To make this easy, lets say I have a movie clip with a diagonal line in it (top left to bottom right).
I want to start my y position at 0, and place it in a loop and increase y by 5 every time. So we start at the top left of the clip, and move to the bottom of the clip.
Now since we have our y value (0, 5, 10, etc.) would you get the x position of the line? such as (x, y):
5, 10
or
42, 15
How could I find the distance between a given point and a drawn line? I was thinking about using the point-slope formula to draw a perpendicular line between the point and the line and measuring it as the distance, but I'm not quite sure how to do this in as3.
View 0 RepliesI've got a grid of sprites.Now I would like to drag an image on a grid-element.Therefore I need to know which x/y of the grid-element is the closest point to the mouse-position.
View 2 Repliesif i have a movieclip or a sprite could i find (via code) where is its registration point compared to its top left point (no rotation included) ?
View 4 RepliesI know some basic trig. but how do i find an angle from an anchor point and a moving point?
adjecent = _root.anchor._x-_root._xmouse;
opposite = _root.anchor._y-_root._ymouse;
hypotenuse = ((adjacent ^ 2)+(opposite ^ 2)) ^ 0.5;
I am using AS2 with Flash 8 Professional So, my problem is that I currently have a man in the middle of the screen, who shoots a line towards the mouse when I click. However, when I use hitTest to see if the line collides with another object, Flash recognizes the line as a large box if it is diagonal, so the hitTest isn't very accurate. The line only satys there for one frame, so I can't have the usual moving-bullet-style. I am either looking for a way to create an imaginary line with AS from the starting point to the mouse and beyond, and tell whether or not this line intersects with an object... or some other way that I haven't thought of to fix my problem. Keep in mind that the line rotates from a center point towards anywhere around it for 360 degrees.
View 1 RepliesHow can I move a point (or circle) slowly and the point let behind a line.When the point turn back the line will be erase.
View 8 RepliesLets say I have two objects, and I want to use action script to draw a line connecting them, which will update automatically as they are moved/ dragged.Can anyone show me how to do that, and also how to control line's parameters like colour, weight etc?
View 1 Replieshow 2 sript: to draw a line ( 4 example ) frm point A to point B and then code it such that the viewer can see the line being drawn ?
View 9 RepliesI am developing a white board application which allows the user to draw line with arrow head (some like Microsoft Word line with arrow feature). I am using graphics property along with lineTo() method to draw a line. Now i have to draw a angular arrow on the last point of line. I am drawing the arrow by connecting the points around last points. As 360 line can pass through this point and each line can have a different angle of arrow.
View 1 Repliesi am trying to find center point of my pie.
var myTween:Tween = new Tween(piechart,"............", Strong.easeOut,1200, 510, 1, true);
i want to animate from the center and it has a xml driven data file.
I'm not a genius in geometry, I'd like to find a point in as3 with the radius and a angle but I don't remember the rule, I know this should be simple!
View 3 RepliesI am currently looking at a way of visualising data and a part of the algorithm needs something 99% the same as A* Path Finding but I cannot wrap my head round it fully. I know it will be a fairly simple modification. (specific cost instead of least cost)
Basically I need to plot a path (2D Grid, no Diagonals) from point A to point B in X number of steps.
E.g. If the start and end points were right next to each other and I needed the path to be 3 steps it would have a tiny loop. (moves: up, right, down). Is there a known name for this algorithm or is the use of this so rare it's uncommon? I am currently looking at this AS3 Librbay for modification as it is apparently very fast and seems clean and simple to me:[URL]..
I once needed to pan a sprite full of movieClips (picture 1) which were draggable with no bounds so you could drag them up and left (higher than sprites registration point).
You could also drag them down and right and expand a sprite that way (as well) but that didnt matter.What did matter was how much is 'up' and 'left' from the sprite registration point so i can include that in my formula for panning.
This was easy to find, i just sorted my array of movieClips, sorted on x and sorted on y after each drag, and thats how i got my up and left points.
Now I am trying to do that same thing, exept my movieclips inside a sprite have a custom rotation (picture 2), so i also have an up and left point, except this time up and left are caused by rotations of movieClips inside a sprite... (and these are dragable as well with no bounds)
Given this grid (tinyurl.com/63dgoja --> This is a trapezium/trapezoid, not a square), how do you find the point clicked by the user? I.e. When the user clicks a point in the grid, it should return the coordinates like A1 or D5.
