I'm implementing some basic annotation draw features, such as arrows. Now I'm a little bit stuck with ellipse.
The methods to draw an ellipse usually address using it's two diameters and eventually a rotation:
However I want to display the ellipse between the point user clicked and the one he's hovering, therefore I need a function that calculates diameters and rotation based on two points:
How would I do that? Can it be achieved with sufficient performance (as it renders during mouse-hovering)?
the steps you shoul follow:
get the angle of the line (from this post: get angle of a line from horizon)
rotate the canvas or at least the part you currently drawing (live demo here: http://www.html5canvastutorials.com/advanced/html5-canvas-transform-rotate-tutorial)
draw an ellipse in canvas (http://www.scienceprimer.com/draw-oval-html5-canvas)
the resulted ellipse will be transformed as described
It can be done in the same way that it is normally done, just using different math to calculate the shape. Without writing the entire code for you, you can start by having an event trigger when the user clicks the mouse button down. The function will copy the users x and y position based on the screen. Then there is a second function which will handle mouse movement. This function will keep track of the x and y coords of the mouse while it is in motion. The final function will be a mouse up event, when a user lifts their finger from the mouse button (assuming this is when the event should be finished). Using the initial and final position of the x and y coordinates, you can calculate the length of the line the user created. That line is the long diameter of the ellipse. Half this number for the large radius. Then use whatever ratio you are using to calculate the smaller radius from the larger one. Then create an ellipse based on these numbers.
For the math: Suppose your first point is x1,y1 and the end point is x2,y2
I'm also assuming that we have a line going from bottom-left to top-right
Distance between two points = sqrt((x2-x1)^2 + (y2-y1)^2) ---> (we will call this d1)
half of this is the length of the large radius ---> (we will call this r1)
Midpoint formula = ((x1+x2)/2 , (y1+y2)/2) ---> axis of rotation (we will call it (m1, m2))
distance from midpoint to end is just the radius
radius is now the hypotenuse of constructed plane, y2-m2 is height of right triangle.
Find the angles between midpoint and one end of larger radius - sin((y2-m2)/r1).
Angle of smaller radius is this angle + pi/4 radians.
calculate length of smaller radius based on ratio.
Related
I have a series of circles which are randomly positioned on the scene (x,y). I was wondering if anyone knew of a solution to make sure that when randomly placed, the circles would not overlap.
In your loop when you place circles, take the randomized (x,y) coordinate, and get the distance to all of the existing circles (another loop) --> √((x1-x2)^2 + (y1-y2)^2), if the distance is greater than the radii of both circles added together for EVERY circle, then you can place the circle, otherwise they overlap.
A statement checking to each circle's co-ordinates plus and minus its radius does not come within the co-ordinates of another circle plus and minus its radius in both the x and y direction could work
This image is what I am trying to achieve. Circles on the edges are clickable. Its structure is made such that each rotation will be multiple of 45 deg.
I am using css transform rotate property for rotation.
How is it supposed to work?
When we click on any circle on the edge it should come to its active position and it should always rotate in clockwise direction.
Here is what I did
I achieved rotation by assigning numbers to the circles.
i.e., if clicked on 7 number then it will rotate by angle (9-7)*45=90 degrees.
(This time I did not change the numbers dynamically. i.e., as the circle is rotating the numbers given to the circles are the same.)
This works fine here. But, lets see this scenario: when we click on 2nd position circle it will rotate by 315 deg and then if you click again on the same (second) position then it will make the angle of 270 and it rotates the div in anti-clockwise direction. I think this is its behavior. But, I don't want this to happen. It should rotate again in clockwise direction and should take the active position.
Now to achieve above I did this:
Adding angle with prev angle.
var prev_degree = prev_degree + current_degree;
(current degree is being calculated using the same formula from above.)
This time I changed numbers dynamically i.e., on each click numbers are given to the desired positions. Starting 1 as from Active position till 8 as shown in Image
But this time, when I rotate circle 1, 2 times it rotates perfect then it starts working strange. If you are constantly clicking on the same number then it will add the same angle and will keep rotating perfect no matter what your angle is. If you click on random circles then it wont work which is totally wrong.
Why clockwise?
Because the circles on edges, they contain icon of which I have to maintain position when whole circle rotates.(I can achieve that rotating circles on the edges by assigning negative angles. But, now this is not a problem.)
There is no case when I am getting negative angle.
Let me know if you are not clear with my question.
Please let me know your suggestions to make this work :-)
I am trying to build a small app where my users can straighten up a tilted face with just 2 clicks
I ask my users to click on the middle of the nose and the middle of the eyebrows of the face within the image.
From there I get 2 points eyebrowMiddle(x1,y1) and noseMiddle (x2,y2).
Is it possible via these 2 points to calculate how much Canvas
rotation I need to have to rotate the image and make the face straight
in relation to the canvas rectangle?
Also, how can I detect and adjust accordingly if the image is tilted
to the left or right?
Here is a more descriptive image to show you what I mean now.
PS:
x1,y1 and x2,y2 are in relation to the canvas perimeter of
course, not the browser window or anything else.
We have tried the line equation such as m = (x2-x1) / (y2-y1) but the
result is always near 1 so I don't think we are following the right
course at the moment.
