I think that first one out of the following states the point in the coordinate system where we wish to place the image. Correct?
What does the other one exactly say in layman's terms?
http://harmattan-dev.nokia.com/docs/platform-api-reference/xml/daily-docs/libqt4/qml-rotation.html#axis.x-prop
axis.x : real
axis.y : real
axis.z : real
The axis to rotate around. For simple (2D) rotation around a point, you do not need to specify an axis, as the default axis is the z axis (axis { x: 0; y: 0; z: 1 }).
and
origin.x : real
origin.y : real
The origin point of the rotation (i.e., the point that stays fixed relative to the parent as the rest of the item rotates). By default the origin is 0, 0.
Not correct.
The first is the axis which will rotated.
X is horizontal axis = Rotation will be to the viewer from top to bottom.
Y is vertical axis = Rotation will be to the viewer form left to right.
Z (the default) is perpendicular to the other 2 axis and is pointing to the viewer = Therefor rotation around Z is rotating on the screen-plane.
In the first parameter you just tell which axes to rotate around. So axis(x:0; y:0; z:1) just means to rotate on the screen plane.
The second are the origin-coordinates. This is the fixpoint where 0/0/0 of the axis coordinate system to rotate around is located. If this is the top-left corner of your object, you will rotate around that corner. You can rotate to any fixpoint, this also means: fixpoints other then the center point will always move your object.
Related
I want an object that rotates on a relative axis (the axis rotates with it) to rotate as if it's axis hadn't moved.
In a game (not my own) I have an object that stays in view of the player at all times (it follows the camera). I also want said object to keep the same rotation relative to the camera so it always looks like the same orientation to the player. It is simple to make it hold an orientation with modified X or Z values relative to the camera yaw, but once I rotate the object on the Y axis, the X axis has moved to the camera pitch makes the object rotate incorrectly (the roll would be affected too).
I am pretty sure it will require either a matrix or a combination of sin/cos, but I have had no luck in finding an answer.
Note: Rotation is in pitch, yaw, and roll using radians on the x, y, and z-axis respectively.
So let's say I have 2 objects. One with the sprite of a circle, other with the sprite of triangle.
My triangle object is set to the position of mouse in every step of the game, while circle is either standing in place or just moving in its own way, whatever.
What I want to do is to have the TRIANGLE move around the circle, but not on it's own, rather on the way your cursor is positioned.
So basically, calculate degree between circle's center and triangle's center. Whenever they are far from each other I just set triangle position to mouse position, BUT when you hover your mouse too close (past some X distance) you can't get any closer (the TRIANGLE is then positioned at maximum that X distance in the direction from circle center to mouse point)
I'll add a picture and hopefully you can get what I mean.
https://dl.dropboxusercontent.com/u/23334107/help2.png
Steps:
1. Calculate the distance between the cursor and the center of the circle. If it is more than the 'limit' then set the triangle's position to the cursor's position and skip to step 4.
2. Obtain the angle formed between the center of the circle and the cursor.
3. Calculate the new Cartesian coordinates (x, y) of the triangle based of off the polar coordinates we have now (angle and radius). The radius will be set to the limit of the circle before we calculate x and y, because we want the triangle to be prevented from entering this limit.
4. Rotate the image of the triangle to 1.5708-angle where angle was found in step 2. (1.5708, or pi/2, in radians is equivalent to 90°)
Details:
1. The distance between two points (x1, y1) and (x2, y2) is sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2))
2. The angle (in radians) can be calculated with
double angle = Math.atan2(circleY-cursorY, cursorX-circleX);
The seemingly mistaken difference in order of circleY-cursorY and cursorX-circleX is an artefact of the coordinate system in most programming languages. (The y coordinate increases downwards instead of upwards, as it does in mathematics, while the x coordinate increases in concord with math - to the right.)
3. To convert polar coordinates to Cartesian coordinates use these conversions:
triangle.setX( cos(angle)*limit );
triangle.setY( sin(angle)*limit );
where limit is the distance you want the triangle to remain from the circle.
4. In order to get your triangle to 'face' the circle (as you illustrated), you have to rotate it using the libgdx Sprite function setRotation. It will rotate around the point set with setOrigin.
Now, you have to rotate by 1.5708-angle – this is because of further differences between angles in mathematics and angles in programming! The atan2 function returns the angle as measured mathematically, with 0° at three o'clock and increasing counterclockwise. The setRotation function (as far as I can tell) has 0° at twelve o'clock and increases clockwise. Also, we have to convert from radians to degrees. In short, this should work, but I haven't tested it:
triangle.setRotation(Math.toDegrees(1.4708-angle));
Hope this helps!
