How to set position for InputDriver.emit() event? - raspberry-pi3

I have Android Things 1.0 Raspberry Pi 3 and 800x480 (width / height) touch pad and need to track user touches on it for Activity layout controls. I'm got touch x and y coordinates from touch pad and try to generate Android touch event with that code:
private InputDriver mInputDriver;
private InputDriverEvent mInputDriverEvent = new InputDriverEvent();
...
int x = getTouchX(); // returns x within [0, 800]
int y = getTouchY(); // returns y within [0, 480]
mInputDriverEvent.clear();
mInputDriverEvent.setPosition(MotionEvent.AXIS_X, x);
mInputDriverEvent.setPosition(MotionEvent.AXIS_Y, y);
mInputDriverEvent.setContact(true);
mInputDriver.emit(mInputDriverEvent);
...
where mInputDriver initialized that way:
mInputDriver = new InputDriver.Builder()
.setName("TouchPad")
.setVendorId(0)
.setProductId(0)
.setAxisConfiguration(MotionEvent.AXIS_X, 0, 800, 0 ,0)
.setAxisConfiguration(MotionEvent.AXIS_Y, 0, 480, 0 ,0)
.build();
UserDriverManager.getInstance().registerInputDriver(mInputDriver);
event was generated for every touch, but pointer (cursor) always is in the screen center:
How to move pointer (cursor) to (x,y) position on screen?

Related

How to more + rotate 180 degree + move a group of shape using Processing?

