I'm designing a simple app to display some Cartesian type graphics using DrawLine() and DrawEllipse() functions into a PictureBox control. To make the coordinate system more "real-world" instead of the picture box I am using a matrix to flip the Y axis, scale everything down and reposition it so that (0,0) is at the center of the screen and (+2,+2) is at the upper right corner. All works well for drawing of graphics. However, in trying to read mouse events it appears that the MouseEventArgs variable (returned by most Mouse events) returns the mouse position X and Y as integers. I am properly using an inverted matrix to retrieve the coordinates at the scaled values, but at the scale I am using, this won't work as integers as I require screen positions in fractional values (1.5, 1.6, etc).
Is there no way to retrieve the mouse values as a floating point or double/decimal value that will give the "resolution" I require?
Some code fragements:
--Globally
Private MyTransform As Matrix
--Within the picturebox
Paint() event
Dim G As Graphics = e.Graphics
Dim mx As New Matrix(1, 0, 0, -1, 0, 0) 'Y-axis orientation flipped to match Cartesian plane
mx.Translate(PictureBox.Width / 2, -PictureBox.Height / 2) 'Move 0,0 to lower left corner
mx.Scale(100, 100)
G.Transform = mx
MyTransform = G.Transform
'All drawing is performed at this point and works fine.
--Within the MouseDown event
MyTransform.Invert()
'Here is the issue--the Mouse points returned, being integers, cannot properly
'show the mouse point if the transformation matrix has scaled up the drawing space at all.
e.Location.x 'is an integer, so it cannot show .01 as the proper mouse location within the transformed viewspace.
e.Location.y 'same issue.
MyTransform.Invert()
I've looked for a cartesian coordinate-based picturebox alternative to no avail, and Charting components won't work because they require the points being drawn be contained in their own proprietary containers/sets. I'm doing all the drawing myself with GDI-type methods. The only alternative seems to be to avoid doing the transformations with vb and doing all the translation/untranslation myself, unless someone has an alternative or example to suggest....?
Store the Scale as a property. Then as you change the scale you apply a function to the mouse coordinates difference. Example will use a 10/1 plane so the Scale will be 10.
Private Property Scale As Single
Function:
Private Function CorrectForScale(coord As Integer) As Single
Return (coord / Scale)
End Function
Now if the difference in distance between mouse position = 3 then the result after the function would be 0.3.
Related
I'm having a fragment shader that draw some stuff. On top of that I want it to draw 1-pixel thick rectangle around the fragment. I have using step function, but the problem is the UV coordinates that is between 0.0 -1.0. How do I know when the fragment is at a specific pixel? For this I want to draw on the edges.
c.r = step(0.99, UV.x);
c.r += step(0.99, 1.0-UV.x);
c.r += step(0.99, UV.y);
c.r += step(0.99, 1.0-UV.y);
The code above just draw a rectangle, but the problem thicknes is 0.01% of total width/hight.
Is there any good description of UX, FRAGCOORD, SCREEN_TEXTURE and SCREEN_UV?
If it is good enough for you to work in screen coordinates (i.e., you want to define position and thickness in terms of screen space) you can use FRAGCOORD. It corresponds to the (x, y) pixel coordinates within the viewport, i.e., with the default viewport of 1024 x 600, the lower left pixel would be (0, 0), and the top right would be (1024, 600).
If you want to map the fragment coordinates back to world space (i.e., you want to define position and thickness in terms of world space), you must follow the work-around mentioned here.
Say, I have an image on an HTML page.
I apply an affine transformation to the image using CSS3 matrix function.
It looks like:
img#myimage {
transform: matrix(a, b, c, d, tx, ty);
/* use -webkit-transform, -moz-transform etc. */
}
The origin of an HTML page is the top-left corner and the y-axis is inverted.
I'm trying to put the same image in an environment (cocos2d) where the origin is the bottom-left corner and the y-axis is upright.
To get the same result in the other environment, I need to transform the origin somehow and reflect that in the resulting CGAffineTransform.
It would be great if I can get some help with the matrix math that goes here. (I'm not so good with matrices.)
The following formula would work,
for converting the position from CSS3 to Cocos2d:
(screen Size - "y" position in CSS3 - height of object)
Explanation:
To make the origin for the Cocos environment same as for the CSS3 environment we would only have to add the screen size to the cocos2d's bodies y co-ordinate.
Eg. The screen size is (100,100) and the body is a point object if you place it at (0,0) in CSS3 it would be at the top left corner. If we add the screen size to the y co-ordinates for cocos2d the object would be placed at (0,100) which is the top-left corner for cocos2d as well
To make the co-ordinates same, since the Y axis is inverted, we have to subtract the "Y" co-ordinate given in CSS3 from the Screen Size for Cocos2d. Suppose we place the same point object in the previous example at (0,10) in CSS3 we would place it at (0, 100 - 10) in cocos2d which would be the same positions on the screen
Since our body would NOT always be a point object we have to take care of its anchor point as well. If suppose the body's height is 20 and we place it at (0,10) in CSS3 then it would be placed at the top-left position and would be coming down because the Y axis is inverted
Hence we would also have to subtract the body's total height from the screen size and "y" co-ordinate to place it at the same position which would be (0, 100 - 10 - 20) putting the body at the same place in cocos2d environment
I hope I am correct and clear :)
I have create a Quartz composition for use in MAC OS program as part of my interface.
I am relying on the fact that when you have composition sprite movement (a text bullet point in my case) is limited both in the X plane and Y plane to minimum -1 and maximum +1.
When I scale up the window / make my window full screen, I find that the horizontal plane (X axis) remains the same, with -1 being my far left point and +1 being my far right point. However the vertical plane (Y axis) changes, in full screen mode it goes from -0.7 to +0.7.
