This problem is in 3D space.
There is a rectangle, defined by 4 vertices. We rotate it around one of its sides.
There is a triangle, defined by 3 vertices.
After a full 360 degree rotation, will the rectangle ever intersect/touch the triangle?
If so, what is the angle of rotation at which intersection first occurs? And what is the point of this first intersection?
After thinking about this for a while, it seems like there are 3 main cases:
triangle vertex touches rectangle surface
triangle surface touches rectangle vertex
triangle edge touches rectangle edge
And there are 2 unlikely cases where the two are perpendicular when the intersect:
rectangle edge hits triangle surface
rectangle surface hits triangle edge
However identifying these cases hasn't really gotten me closer to a solution. I'm hoping someone can point me in the right direction for how to solve this problem. I want to solve it fast for a small number of rectangles x a large number of triangles.
Context: the larger problem I'm trying to solve is I want to wrap a rectangle around a closed polygonal mesh. I wish to do this step by step by rotating the rectangle until it intersects, then rotating the remaining rectangle around the intersection point, etc.
When you rotate a rectangle around one of its sides, you get a cylinder. Intersect each of the lines with the cylinder. The position of the intersection points gives you the rotation angles. Since this doesn't catch the case where the triangle is completely contained within the cylinder, test whether the vertices' distance to the cylinder's axis is smaller than the cylinder's radius, too.
Say your rectangle has the vertices A to D. You want to rotate around the side AB. The radius of your cylinder is then r = |AD|.
First, transform the coordinates so that the rectangle is placed with the side that you want to rotate about along the z axis and the adjacent side along the x axis.
A′ = {M} · A = {0, 0, 0}
B′ = {M} · B = {0, 0, |AB|}
C′ = {M} · C = {r, 0, 0}
Apply the same transformation {M} to the vertices of the triangle.
Now find the intersections of all three sides of the triangle with the cylinder. Because the cylinder is aligned to the z axis, the problem can be separated into two subproblems: (1) Find any intersections with the top and bottom surfaces a z == 0 and z == |AB|. (2) Find the intersections with the "coat" of the cylinder; this is the intersection of a line with a circle in the xy plane.
You can then calculate the rotation angles with the tangent function of the y and x coordinates of these points as atan2(y, x).
If you need the coordinates of the intersection points in the original coordinates, don't forget to undo the transformation.
Related
I have to find the exact centroid of multiple rectangles. The coordinates of each rectangle are as follows:
coord = (0.294792, 0.474537, 0.0989583, 0.347222) ## (xcenter, ycenter, width, height)
I have around 200 rectangles, how can I compute the centroid of them?
I already tried to implement it, but the code did not work well.
My code:
for i in range(len(xCenter)):
center = np.array((xCenter[i]+(Width[i]/2), yCenter[i]+(Height[i]/2)))
This is a somewhat vague question, but if you mean the centroid of all rectangles by area, then each center of a rectangle is weighted by the area of the rectangle. Think of it as the all the mass of the rectangle being compressed into the center, and then having to take the centroid of several weighted points. The formula for that would be the sum of 1 through n (assuming rectangles are numbered 1 to n) of Area(Rec(i)) * vec(center(i)) all divided by the total mass of the system (the sum of all the areas). If you are referring to the centroid of the area in general, ignoring rectangle overlap, that is a little more tricky. One thing you could do is for each rectangle, check it against all other rectangles, and if a pair of rectangles overlap, split them up into a set of non-overlapping rectangles and put them back into the set of rectangles. Once all rectangles are non-overlapping, find the centroid by mass.
Have to make a program that given the option to input square or circle, user inputs width and a center x,y coordinate.
What I don't understand is how to write code for if there are two shapes on a plane and how to identify if one is inside the other
I'm super helpless, and have no background in computer science. Thank you!
You have to address Circle in Circle, Circle in Square, Square in Circle, and Square in Square. I suggest drawing some pictures with the centers marked, and observe their relationships.
Circle in Circle: the distance between the centers has to be less than the difference in radius's.
Circle in Square: Same as circle in circle I think
Square in Circle: The radius of the circle must be larger than the distance from the center of the square to a corner PLUS the distance between centers
Square in Square: you got this! solve yourself ;)
I'm trying to do draw widgets on the circle, for this I need to paint the widgets as a arc. I know the number of widgets (let's say), then each widget is at 36 degrees from the origin to the circumference. The information I have is the center of circle, radius and I know the starting and end point on the circumference for each such widget.
This is computed by doing
dx = int(round(400 + 300 * np.cos(angle)))
dy = int(round(400 + 300 * np.sin(angle)))
where angle = 2 * np.pi / 15
I go over a for loop computing the new value for angle which is basically angle * i where i = (1, 10)
I don't understand the start angle and span angle for the arcs function in QPainter.QPainter Arc. I googled and not many terms came up. Maybe there is a different term for them.
So the problem is I have a starting point and ending point on the circumference and center and radius, how do I use them to draw Arcs such that I get something that looks like :
circos
What I have tried is, I can compute the center point (cx) of the two end points, if I draw a line from the center of the circle to this point cx, then I can compute how far this point circumference which essentially is my width, but how to get the orientation correct to represent them as circles.
Instead of circular I do have a layout with just lines like this, but would like to be like the circos one.
