Rays, Planes and Normals

While the player holds the drop button, Harbour Hopper shows a small target ring where the crate will land: on a deck, on the sloping boat ramp, or in the water. To place it, the game fires an invisible ray straight down from the drone and asks where it first meets a surface. Two dot products and one division answer that.

The rule

Signed distance to a plane: normal dot point, minus offset.

A ray is a start point plus a direction that can be stretched: $\mathbf{p}(t) = \mathbf{o} + t\,\mathbf{u}$ with $t \ge 0$. When $\mathbf{u}$ is a unit vector, $t$ is simply the distance travelled.

A plane — in our side view a line, in 3D a flat sheet — is described by a unit normal $\mathbf{n}$, the arrow sticking straight out of the surface, and an offset $d$. The plane is every point $\mathbf{p}$ with $\mathbf{n} \cdot \mathbf{p} = d$. The water is $\mathbf{n} = (0, 1),\ d = 0$; the deck is $\mathbf{n} = (0, 1),\ d = 2$.

The signed distance of any point from the plane is $s = \mathbf{n} \cdot \mathbf{p} - d$. It is positive on the side the normal points to, zero on the plane and negative behind it. A crate at $(7, -0.4)$ has $s = -0.4$ against the water: it is 0.4 m under.

To hit a plane with a ray, put $\mathbf{p}(t)$ into the plane's rule and solve for $t$:

$t = \dfrac{d - \mathbf{n} \cdot \mathbf{o}}{\mathbf{n} \cdot \mathbf{u}}$

If $\mathbf{n} \cdot \mathbf{u} = 0$, the ray runs parallel to the plane and never crosses it. If $t < 0$, the plane is behind the ray's start.

The shortcut

Two dots and a divide. Compute $\mathbf{n} \cdot \mathbf{o}$ and $\mathbf{n} \cdot \mathbf{u}$; then $t$ = (offset − first dot) ÷ second dot. The first dot hands you the signed distance for free.

The boat ramp slopes down into the water along $0.6x + 0.8y = 6$, so $\mathbf{n} = (0.6, 0.8)$ and $d = 6$. The drone is at $(4, 9)$ and the ray points straight down, $\mathbf{u} = (0, -1)$.

Notice that 4.5 m straight down is more than the 3.6 m signed distance: on a slope, "straight down" is not the shortest way to the surface.

The trap is a normal that isn't unit length. Use $(3, 4)$ with $d = 30$ instead of $(0.6, 0.8)$ with $d = 6$ and it's the same ramp, but every signed distance comes out five times too big.

Drills

Answer fast — trust the rule.

Remember

• A ray is o + t·u; a plane is every point with n · p = d, using a unit normal.
• Signed distance = n · p − d: positive in front, zero on, negative behind.
• Ray hit: t = (d − n · o) ÷ (n · u); a zero divisor means parallel, and a negative t means behind.