I have built a small custom qml item that is used as a selection area (something like the QRubberBand component provided in Qt Widgets). The item also give the ability to user to resize the content of the selection, so by grabbing the bottom corner of the selection rectangle it is possible to drag to enlarge the content. After the user has done resizing I would like to compute the QTransform matrix of the transformation. QTransform provides a convenient QTransform::scale method to get a scale transformation matrix (which I can use by comparing the width and height ratio with the previous size of the selection). The problem is that QTransform::scale assumes that the center point of the transformation is the center of the object, but I would like my transformation origin to be the top left of the selection (since the user is dragging from the bottom-right).
So for example, if I have the following code:
QRectF selectionRect = QRectF(QPointF(10,10), QPointF(200,100));
// let's resize the rectangle by changing its bottom-right corner
auto newSelectionRect = selectionRect;
newSelectionRect.setBottomRight(QPointF(250, 120));
QTransform t;
t.scale(newSelectionRect.width()/selectionRect.width(), newSelectionRect.height()/selectionRect.height());
The problem here is that if I apply the transformation t to my original selectionRect I don't get my new rectangle newSelectionRect back, but I get the following:
QRectF selectionRect = QRectF(QPointF(10,10)*sx, QPointF(200,100)*sy);
where sx and sy are the scale factors of the transform. I would like a way to compute the QTransform of my transformation that gives back newSelectionRect when applied to selectionRect.
The problem lies in this assumption:
QTransform::scale assumes that the center point of the transformation is the center of the object
All transformations performed by QTransform are referred to the origin of the axis, is just an application of various tranformation matrixes (https://en.wikipedia.org/wiki/Transformation_matrix):
Also, QTransform::translate (https://doc.qt.io/qt-5/qtransform.html#translate) states:
Moves the coordinate system dx along the x axis and dy along the y axis, and returns a reference to the matrix.
Thereby, what you are looking for is:
QTransform t;
t.translate(+10, +10); // Move the origin to the top left corner of the rectangle
t.scale(newSelectionRect.width()/selectionRect.width(), newSelectionRect.height()/selectionRect.height()); // scale
t.translate(-10, -10); // move the origin back to where it was
QRectF resultRect = t.mapRect(selectionRect); // resultRect == newSelectionRect!
Related
I want rotate a QGraphicsPixmapItem around a point according to mouse position.
So i tried this:
void Game::mouseMoveEvent(QMouseEvent* e){
setMouseTracking(true);
QPoint midPos((sceneRect().width() / 2), 0), currPos;
currPos = QPoint(mapToScene(e->pos()).x(), mapToScene(e->pos()).y());
QPoint itemPos((midPos.x() - cannon->scenePos().x()), (midPos.y() - cannon->scenePos().y()));
double angle = atan2(currPos.y(), midPos.x()) - atan2(midPos.y(), currPos.x());
cannon->setTransformOriginPoint(itemPos);
cannon->setRotation(angle); }
But the pixmap moves a few of pixels.
I want a result like this:
Besides the mixup of degrees and radians that #rafix07 pointed out there is a bug in the angle calculation. You basically need the angle of the line from midPos to currPos which you calculate by
double angle = atan2(currPos.y() - midPos.y(), currPos.x() - midPos.x());
Additionally the calculation of the transformation origin assumes the wrong coordinate system. The origin must be given in the coordinate system of the item in question (see QGraphicsItem::setTransformOriginPoint), not in scene coordinates. Since you want to rotate around the center of that item it would just be:
QPointF itemPos(cannon->boundingRect().center());
Then there is the question whether midPos is actually the point highlighted in your image in the middle of the canon. The y-coordinate is set to 0 which would normally be the edge of the screen, but your coordinate system may be different.
I would assume the itemPos calculated above is just the right point, you only need to map it to scene coordinates (cannon->mapToScene(itemPos)).
Lastly I would strongly advise against rounding scene coordinates (which are doubles) to ints as it is done in the code by forcing it to QPoints instead of QPointFs. Just use QPointF whenever you are dealing with scene coordinates.