We don't care if the image looks wrong in the canvas as long as the
face features are parallel in relation to the bottom of the canvas
(they should be looking straight).
To perform such a rotation, you need to decide of the pivot point. Here i choose the eyebrow.
Then you have to choose a point in the target canvas where this pivot point will be hooked. I decided to choose the point at middle x coordinates, and at fourth of the screen in y.
To compute the rotation angle, you have to use atan2, which will nicely give you the angle for a given deltaY / deltaX in between two points ( angle = Math.atan2 ( delta y , delta x ) ) .
Then to draw :
- Translate to the target point.
- rotate by right angle.
- draw the image centering on its pivot.
ET VOILA, it works :-)
function rotate() {
ctx.save();
// go to default center position
ctx.translate(eyeBrowTargetPosition.x, eyeBrowTargetPosition.y);
// compute angle
var yDelta = noseMiddle.y - eyebrowMiddle.y;
var xDelta = noseMiddle.x - eyebrowMiddle.x ;
var angle = Math.atan2 (yDelta ,xDelta);
// compensate for angle
ctx.rotate(angle);
//draw image centering input on eyebrow
ctx.drawImage(face, -eyebrowMiddle.x, -eyebrowMiddle.y);
ctx.restore();
};
jsbin is here :
http://jsbin.com/wavokaku/2/edit?js,output
result with an approximation of the existing green dots :
I've searched for the answer to this and have tried many proposed solutions, but none seem to work. I've been struggling with this forever so any insight is greatly appreciated.
I have 3 shapes (vectors I suppose) on a JS canvas, each with an orientation represented as degrees off of 0 and a width. I need to drag one of these shapes "straight out" from its orientation. This is difficult to explain in words so please view the graphic I created:
The middle (diagonal) shape is at 45 degrees. It's origin is the red dot, (x1,y1). The user drags the shape and their mouse lies at the green dot, (x2,y2). Since the shape's origin is in the lower left, I need to position the shape at the position of the lighter blue shape as if the user has dragged straight outward from the shape's origin.
I don't think it matters, but the library I'm using to do this is KineticJS. Here's the code and some information I have available which may help solve the problem. This code positions the shape on top of the mouse, which isn't what I want:
var rotationDeg = this.model.get("DisplayOri"), // rotation in degrees
rotationRadians = rotationDeg * Math.PI / 180, // rotation in rads
unchanged = this.content.getAbsolutePosition(), // {x,y} of the shape before any dragging
dragBoundFunc = function (changed) {
// called on a mouseMove event, so changed is always different and is the x,y of mouse on stage
var delta = {
x: changed.x - unchanged.x,
y: changed.y - unchanged.y
};
return changed; // go to the mouse position
};
[edit] I should mention that the obvious of "return delta" doesn't work.
It sounds like you want to constrain the movement of the object.
Determine the vector representing the constraint axis : that is, we only want motion to occur along this line. It appears from your drawing that this is in the direction of the short line from the red dot out to the left. That vector has a direction of -1/m where m is the slope of the line we are moving.
Constrain the movement. The movement is represented by the mouse move delta - but we only want the portion of that movement in the direction of the constraint axis. This is done with a dot product of the two vectors.
So in pseudo code
m = (line.y2 - line.y1)/(line.x2 - line.x1)
constraintSlope = -1/m
contraintVector = {1, constraintSlope} //unit vector in that direction
userMove = {x2-x1, y2-y1} //vector of mouse move direction
projection = userMove.x * constraintVector.x + userMove.y * constraintVector.y
translation = projection * constraintVector //scaled vector
OH Great and Knowledgeable Stack Overflow, I humbly request your great minds assistance...
I'm using the three js library, and I need to implement a 'show extents' button. It will move the camera to a position such that all the objects in the world are visible in the camera view (given they are not blocked of course).
I can find the bounding box of all the objects in the world, say they are w0x,w0y,w0z and w1x,w1y,w1z
How can I, given these to bounds, place the camera such that it will have a clear view of the edges of the box?
Obviously there will have to be a 'side' chosen to view from...I've googled for an algorithm to no avail!
Thanks!
So Let's say that you have picked a face. and that you are picking a camera position so that the camera's line-of-sight is parallel to one of the axes.
Let's say that the face has a certain width, "w", and let's say that your camera has a horizontal field-of-view "a". What you want to figure out is what is the distance, "d" from the center of the face that the camera should be to see the whole width.
If you draw it out you will see that you basically have an isosceles triangle whose base is length w and with the angle a at the apex.
Not only that but the angle bisector of the apex angle forms two identical right triangles and it's length (to the base) is the distance we need to figure out.
Trig tells us that the tangent of an angle is the ratio of the oposite and adjacent sides of the triangle. So
tan(a/2) = (w/2) / d
simplifying to:
d = w / 2*tan(a/2)
So if you are placing the camera some axis-aligned distance from one of your bounding box faces then you just need to move d distance along the axis of choice.
Some caveats, make sure you are using radians for the javascript trig function input. Also you may have to compute this again for your face height and camera's vertical field-of-view and pick the farther distance if your face is not square.
If you want to fit the bounding box from an arbitrary angle you can use the same ideas - but first you have to find the (aligned) bounding box of the scene projected onto a plane perpendicular to the camera's line of sight