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.
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 created a custom venn chart for two sets an their intersection in JavaFX.
Now I want to show the number of elements in this sets like in the following picture:
The sets are ellipses and I used masks to give the intersection a different color.
Now I want to show the number with size and position related to the available space so that the numbers are always inside the according area.
The width for the text element is easy to calculate, but I don't know how to get available height for the text elements.
Maybe it could be helpful to have the pixels of the ellipses as path, but I have no idea to how I can get this.
Does anyone know how to implement that?
Edit:
I developed an algorithm to find the size of a rectangle, which fits in the required areas of the ellipses. The text are scaled to the size of the rectangle and it works, but now I have another problem.
I need to center the scaled text in the rectangle an I used a StackPane for that. But I can't position the StackPane in the chart parent. If I set the layoutX and layoutY the bounds in parent are different.
For example: stackPane with text = sp;
sp.setLayoutY(122.1662320906945);
The Result for getBoundsInParent().getMinY() is 97.16622924804688;
How can I set the StackPane position, if I use it in a chart class as chart children?
I'm experimenting with a vector based graphics style with objects represented as series of line segments with a given width(it would probably be easier to think of these as rectangles). The problem is that these segments are connected at the center and leave a gap (shown below). I've determined that the most efficient way to cover this gap is simply to cover it with a triangle, and since I'm working in OpenGL, all I need are the points of the two points that don't overlap with the other rectangle, the third point being the center point where the two line segments(rectangles) are connected. How can I determine which points I need to use for the triangle, given that I have all of the points from both rectangles?
EDIT: I will also accept alternative solutions, as long as they cover up that gap.
EDIT 2: Nevermind, I solved it. I'll post code once I have better Internet connection.
Maybe I'm misunderstanding the question... but if you zoom in on the top corner of your red pentagon, you get something like this, am I right?
where A and B are nodes on the rectangle for edge1 and C and D are nodes on the rectangle for edge2. You say you already know these coordinates. And from what you say, the edges meet at the centre, which is halfway between A and B, and also halfway between C and D. So call this point X, and you can calculate its coordinates easily I guess.
So all you need to do is draw the missing triangle AXC, right? So one way would be to determine that A and C are on the "outside" of the polygon (and therefore need filling) and B and D are on the "inside" and therefore don't. But it's probably easier to just draw both, as it doesn't hurt. So if you fill AXC and BXD, you'd get this:
The solution I found assumes that there are 3 basic cases:
First, the three unique center points for the two rectangle proceed upward (positive y direction) so the gap is either on the left or right of the connection. In my code, I had the corner points of the rectangle organized by their orientation to the left or right of the center point, so if the bottom rectangle's left point is below the top rectangle's left point, then the gap is between the left points of the two rectangles, otherwise the gap is between the right points.
Second, the three unique center points have a maximum at the center most of the center points, so the gap is on the top. The gap is then between the two points with the maximum y values.
Third, the three unique center points have a minimum at the center most of the center points, so the gap is on the bottom. The gap is then between the two points with the minimum y values.
[I'll post pictures of the example cases if it is requested]
I'm trying to make a square move by rotating it sideways with its reference point set to the bottom side of the direction it is moving to.
For example: if I would move the square to the right, I would set its reference point to the bottom right of it and animate a rotation of 90 degrees, after the movement is done, I increase the square X by the width of it and centrally set its rotation to 90 degrees (so that I can keep track of its position)
The thing is, how should I proceed to keep repeating it? 'cause if I try to rotate by another 90 degrees using bottom right reference point, it won't be using the right position. What should I do to get the new bottom right relative position?
Thanks!
Create a transparent image twice as wide and high as the square.
Place the square in the top left corner.
Rotating the image around its center will now rotate the square in the way you describe.
You will need some maths to track where the new bottom is, using some simple formula for 2D rotation, so you calculate BEFORE rotating, where the point you wish to modify.
Then you use xReference and yReference variables :)
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.