Shadow Length Calculation Explained

The length of a shadow cast by an object, like a tower, depends on the angle of elevation of the sun. When the sun is directly overhead (at its highest point), its angle of elevation is the maximum (90 degrees), and the shadow is shortest (ideally zero for a point source, but practically very short). As the sun moves lower in the sky towards sunrise or sunset, the angle of elevation decreases, and the shadow becomes longer.

The relationship between the height of the object (h), the length of the shadow (L), and the sun's angle of elevation ($\alpha$) is given by trigonometry:

$\tan(\alpha) = \frac{\text{height}}{\text{shadow length}} = \frac{h}{L}$

Rearranging this formula to find the shadow length:

$L = \frac{h}{\tan(\alpha)}$

The question states that the sun is overhead at noon at this location. This happens at a place on the Earth where the latitude is equal to the sun's declination angle. A common scenario where the sun is directly overhead (at the zenith, angle 90 degrees) at noon is at the equator during the time of the equinox (around March 20/21 and September 22/23), where both the latitude and the sun's declination are 0 degrees.

Let's assume this scenario: Latitude ($\phi$) = 0°, Sun's Declination ($\delta$) = 0°. The angle of elevation ($\alpha$) of the sun at a given hour angle ($h$) is approximately given by:

$\sin \alpha = \sin \phi \sin \delta + \cos \phi \cos \delta \cos h$

If $\phi = 0$ and $\delta = 0$, then $\sin 0 = 0$ and $\cos 0 = 1$. The formula simplifies to:

$\sin \alpha = 0 \times 0 + 1 \times 1 \times \cos h = \cos h$

The hour angle ($h$) is the angle through which the Earth has rotated since solar noon. It changes by 15 degrees every hour (360 degrees / 24 hours = 15 degrees/hour). Noon is at 12:00 PM, where the hour angle is 0 degrees.

Shadow Length at 10:00 AM

10:00 AM is 2 hours before noon (12:00 PM). So the hour angle is $2 \times 15^\circ = 30^\circ$ (West).

The sun's angle of elevation ($\alpha_{10}$) at 10:00 AM is found using $\sin \alpha_{10} = \cos 30^\circ$:

$\sin \alpha_{10} = \cos 30^\circ = \frac{\sqrt{3}}{2}$

$\alpha_{10} = \arcsin\left(\frac{\sqrt{3}}{2}\right) = 60^\circ$

The shadow length at 10:00 AM ($L_{10}$) is:

$L_{10} = \frac{h}{\tan(\alpha_{10})} = \frac{h}{\tan(60^\circ)} = \frac{h}{\sqrt{3}}$

Shadow Length at 4:00 PM

4:00 PM is 4 hours after noon (12:00 PM). So the hour angle is $4 \times 15^\circ = 60^\circ$ (East).

The sun's angle of elevation ($\alpha_4$) at 4:00 PM is found using $\sin \alpha_4 = \cos 60^\circ$:

$\sin \alpha_4 = \cos 60^\circ = \frac{1}{2}$

$\alpha_4 = \arcsin\left(\frac{1}{2}\right) = 30^\circ$

The shadow length at 4:00 PM ($L_4$) is:

$L_4 = \frac{h}{\tan(\alpha_4)} = \frac{h}{\tan(30^\circ)} = \frac{h}{1/\sqrt{3}} = h\sqrt{3}$

Comparing Shadow Lengths

To compare the shadow at 4:00 PM to the shadow at 10:00 AM, we find the ratio $L_4 / L_{10}$:

$\frac{L_4}{L_{10}} = \frac{h\sqrt{3}}{\frac{h}{\sqrt{3}}} = h\sqrt{3} \times \frac{\sqrt{3}}{h} = \sqrt{3} \times \sqrt{3} = 3$

This shows that the shadow length at 4:00 PM is 3 times the shadow length at 10:00 AM under these conditions.

The comparison confirms that the shadow at 4:00 PM is three times longer than the shadow at 10:00 AM, based on the assumption of being at the equator during an equinox where the sun is overhead at noon.