Consider a lit candle placed on a cylinder. If the candle is placed at the center of the top surface, let the distance from the origin (center of the surface) to the end of the shadow be $r$. In this case the area of the shadow can easily be calculated by the difference of 2 circular areas ie:- $\pi r^2-$ Area of cylinder's base.

Now suppose we shift the candle from the origin & place it at some point $(x,y)$ on the circular surface, how do we calculate the area of the shadow? The dimensions of the cylinder are known. The length of the candle at the particular instant is known.

Further, what is the equation the shape of the shadow?

Thanks in advance.

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    $\begingroup$ I think this may be interesting. math.utah.edu/~treiberg/Perspect/Perspect.htm $\endgroup$ – Siminore Jul 6 '12 at 9:22
  • $\begingroup$ I think the shape of the shadow doesn't change at all, but I don't have the time to go through a proper working. $\endgroup$ – Ben Millwood Jul 6 '12 at 10:14
  • $\begingroup$ @BenMillwood I'm sure the shape changes but the area might be constant. $\endgroup$ – Green Noob Jul 6 '12 at 10:18
  • $\begingroup$ Why are you sure? It's possible I've misunderstood the question, but the mental picture I have in my head is just a cone of light with a shear (i.e. a linear transformation) applied. $\endgroup$ – Ben Millwood Jul 6 '12 at 10:20

My answer may not be as complete as desired -- so far I've just done the case of the candle lying somewhere above the cylinder, at a height $d$ above the top of the cylinder, with horizontal radial distance $s$ from the center.

Let $h$ be the height of the cylinder, and $r \ge s$ be the radius of the cylinder. Parametrize the circular top of the cylinder with $\theta \in [0,2\pi)$ . Note that the segment connecting the candle's projection onto the cylinder, with a point $\theta$ on the circle, has length $\sqrt{r^2+s^2-2rs\cos\theta}$. Let the length of the shadow on the ground, considered radially from the candle, be $R(\theta)$. Then seeing that the triangles formed by the light ray extending from the candle to the edge of the cylinder's top, and that to the ground, are similar, comparing tangents we find that $$ R(\theta) = \left(1+\frac{h}{d}\right)\sqrt{r^2+s^2-2rs\cos\theta} ~~. $$

To calculate the area of this shadow, we note that the area of a differential sector is $\frac{1}{2}R^2d\theta$ , and compute $$ A_s = \int_0^{2\pi} \frac{1}{2}\left(1+\frac{h}{d}\right)^2(r^2+s^2-2rs\cos\theta)d\theta = \pi(s^2+r^2)\left(1+\frac{h}{d}\right)^2 ~~. $$ Finally, removing the cylinder yields the desired area, $A_s - \pi r^2$.

When $s=0$, we retrieve the correct candle-above-the-center case, $\pi r^2 \left(1+\frac{h}{d}\right)^2 - \pi r^2$.

  • $\begingroup$ I like how your original version managed to combine "candle" and "cylinder" and get "calendar". Linguistics is a funny thing :) $\endgroup$ – Ben Millwood Jul 6 '12 at 10:39
  • $\begingroup$ Haha, you caught that ;D I actually caught myself typing it several times during the original draft. $\endgroup$ – Eugene Shvarts Jul 6 '12 at 10:39


$HJ$ is the candle, $BAIF$ is cylinder (vertical section of cylinder through its axis),$ED$ forms the shadow.

The light from candle to edges of cylinder's surface $FI$ forms an oblique cone which is similar to oblique cone formed from $J$ to the ground $ED$.

It is easy to see that the shadow is circle with center at $C$ and radius $EC$ or $CD$. (which can be calculated by observing that triangle $JFG$ and $ECJ$ are similar)

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    $\begingroup$ Yes! This is essentially the same idea in my answer, but the picture makes it a billion times easier to understand :) $\endgroup$ – Ben Millwood Jul 6 '12 at 11:31
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    $\begingroup$ In this type of question, how can you tell that the shadow is not an oval instead? $\endgroup$ – wjm Jul 6 '12 at 13:11
  • $\begingroup$ It is easy to see that the shadow is circle with center at C and radius EC or CD I'm sorry I do not see how this is the case. As @JosuéMolina mentioned, why can't it be an oval? $\endgroup$ – Green Noob Jul 6 '12 at 13:22
  • $\begingroup$ @Josué_Molina,@Green_Noob If you try to visualize the light from candle to the edges of cylinder's top FI it forms an oblique cone with circle($FGI$ diameter) as the base and J the top of candle as the vertex. And because light rays travel in straight line(in this case) the oblique cone formed on ground is similar to the one formed above which has circle as its base. So, shadow is circular with $ECD$ as diameter. $\endgroup$ – sabertooth Jul 6 '12 at 20:15

I'm not absolutely sure I've understood your problem correctly, but if I'm right then it's equivalent to the following setup:

  • An opaque plane - the ground - stretches infinitely in two directions.
  • A small opaque disc parallel to the ground floats some distance above it.
  • There is light source floating over the disc.

We wish to find the shadow of the disc on the ground.

Essentially, I'm ignoring most of the cylinder because if the light source is directly above the top of it then it can't "see" the sides of the cylinder, so the shadow can't be affected by them.

Now draw a line from the point light source through the centre of the disc, and continue that line downwards until it hits the ground. Let $\gamma:\mathbb R\to\mathbb R^3$ be a constant-speed parametrisation of this line with $\gamma(0)$ the light source and $\gamma(1)$ the centre of the disc. Let $c$ be such that $\gamma(c)$ is the point at which the line reaches the ground. Let $r$ be the radius of the disc.

Now for each $t$ imagine a circle with centre $\gamma(t)$ and radius $rt$. At $t=0$ you have the light source, and at $t = 1$ you have the disc. The circles for all $t>0$ form a cone, the sides are straight, so in particular the circle at $t=c$ is the shadow on the ground. By definition, that's a circle with centre $\gamma(c)$ and radius $rc$.

This argument is simpler than it sounds, but the terminology necessary to describe it clearly is a bit of a nuisance. Leave a comment if it's unclear (or if it's outright wrong; I've not yet checked it's consistent with the other answer posted so far).

  • $\begingroup$ When $t=1$, you'll get a translation of the disk with center beneath the light source, which won't be the disk unless the source is above the disk's center, right? $\endgroup$ – Eugene Shvarts Jul 6 '12 at 10:48
  • $\begingroup$ @EugeneShvarts: no, $\gamma$ doesn't go directly downwards, in fact I defined $\gamma(1)$ to be the centre of the disc. $\endgroup$ – Ben Millwood Jul 6 '12 at 10:56
  • $\begingroup$ Oh I apologize, totally misread it. That said, $1+h/d$ is the factor $c$ for when the light is straight above, so if you can show that $(rc)^2 = (r^2+s^2)(1+h/d)^2$ , mission accomplished. $\endgroup$ – Eugene Shvarts Jul 6 '12 at 10:58

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