I'm having this Processing issue where I'm trying to make my ariplane do a 180 turn-over right before it gets to the end of the drawing... It looks fairly simple but I just can't get it to work. Help would be much appreciated.
As I have other shapes (not showned in code below) I finally managed to use pushMatrix(); & popMatrix(); to only impact the airplane. Although rotating it brings me sideways.
int globalX = 0;
int globalY = 600;
int vitesse = 3;
void setup() {
size(800, 600);
rectMode(CENTER);
}
void draw() {
background(255);
pushMatrix();
bouger();
tourner();
dessinerAvion(globalX, globalY);
popMatrix();
}
void bouger() {
// Change the x location by vitesse
//globalX = globalX + vitesse;
globalY = globalY - vitesse;
}
void tourner() {
if (globalY < 0) {
for (int i = 10; i > 0; i = i-1) {
rotate(PI/3.0);
//rotate(radians(10));
}
for (int i = 10; i > 0; i = i-1) {
rotate(PI/3.0);
//rotate(radians(10));
vitesse = vitesse * -1;
}
}
}
void dessinerAvion(int x, int y) {
rectMode(CENTER);
// corps
fill(156,21,21);
ellipse(x + 250, y + 100, 50, 100);
rect(x + 250, y + 200, 50, 250,60);
// reacteur gauche
rect(x + 125, y + 225, 25, 50, 75);
// reacteur droite
rect(x + 375, y + 225, 25, 50, 75);
fill(210,14,14);
// aile gauche
quad(x + 75, y + 275, x + 75, y + 250, x + 225, y + 175, x + 225, y + 250);
// aile droite
quad(x + 425, y + 275, x + 425, y + 250, x + 275, y + 175, x + 275, y + 250);
fill(112,14,14);
// ailette gauche
triangle(x + 225, y + 290, x + 226, y + 310, x + 175, y + 310);
// ailette droites
triangle(x + 275, y + 290, x + 274, y + 310, x + 325, y + 310);
// aile verticale arrière
triangle(x + 250, y + 285, x + 248, y + 310, x + 252, y + 310);
}
Matrix are weird animals. They let you take a bunch of shapes and translate, rotate or scale them all at once, but only through playing with the coordinate system, which implies that the way you draw the shapes that you'll process will have a huge impact on how they react.
This is your plane on a neutral grid (by which I mean that globalX and globalY are 0):
Every square on this grid has a 20 pixels side. As you can see, your plane is far from the [0, 0] point, which is the start of the coordinate system it's drawn into. Also, when you modify it's coordinates, you're doing it through modifying it's coordinates, as we can see on this line of code:
void dessinerAvion(int x, int y)
The problem here is that applying a geometric transformation to an object with these coordinates can be counterintuitive. To simplify the matter, imagine the whole coordinate system that the plane use as if it was an image, like a png. The top left of the image is the [0, 0] point. When you change the plane's y coordinates for a higher number, your png image gets taller. If you change the x coordinate of the plane for a bigger number, the png image gets wider. You get the general idea.
When you apply a rotation to the plane, it's the entire 'png image', the entire coordinate system, that gets rotated. It's the same for any other transformation, too. Now, experimentally, look at what happens if I do this:
for(float i=0; i<20; i+=1) {
pushMatrix();
rotate(PI/i);
dessinerAvion(0, 0);
popMatrix();
}
As you can see, the plane isn't rotating on itself. It's the whole coordinate system that rotates, pivoting on the [0, 0] point, and it brings the plane with itself. That's the deal with matrix: they keep a lot of geometry waaay more simple, but you have to work accordingly.
Which brings me back to the previous point: you're updating the coordinates where you draw the plane. Then you use a matrix. Your updated plane will rotate in an unexpected way because you apply 2 different logic to your transformations: on one hand you modify coordinates (which is fine), while on the other you apply a transformation on a matrix (which is fine too). The issue is to mix those without acknowledging their differences.
Honestly, unless you have a specific thing in mind, you shouldn't mix those to move a single object.
The obvious solution to the first part of this issue would be to translate the plane instead of moving it's coordinates:
void draw() {
background(255);
bouger(); // updating coordinates, this doesn't need to be in the matrix block
pushMatrix();
//tourner();
translate(globalX, globalY);
dessinerAvion(0, 0); // zero here as the translation is doing that part of the job now
popMatrix();
}
Now you should see... well, you should see no difference at all. That's normal. The real difference is not in how the sketch appears, but how it's processed behind the scenes. Now your plane is a static object on which we apply geometric transformations.
Now to the other part of your problem: animating an object. Well, to be honest the animation part is quite easy, but there's a lot of work to do beforehand, more precisely to fix the plane's original coordinates so they are easier to manage. Let me explain:
If you look at the first image where I show the plane without any geometric transformation, you'll notice that the left wing starts at about ~75 pixels from the 0 point, and that the nose of the plane is ~45 pixels away from the 0 on the y axis. Currently, that's the rotation's anchor point. If you want to rotate the plane in a gracious manner, you'll have to fix this. There are many ways to proceed, follow me.
Take notes, because for some reason very few people will explain in simple words how to rotate on an anchor point using a matrix.
To rotate a shape on a specific anchor point, you have to proceed as follow:
pushMatrix
translate to the shape's destination
rotate
draw the shape using [0, 0] as it's anchor point
popMatrix
So if you want to rotate your plane on itself, thus deciding on an angle then moving it to it's destination before you draw the plane, you act as follow:
void draw() {
background(255);
// grid
stroke(0);
for (int i=0; i<height; i+=20) {
line(0, i, width, i);
}
for (int i=0; i<width; i+=20) {
line(i, 0, i, height);
}
pushMatrix();
translate(200, 200); // for the demonstration only
globalAngle--;
rotate(radians(globalAngle)); // I'm lazy and I don't want to do this in radians so I use degrees and convert them
dessinerAvion(-250, -200); // the center of the plane is the [250, 200] point. That's my anchor point.
popMatrix();
}
Which gives us this result:
Now, using the first translate, you can draw the plane wherever you want, and you can use the rotation to make it face any direction you want it to face.
Using these, it would be real easy to make the plane cross the screen back and forth:
void draw() {
background(255);
drawGrid();
// implementing the "u-turn"
globalY -= vitesse;
if (globalY < -200) {
globalX -= 200;
globalAngle = 180;
vitesse = -vitesse;
}
// drawing the sketch
pushMatrix();
translate(globalX, globalY);
rotate(radians(globalAngle));
dessinerAvion(-250, -200);
popMatrix();
}
Now, if we want to watch the plane turn, we have several options. You can rotate the plane frame by frame while plotting it's course. That's a good option. But I'm lazy, as previously stated, so instead I'll move the anchor point somewhere left of the plane and use it to rotate around it in a gracious manner:
void draw() {
background(255);
drawGrid();
pushMatrix();
translate(globalX, globalY);
tourner();
bouger();
dessinerAvion(0, -200); // this coordinate sets my anchor point
popMatrix();
}
void tourner() {
if (globalY < 300) { // starting the u-turn
vitesse = 0; // The plane won't move while it's rotating... except for the rotation itself
globalAngle--; // rotating counterclockwise 1 degree per frame
if (globalAngle<-180) { // u-turn completed
globalAngle = -180;
vitesse = -3;
}
}
rotate(radians(globalAngle));
}
I think that does it. I hope this helps. I'll lurk around in case you have questions about this answer.
Have fun!