This scaling is screwing with my calculations. Is there anyway to get the application to keep the scale as -1 to +1 for both horizontal and vertical planes? Or is there a way of determining the upper and lower limits?
Appreciate any help/pointers
Quartz Composer viewer Y limits are usually -0.75 -> 0.75 but it's only a matter of aspect ratio. X limits are allways -1 -> 1, Y ones are dependents on them.
You might want to assign dynamically customs width and heigth variables, capturing the context bounds size. For example :
double myWidth = context.bounds.size.width;
double myHeight = context.bounds.size.height;
Where "context" is your viewer context object.
If you're working directly with the QC viewer : you should use the Rendering Destination Dimensions patch that will give you the width and the height. Divide Height by 2, then multiply the result by -1 to have the other side.
I am new to GLUT and opengl. I need to draw a scatterplot matrix for n dimensional array.
I have saved the data from csv to a vector of vectors and each vector corresponds to a row. I have plotted just one scatterplot. And used GL_LINES to draw the grid. My questions
1. How do I draw points in a particular grid? Using GL_POINTS I can only draw points in the entire window.
Please let me know need any further info to answer this question
Thanks
What you need to do is be able to transform your data's (x,y) coordinates into screen coordinates. The most straightforward way to do it actually does not rely on OpenGL or GLUT. All you have to do is use a little math. Determine the screen (x,y) coordinates of the place where you want a datapoint for (0,0) to be on the screen, and then determine how far apart you want one increment to be on the screen. Simply take your original data points, apply the offset, and then scale them, to get your screen coordinates, which you then pass into glVertex2f() (or whatever function you are using to specify points in your API).
For instance, you might decide you want point (0,0) in your data to be at location (200,0) on your screen, and the distance between 0 and 1 in your data to be 30 pixels on the screen. This operation will look like this:
int x = 0, y = 0; //Original data points
int scaleX = 30, scaleY = 30; //Scaling values for each component
int offsetX = 100, offsetY = 100; //Where you want the origin of your graph to be
// Apply the scaling values and offsets:
int screenX = x * scaleX + offsetX;
int screenY = y * scaleY + offsetY;
// Calls to your drawing functions using screenX and screenY as your coordinates
You will have to determine values that make sense for the scalaing and offsets. You can also have your program use different values for different sets of data, so you can display multiple graphs on the same screen. But this is a simple way to do it.
There are also other ways you can go about this. OpenGL has very powerful coordinate transformation functions and matrix math capabilities. Those may become more useful when you develop increasingly elaborate programs. They're most useful if you're going to be moving things around the screen in real-time, or operating on incredibly large data sets, as they allow you to perform these mathematical calculations very quickly using your graphics hardware (which is able to do them much faster than the CPU). However, the time it takes for the CPU to do simple calculations like those where you only are going to do them once or very infrequently on limited sets of data is not a problem for computers today.
Let's say I have circle bouncing around inside a rectangular area. At some point this circle will collide with one of the surfaces of the rectangle and reflect back. The usual way I'd do this would be to let the circle overlap that boundary and then reflect the velocity vector. The fact that the circle actually overlaps the boundary isn't usually a problem, nor really noticeable at low velocity. At high velocity it becomes quite clear that the circle is doing something it shouldn't.
What I'd like to do is to programmatically take reflection into account and place the circle at it's proper position before displaying it on the screen. This means that I have to calculate the point where it hits the boundary between it's current position and it's future position -- rather than calculating it's new position and then checking if it has hit the boundary.
This is a little bit more complicated than the usual circle/rectangle collision problem. I have a vague idea of how I should do it -- basically create a bounding rectangle between the current position and the new position, which brings up a slew of problems of it's own (Since the rectangle is rotated according to the direction of the circle's velocity). However, I'm thinking that this is a common problem, and that a common solution already exists.
Is there a common solution to this kind of problem? Perhaps some basic theories which I should look into?
Since you just have a circle and a rectangle, it's actually pretty simple. A circle of radius r bouncing around inside a rectangle of dimensions w, h can be treated the same as a point p at the circle's center, inside a rectangle (w-r), (h-r).
Now position update becomes simple. Given your point at position x, y and a per-frame velocity of dx, dy, the updated position is x+dx, y+dy - except when you cross a boundary. If, say, you end up with x+dx > W (letting W = w-r), then you do the following:
crossover = (x+dx) - W // this is how far "past" the edge your ball went
x = W - crossover // so you bring it back the same amount on the correct side
dx = -dx // and flip the velocity to the opposite direction
And similarly for y. You'll have to set up a similar (reflected) check for the opposite boundaries in each dimension.
At each step, you can calculate the projected/expected position of the circle for the next frame.
If this lies outside the rectangle, then you can then use the distance from the old circle position to the rectangle's edge and the amount "past" the rectangle's edge that the next position lies at (the interpenetration) to linearly interpolate and determine the precise time when the circle "hits" the rectangle edge.
For example, if the circle is 10 pixels away from the rectangle's edge, then is predicted to move to 5 pixels beyond it, you know that for 2/3rds of the timestep (10/15ths) it moves on its orginal path, then is reflected and continues on its new path for the remaining 1/3rd of the timestep (5/15ths). By calculating these two parts of the motion and "adding" the translations together, you can find the correct new position.
(Of course, it gets more complicated if you hit near a corner, as there may be several collisions during the timestep, off different edges. And if you have more than one circle moving, things get a lot more complex. But that's where you can start for the case you've asked about)
Reflection across a rectangular boundary is incredibly simple. Just take the amount that the object passed the boundary and subtract it from the boundary position. If the position without reflecting would be (-0.8,-0.2) for example and the upper left corner is at (0,0), the reflected position would be (0.8,0.2).