# [SOLVED] calculating the length of a shadow

• May 10th 2006, 09:06 AM
romed
[SOLVED] calculating the length of a shadow
How does one calculate the lenth of a shadow at any given time
lets say my latitute is 36 degrees.

And the height of the object is 1.2 meteres

also how do i claculate when the length of the shadow is the same as the height thats is 1.2 meters

Thanks Rod
• May 10th 2006, 11:23 AM
JakeD
A real calculation of the length of shadow is non-trivial and is in the domain of celestial navigation. See here for calculators for educational purposes.
• May 10th 2006, 11:51 AM
earboth
Quote:

Originally Posted by romed
How does one calculate the lenth of a shadow at any given time
lets say my latitute is 36 degrees.
And the height of the object is 1.2 meteres
also how do i claculate when the length of the shadow is the same as the height thats is 1.2 meters
Thanks Rod

Hello,

I can give you only a few informations:

1. When the sun is perpendicular over the tropic it has it's maximum altitude over the horizon at 12 o'clock local time . At a latitude of 36°N the maximum altitude of the sun is approximately 77.5° over the horizon. Then you get the shortest shadow:
$s=1.2 m \cdot \cot(77.5^\circ) \approx 26.6 cm$

The longest shadow you get when the sun touches the horizon: The shadow has an infinite length.

2. When the shadow is as long as the stick itself, then the sun has a altitude of 45° over the horizon. This situation will happen twice a day in the summer.
In winter the maximum altitude of the sun reaches only approximately 30.5° (at 36°N, of course!). That means the shortest shadow will have a length of

$s=1.2 m \cdot \cot(30.5^\circ) \approx 2.04 m$

I hope these informations are of some use to you.

Greetings

EB
• May 10th 2006, 12:10 PM
earboth
Quote:

Originally Posted by romed
How does one calculate the lenth of a shadow at any given time
lets say my latitute is 36 degrees....

Thanks Rod

Hello,

it's me again.

I've attached a diagram, to demonstrate where all my numbers came from.

Greetings

EB
• May 10th 2006, 01:03 PM
JakeD
Quote:

Originally Posted by earboth
Hello,

I can give you only a few informations:

1. When the sun is perpendicular over the tropic it has it's maximum altitude over the horizon at 12 o'clock local time . At a latitude of 36°N the maximum altitude of the sun is approximately 81° over the horizon. Then you get the shortest shadow:
$s=1.2 m \cdot \cot(81^\circ) \approx 19 cm$

The longest shadow you get when the sun touches the horizon: The shadow has an infinite length.

2. When the shadow is as long as the stick itself, then the sun has a altitude of 45° over the horizon. This situation will happen twice a day in the summer.
In winter the maximum altitude of the sun reaches only approximately 27° (at 36°N, of course!). That means the shortest shadow will have a length of

$s=1.2 m \cdot \cot(27^\circ) \approx 2.36 m$

I hope these informations are of some use to you.

Greetings

EB

Hi. According to this page from NASA, the maximum and minimum altitudes of the sun at latitude $L$ are given by $90 - L \pm 23.5,$ which for $L = 36$ yields 77.5 and 30.5. How did you calculate your maximum and minimum?
• May 10th 2006, 09:48 PM
earboth
Quote:

Originally Posted by JakeD
Hi. According to this page from NASA, the maximum and minimum altitudes of the sun at latitude $L$ are given by $90 - L \pm 23.5,$ which for $L = 36$ yields 77.5 and 30.5. How did you calculate your maximum and minimum?

Hello,

you are right. Instead of 23.5° I took 27°. I can only guess why I use this number: The polar circle has a latitude of nearly 67° (and I'm living nearer to the polar circle than to the 36°N regions) and so I mixed up those ciphers. So sorry!

Greetings

EB