recalibrate ball in pong so the edge hits first, not the centre in Processing

I've recently built a game of pong for a UNI assignment in Processing, but whenever the ball hits the top, bottom, side of the screen or the 'paddle' it only bounces back once half the ball is off the screen. I just want the edge of the ball to hit first rather than the centre, but am unsure where my code is going wrong. I hope this makes sense, I'm a definite beginner.
Here is my code for reference
//underwater pong
float x, y, speedX, speedY;
float diam = 10;
float rectSize = 200;
float diamHit;
PImage bg;
PImage img;
int z;
void setup() {
size(920, 500);
smooth();
fill(255);
stroke(255);
imageMode(CENTER);
bg = loadImage("underthesea.jpg");
img = loadImage("plastic.png");
reset();
}
void reset() {
x = width/2;
y = height/2;
//allows plastic to bounce
speedX = random(5, 5);
speedY = random(5, 5);
}
void draw() {
background(bg);
image(img, x/2, y);
rect(0, 0, 20, height);
rect(width/2, mouseY-rectSize/2, 50, rectSize);
//allows plastic to bounce
x += speedX;
y += speedY;
// if plastic hits movable bar, invert X direction
if ( x > width-30 && x < width -20 && y > mouseY-rectSize/2 && y < mouseY+rectSize/2 ) {
speedX = speedX * -1;
}
// if plastic hits wall, change direction of X
if (x < 25) {
speedX *= -1.1;
speedY *= 1.1;
x += speedX;
}
// if plastic hits up or down, change direction of Y
if ( y > height || y < 0 ) {
speedY *= -1;
}
}
void mousePressed() {
reset();
}
I wasn't able to run your code because I am missing the background and plastic images, but here's what's probably going wrong. I'm not 100% since I do not know the dimensions of your images either.
You are using imageMode(CENTER). See the documentation for details.
From the docs:
imageMode(CENTER) interprets the second and third parameters of image() as the image's center point. If two additional parameters are specified, they are used to set the image's width and height.
This treats the coordinates you input into the image function as the center of the image.
Your first issue is that you are placing your image at x/2 but doing all your collision checks with x in mind. x does not represent the middle of your image, because you're drawing it at x/2.
Then I'm not sure if you are doing your horizontal collision checks right, as you are checking against hardcoded values. I do know you are doing your vertical collision checks wrong. You are checking if the center of the image is at the top of the canvas, 0, or the bottom of the canvas, height. This means that your image will already have moved out of the screen halfway.
If you want to treat the image coordinates as the center of your image, you need to check the left edge of the image at x - imageWidth / 2, the right edge at X+ imageWidth / 2, the top edge at y - imageWidth / 2 (remember the y coordinates are 0 at the top of the canvas and increase towards the bottom) and the bottom at y - imageWidth / 2. Here's a great website which goes into more detail on 2d collision detection, i'd highly recommend you give it a read. It's an awesome website.

