# Ray vs Surface Intersection Calculations

For all equations below, assume the ray is defined as:

$\vec{T}(t) = \vec{S} + t \vec{V}$

### Ray vs Triangle

A triangle is defined by three points $$\vec{P_0}, \vec{P_1}, \vec{P_2}$$.

We need to find the plane that these points create. A plane can be represented as a normal vector and a distance from the origin.

First, we get the normal. $\vec{N} = (\vec{P_1} - \vec{P_0}) \times (\vec{P_2} - \vec{P_0})$

Second, to get the signed distance, $$d$$, we can take the negative dot product of $$\vec{N}$$ and any arbitrary point in the plane, so we'll just pick $$\vec{P_0}$$.

$d = -\vec{N} \cdot \vec{P_0}$

Finally, the 4D plane vector is given as:

$\vec{L} = \langle \vec{N}, d \rangle$

The value of $$t$$ that corresponds to where the ray intersects this plane is given as:

$t = -\frac{\vec{L} \cdot \vec{S}}{\vec{L} \cdot \vec{V}}$

If the denominator above is 0, then there is no intersection, otherwise you can plug this $$t$$ back into the ray equation to get the intersection point, $$\vec{P_{\text{intersection}}}$$.

But now you have to determine whether this intersection point falls inside of the triangle, and not just along the plane this triangle creates.

$\vec{R} = w_1 \vec{Q_1} + w_2 \vec{Q_2}$

Where: $\vec{R} = \vec{P_{\text{intersection}}} - \vec{P_0}$ $\vec{Q_1} = \vec{P_1} - \vec{P_0}$ $\vec{Q_2} = \vec{P_2} - \vec{P_0}$

Taking the dot product of both sides of the equation above by $$\vec{Q_1}$$ and then $$\vec{Q_2}$$ creates a system of equations that can help us solve for $$w_1$$ and $$w_2$$.

$\begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} \vec{Q_1} \cdot \vec{Q_1} & \vec{Q_1} \cdot \vec{Q_2} \\ \vec{Q_1} \cdot \vec{Q_2} & \vec{Q_2} \cdot \vec{Q_2} \\ \end{bmatrix}^{-1} \begin{bmatrix} \vec{R} \cdot \vec{Q_1} \\ \vec{R} \cdot \vec{Q_2} \end{bmatrix}$ $= \frac{1}{ ( (\vec{Q_1} \cdot \vec{Q_1}) (\vec{Q_2} \cdot \vec{Q_2}) ) - (\vec{Q_1} \cdot \vec{Q_2})^2 } \begin{bmatrix} \vec{Q_2} \cdot \vec{Q_2} & -\vec{Q_1} \cdot \vec{Q_2} \\ -\vec{Q_1} \cdot \vec{Q_2} & \vec{Q_1} \cdot \vec{Q_1} \\ \end{bmatrix} \begin{bmatrix} \vec{R} \cdot \vec{Q_1} \\ \vec{R} \cdot \vec{Q_2} \end{bmatrix}$

The point $$\vec{R}$$ lies inside the triangle iff all three weights $$w_0$$, $$w_1$$, $$w_2$$ are nonnegative. Note that $$w_0 = 1 - w_1 - w_2$$.

### Ray vs Box

A box is defined with six plane equations.

$x = 0$ $y = 0$ $z = 0$ $x = r_x$ $y = r_y$ $z = r_z$

Where $$r_x$$, $$r_y$$, $$r_z$$ are the dimensions of the box.

We need to consider at most three planes to determine whether the ray intersects the box, since the other three planes would be facing away from the ray's direction $$\vec{V}$$.

We can do this with a little bit of logic, using each component of $$\vec{V}$$ one-by-one.

• If $$\vec{V_x} = 0$$ , then the ray cannot intersect the planes at $$x = 0$$ and $$x = r_x$$ since it is parallel to those planes.
• If $$\vec{V_x} \gt 0$$, then the ray would intersect the plane at $$x = 0$$ before the plane at $$x = r_x$$, and therefore we'd only need to consider the first plane.
• Alternatively, if $$\vec{V_x} \lt 0$$, then the ray would intersect the plane at $$x = r_x$$ before the plane at $$x = 0$$, and therefore we'd only need to consider the first plane.

This reduces the problem to at most three Ray vs Surface intersections for each component of the $$\vec{V}$$.

The value of $$t$$ which intersects the plane $$x = r_x$$ is: $t = \frac{r_x - \vec{S_x}}{\vec{V_x}}$

Once we have the intersection point on the plane, we need to determine whether the point is inside the face of the box along that plane, which is pretty easy. Both of the following conditions must be true:

$0 \ge \vec{T(t)}_y \ge r_y$ $0 \ge \vec{T(t)}_z \ge r_z$

### Ray vs Sphere

A sphere of radius $$r$$ is defined by this equation: $x^2 + y^2 + z^2 = r^2$

Using the components of a ray in place of $$x$$, $$y$$, and $$z$$: $\left(\vec{S_x} + t \vec{V_x}\right)^2 + \left(\vec{S_y} + t \vec{V_y}\right)^2 + \left(\vec{S_z} + t \vec{V_z}\right)^2 = r^2$

This equation expands to a quadratic polynomial:

$\left(\vec{V_x}^2 + \vec{V_y}^2 + \vec{V_z}^2\right) t^2 + 2 \left(\vec{S_x}\vec{V_x} + \vec{S_y}\vec{V_y} + \vec{S_z}\vec{V_z}\right) t + \vec{S_x}^2 + \vec{S_y}^2 + \vec{S_z}^2 - r^2 = 0$

This might not look like a quadratic polynomial, but it is! It is in the form:

$a t^2 + bt + c = 0$

Where

• $$a = \vec{V}^2$$
• $$b = 2 \left(\vec{S}\vec{V}\right)$$
• $$c = \vec{S}^2 - r^2$$

Since this is a quadratic polynomial, we can use the discriminant, $$D = b^2 - 4ac$$ to tell us whether the ray intersects the sphere at all.

• If $$D \lt 0$$, it does not intersect the sphere.
• If $$D = 0$$, it is tangential to the sphere.
• If $$D \gt 0$$, it intersects the sphere at two points.

To find the intersection point closer to the ray's origin $$\vec{S}$$, which corresponds to a smaller value of $$t$$, it is given by:

$t = \frac{-b - \sqrt{D}}{2a}$

### Ray vs Cylinder

Imagine a cylinder placed flat and centered on the origin, such that its flat bottom face is at $$z = 0$$.

The equations that define a cylinder is:

$x^2 + m^2 y^2 = r^2$ $0 \ge z \ge h$

Where

• The cylinder's height is $$h$$.
• $$m = \frac{r}{s}$$
• $$r$$ is the radius in the x direction.
• $$s$$ is the radius in the y direction.
• When $$r = s$$, the cylinder is circular and $$m = 1$$.

Taking the above function, substituting the components of the ray, then expanding the exponent terms yields yet another quadratic polynomial:

$\left(\vec{V_x}^2 + m^2 \vec{V_y}^2\right) t^2 + 2 \left(\vec{S_x}\vec{V_x} + m^2 \vec{S_y}\vec{V_y}\right) t + \vec{S_x}^2 + m^2 \vec{S_y}^2 - r^2 = 0$

See the discussion above with calculating this for a sphere. The solution to the above equation yields ray intersections against a cylinder infinite along the z axis. Once you have a point, you need only compare its z-coordinate to be $$0 \ge z \ge h$$.

TBD