My image
I don't understand the start angle and span angle for the arcs function in QPainter.QPainter Arc.
Why? The documentation is IMHO very clear:
The startAngle and spanAngle must be specified in 1/16th of a degree, i.e. a full circle equals 5760 (16 * 360).
This means that your units are 1/16º. E.g. 45º is 45*16 units.
Positive values for the angles mean counter-clockwise while negative values mean the clockwise direction. Zero degrees is at the 3 o’clock position."
This means that 90*16 points at 12 o'clock (goes 90º counter-clockwise from 3 o'clock), and -90*16 points at 6 o'clock.
Of course the "zero" degrees only has sense for the start angle. The span angle states how much further does the arc go, and in which direction.
For example, to draw an arc from 3 o'clock to 12 o'clock, you'd do
painter.drawArc(rect, 0, 90*16)
*or*
painter.drawArc(rect, 90*16, -90*16)
But to draw an arc from 3 o'clock to 6 o'clock, you'd do
painter.drawArc(rect, 0, -90*16)
*or*
painter.drawArc(rect, -90*16, 90*16)
The arcs are not specified using center and radius, but rather using a bounding rectangle. If the arc was a full ellipse, it would be inscribed in the rectangle - the arcs are implicitly elliptical arcs.
So, given x and y centerpoint, and r for circular radius, the bounding rectangle is
rect = QRect(x-r, y-r, 2*r, 2*r)
I am trying to rotate a vtk camera around its focal point. The aim being to 'orbit' the model.
I'm using the camera.SetPosiiton(x, y, z) call to set the camera location, and I know I can do the same at each update period in my render window.
The focal point has the location (0, 0, 0), and some other bounding box getting gives me my initial camera (x, y, z) location. The distance from the focal point (0, 0, 0) to the camera location 9x, y, z) describes the radius the of the sphere.
In my head, this essentially moving the camera in steps around the point (0, 0, 0) and I am presuming there is a maths function I could use to feed it my starting camera point, and work out my next camera location.
This should result in the model appearing to spin in space. My camera view is offset from all x, y, z, planes, making it a 3d problem, not a 2d problem. However, I do want my camera to remain the same distance from the model (focal point)
What I am trying to achieve is like this:- take a pencil (my model is long and narrow). Hold it in your finger tips at arms length, tip pointing to the ceiling. Tilt the pencil by ~30 degrees. This is the camera start position. Rotate the pencil body in your fingers, maintaining tilt angle, and the distance from your eye.
THis post looks helpful: Plotting a point on the edge of a sphere however, this assumes you know the radius to get to the initial x, y location.
Could some one point me towards the maths I need to do this, my maths is horribly rusty.
It seems what you want is to rotate a vector about an axis, this can be most easily done using a rotation matrix
So, if your desired axis of rotation is tilted 30 degrees from the z axis on the zx plane, your axis of rotation is (cos(pi/6),0,sin(pi/6)), increment the rotation angle, plug that into the rotation matrix to get matrix R, the new camera position vector will be R*(x,y,z)'
Start off with the points (+-1,0,0), (0,+-1,0), (0,0,+-1). These form two Pyramids with all the points on the unit sphere.
Now you can take the midpoints of each triangle, and project out so they lie on the unit sphere too. For each triangle this now gives you 3 new triangles, and you can repeat the process.
The alternative to the midpoint of the triangle is to take the midpoints of each side, and join them up. That gives 3 new points that can be projected out to the unit circle. This gives you 4 triangles for each sub division.
Repeat as many times as you need.
I have been working around extracting the time series from shapes based on distances to center of mass clockwise starting from angle 0 to 360.
My Implementation that arranges contour points based on their angle to the [1,0], vector might be good for some shapes but is not useful for shapes that has much articulation. Consider the following code:
im = Image.open(os.path.join(path,filename))
im = im.filter(ifilter.MedianFilter)
contim = im.filter(ifilter.CONTOUR)
contim = contim[1:-1,1:-1] # this is because borders are extracted here as contours
contpts = np.where(contim ==0)
contpts = np.vstack((contpts[0],contpts[1])) # Just need to arrange these points clockwise with respect to the center of mass of the shape
Can anyone direct me to how I can extract that feature from any shape where I can start from a point and keep going along the contour to extract all the distances to the center of mass of the shape.
For more information about the feature, please view this paper: "LB_Keogh Supports Exact Indexing of Shapes under Rotation Invariance with Arbitrary Representations and Distance Measures"
If I understood, there's a geometrical figure in a discretized plane, represented as a matrix. If the entry is 1, you're inside the figure. If it's 0, you're outside. He wants to determine de distance between the edge of the figure and the center of the figure for all points in the edge. He parametrized it with a polar coordinate system. The center of the figure is the origin and now he wants to get the distance to the border as a function of the angle. This is what he calls his "time series".
Is this correct?
If yes, couldn't you just:
1. determine the center of mass,
2. reposition the origin to match the center of mass.
3. start angle at 0
4. r = 0
5. for each angle in [0,1,...,360]
1. If you're in inside the figure, increase r until you reach the border.
2. If you're outside the figure, decrease r until you reach the border.
3. When you reach the border, d(angle) = r
It the figure have a more or less continuous border, this will follow the contour.
Would this work?