Lampshade geometry and moving the source of light

Question

I've been thinking about lampshade geometry and the hyperbolic outline of light formed when the lamp is against the wall. This happens when the source of light is at the centre between the lampshade. However when you change where the source of light is - so move it to the side a little - will that change the shape formed by light on the wall at all? If it does, is there a certain pattern to it?

Answer 1

To expand on Blue’s comments, moving the light source around is tantamount to moving and tilting the plane in which the aperture lies. The resulting cone nappe is right elliptical and is an affine image of the original circular cone. The plane on which the shadow is cast is also moved and tilted relative to the light source. The intersection of the new cone and shadow plane is also a conic of some sort.

Since we’re basically computing a central projection of the aperture curve onto the shadow plane, it seems natural to use projective geometry to formulate the above in a more precise way. The projection can be constructed in stages. Treating the aperture plane as the image plane of a standard pinhole camera at the light source location $\mathbf L$, we can construct the corresponding projection matrix $\mathtt P$. If $\mathtt C$ is the matrix of a conic on this plane, then points $\mathbf X$ in space that project to points on $\mathtt C$ satisfy $(\mathtt P\mathbf X)^T\mathtt C(\mathtt P\mathbf X)=0$, i.e., the back-projection of $\mathtt C$ is the quadric $\mathtt Q = \mathtt P^T\mathtt C\mathtt P$. By construction, $\mathtt P\mathbf L=0$, so also $\mathtt Q\mathbf L = 0$, therefore $\mathtt Q$ is rank-deficient. We can specify a coordinate system for the shadow plane by choosing a full-rank $4\times3$ matrix $\mathtt M$ such that all points on the plane are of the form $\mathtt M\mathbf x$. In this coordinate system, the intersection of $\mathtt Q$ and the shadow plane is $\mathtt C' = \mathtt M^T\mathtt Q\mathtt M = (\mathtt P\mathtt M)^T\mathtt C(\mathtt P\mathtt M).$ To put it another way, the shadow outline is the image of $\mathtt C$ under the homography $\mathtt H = (\mathtt P\mathtt M)^{-1}$.

As a concrete example, let the aperture for the upper light cone be a circle on the $z=h\gt0$ plane, centered on the $z$-axis and with radius $r$. Using the natural coordinate system for this plane, $\mathtt C = \operatorname{diag}(1,1,-r^2)$. If the light source is at $\mathbf L = (x_l,y_l,z_l,1)^T$, the corresponding projection matrix is $$\mathtt P = \begin{bmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\end{bmatrix} \begin{bmatrix} h-z_l & 0 & x_l & -h x_l \\ 0 & h-z_l & y_l & -h y_l \\ 0 & 0 & h & -h z_l \\ 0&0&1&-z_l \end{bmatrix} = \begin{bmatrix} 0 & h-z_l & y_l & -h y_l \\ h-z_l & 0 & x_l & -h x_l \\ 0&0&1&-z_l \end{bmatrix}.$$ (I’ve swapped the $x$- and $y$- axes in the projection so that they’ll back-project to the corresponding axes on the shadow plane.) Taking $x=-d$, $d>r$ as the shadow plane, a natural choice of coordinates for it is $$\mathtt M = \begin{bmatrix}0&0&-d\\1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$ with which $$\mathtt H^{-1} = \begin{bmatrix} h - z_l & y_l & -h y_l \\ 0 & x_l & -h x_l - d(h-z_l) \\ 0 & 1 & -z_l \end{bmatrix}.$$ We also require $z_l\lt h$ so that this light cone nappe opens upward and $x_l\gt-r$ so that this nappe intersects the shadow plane.

As a sanity check, when $\mathbf L$ is the origin, we obtain $\mathtt Q = \operatorname{diag}(h^2,h^2,-r^2,0)$, a circular cone, and the corresponding shadow edge is $\mathtt C' = \operatorname{diag}(h^2,-r^2,h^2d^2)$, which is the expected hyperbola opening upward. With $h=r=1$, this is a right circular cone and the shadow boundary is a right hyperbola.

The general form of $\mathtt Q$ that results is a bit messy, so I’ll omit it here, but its principal $3\times3$ minor is equal to $-r^2(h-z_l)^4$, which is negative, so this is an elliptic cone with apex $\mathbf L$. The type of conic that forms the shadow boundary can be determined from $\mathtt H$: with a little bit of work one can find that it maps the line $x=x_l$ to the line at infinity. This line bounds the rays that intersect the shadow plane: the plane through this line and $\mathbf L$ is parallel to the shadow plane. When $-r\lt x_l\lt r$, this line intersects the aperture circle $x^2+y^2=r^2$ in two points and the image of the aperture is a hyperbola; when $x_l=r$ it is tangent to the aperture and the image is a parabola; and when $x_l\gt r$ it doesn’t intersect the circle at all and the image is an ellipse.

With the above basics you can explore this model further. For instance, it might be interesting to compute the homography that relates the “standard” hyperbolic shadow with the light source at the origin to the shadow produced by the displaced light source. This homography in a sense captures the effects on the shadow of moving the light source around.