Different Processing rendering between native and online sketch

I get different results when running this sample with Processing directly, and with Processing.js in a browser. Why?
I was happy about my result and wanted to share it on open Processing, but the rendering was totally different and I don't see why. Below is a minimal working example.
/* Program that rotates a triange and draws an ellipse when the third vertex is on top of the screen*/
float y = 3*height/2;
float x = 3*width/2;
float previous_1 = 0.0;
float previous_2 = 0.0;
float current;
float angle = 0.0;
void setup() {
size(1100, 500);
}
void draw() {
fill(0, 30);
// rotate triangle
angle = angle - 0.02;
translate(x, y);
rotate(angle);
// display triangle
triangle(-50, -50, -30, 30, -90, -60);
// detect whether third vertex is on top by comparing its 3 successive positions
current = screenY(-90, -60); // current position of the third vertex
if (previous_1 < previous_2 && previous_1 < current) {
// draw ellipse at the extrema position
fill(128, 9, 9);
ellipse(-90, -60, 7, 10);
}
// update the 2 previous positions of the third vertex
previous_2 = previous_1;
previous_1 = current;
}
In processing, the ellipse is drawn when a triangle vertex is on top, which is my goal.
In online sketching, the ellipse is drawn during the whole time :/
In order to get the same results online as you get by running Processing locally you will need to specify the rendering mode as 3d when calling size
For example:
void setup() {
size(1100, 500, P3D);
}
You will also need to specify the z coordinate in the call to screenY()
current = screenY(-90, -60, 0);
With these two changes you should get the same results online as you get running locally.
Online:
Triangle Ellipse Example
Local:
The problem lies in the screenY function. Print out the current variable in your processing sketch locally and online. In OpenProcessing, the variable current grows quickly above multiple thousands, while it stays between 0 and ~260 locally.
It seems like OpenProcessing has a bug inside this function.
To fix this however, I would recommend you to register differently when you drew a triangle at the top of the circle, for example by using your angle variable:
// Calculate angle and modulo it by 2 * PI
angle = (angle - 0.02) % (2 * PI);
// If the sketch has made a full revolution
if (previous_1 < previous_2 && previous_1 < angle) {
// draw ellipse at the extrema position
fill(128, 9, 9);
ellipse(-90, -60, 7, 10);
}
// update the 2 previous angles of the third vertex
previous_2 = previous_1;
previous_1 = angle;
However, because of how you draw the triangles, the ellipse is at an angle of about PI / 3. To fix this, one option would be to rotate the screen by angle + PI / 3 like so:
rotate(angle + PI / 3);
You might have to experiment with the angle offset a bit more to draw the ellipse perfectly at the top of the circle.

OpenSceneGraph osg::Quat: shape not rotating

I have a small function to create a new instance of a WorldObject.
I want to use osg::ref_ptr<osg::PositionAttitudeTransform> for translation and rotation but there is a problem I can't figure out.
I use setTranslation() with a Vec3 which works very well. But the Quat with makeRotation() just does nothing.
Here is the code:
osg::ref_ptr <osg::PositionAttitudeTransform> getWorldObjectClone(const char* name, osg::Vec3 position = osg::Vec3(0, 0, 0), osg::Vec3 rotation = osg::Vec3(0, 0, 0))
{
osg::ref_ptr <osg::PositionAttitudeTransform> tmp = new osg::PositionAttitudeTransform;
osg::Quat q(0, osg::Vec3(0, 0, 0));
tmp = dynamic_cast<osg::PositionAttitudeTransform*>(WorldObjects[name]->clone(osg::CopyOp::DEEP_COPY_ALL));
tmp->setPosition(position);
q.makeRotate(rotation.x(), 1, 0, 0);
q.makeRotate(rotation.y(), 0, 1, 0);
q.makeRotate(rotation.z(), 0, 0, 1);
tmp->setAttitude(q);
return tmp;
}
I tried rotation = {90,0,0} (degrees) and rotation = {1,0,0} (radians) but both have no effect. Is there an mistake in how the code is using the Quat?
The rotation method you are using works with radians.
If you want to rotate 90 degrees around the X axis, you need to call:
q.makeRotate(osg::PI_2, 1, 0, 0 );
// or the equivalent
q.makeRotate(osg::PI_2, osg::X_AXIS);
Keep in mind that every call to makeRotate will reset the full quaternion to the given rotation. If you're trying to concatenate several rotations, you have to multiply the corresponding quaternions.
For instance:
osg::Quar xRot, yRot;
// rotate 90 degrees around x
xRot.makeRotate(osg::PI_2, osg::X_AXIS);
// rotate 90 degrees around y
yRot.makeRotate(osg::PI_2, osg::Y_AXIS);
// concatenate the 2 into a resulting quat
osg::Quat fullRot = xRot * yRot;

