Assume the following three things are known:
The (x,y) coordinate of point A (the top left point in each of the example triangles shown below),
The slope and y-intercept of the line shown in green,
An aspect ratio for the target rectangle.
...I would like to find out how to determine the location of the point shown in yellow--which intersects the green line--such that the rectangle shown in black matches the required target aspect ratio.
Thank you.
-lagouyn
Let's A point has coordinates (ax, ay), and line equation is y=y0+k*x
Then
(y0 + k*X - ay) * aspect = X - ax
X * (k*aspect - 1) = ay*aspect - ax - y0*aspect
X = (ay*aspect - ax - y0*aspect) / (k*aspect - 1)
Y = y0+ k*X
Related
i have some X/Y coordinates that represent the center of a circle somewhere in an image. From that circle, i want to compoute the mean of all the point contained inside the circle.
currently, i compute the mean of a square patche as follow, but a square is not relevent for the project. Ideally, i would like to do it only with numpy. but if it is not possible, i would concidere something else.
mean = np.mean(image[Y - margin : Y + margin, X - margin, X + margin])
As I understood,
YOU HAVE: (x,y) of the center of the circle
YOU WANT: mean of all the points contained in the circle
Since all the points on the right side should be equal to the number of points on the left side of the center, Wouldn't the mean be the same as the center of the circle !?
i found a solution where I compute all indexes contained in a centered disk.
I calculate the squarred euclidean distance of each X/Y coordinate of a squarre array.
I compare it to the squarred radius of the circle.
If it is superior, the point is not contained in the circle. Mark it as 0, 1 otherwise
I extract indexes where the array is equal to one (disk-shaped)
I center the computed indexes. To use them, i add the X/Y coordinate of a specific image point.
NOTE: I used the squarred euclidean distance because the square root function is monotonic (i.e constantly increasing). So, it saves compational power to keep with the squarred version.
radius = 4
size = 2 * radius + 1
radiusSquarred= radius**2
mask = np.zeros((size, size))
distance = lambda x, y: (x-radius)**2 + (y-radius)**2
for i in range(size):
for j in range (2 * radius+ 1):
if distance(i, j) <= radiusSquarred:
mask[i, j] = 1
index = np.where(mask == 1)
diskIndexes = (index[0] - radius, index[1] - radius)
X, Y = 100, 150
np.mean(image[diskIndexes[0] + Y, diskIndexes[1] + X])
Here some examples of twisted triangle prisms.
I want to know if a moving triangle will hit a certain point. That's why I need to solve this problem.
The idea is that a triangle with random coordinates becomes the other random triangle whose vertices all move between then
related: How to determine point/time of intersection for ray hitting a moving triangle?
One of my students made this little animation in Mathematica.
It shows the twisting of a prism to the Schönhardt polyhedron.
See the Wikipedia page for its significance.
It would be easy to determine if a particular point is inside the polyhedron.
But whether it is inside a particular smooth twisting, as in your image, depends on the details (the rate) of the twisting.
Let's bottom triangle lies in plane z=0, it has rotation angle 0, top triangle has rotation angle Fi. Height of twisted prism is Hgt.
Rotation angle linearly depends on height, so layer at height h has rotation angle
a(h) = Fi * h / Hgt
If point coordinates are (x,y,z), then shift point to z=0 and rotate (x,y) coordinates about rotation axis (rx, ry) by -a(z) angle
t = -a(z) = - Fi * z / Hgt
xn = rx + (x-rx) * Cos(t) - (y-ry) * Sin(t)
yn = ry + (x-rx) * Sin(t) - (y-ry) * Cos(t)
Then check whether (xn, yn) lies inside bottom triangle
I have an infinite grid of hexagons, defined by a cubic (x y z) coordinate system like so:
I also have a viewport -- a rectangular canvas where I will draw the hexagons.
My issue is this. Because the grid of hexagons is infinite in all directions, I can't feasibly draw all of them at once. Therefore, I need to draw all the hexagons that are in the viewport, and ONLY those hexagons.
This image summarizes what I want to do:
In this image, purple-colored hexagons are those I want to render, while white-colored hexagons are those I don't want to render. The black rectangle is hte viewport -- all the hexagons that intersect with it are to be drawn. How would I find which hexagons to render (IE their xyz coordinates)?
Some other info:
I have a function that can recall a hexagon tile and draw it centered at position(x,y) in the viewport, given its cubic xyz coordinates. Therefore, all I should need is the xyz coords of each rectangle to draw, and I can draw them. This might simplify the problem.
I have formulas to convert from cubic hexagon coordinates to x/y coordinates, and back. Given the above diagram, r/g/b being the axes for the cubic coords with the image above, x and y being the cartesian coordinates, and s being the length of a hexagon's edge...
y = 3/2 * s * b
b = 2/3 * y / s
x = sqrt(3) * s * ( b/2 + r)
x = - sqrt(3) * s * ( b/2 + g )
r = (sqrt(3)/3 * x - y/3 ) / s
g = -(sqrt(3)/3 * x + y/3 ) / s
r + b + g = 0
Let's X0, Y0 are coordinates of top left corner, RectWidth is rectangle width, HexWidth = s * Sqrt(3/2) is hexagon width.
Find center of the closest hexagon r0, g0, b0, HX0, HY0. (Rect corner lies in this hexagon, because hexagons are Voronoy diagram cells). Remember horizontal and vertical shift DX = X0 - HX0, DY = Y0 - HY0
Draw horizontal row of Ceil(RectWidth/HexWidth) hexagons, incrementing r coordinate, decrementing f, and keeping b the same, ROWINC=(1,-1,0).
