I have a 360° panorama picture (texture) that I map around a sphere and I place my camera inside the sphere to have a spherical projection. Now the texture is projected on the outside of the sphere.
I want to 'turn the texture inside-out' and i read it could be done by applying a matrix.
I'd like to apply a matrix to a sphere (Object3D) in Rajawali/OpenGL.
This can be done in ThreeJS like this:
var sphere = new THREE.SphereGeometry(100, 100, 40);
sphere.applyMatrix(new THREE.Matrix4().makeScale(-1, 1, 1));
three.js documentation for applyMatrix:
this updates the position, rotation and scale with the matrix.
and I'd like to do this in Rajawali:
mSphere = new Sphere(100, 100, 40);
//Apply the matrix here
Documentation for Matrix/Object3D can be found here
Matrix4
Object3D
Or is there any other way it could be done?
This has been fixed!
mSphere = new Sphere(100, 100, 40);
mSphere.setScaleX(-1);
mSphere.setScaleY(1);
mSphere.setScaleZ(1);
Isn't that really a job for material.side?
Quote from three.js doc for Material:
.side
Defines which of the face sides will be rendered - front, back or both.
Default is THREE.FrontSide. Other options are THREE.BackSide and THREE.DoubleSide.
So looking at a sphere surface from the inside should be using THREE.BackSide.
Related
I have an XYLineChart and I want to customize the shape of nodes into rings.
Like this:
Here is the code how I try to done this:
Ellipse2D bigCircle = new Ellipse2D.Double(-4, -4, 8, 8);
Ellipse2D smallCircle = new Ellipse2D.Double(-2,-2,4,4);
Area ring = new Area(bigCircle);
ring.subtract(new Area(smallCircle));
Shape ringShape = (Shape)ring;
renderer.setSeriesShape(0, ringShape);
But the result is like in the left image. It is like this renderer.setSeriesShape(0, bigCircle);
What I am doing wrong?
or
Exist a factory with shapes which include my shape?
My code is fine, but the problem is from line. How #samabcde said.
I made a bigger ring to see what exactly is.
So I have this program to test the possibility of an object to slide down in a ramp given its friction, object mass and ramp angle. However I need to animate the box if the force is positive. Just a simple animation moving the box from that point to the end of the ramp. But I can't. Please help
private void drawTransform(Graphics g, double modifier) {
// redtowhite = new GradientPaint(0,0,color.RED,100, 0,color.WHITE);
Rectangle rect = new Rectangle(130,350, 350, 15);
Rectangle box = new Rectangle((int) (rect.getX()+300), 300, 50, 50);
AffineTransform at = new AffineTransform();
at.rotate(-Math.toRadians(modifier), rect.getX(), rect.getY() + rect.height);
// Transform the shape and draw it to screen
Graphics2D g2d = (Graphics2D) g;
g2d.setColor(Color.DARK_GRAY);
// g2d.fillRect(0, 0, 350, 600);
g2d.fill(at.createTransformedShape(rect));
g2d.draw(at.createTransformedShape(box));
}
Screenshot:
If all you want to do is move the box, this can be done by simply updating it's X position. You should be able to manipulate the rectangle's X position directly using something like "box.x++". Alternatively you could create a variable and reference that to provide the initial X co-ordinate, then updating that variable will "move" the box. One issue is this will only move the box along the X axis, hence you will also need some kind of constant downward force acting as gravity. This is easy to achieve, just minus the box's Y position value when it is not colliding with the ground, or your ramp.
Another approach is velocity based movement using vectors, however you mentioned that the animation should be simple. If you do want a smoother animation velocity based movement will provide this but you will need to perform a little research first.
I need to plot a polynomial given a start point and an end point. I've looked at using the paincomponent to do do it, but I'm getting confused because the origin is in the top left corner and I need my grid lines to be dynamic.
How do I make it so I can create a dynamic scale and what java class is best used for plotting a function?
For example if the user were to enter -2 2 x^2 the plotter would need to create a grid that goes from -2 has 0 in the middle and 2 on the right for the x coordinates. However if the user enters x^3 we have negative coordinates in the y plane and I'll need to slide the line containing the x coordinates up to accomidate.
Thanks!
The way I think you could do this(and I may be wrong) is to draw the y-line and x-line of your graph, and label the values for x and y using drawString at the correct position.
