I would like to draw those black dots around the whole circle with regular spacing in between each other (rough image below). Those black dots should sit directly on the circle. Is there an easy way to do it somehow with painter.drawArc() function or something like that?
No, there isn't. I am afraid you will have to use... math!
Actually, you could get away without that. For sweeping around a circle in particular there is an easy way to do it by transforming the painter. You initially transform the painter to the center of the circle, and draw the first dot at the circle's 12 o'clock position. That means drawing it into negative y space. Then you simply rotate the painter by 360 / numOfObjects degree and draw the same circle again.
They key thing here is to transform the painter to the center and draw the object offset, because otherwise it will be more complicated to calculate the position and angle yourself. The angle doesn't play a role here, since you are drawing dots, but it will make a huge difference if you draw something other than dots. This way you can easily sweep the circle by rotating the painter around its origin.
To put it in pseudo code:
draw big circle
move painter to big circle center
degree = 360 / numOfObjects
while numOfObjects--
draw dot at 12 o'clock
rotate painter by degree
Related
I have a program displaying planes of cubes, like levels in a house, I have the planes displayed so that the display angle is consistent to the viewport projection plane. I would like to be able to allow the user to select them.
First I draw them relative to each other with the first square drawn at {0,0,0}
then I translate and rotate them, each plane has it's own rotate and translate.
Thanks to this this page I have code that can cast a ray using the user's last touch. If you notice in the picture above, there is a green square and blue square, this is debug graphic displaying the ray intersecting the near and far planes in the projection matrix after clicking in the centre (with z of zero in order to display them), so it appears to be working.
I can get a bounding box of the cube, but it's coordinates will think they are still up in the left corner.
My question is how do I use my ray to check intersections with the objects after they have been rotated and translated? I'm very confused as I once had this working when I was translating and rotating the whole grid as one, now each plane is being moved separately I can't work out how to do it.
I plan on making a game, and I want to create some background animations for said game. One of these animations is a rotating rectangle. I've looked all over, and I cannot find any form of math or logic that allows me to rotate a rectangle (SDL_Rect to be specific, but you might have already known that).
I can't figure out the math for myself, I really don't have any working code for this, so I can't show anything.
Essentially I'm looking for some type of logic that I can apply the rectangle's coordinates so that whenever the main game loop loops, it will rotate the rectangle some amount of degrees.
You can't rotate an SDL_Rect. If you look at its definition, it's made of coordinates for the top-left corner, the width and the height. There's no way to represent a rectangle with sides that aren't parallel to the coordinate system's axes.
SDL_RenderCopyEx supports drawing rotated textures, though.
I have a sprite, a square, just for orthogonal projection. Now I want to project it in a very basic, simple isometric way. (I know this might not be pretty, but I just want to figure this out)
Given my square, I rotate it 45 degrees. Now if I understand correctly, I should still divide my height by 2. This has been impossible for me in SFML. There is a scale function but if I scale with a factor 0.5 in the y-axis direction, my cube just gets stretched, instead of a diamond shape. It looks as though SFML transforms the sprite according to it's own relative axes (that were rotated before..).
Since you cannot access the height of a sprite, I was wondering if this was even possible?
Can I convert a square sprite to a diamond shape in SFML?
Using a sf::RenderTexture is an option (see other answer). Another option is to fiddle with the sf::View. Double the view's height, and adjust coordinates. It would go something like this:
my_sprite.setRotation(45.f);
//adjust the position for new screen coordinates (once)
my_sprite.setPosition(my_sprite.getPosition().x, my_sprite.getPosition().y * 2);
//...
//when drawing:
sf::View v = my_render_window.getDefaultView();
v.setSize(v.getSize().x, v.getSize().y * 2);
v.setCenter(v.getSize() *.5f);
my_render_window.setView(v);
my_render_window.draw(my_sprite);
my_render_window.setView(my_render_window.getDefaultView());
Rotate your sprite as you are doing now. Render it to an sf::RenderTexture. Use the member function getTexture, and make a new sprite from it, or reuse the old sprite. Scale the sprite along the y-axis. Draw it to the render window.
Some math on your part may be required in order to set the RenderTexture to the right size and to draw the original sprite in the correct location on it.
original_sprite.setRotation(45);
sf::RenderTexture rt;
rt.create(FigureOutWidth(),FigureOutHeight());
original_sprite.setPosition(MoreMathHere());
rt.draw(original_sprite);
sf::Sprite new_sprite(rt.getTexture());
new_sprite.setScale(1.0,0.5);
It should go without saying, but do this once in initialization, not every frame.
first thing I want to do is translating to the center of the screen and draw all of the objects from there.
then I would like to apply tranlsate for panning and scale for zoom. I want to zoom relative to a center point ! so how should be the order of them so that it works ?
glTranslatef(width/2, height/2, 0);
gltranslate(centerX,centerY); // go to center point
glscale(zoom);
glTranslatef(offset.x/zoom, offset.y/zoom, offset.z/zoom); // pan
I tried the above order but it doesn't go to the center point and it always zoom relative to (0,0).
I suppose you are drawing a square with both x and y between 0,1.
first you have to translate to the point the scaled object should be:
glTranslate3f(centerX,centerY,0);
glScale(zoom);
glTranslatef(-0.5f, -0.5f,0); // to the middle
draw stuff
opengl executes the transformations in reverse order since it's a pipeline.
reading the above sequence in the bottom-up direction will give the key.
I am looking for a way to get all the vertices of a rectangle whose center, normal, length and height I know. I am a little weak in maths so please help me.
Edit : the plane is in 3D space.
You can easily calculate the x and y coordinates of the vertices of a rectangle in 2D space given the center, width and height by subtracting/adding half the width/height from the x/y position of the center point.
If you need this in 3D space, this becomes a little more tricky and relies on a bit of trigonometry, but still follows the same principle. You'll need one extra piece of information. You need some way of fixing the orientation of the square in some direction; ie, which direction is the rectangle 'facing'. The normal will allow you to work out what plane the rectangle is on, but without some orientation on that plane, the best you can do is work out a set of possible values in a circle around the center for each of the vertices.