Note that if DY > HexWidth/2, you need extra top row with initial coordinates shifted up (r0, g0-1, b0+1)
Shift starting point by L=(0, 1, -1) if the DX < 0, or by R=(1, 0, -1) otherwise. Draw another horizontal row with the same ROWINC
Shift row starting point by alternative way (L after R, R after L). Draw horizontal rows until bottom edge is reached.
Check whether extra row is needed in the bottom.
You can think of the rectangular box in terms of constraints on an axis.
In the diagram, the horizontal lines correspond to b and your constraints will be of the form somenumber ≤ b and b ≤ somenumber. For example the rectangle might be in the range 3 ≤ b ≤ 7.
The vertical lines are a little trickier, but they are a “diagonal” that corresponds to r-g. Your constraints will be of the form somenumber ≤ r-g and r-g ≤ somenumber. For example it might be the range -4 ≤ r-g ≤ 5.
Now you have two axes with constraints on them, and you can form a loop. The easiest thing will be to have the outer loop use b:
for (b = 3; b ≤ 7; b++) {
…
}
The inner loop is a little trickier, because that's the diagonal constraint. Since we know r+g+b=0, and we know the value of b from the outer loop, we can rewrite the two-variable constraint on r-g. Express r+g+b=0 as g=0-r-b. Now substitute into r-g and get r-(0-r-b). Simplify r-(0-r-b) to 2*r-b. Instead of -4 ≤ r-g we can say -4 ≤ 2*r-b or -4+b ≤ 2*r or (-4+b)/2 ≤ r. Similarly, we can rearrange r-g ≤ 5 to 2*r-b ≤ 5 to r ≤ (5+b)/2. This gives us our inner loop:
for (b = 3; b ≤ 7; b++) {
for (r = (-4+b)/2; r ≤ (5+b)/2; r++) {
g = 0-b-r;
…
}
}
The last bit is to generalize, replacing the constants 3,7,-4,5 with the actual bounds for your rectangle.
My trigonometry needs a little help.
How would I go about calculating the point of the nearest possible intersection with a line along a rounded corner?
Take this image:
What I would like to know is, given that I know point a, and the dimensions of the rectangle, how would I find point b when the edges of the rectangle are curved?
So far, as you can see, I've only managed to calculate the nearest edge of the rectangle as if it had right-angled corners.
If it matters, I'm doing this in ActionScript 3. But example sudo-code will suffice.
Calculate the vector from the midpoint M of the corner to A:
v_x = a_x - m_x
v_y = a_y - m_y
then go radius of the corner r times towards A to get to the intersection point I
i_x = m_x + r*v_x
i_y = m_y + r*v_y
This obviously only works if the nearest intersection is on the rounded corner. Just calculate the other intersections with the edges, too, and then check which has the nearest distance to A.
You need to know the radius R of the circle that generates the round corner and the coordinates (Xr,Yr) of the point where the two sides of a non rounded rectangle cross each other.
Then the coordinates for the center of the circle that generates the round corner are (Xc, Yc) = (Xr-R, Yr-R)
From here, it's a matter of solving the equation of the cross point between the segment line defined by point A=(Xa, Ya) and point (Xc, Yc), whose parametric equation is:
x = Xa + p*(Xc-Xa)
y = Ya + p*(Yc-Ya)
and the circle whose equation is
(x-Xc)^2 + (y-Yc)^2 = R^2
Substitute values for x and y from the parametric euation of the line in the equation of the circle, and you will have an equation with only one unkown: p. Solve the equation, and if there are more than one solution, choose the one that is in the range [0,1]. Substitute the found value of p in the parametric equation of the line to get the point of intersection.
Graphically:
If you know the radius and center of the corner as R and C=(Xc, Yc), then the nearest point on the corner to the given point A=(Xa, Ya) is the intersection point of the corner and the line defined by the given point and the center. This point can be directly expressed as
X = Xc + R*(Xa-Xc)/|AC|
Y = Yc + R*(Ya-Yc)/|AC|
where |AC| = Sqrt((Xa-Xc)^2 + (Ya-Yc)^2)
Imagine a circle. Imagine a pie. Imagine trying to return a bool that determines whether the provided parameters of X, Y are contained within one of those pie pieces.
What I know about the arc:
I have the CenterX, CenterY, Radius, StartingAngle, EndingAngle, StartingPoint (point on circumference), EndingPoint (point on circumference).
Given a coordinate of X,Y, I'd like to determine if this coordinate is contained anywhere within the pie slide.
Check:
The angle from the centerX,centerY through X,Y should be between start&endangle.
The distance from centerX,centerY to X,Y should be less then the Radius
And you'll have your answer.
I know this question is old but none of the answers consider the placement of the arc on the circle.
This algorithm considers that all angles are between 0 and 360, and the arcs are drawn in positive mathematical direction (counter-clockwise)
First you can transform to polar coordinates: radius (R) and angle (A). Note: use Atan2 function if available. wiki
R = sqrt ((X - CenterX)^2 + (Y - CenterY)^2)
A = atan2 (Y - CenterY, X - CenterX)
Now if R < Radius the point is inside the circle.
To check if the angle is between StartingAngle (S) and EndingAngle (E) you need to consider two possibilities:
1) if S < E then if S < A < E the point lies inside the slice
2) if S > E then there are 2 possible scenarios
if A > S
then the point lies inside the slice
if A < E
then the point lies inside the slice
In all other cases the point lies outside the slice.
Convert X,Y to polar coordinates using this:
Angle = arctan(y/x);
Radius = sqrt(x * x + y * y);
Then Angle must be between StartingAngle and EndingAngle, and Radius between 0 and your Radius.
You have to convert atan2() to into 0-360 before making comparisons with starting and ending angles.
(A > 0 ? A : (2PI + A)) * 360 / (2PI)