Window positioning results in space around windows on Windows 10

I have some code that positions windows to screen quadrants. It works fine on Windows XP, 7, and 8/8.1. However, on Windows 10, there is a weird gap between windows. The extra space surrounds the window on all 4 sides. I presume it has something to do with window borders, but can't figure out how to correct the problem. Any input would be highly appreciated. The code is as follows:
// Get monitor info
HMONITOR hm = MonitorFromWindow(hWnd, MONITOR_DEFAULTTONEAREST);
MONITORINFO mi;
mi.cbSize = sizeof(mi);
GetMonitorInfo(hm, &mi);
// Set screen coordinates and dimensions of monitor's work area
DWORD x = mi.rcWork.left;
DWORD y = mi.rcWork.top;
DWORD w = mi.rcWork.right - x;
DWORD h = mi.rcWork.bottom - y;
switch (corner) {
case 0: // Left top
SetWindowPos(hWnd, HWND_TOP, x, y, w / 2, h / 2, SWP_NOZORDER);
break;
case 1: // Right top
SetWindowPos(hWnd, HWND_TOP, x + w / 2, y, w / 2, h / 2, SWP_NOZORDER);
break;
case 2: // Right bottom
SetWindowPos(hWnd, HWND_TOP, x + w / 2, y + h / 2, w / 2, h / 2, SWP_NOZORDER);
break;
case 3: // Left bottom
SetWindowPos(hWnd, HWND_TOP, x, y + h / 2, w / 2, h / 2, SWP_NOZORDER);
break;
}
I managed to correct this effect by inflating target rectangle by a calculated margin like this:
static RECT GetSystemMargin(IntPtr handle) {
HResult success = DwmGetWindowAttribute(handle, DwmApi.DWMWINDOWATTRIBUTE.DWMWA_EXTENDED_FRAME_BOUNDS,
out var withMargin, Marshal.SizeOf<RECT>());
if (!success.Succeeded) {
Debug.WriteLine($"DwmGetWindowAttribute: {success.GetException()}");
return new RECT();
}
if (!GetWindowRect(handle, out var noMargin)) {
Debug.WriteLine($"GetWindowRect: {new Win32Exception()}");
return new RECT();
}
return new RECT {
left = withMargin.left - noMargin.left,
top = withMargin.top - noMargin.top,
right = noMargin.right - withMargin.right,
bottom = noMargin.bottom - withMargin.bottom,
};
}
And then doing
RECT systemMargin = GetSystemMargin(this.Handle);
targetBounds.X -= systemMargin.left;
targetBounds.Y -= systemMargin.top;
targetBounds.Width += systemMargin.left + systemMargin.right;
targetBounds.Height += systemMargin.top + systemMargin.bottom;
That worked for all windows I could test it with, except Explorer windows, which I hardcoded to exclude. If I'd do that expansion on Explorer near the screen edge, window ends up spilling a large area past it to the adjacent monitor.
The default font size of windows XP/7/8/8.1 is 100%, but in windows 10 the default is to display text and items in 125%. That affects directly all the window sizes.
Go to settings, display and you will find a scroller, move it to 100% and everything should display the same way as it did in Windows 8/7/XP

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