I have a perspective projection. I want to have an object follow the mouse. It works fine when I set the object to be almost on a near clipping plane. But as the object goes beyond the near clipping plane, its movement is more and more distorted in a comparison to the mouse position. I know I need to change X and Y coordinates to reflect modified Z, but I don't know exact equation.
The viewport limits map to the near plane, so close to the near plane the scaling factor is ~1. So all you have to do is to scale by the distance of the object in view coordinates in relation to the distance of the near clipping plane:
scale = Z_object / Z_near
I have the coordinates of a centre point . I also have an array called the asteroid normal which I assume is the relative rotation of the axis (its 3 numbers between zero and one).
How can I make an object revolve around this object? I haven't been able to find any formula that does this.
Try this:
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
glTranslate(-x,-y,-z);
glRotate(angle,nx,ny,nz);
glTranslate(x,y,z);
Use the rotation matrix for an axis and angle. The new position p' of a point p on the object is
p' = center + R(angle, axis) * (p - center)
where R(angle, axis) is the matrix that rotates by angle about axis, and center is a point that the axis passes through. Tal Darom's answer is the same, only in OpenGL notation.
let me begin by stating that's i'm dreadful at math.
i'm attempting to reposition and rotate a rectangle. however, i need to rotate the rectangle from a point that is not 0,0 but according to how far its coordinates has shifted. i'm sure that doesn't make much sense, so i've made some sketches to help explain what i need.
the image above shows 3 stages of the red rectangle moving from 0% to 100%. the red rectangle's X and Y coordinates (top left of the red rectangle) only moves a percentage of the blue rectangle's height.
the red rectangle can rotate. focusing only on the middle example ("Distance -50%") from above, where the red rectangle is repositioned at -50 of the blue rectangle's height, its new angle in the above image is now -45º. it has been rotated from its 0, 0 point.
now, my problem is that i want its rotational point to reflect its position.
the red and blue rectangles are the same size, but have opposite widths and heights. since the red rectangle's 0,0 coordinates are now -50% of the blue rectangle's height, and since they have opposite widths and heights, i want the rotational point to be 50% of the red rectangle's width (or 50% of the blue rectangle's height, which is the same thing).
rather than specifically telling the red rectangle to rotate at 50% of its width, in order to do what i want, i need to emulate doing so by using a formula that will position the red rectangle's X and Y coordinates so that its rotational point reflects its position.
Here's an illustrated solution to your problem:
I don't exactly understand what you need, but it seems that a procedure to rotate a rectangle around an arbitrary point may help.
Suppose we want to rotate a point (x,y) d radians around the origin (0,0). The formula for the location of the rotated point is:
x' = x*cos(d) - y*sin(d)
y' = x*sin(d) + y*cos(d)
Now we don't want to rotate around the origin, but around a given point (a,b). What we do is first move the origin to (a,b), then apply the rotation formula above, and then move the origin back to (0,0).
x' = (x-a)*cos(d) - (y-b)*sin(d) + a
y' = (x-a)*sin(d) + (y-b)*cos(d) + b
This is your formula for rotating a point (x,y) d radians around the point (a,b).
For your problem (a,b) would be the point halfway on the right side of the blue rectangle, and (x,y) would be every corner of the red rectangle. The formula gives (x',y') for the coordinates of the corners of rotated red rectangle.
It's quite simple really.
1. Let's settle on your point you want to rotate the rectangle about, i.e. the point of rotation (RP) which does not move when you swivel your rectangle around. Let's assume that the point is represented by the diamond in the figure below.
2. Translate the 4 points so that RP is at (0,0). Suppose the coordinates of that point is (RPx,RPy), therefore subtract all 4 corners of the rectangle by those coordinates.
3. Rotate the points with a rotation matrix (which rotates a point anticlockwise around the origin through some angle which is now the point of rotation thanks to the previous translation):
The following figure shows the rectangle rotated by 45° anticlockwise.
4. Translate the rectangle back (by adding RP to all 4 points):
I assume this is what you want :)
It seems like you could avoid a more complex rotation by more crafty positioning initially? For example, in the last example, position the red box at "-25% Blue Height" and "-25% Red Height" -- if I follow your referencing scheme -- then perform the rotation you want.
If you know the origin O and a point P on the side of rotated rectangle, you can calculate the vector between the two:
(source: equationsheet.com)
You can get the angle between the vector and the x-axis by taking the dot product with this vector:
(source: equationsheet.com)
Given this, you can transform any point on the rectangle by multiplying it by a rotation matrix:
(source: equationsheet.com)