I am making a 2D board game. the game board grid is 8x8 and each cell of the grid is an object. So a board consists of 64 cell objects. My aim is to work out which cell the mouse is in. I am attempting this by tracking the mouse coordinates and comparing it to the grid coordinates.
my coordinate system is as follows:
gluOrtho2D(-4,4,-4,4);
I am trying to get the current mouse position by using the following in my update function:
POINT p
if (GetCursorPos(&p)){
}
if (ScreenToClient(hWnd, &p))
{
}
However although this is tracking the coordinates of the mouse it is not correctly tracking the world coordinates that I set with gluOrtho2D. How can I achieve this?
It depends on your glViewPort
Let's say you have:
glViewport(0,0, 640, 640);
The mouse position is (mousePos.x,mousePos.y) and the world position you want to know is (world.x, world.y)
And, give that the top/left corner of your screen is the (0, 0) coordinate
Then we can make the following:
world.x = -4.0 + (mousePos.x / 640.0) * (4*2)
world.y = 4.0 - (mousePos.y / 640.0) * (4*2)
What we are doing here is a linear interpolation using the normalize position of the mouse within the screen (mousePos.x / 640) and then multiplying this value to the width of the word (4*2).
Given that the top/left corner of the grid start at (-4, 4), we add the offset of the world position.
I'm creating a simple game that involves swinging a club using the bullet library in c++. However, I'm having trouble rotating the club (rigid body) the way I want.
For reference, here is the club structure:
struct Club {
btRigidBody *btPhys;
btScalar rotateAmount;
btVector3 axis;
}
Here is how I rotate the club:
//Create a rotation matrix from the club's world transform
btMatrix3x3 orn = club.btPhys->getWorldTransform().getBasis();
//Apply a rotation to the matrix
orn *= btMatrix3x3(btQuaternion( btVector3(0, 0, 1), btScalar(degreesToRads(club.rotationAmount))));
//Set the rotation of the club
club.btPhys->getWorldTransform().setBasis(orn);
However, this just rotates the rigid body on the center of the club. I want the rigid body to be rotated near it's edge (the club's handle).
So I added a transform to move the club upward above the pivot point:
(this code follows directly after the previous code)
btTransform trans;
//Get the club's trasnform
club.btPhys->getMotionState()->getWorldTransform(trans);
float x, y, z;
x = trans.getOrigin().getX();
y = trans.getOrigin().getY();
z = trans.getOrigin().getZ();
//Set the transform to be 2 units above the club's origin (the club is 4 units long)
trans.setOrigin(btVector3(btScalar(x), btScalar(y + 2.0f), btScalar(z)));
//Apply the transform to the club
club.btPhys->getMotionState()->setWorldTransform(trans);
However, this doesn't change the pivot point, it just moves the club above the pivot point (creating an unwanted space).
In short, the rotation matrix just rotates the rigidbody around it's center. I just want to move the pivot point from the center of the rigid body to the edge of the rigidbody.
SetWorldTransform of MotionState always transforms the body around its center of mass.
There are two ways in which you can solve your problem. If you're using btDefaultMotionState as your motion state, you can specify the offset of center of mass in the constructor. This way the body will rotate around another point. There is however a serious side effect of this approach. All physical interaction with this object will treat the new pivot point as the center of mass. If your body is kinematic, it is perfectly fine but if it is dynamic, the effects may differ from desired effects.
Another probably more safe approach is to stack 3 transformations together to achieve the rotation you're looking for. First, translate the body to the location of desired pivot. Next, rotate it. Finally, translate it back where the center of mass should be.
I'm trying to detect if my mouse is hovering over a rectangle that I drew(with VBOs), but when I get the mouse coordinates with Mouse.getX() & Mouse.getY(), it returns the window coordinates(i.e. (480, 200)). How can I get the mouse coordinates from the range of [-1, 1]?
trivial approach
You can do this by knowing only the viewport or if drawing in the whole window its inner size. Assuming mouse coordinates are 0,0 in the top left corner.
