I have a canvas which display a picture that The user has uploaded.What I want to do is to ask the client to give me the width and height between two points that he/she has selected on canvas in cm .For example,The client click on top left on canvas and then select top right .after that he/she should enter the width between those two selected points.
How should I do that?
Sorry for my bad English.
Thanks
Just get the coordinates of your two points.
Let us assume the first point is called A and the second point is called B. So their coordinates would be:
A = (xa, ya)
B = (xb, xb)
So the horizontal distance between A and B would be, |xa-xb|1. And the vertical distance between them would be |ya-yb|1.
If we draw a rectangle using A and B as diagonally opposite points of it, that rectangle's width would be |xa-xb|, and it's height would be |ya-yb|.
|| represents "Absolute Value". See http://en.wikipedia.org/wiki/Absolute_value.
Related
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 have a situation where I need to output some code that simultaneously zooms a div from 0 to it's "final" size whilst moving it down from the top of the screen.
The x and y final sizes could be different every time but my code WILL know them (the div will contain a user-supplied image so I can just read the image size).
Now, I can do the code to "zoom" the div and also move it down the screen. I know how to find the y-center of the browser window etc...
I am moving it down the screen by adding "yvalue" to the topMargin.
Here is where I am stuck and I guess it is more of a math question rather than code??
Q: How do I calculate what "yvalue" (the y position increment) should be so that the div arrives in the center of the screen at the same time (roughly) that the div zooms to maximum size. Also I need to find out what the corresponding div_xvalue and div_yvalue should be to make the "zoom" happen at the same time (if that makes sense)
Please assume the following variables :
yvalue = increment to add to div topMargin each iteration
zoom_x_final = final width of div
zoom_y_final = final height of div
browser_center_y = center position of browser window (y direction)
x = current x size of div
y = current y size of div
div_xvalue = value to add to x for increasing div size per iteration
div_yvalue = value to add to y for increasing div size per iteration
stop_y = center screen position stop for slide down calc'd from browser_center_y-(zoom_y_final/2)
I think this might help you.
I have tried to find y_increment_per_iteration and got a very simple answer:
y_increment_per_iteration
= [{(browser_width) or (working_area_width)} / 2 ] / zoom_y_final
Please note that this answer might not be correct as it is not tested.
Need some inspiration. I've got a picture (blue) and want it to move proportional to the mouse position inside an invisible area (orange). So, if the mouse moves in top-left direction, the image should follow the movement.
I don't want to simply copy the mouse position, rather create an Joystick like behaviour, so if the mouse moves, the image should move stepwise in the desired direction.
But how? I've no idea how to set the right x+y coordinates for the image or how to establish a formula to calculate them.
Incremental (vectored) positions. Consider:
Let's call the dead center of your invisible circle the vector reference point (0,0) or VRP.
You move your mouse away form the VRP. Let's use your image as a reference and say that your mouse is at (-3x -2y) relative to the VRP. You keep it there; It creates a -3 X vector and a -2 Y vector.
For as long as you keep your mouse there, those vectors will be applied to the square's current coordinate at each cycle, like this:
Assume Picture starter position is absolute 100,100.
Cycle 1: [x]:100 -3 = 97;[Y]:100 -2 = 97. New picture position = 97x98y.
Cycle 2: [x]:97 -3 = 94;[Y]:98 -2 = 96. New picture position = 94x96y.
And so forth. If you want to stop the movement, just bring the cursor back to the VRP.
You may sophisticate the mechanism creating acceleration intermediate vectors, or a dead zone around the vector reference point.
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
I'm using canvas for a project and I have a number of elements that I'm skewing. I'm only skewing on the y value and just want to know what the new width of the image is after skewing (so I can align it with another canvas element). Check out the code below to see what I mean
ctx.save();
//skew the context
ctx.transform(1,0,1.3,0,0,0);
//draw two images with different heights/widths
ctx.drawImage(image,0,0,42,60);
ctx.drawImage(image,0,0,32,25);
The goal would be to know that the 42 by 60 image was now a X by 60 image so I could do some translating before drawing it at 0,0. It's easy enough to measure each image individually, but I have different skew values and heights/widths throughout the project that need to be align. Currently I use this code (works decently for images between 25 and 42 widths):
var skewModifier = imageWidth*(8/6)+(19/3);
var skewAmount = 1.3; //this is dynamic in my app
var width = (skewModifier*skewAmount)+imageWidth;
As images get wider though this formula quickly falls apart (I think it's a sloping formula not a straight value like this one). Any ideas on what canvas does for skews?
You should be able to derive it mathematically. I believe:
Math.atan(skewAmount) is the angle, in radians, that something is skewed with respect to the origin.
So 1.3 would skew the object by 0.915 radians or 52 degrees.
So here's a red unskewed object next to the same object skewed (painted green). So you have a right triangle:
We know the origin angle (0.915 rads) and we know the adjacent side length, which is 60 and 25 for your two images. (red's height).
The hypotenuse is the long side thats being skewed.
And the opposite side is the triangle bottom - how much its been skewed!
Tangent gets us opposite / adjacent if I recall, so for the first one:
tan(0.915) = opposite / 60, solving for the opposite in JavaScript code we have:
opposite = Math.tan(0.915)*60
So the bottom side of the skewed object starts about 77 pixels away from the origin. Lets check our work in the canvas:
http://jsfiddle.net/LBzUt/
Looks good to me!
The triangle in question of course is the canvas origin, that black dot I painted, and the bottom-left of the red rectangle, which is the original position that we're searching for before skewing.
That was a bit of a haphazard explanation. Any questions?
Taking Simon's fiddle example one step further, so you can simply enter the degrees:
Here's the fiddle
http://jsfiddle.net/LBzUt/33/