To illustrate this I did this example for you:
public void paintComponent(Graphics g)
{
Graphics2D g2 = (Graphics2D) g;
Point2D.Double from = new Point2D.Double(300, 100);
Point2D.Double to = new Point2D.Double(300, 300);
Line2D.Double line = new Line2D.Double(from, to);
g2.draw(line);
Point2D.Double from2 = new Point2D.Double(150, 200);
Point2D.Double to2 = new Point2D.Double(450, 200);
Line2D.Double line2 = new Line2D.Double(from2, to2);
g2.draw(line2);
g2.drawString("-1", 290, 215);
}
The drawString method would draw "-1" below 0. So you could use the same thing to draw -2 below -1, etc.
I know this sounds inefficient, but it's the only way I can think of.
Hope someone else will have a better solution for you.
It shouldn't be too difficult to transform the y-coordinate if you know the size of area you want to paint to. But to be honest, why don't you use one of the plotting libs like JFreeChart, ...?
I'm trying to detect the center of a circle. I try to do this with cvHoughCircle. But I can't seem to get it working properly .
The only thing that can vary is the size of the circle.
I try detecting the circle by doing :
circle = cvHoughCircles(imgThreshold, storage, CV_HOUGH_GRADIENT, 1,
(double)imgThreshold.height()/20, 200, 20, 0, 0);
imgThreshold is the b/w image you can see here. The resolution of the image is in fact 1280*1024.
Can anyone tell me what I am doing wrong.
Instead of using cvHoughCircle it is possible to solve this problem with a bit of math:
CvMoments moments = new CvMoments();
cvMoments(imgThreshold, moments, 1);
double moment10 = cvGetSpatialMoment(moments, 1, 0);
double moment01 = cvGetSpatialMoment(moments,0,1);
double area = cvGetCentralMoment(moments, 0, 0);
int posX = 0;
int posY = 0;
int lastX = posX;
int lastY = posY;
posX = (int) (moment10/area);
posY = (int) (moment01/area);
cvCircle(iplRgbImage, new CvPoint(posX,posY), 3, CvScalar.GREEN, -1, 8, 0);
source = http://aishack.in/tutorials/tracking-colored-objects-in-opencv/
If the circle is complete and filled and not occluded by other shapes, you can use findContours() and then find the center of the contour.
use cvBlob
https://code.google.com/p/cvblob/
Concerning Hough transform it can detect circles by identifying pixels that belongs to a circle periphery. More precisely given a binary (thresholded) image containing ie white pixels along a cyclic path, the hough circle transform will detect the circle. So the image to feed the algorithm should be binary and thresholded but in your example must be the thresholded example of an edge filter (ex Sobel) rather than a solid filled circle.
I can not tell a right way of "fitting" a circle on the above image, but the centroid of the blob extracted with connected components is a good and fast way to go.
If I have a canvas, on which I draw a Bitmap like this:
canvas.drawBitmap(bmLargeImage, srcRect, destRect, paint);
and I scale the bitmap:
canvas.scale(1.5f, 1.5f, 450, 250);
I want to get the position of the Bitmap after the scale. If the position before scale was (0, 0), after scale there is a offset and I need that offset.. how can I get it?
Thanks and sorry for the simple question, newbie here...
Ok lets try to work out the best formula for this
canvas.scale(scaleX, scaleY, pivotX, pivotY);
if (scaleX >= 1){
objectNewX = objectOldX + (objectOldX - pivotX)*(scaleX - 1);
}else{
objectNewX = objectOldX - (objectOldX - pivotX)*(1 - scaleX);
}
The same for objectNewY. The new width and height of the bitmap would of course be the multiple of the old size and scale.
I believe the cleanest Solution would be to use the underlying transformation Matrix of the Canvas you are manipulating.
In Android there is the canvas.getMatrix(Matrix cmt) method available which will yield it. The transformation matrix will transform any point in world space you throw at into screen coordinates. Just use the matrix.mapPoints(float[] points) and you will be fine.
FYI, you can easily do it the other way around too. If you want to know what screen coordinate maps to which point in world space, e.g. for tapping; the inverse matrix can be used for that. It can be obtained via the matrix.invert(Matrix out) method. Use its mapPoints() for the coordinate mapping then.
Here are the official docs:
mapPoints(), invert(), getMatrix()
If you'd like know the corners of your screen relative to your original canvas, you can use canvas.getClipBounds(). This returns a Rect with edge coordinates relative to your original canvas. For instance, if you start off with a canvas size of 320 x 480 and call
canvas.scale(2, 2, getWidth()/2, getHeight()/2);
and then
canvas.getClipBounds();
you will have a Rect (call this rect) where
rect.top == 120
rect.bottom == 360
rect.left == 80
rect.right == 240