Line segment circle minimal vector before collision

collision detectiontrigonometry

Let this problem be in 2D cartesian plane.
- Let 'Collision' of two objects be $<=>$ they are not touching and they have atleast one intersection point.
- Let $C$ be a circle with centre $O$ and radius $R$.
- Let $L$ be a line segment defined by two points, $A$, $B$.
- Let $Vm$ be vector defining the movement of $C$.

Suppose, that we move $C$ using vector Vm and on the path, the circle will collide $L$.

Let Vc be vector.
Let $C'$ be circle $C$ moved using $Vc$. (definition and explanation below)

I am tying to find $Vc$.

If we move $C$ by $Vm$ and on the movement path, and $C$ would collide $L$, then $Vc$ is vector such as $C'$ will touch L and length of $Vc$ $\lt$ $Vm$.

If $C$ during movement mentioned above would not collide $L$, then $Vc=Vm$

Illustration of the problem

On paper this can be solved by look and see, however, I am looking for idea on how to solve this algorithmically.
Result will be used in computer code.

I solved most of cases by:

finding $L2$ orthogonal to $L$ such as $O$ lies on $L2$
calculating $X$ intersection point of $L$ and $L2$
calculating distance $|XO|$
subracting $|XO|$ from length of $Vm$ to get $Vc$ (reducing its length by $|XO|$)

This, however, does deal only with some cases of the problem. Not, for example, the first one displayed in illustration.

Thank you for any help

Node: I would like to avoid having for loop and checking for collision during all the movement steps

Best Answer

So finally I am answering my own question after months of work. The entire problem is quite complicated, but worth solving, since the final collision handling is really worth it.

This explanation is copied here from my webpage, where the math solution is more detailed and supported with more images plus wolfram mathematica notebooks.

This describes how the collision between the objects in the game is being handled.

Circle - Line collision

If collision between circle and line segment (which is line limited on interval <a,b> is resolved, then, using line segment, other geometrical shapes can be created (such as triangle or polygon), and thus, the collision between them and circle is already solved.

Definitions

Term line reffers to infinite line, not line segment

Let $C$ be circle with centre $C_s = (c_x, c_y)$ that we want to move

Let $L_s$ be Line Segment defined by 2 points, $A, B$. $L$ might collide circle $C$

Let $L$ be Line defined also by 2 points, $A, B$

Let $v=(v_x, v_y)$ be vector, defining the movement of the circle $C$

Let $v'$ be the final movement vector (the result)

$L_s$ touches $C \Leftrightarrow C$ and $L_s$ have exactly $1$ intersection point

$L_s$ collides $C \Leftrightarrow C$ and $L_s$ have exactly $2$ intersection points

Objective

Imagine, that being given movement vector $v$, the circle $C$ would collide $L_s$ during movement, if we move $C$ by $v$.

The objective is to determine vector $v'$ such as, when moving $C$ by vector $v'$, then $C$ will collide $L_s$ in exactly one point.

Mathematical solution

The problem is divided to 6 cases, all of them will be discussed below.

If $C$ does not touch nor collides line $L_s$ before movement begins

$C$

$L_s$

$A$

$B$

$A$

$B$

$L_s$

$C$

$L_s$

$P$

$M$

$L_s$

$P$

$P \in \{M\} / \{A, B\} $

$C$

$L_s$

If $C$ touches $L_s$ before movement begins

$C$

$L_s$

$C$

$L_s$

$C$

$L_s$

Circle - Line segment intersection relationship

To solve all the cases, we need to first check for relation between $C$ and $L$ (do they collide each other? Or touch?)

Circle touches Line segment

For this, following statements apply:

Let $P$ be set of all intersection points between $C$ and $L$. Applies $|P| \in \{0,1,2\}$

If $|P|=0 $, then $C$ does not touch $L_s$

If $|P|=1 \wedge P \in L $, then $C$ touches $L_s$

If $ |P|=2 \wedge P_x,P_y \in P: P_x = B \wedge P_y \notin L_s \wedge |A C_s| >= r $, then $C$ touches $L_s$

If $|P|=2 \wedge P_x,P_y \in P: P_x = A \wedge P_y \notin L_s \wedge |B C_s| >= r $, then $C$ touches $L_s$

To explain the math above in plain words:

If $C$ and $L$ have 0 intersection points, then they do not touch.

If $C$ and $L$ have 1 intersection point, then they touch.

If $C$ and $L$ have 2 intersection points:

If one of the intersection points is $A$, and the other intersection point does not lie on the line segment $L_s$ and the B point lies outside the circle, then they touch
If one of the intersection points is $B$, and the other intersection point does not lie on the line segment $L_s$ and the A point lies outside the circle, then they touch

Circle collides Line segment

For this, following statements apply:

Let $P$ be set of all intersection points between $C$ and $L$. Applies $|P| \in \{0,1,2\}$

If $|P| \in \{0,1\} $, then $C$ does not collide $L_s$

If $|P|=2 \wedge P_x,P_y \in P: P_x \in L_s \wedge P_y \in L_s $, then $C$ collides $L_s$

If $|P|=2 \wedge P_x,P_y \in P: P_x = A \wedge |B C_s| >= r $, then $C$ collides $L_s$

If $|P|=2 \wedge P_x,P_y \in P: P_x = B \wedge |A C_s| >= r $, then $C$ collides $L_s$

To explain the math above in plain words:

If $C$ and $L$ have 0 r 1 intersection points, then they do not collide.

