor(px, newColor)
show(picture)
repaint(picture)