Shadow cast by a wall; coordinate geometry/ calculus problem

Question

Out for a walk yesterday, I noticed something curious about the shadow cast by a certain wall.

This wall had a height that grew approximately linearly. The base of the wall followed a smooth curve (hugging a bend in the road).

The sun was fairly high in the sky behind the wall, causing the wall to cast a shadow. The extent of this shadow remained at approximately a constant perpendicular distance from the wall.

In some sense, the curve in the wall "cancelled out" the increasing height of the wall.

My question: what curve causes this phenomenon?

I have managed to sketch out the setup in Geogebra, at the following link:

https://ggbm.at/jtrnv9ds

... and here’s a pencil sketch of the setup:

... but I haven't got very far analytically. Can anybody help me? Thanks a lot!

Answer 1

This is a comment but not yet an answer.Trying to find traces of upper and lower ends of a constant ascent spiral wall shadow. Please comment for further 3D geometry understanding, I do not get how constant width comes about.

EDIT1:

Please ignore the above. In the following sketch

Constant normal curves/surfaces are parallel Bertrand curves/surfaces. The mutual common normal (red to black) is of constant length.

Translational or surfaces dragged/extruded parallel to themselves can be defined.

In your case it is a combination of these two types. Ground intersection of canal coincides with Bertrand curve.

There is no unique upper edge to the wall. Any line can be drawn on the canal surface. Wall height can vary arbitrarily with respect to arc so long it is on the canal surface.

Differential equations with two independent constants can be written to describe the situation.

Answer 2

If a wall is plane and of fixed height, the edge of its shadow is straight and parallel to the wall's base.

If the wall is plane but of uniformly increasing height, the edge of the shadow is straight but diverges from the wall's base as the wall's height increases.

What curve in the base of a wall of uniformly increasing height produces a shadow whose edge is a parallel curve, i.e. a curve whose minimum distance from the wall's base is the same at every point? Not the arc of a circle it seems.

In the figure, $BJHKE$ is the wall viewed directly from above, with $BJ$, $JH$, $HK$, $KE$ equal arcs along the wall's base. From its height at $B$, the wall is higher by equal amounts at $J$, $H$, $K$, $E$. Denote the corresponding points at the top of the wall $B'$, $J'$, $H'$, $K'$, $E'$. From its great distance, light from the sun arrives at the wall in parallel rays, casting shadows of $B'$, $J'$, $H'$, $K'$, $E'$ at $C$, $N$, $O$, $P$, $F$ lying on curve $CD$. Since the shadow falls farther out from the base of the wall in proportion to the wall's height at the point of incidence, distances $JN$, $HO$, $KP$, $EF$ increase by equal amounts from $BC$. But in concentric circles, segments of chords intercepted between parallels drawn through points marking equal arcs on one of the circles do not differ by equal amounts.

Though not a formal proof, the following may be useful:

$$JN>BC$$and$$HO>JN$$but$$JN-BC<HO-JN$$ because $J$ rises higher in its run from $B$ than $N$ does in its run from $C$, but $H$ rises even higher from $J$ than $O$ does from $N$, and so on. The slopes where the parallels cross the circle at $J$, $H$, $K$, ... are greater, and increase faster, than the slopes where those parallels cross the arc of a concentric circle which is larger, and hence of smaller curvature, at $N$, $O$, $P$, .... In short, there is increasingly more rise for the run along the arc of the smaller circle than along the arc of the larger, so that $JN$, $HO$, $KP$, ... do not increase by equal amounts for equal arc increases $BJ$, $JH$, $HK$, ....

In the above photo of a handsaw blade clamped to a circular trashcan lid, the sunlight is perpendicular at the narrow end of the blade as in the previous diagram. The edge of the shadow is evidently not the arc of a circle concentric with the can lid; the length of a perpendicular drawn from a point on the shadow's edge to the blade increases for a while with the increasing height and then decreases to nothing.

I conclude that under the given conditions the base of the wall cannot trace a circular arc. The proof sketched above relied on the fact that an arc parallel to a circular arc is itself circular. This is not true for curves generally, however. E.g. a curve parallel to a parabola is not parabolic.

Note too that, if we assume as I have in the figure for simplicity, that the sunlight is perpendicular to the wall at its lowest point, then the base of the wall can bend at most through $90^o$. Shadows of any extension beyond that will eclipse portions of the shadow preceding (or fall to the other side of the wall?), as the wall rises ever higher. So we are looking for a finite portion of a non-circular curve, if it is a closed curve. Then again, from its principal vertex to any point however far out on a hyperbola, the change in direction is less than half the angle of the asymptotes, i.e. less than some acute angle; and from its principal vertex to any point however far out on a parabola, the change in direction is less than, although arbitrarily close to, $90^o$. So if either of these curves is the solution, the shadow edge could be parallel to the wall base for an unlimited arc of the curve.

Finally, it appears that to compensate for the non-uniform width of the shadow evident in the photo above, the curvature of the wall's base must diminish in some way as the wall's height increases, as in OP's sketch and in the figure below.

$EHKG$ is the top of the curved wall, rising linearly from $E$ to $G$, i.e. the height of the wall increases by equal amounts for equal distances along the wall's base $CILSF$. Area $DMFC$ is in shadow. $DOJPNM$ is the curved edge of the shadow. $OC=PI$ are normals from points on the shadow's edge to the base of the wall.

$D$, $J$, $N$, $M$ are the shadows of points $E$, $H$, $K$, $G$. And since the sun's rays $ED$, $HJ$, $KN$, $GM$ are virtually parallel, then right triangles $ECD$, $HIJ$, $KLN$, $GFM$ are similar, and $CD$, $IJ$, $LN$, $FM$ increase linearly with the height of the wall at points $E$, $H$, $K$, $G$, respectively. Therefore, $JI$, $NL$, increase linearly with distances $CI$, $CL$ along the base of the wall. More precisely, supposing $CE=1$, and that $m$ is the constant steepness of the rising curve $EHKG$, then$$HI=1+m\cdot CI$$and$$KL=1+m\cdot CL$$Therefore$$\frac{JI}{NL}=\frac{HI}{KL}=\frac{1+m\cdot CI}{1+m\cdot CL}$$We are looking for a curve whose distance from a parallel curve, measured along parallel intercepts, increases linearly with its arc length.

The saw blade proved well suited to simulate the wall, because of its temper as well as its taper. Since under a given force it bends more at the narrow end than at the broad end, it was possible to verify, as the second photo indicates, that the shadow's width can indeed be more nearly constant when the curvature of the wall diminishes as its height increases.

But even for a given curve, variation or constancy of shadow width seems to depend on the wall's compass-orientation to the sun. To achieve the near-constant shadow width of the second photo I had sunlight striking more directly at the narrow end of the blade than at the broad end, but nowhere perpendicular. But that near-constant width was not seen in other orientations. It is easier to move a saw than a wall, if you have to do it before the sun moves.

The above is of course an incomplete answer to the posted question, and may contain no more than the OP already knows, but perhaps it may be useful to someone.