If $C$ and $L$ have 2 intersection points:

If both intersection points lie on $L_s$, then they collide
If one of the intersection points is $A$, and the other intersection point does not lie on the line segment $L_s$ and the $B$ point lies inside or on the circle, then they collide
If one of the intersection points is $B$, and the other intersection point does not lie on the line segment $L_s$ and the $A$ point lies inside or on the circle, then they collide

Case 1: $C$ will collide $L_s$ during movement at one of points $A$ or $B$, where $A$, $B$ are points defining the line $L_s$

To solve this, similar mathematical approach is used as in case2.
It is strongly recomended to read that first, and then come back here for better understanding...

By taking circle equation and line equation and combining them together, the point A (later also B) is plugged into the equations. If there is solution, and the final movement vector $v$ matches criteria (smaller than original movement vector, etc...for more details, read through case2), then the circle is considered to be colliding line segment at point A (or B, respectivelly).

Case 2: $C$ will collide $L_s$ during movement at point $P$. Let $M$ be, set of all points on line $L_s$. Then, for $P$ applies: $P \in \{M\} / \{A, B\} $

Let's reiterate, that $L$ is line, not a line segment. For simplicity, I will explain the solution on LINE, understanding that narrowing the solution down for line segment involves limitation on interval and adding few other edge cases to entire solution.

Let's start by having the equation of circle $C$:

$$(x-c_x)^2+(y-c_y)^2=r^2$$

Now, we perform translation of the centre by vector $v$:

$$(x-(c_x+v_x))^2+(y-(c_y+v_y))^2=r^2$$

However, we do not know how far to move the circle to get exactly one intersection point with $L$. Therefore, we use parameter $t$:

$$(x-(c_x+v_x*t))^2+(y-(c_y+v_y*t))^2=r^2$$
The equation above describes all circles, that are moved on the line defined by movement vector $v'$

Now, we will solve system of 2 equations:

$$(x-(c_x+v_x*t))^2+(y-(c_y+v_y*t))^2=r^2$$ $$a*x+b*y+c=0$$

Where the second equation is the equation of $L$ in general form

We express $x$ from the line equation (alternativelly, $y$ must be expressed in some cases to avoid diving by 0):

$$x=\frac{-b*y-c}{a}$$

That value is plugged into the circle equation to get:

$$(\frac{-b*y-c}{a}-(c_x+v_x*t))^2+(y-(c_y+v_y*t))^2=r^2$$

Now, considering $t$ as the polynom variable, if discriminant equals 0, then there is exactly one solution (circle collides line in one point only)

$$Discriminant[(\frac{-b*y-c}{a}-(c_x+v_x*t))^2+(y-(c_y+v_y*t))^2-r^2,y]=0$$

The discriminant is then:

$$D=-(\frac{1}{a^2})* 4 *(c^2 + 2 a c c_x + a^2 c_x^2 + 2 b c c_y + 2 a b c_x c_y + b^2 c_y^2 - a^2 r^2 - b^2 r^2 + 2 a c t v_x + 2 a^2 c_x t v_x + 2 a b c_y t v_x + a^2 t^2 v_x^2 + 2 b c t v_y + 2 a b c_x t v_y + 2 b^2 c_y t v_y + 2 a b t^2 v_x v_y + b^2 t^2 v_y^2)$$

Considering some input values for circle centre, movement vector, radius, etc., this would be more readable result:

$$D=-4(28-36*t+9t^2$$

The thing left is to compute the $t$, which is the movement vector multiplier.
Final solution vectors (if the solutions exist) will be:

$v'_1 = (v_x * t_1, v_y * t_1)$
$v'_2 = (v_x * t_2, v_y * t_2)$

Of course, the final vectors $v'_1, v'_2$ might not face in same direction as $v'$.
In that case, the solution is discarted.
Moreover, if $|t_1| > 1 \lor |t_2| > 1)$, then $t_1$, respectivelly $t_2$ is discarted.
That is because we require final vector to have same or smaller length then the original move vector

Case 3: $C$ will not collide $L_s$ during movement at all

If none of Case1 or Case2 holds and yields a solution, then the circle will not collide line segment at all during movement.

Case 4 and 5: $C$ touches $L_s$ before movement begins, and $C$ will (or not) collide $L_s$ during movement

For this one, imagine following figure:

touchMoveIllustration1

Let $t$ be tangent to $C$ in point $P$, which is touch point of $C$ and $L_s$ (in this example, $P=A$)
Let $u=|AE|$, $v=|AD|$ be two movement vectors, where we position them to begin in touch point $P=A$
$u=|AE|$ represents movement that would be allowed, whereas $v=|AD|$ represents movement that would NOT be allowed

In general case, imagining vector $u=|AD|$, then movement of $C$ using vector $u$ is valid $\iff D$ is on same side of line $t$ as centre of circle $C$ or $D \in t$

Case 6: $C$ collides $L_s$ before movement begins

Trivial case, when is done check for whether $C$ collides $L_s$, and if so, then no movement is done...

Best Answer

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