Need some math - Projecting Slope

Question

I have a rectangle. Its height (RH) is 400. Its width (RW) is 500.

I have circle. Its height (CH) is 10. Its width (CW) is 10. Its starting location (CX1, CY1) is 20, 20.

The circle has moved. Its new location (CX2, CY2) is 30, 35.

Assuming my circle continues to move in a straight line. What is the circle's location when its edge reaches the boundary?

enter image description here

Hopefully you can provide a reusable formula.

Perhaps some C# method with a signature like this?

point GetDest(size itemSize, point itemPos1, point itemPos2, size boundarySize)

I need to calculate what that location WILL be once it arrives - knowing that it is not there yet.

Thank you.

PS: I need this because my application is watching the accelerometer on my Windows Phone. I am calculating the target necessary to animate the motion of the circle inside the rectangle as the user is tilting their device.

I don't know about phone but in winforws should be like this: if it is going to right would be like `while(circle.Location.X < (this.Width - circle.Width)) { /* DO STUFF */};` — Dumbo, Jan 23 '12 at 23:37
No, you are attempting to test if it has reached the final location. I am attempting to calc the final location before it reaches it. — Jerry Nixon, Jan 23 '12 at 23:44
You know the radii of the two circles and the position of their centers, so use the line equation: `y = m*x + b` , no calculus, just algebra. — Chris O, Jan 23 '12 at 23:48
@ChrisO, yes. I just worked it out as an answer. I hope you spot any error. — Johan Lundberg, Jan 24 '12 at 00:43

score 4 · Answer 1 · answered Jan 23 '12 at 23:36

4

It is 1 radius away from the boundar(y/ies).

answered Jan 23 '12 at 23:36

Ignacio Vazquez-Abrams

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My answer still stands. Think normals. – Ignacio Vazquez-Abrams Jan 23 '12 at 23:48
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Your answer is too cryptic to be useful. Sorry. – Jerry Nixon Jan 24 '12 at 05:09
@IgnacioVazquez-Abrams. Your answer is true but it's just saying it's at the edge. He would like to know where at the edge? – Johan Lundberg Jan 25 '12 at 20:50

Johan Lundberg · Accepted Answer · 2012-01-24T08:27:21.267

The answer is X=270 Y=395

first define the slope V as dy/dx =(y2-y1)/(x2-x1). In your example: (35-20)/(30-20)=1.5

the line equation is y = V * (x-x1) + y1. You are interested in the horizontal locations x at: y= CH/2 OR y= H-CH/2 so (not code, just math)

if (y2-y1)<0:
 x=(CH/2 -y1)/V +x1     10 for your example. OR
if (y2-y1)>0:
 x=(H-CH/2 -y1)/V +x1   270 for your example
else (that is: y2==y1)
 the upper or lower lines were not hit.

if CH/2 <= x <= W-CH/2 the circle did hit the that upper or lower side: since V>0, we use x=270 and that is within CH/2 and W-CH/2.

So the answer to your question is y=H-CH/2 = 395 , X=270

For the side lines it's similar:

(if (x2-x1)<0)
 y=(CH/2 -x1)*V +y1   
(if (x2-x1)>0)
 y=(W-CH/2 -x1)*V +y1 
else (that is: x2==x1)
 the side lines were not hit.

if CH/2 <= y <= H-CH/2 the circle did hit that side at that y.

be careful with the trivial cases of completely horizontal or vertical movement so that you don't divide by zero. when calculating V or 1/V. Also deal with the case where the circle did not move at all.

Since you now asked, here's metacode which you should easily be able to convert to a real method. It deals with the special cases too. The input is all the variables you listed in your example. I here use just one symbol for the circle size, since it's a circle not an ellipse.

method returning a pair of doubles    getzy(x1,y1,W,H,CH){

  if (y2!=y1){ // test for hitting upper or lower edges
    Vinverse=(x2-x1)/(y2-y1)
    if ((y2-y1)<0){
       xout=(CH/2 -y1)*Vinverse +x1 
       if (CH/2 <= y <= H-CH/2) {
           yout=CH/2
           return xout,yout
       }
     }
    if ((y2-y1)>0){
       xout=(H-CH/2 -y1)*Vinverse +x1 
       if (CH/2 <= y <= H-CH/2) {
           yout=H-CH/2
           return xout,yout
       }
    }
  }     
  // reaching here means upper or lower lines were not hit.
  if (x2!=x1){ // test for hitting upper or lower edges
    V=(y2-y1)/(x2-x1)
    if ((x2-x1)<0){
       yout=(CH/2 -x1)*V +y1 
       if (CH/2 <= x <= W-CH/2) {
           xout=CH/2
           return xout,yout
       }
     }
    if ((x2-x1)>0){
       yout=(H-CH/2 -x1)*V +y1 
       if (CH/2 <= x <= W-CH/2) {
           xout=H-CH/2
           return xout,yout
       }
    }
  }     
  // if you reach here that means the circle does not move...
   deal with using exceptions or some other way.
}

@JerryNixon I added C-like metacode which you should be able to just use. — Johan Lundberg, Jan 24 '12 at 08:28

score 1 · Answer 3 · answered Jan 23 '12 at 23:48

It's easy; no calculus required.

Your circle has radius R = CW/2 = CH/2, since the diameter of the circle D = CW = CH.

In order to have the circle touch the vertical edge of the rectangle at a tangent point, you have to move the circle to the right by a distance (W - (CX1 + CW/2))

Likewise, the circle will touch the bottom edge of the rectangle at a tangent point when you move it down by a distance (H - (CY1 + CH/2)).

If you do this in two separate translations (e.g., first to the right by the amount given, then down by the amount given or visa versa), you'll see that the circle will touch both the right hand vertical and the bottom horizontal edges at tangent points.

Glenn · Answer 4 · 2012-01-24T00:22:01.190

When the moving circle arrives at a wall (boundary) then it will be tangent at one of four points on the circle, call them N, S, E, and W. You know their initial coordinates.

The points travel in a line with a slope known to you: m=(y2-y1)/(x2-x1); where in your example (x1, y1) - (20,20) and (x2, y2)= (30, 35).

Your problem is to find the trajectory of the first point N, S, E, or W which reaches any wall. The trajectory will be a line with slope m.

You can do this by adding (or subtracting) the direction vector for the line to the point N, S, E, or W, scaled by some t.

For example, N is (20, 15). The direction vector is (x2-x1,y2-y1) = (10, 15). Then (20, 15) + t * (10, 15) will hit the boundary lines at different t's. You can solve for these; for example 20 + t*10 = 0, and 20 + t*10 = 400, etc.

The t that is smallest in magnitude, over all four trajectories, gives you your tangent point.

score 0 · Answer 5 · answered Jan 23 '12 at 23:39

0

Not sure its calculus..wouldn't it just be the following:

if y >= 390 then it reached the top edge of the rectangle

if x >= 490 then it reached the right edge of the rectangle

if y <= 0 then it reached the bottom edge of the rectangle

if x <= 0 then it reached the left edge of the rectangle

answered Jan 23 '12 at 23:39

Justin Self

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No, this is calculation to determine when it has reached. But it is not there yet. I need to calculate the final location BEFORE it reaches the location. – Jerry Nixon Jan 23 '12 at 23:43

Need some math - Projecting Slope

5 Answers5