Selected Questions -Answers From All Experts Astronomy Forum (Shadow Lengths, Angles And Times)

Question:  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?

Answer:

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 maximum 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 shadow
measured.

By example, say H = 0.5 meters, and L(m) 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 LAT = tan ^-1(2) (e.g. LAT = 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 = -23.5 degrees.

Then the latitude of the observer would be:

LAT  =  90 + ( d  - 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 corresponding 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 mean time)