The following will normalize the input to [-1, 1].
double normalizedX = -1.0 + 2.0 * (double)Mouse.getX() / window.width;
double normalizedY = 1.0 - 2.0 * (double)Mouse.getY() / window.height;
You can also use a more intricate solution by creating an inverse matrix for the viewport and multiply out the mouse input vector.
Hey so I'm integrating box2d and SFML, and box2D has the same odd, mirrored Y-axis coordinate system as SFML, meaning everything is rendered upside down. Is there some kind of function or short amount of code I can put that simply mirrors the window's render contents?
I'm thinking I can put something in sf::view to help with this...
How can i easily flip the Y-axis easily, for rendering purposes, not effecting the bodies dimensions/locations?
I don't know what is box2d but when I wanted to flip Y axis using openGL, I just applied negative scaling factor to projection matrix, like:
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
glScalef(1.0f, -1.0f, 1.0f);
If you want to do it independent of openGL simply apply a sf::View with a negative x value.
It sounds like your model uses a conventional coordinate system (positive y points up), and you need to translate that to the screen coordinate system (positive y points down).
When copying model/Box2D position data to any sf::Drawable, manually transform between the model and screen coordinate systems:
b2Vec2 position = body->GetPosition();
sprite.SetPosition( position.x, window.GetHeight() - position.y )
You can hide this in a wrapper class or function, but it needs to sit between the model and renderer as a pre-render transform. I don't see a place to set that in SFML.
I think Box2D has the coordinate system you want; just set the gravity vector based on your model (0, -10) instead of the screen.
How can i easily flip the Y-axis easily, for rendering purposes, not effecting the bodies dimensions/locations?
By properly applying transforms. First, you can apply a transform that sets the window's bottom-left corner as the origin. Then, scale the Y axis by a factor of -1 to flip it as the second transform.
For this, you can use sf::Transformable to specify each transformation individually (i.e., the setting of the origin and the scaling) and then – by calling sf::Transformable::getTransform() – obtain an sf::Transform object that corresponds to the composed transform.
Finally, when rendering the corresponding object, pass this transform object to the sf::RenderTarget::draw() member function as its second argument. An sf::Transform object implicitly converts to a sf::RenderStates which is the second parameter type of the corresponding sf::RenderTarget::draw() overload.
As an example:
#include <SFML/Graphics.hpp>
auto main() -> int {
auto const width = 300, height = 300;
sf::RenderWindow win(sf::VideoMode(width, height), "Transformation");
win.setFramerateLimit(60);
// create the composed transform object
const sf::Transform transform = [height]{
sf::Transformable transformation;
transformation.setOrigin(0, height); // 1st transform
transformation.setScale(1.f, -1.f); // 2nd transform
return transformation.getTransform();
}();
sf::RectangleShape rect({30, 30});
while (win.isOpen()) {
sf::Event event;
while (win.pollEvent(event))
if (event.type == sf::Event::Closed)
win.close();
// update rectangle's position
rect.move(0, 1);
win.clear();
rect.setFillColor(sf::Color::Blue);
win.draw(rect); // no transformation applied
rect.setFillColor(sf::Color::Red);
win.draw(rect, transform); // transformation applied
win.display();
}
}
There is a single sf::RectangleShape object that is rendered twice with different colors:
Blue: no transform was applied.
Red: the composed transform was applied.
They move in opposite directions as a result of flipping the Y axis.
Note that the object space position coordinates remain the same. Both rendered rectangles correspond to the same object, i.e., there is just a single sf::RectangleShape object, rect – only the color is changed. The object space position is rect.getPosition().
What is different for these two rendered rectangles is the coordinate reference system. Therefore, the absolute space position coordinates of these two rendered rectangles also differ.
You can use this approach in a scene tree. In such a tree, the transforms are applied in a top-down manner from the parents to their children, starting from the root. The net effect is that children's coordinates are relative to their parent's absolute position.