I am currently drawing a LineString via clicking points in my UI.
Before adding a new point to the LineString I check via oldLinestring.intersection(newsegment) if it intersects the previous lines.
This gives me the Point of intersection and leaves me with a LineString like this (note that he intersection point does not split the line segment that it lays on):
I now want to create a Polygon from the enclosed Area but using a Poligonizer wont work because the LineString does not form a loop with matching start and end points.
What is the intended way to approach this?
Related
Given a number of points on a 2d surface and radiuses for these points I can easily paint circles for them. What I need is an algorithm that only paints the envelope (right word for what I am looking for?) or outer bound of these combined circles. Additionally a second set of circles can 'encroach' on these circles, resulting in a kind of 'border'.
Image of what I am looking for
A quick way to draw the outline of the union of the disks is to
fill all disks in yellow, then
fill all disks in white with a smaller radius.
This can be adapted to the "encroached" circles, provided you only fill the remaining portions of the disks. Unfortunately, in a general setting finding the remaining portions can be an uneasy geometric problem.
There is an alternative approach which can work in all cases:
fill an image with zeroes, then for all disks fill every pixel with the value of the distance to the circumference (maximum at the center), but only keep the highest value so far.
while you do this, fill a second image with the color of the disk that achieved that highest value. (Initialize the image with the background color.)
At the end of this process, the first image will represent a "terrain" made of intersecting cones; and for every point of the terrain, you will know the color.
finally, color the pixels that have a height smaller than the desired stroke width, using the color map.
You can do the drawing in two steps.
1) Draw the outline using the following method: For each point, draw a circle using your favorite circle-drawing method, but before drawing a pixel, ensure that it is not contained inside any other circle. Do this for every point and you will get your outline.
2) Draw the borders between different sets using the following method: For each pair of points from different sets, calculate the two intersection points of the circles. If there is an intersection, the border can be drawn as a segment joining these two points. However, you have to make two lines, one for circle A, and another for circle B. To draw the line for circle A, slightly offset the segment towards point A. Then, use your favorite line-drawing method, but before drawing a pixel, ensure that it is closer to point A that any other point of the opposite set. After drawing the line, repeat the process for circle B. Note that both segment are not guaranteed to be the same length since the asymmetry of the points of the different sets. It will, however, always form a closed shape when all outlines and borders are drawn.
I need help with this : I have Freeform area and shapes in form of circles named from A to G. Some of this points are inside of Freeform area and some of them no. I need macro which will check each of this points - if this point is inside/ outside of freeform area. How can i get result in form of table where in Column A are Names of shapes (circles-points) and in column B result (if point is inside/outside of freeform).
Thanks
Victor
Picture of what I need is here:
There are several ways to do that. Either over calculation of the Angles or by creating a ray trough the point and find the intersections To the polygons. If the count are even, it's inside; if it's odd it's outside. Those algorithms are not that trivial and numerical a challenge. Here are some links (They have vb6/vba source):
Paul Bourke Point inside Polygon
VB-Helper Point inside Polygon
If it has not that exact you may draw the points in a picturebox and determine the background color.
To find more search the web for "points inside polygon".
How can I calculate object bottom position?
I created an object and a plane, both in (0,0,0) position, but the model geometrical center is placed in that point, not the bottom face of it, in result, the plane is clipping the half of it, see the demo: http://www.edar.com.pl/test-3d-3/
I'd like my models to have the same height when I place them on the screen. Also I want them all to intersect the shadow plane with their bottom parts, so I guess I have to calculate their heights and move them via object.position.y = obj_height/2; ?
Check Three.GeometryUtils.Center() - function which centers the geometry of an object.
So you could copy and modify the function to your liking to center the geometry at the bottom center.
Also you could also iterate over all vertices and just add or remove an offset-vector.
In an infinite 2D space there are a set of lines, each line having a start and end point, and a time of creation: Line(p0, p1, t).
I want to find the lines that should be rendered in a top-down view of this 2D space (higher values of t show up closer to the viewport, not that it should be relevant.)
The intuitive answer is "check if either point is within the viewport coordinates," but this falls down when the points are further apart than the viewport area covers.
The other idea I had was using something like geohash, this would limit precision i.e. maximum zoom level of the viewport. The idea is enumerating the hashes of the cells intersected and storing them. This way querying is a matter of asking the right question.
Are there any ideal solutions? Has this been solved before?
I think you need to check two conditions: one that the rectangle of viewport overlaps the rectangle with corners (p0,p1) and the second that some corners of viewport rectangle are on the different sides of the whole line which contains line segment (p0,p1).
The task of finding rectangle overlap can be solved very effectively for very large number of rectangles using R-trees (http://en.wikipedia.org/wiki/R-tree).
The second task can be reduced to checking signs of the cross product of (p1-p0) x (corner_coordinate-p0)
(all three quantities taken as 3-d vectors with third coordinate equal to zero, the result will be vector along the perpendicular direction). There should be corners with the opposite sign of this cross product.
Given this grid ( http://i.stack.imgur.com/Nz39I.jpg is a trapezium/trapezoid, not a square), how do you find the point clicked by the user? I.e. When the user clicks a point in the grid, it should return the coordinates like A1 or D5.
I am trying to write pseudo code for this and I am stuck. Can anyone help me? Thanks!
EDIT: I am still stuck... Does anyone know of any way to find the height of the grid?
If it is a true perspective projection, you can run the click-point through the inverse projection to find it's X,Z coordinates in the 3D world. That grid has regular spacing and you can use simple math to get the A1,D5,etc.
If it's just something you drew, then you'll have to compare the Y coordinates to the positions of the horizontal lines to figure out which row. Then you'll need to check its position (left/right) relative to the angled lines to get the column - for that, you'll need either coordinates of the end-points, or equations for the lines.
Yet another option is to store an identical image where each "square" is flood-filled with a different color. You then check the color of the pixel where the user clicked but in this alternate image. This method assumes that it's a fixed image and is the least flexible.
If you have the coordinates of end points of the grid lines then
Try using the inside-outside test for each grid line and find the position
Since this grid is just a 3D view of a 2D grid plane, there is a projective transform that transforms the coordinates on the grid into coordinates on the 2D plane. To find this transform, it is sufficient to mark 4 different points on the plane (say, the edges), assign them coordinates on the 2D plane and solve the resulting linear equation system.