View 2 RepliesI am looking for a way to find a vector if I know one point and 1 angle. Like I have A(x, y) and a vector starting from A(x, y) with an angle. I would like to know the position on a point X (x, y) that can be anywhere on that vector. I only know the coordinates of A and the angle. Possible?? I guess it has to be a function...
View 5 Repliesi'm triying to make a dinamic generated plane, that plane has a grid with lines in X and Y, by now i'm able to generate a grid clicking and dragging points to make 4 lines and i generated 2 lines between the midles of the opposites lines like this:
[Code]...
i want to find third Point of Triangle i Have two point p1,p2 with respect to it i want to find third point of triangle
Note one thing triangle is 45 - 90 - 45 degree so the point i want to find is at 90 degree,
so as per my requirment p1 = 45 degree p2 = -45 degree and p3 = 90 degree so i want to find p3
I have a number of boxes on screen - each box is a rectangular movieclip, and they fall to the bottom of the screen where they pile up to make a stack of boxes. I'm using box2d for the physics. Oh, and each box may be rotated. I need to find the highest point - ie the smallest y value, of the stack of boxes.
Obviously I can find the position of the highest box's registration point easily (which is the centre of each box) - but that may not be the highest point of the pile of boxes. The highest point may not even be the highest box - it could be a lower box that is rotated on end, for example.
I want to find out if the user is currently using more than one finger to draw on the screen.currently, the TouchEvent object doesn't have any property like touchCounts, which tells the number of touches on the screen. it just has a touchPointID, which helps to identify a particular touch.
do i need to handle this on my own ? for eg: in the TOUCH_BEGIN event handler i could see how many different unique touchPointID i have received and update the count myself.
I have two circles, an inner circle and an outter circle. I'm trying to find points that reside within the area between the edge of the inner circle and the edge of the outer circle.
View 1 RepliesI have a path drawn in the flash IDE, accessible as a MovieClip.
With AS3 I need to find the closest point lying on this path to another point.
I'm having trouble with this code
var imageMap:ImageSnapshot= ImageSnapshot.captureImage(object);
var pixelValue:uint = imageMap.bitmapData.getPixel(x, y);
How to find out different circle's center point of a bitmap?
just like photo above, how can i locate the 2 center points of circles?
i only know using .getColorBoundsRect to get a point, but this only apply to one circle, in fact i wanna get different points according to the bitmap.
I have a MovieClip that has four circles on the stage. Each circle draws a line to the other three circles. I'm trying to figure out how to find the intersect point between all of the lines. If there is no intersect point, then the circle in the middle should become the intersect point. I've uploaded the SWF here: [URL].
Here is the code for generating the lines:
ActionScript Code:
package {
import flash.display.*;
import flash.events.*;
import flash.geom.*;
import flash.sensors.Accelerometer;
public class IntersectTest extends MovieClip {
[Code] .....
I'm trying to add some zooming functionality. The problem is that when you zoom in and out, it scales the MC up and down from where the registration point is. This means That when I'm far from the position of the registration point instead of zooming the in straight it also makes it appear as if you are travelling towards the registration point.
To offset this effect I want to calculate the amount of pixels I need to offset the Map by so that it appears as if you are zooming in directly.
So just as a test I've put a little ball inside the map(called "center") which I'm trying to keep in the center of the map.[code]...
I have a rectangle of any arbitrary width and height. I know X,Y, width, and height. How do I solve the upper right hand coordinates when the rectangle is rotated N degrees? I realized if it were axis aligned I would simply solve for (x,y+width). Unforunatly this doesn't hold true when I apply a transform matrix on the rectangle to rotate it around its center.
View 2 RepliesI know the length of the triangle segments and the xy coordinates of two points. How do I figure out the xy of the 3rd point?
View 3 RepliesSo I have a ground that's supposed to be, for the most part, curved. How can I find the utmost y point of the ground at a given x? The ground looks like so: The function I have right now sort-of works, but seems very inefficient. You input an x parameter, and the function loops using hitTestPoint with a constant x and a y that increases until it hits the ground. Then it returns that y value. Is there a better way to do this? This method seems like it would get slow.
View 4 RepliesIs there a way to set the x and y at the same time using the Point object?
View 2 Replies
