Expert: Philip Stahl - 12/1/2005

Question
Dear sir

I would like to know at what time/sun angle over a given place on earth does the shadow of an object becomes the same length of it? How can I calculate that? Thank you.

Answer
Hello, and please note the *correction of notation* (LAT to ALT)  note - for my previous reply.


A very basic relation (along with some simple observations) allows one to make the calculation you reference, and also determine one's latitude.

That is:

 tan (ALT)  =  H/ L_m

where 'ALT' is the max. altitude of the Sun at the location of the observer, H is the height of a vertical object (assumed to be planted on flat ground), and L_m is the (minimum) length of the object shadow measured.

By example, say H = 0.5 meters, and L is found to be 0.25 meters (which would occur at or near astronomical NOON or when the Sun is on the observer's meridian), then:

tan (ALT) = 0.5/ 0.25 =  2

and ALT = tan ^-1(2) (e.g. ALT = arc tan (2) = 63.4 degrees

[Side note: we can find the observer's LATITUDE from this IF we know the Sun's declination on the given day. For example, if we are talking about the day of the winter solstice (Dec. 23), then the Sun's declination angle D(S) = -23.5 degrees.

Then the latitude of the observer would be:

LAT  =  90 + (D(S) - ALT)  = 90 + (-23.5 - 63.4)= 3.1 degrees

That is, the observer is at 3.1 deg North latitude or just above the equator.

Now, when the shadow of the object *equals* its height at the (same) observer's location, one will have:

tan (X) =  0.5/ 0.5 =  1

and X = arc tan (1)=  45 degrees

The difference in angles (ALT - X) =  

63.4  - 45 =   18.4 degrees

And this will give a rough idea of the time at the location, say if one divides by 15 degrees (the number of degrees correponding to one hour of time, e.g. the Earth rotates through 15 degrees in one hour).

For this case:  18.4/ 15  = 1.227 or about 1 hour and 14 minutes past noon. (e.g. 1:14 p.m. local time)

Hope this info proves useful.