Hello Marcelo,
Many on this list empathize with your problem. We know what we want to do
but the math is unfamiliar. In reality the trigonometry here is very simple,
as you have laid out the problem. The ratio of the height of the shadow
caster, G to the shadow length, L is the Tangent of the altitude, H. Tan H
= G/L. Or rearranging L = G/Tan H. The shadow length is equal to the height
of the shadow caster divided by a simple number. the Tangent of the solar
altitude angle H.
This assumes you know the altitude angle. At solar noon when the sun is on
the meridian, this is an easy calculation as the Noon altitude equals the
co-latitude minus the solar declination or H = 90-Lat-Dec.
This assumes you know your latitude and solar declination. Latitude is easy
from maps, websites, GPS etc. Solar declination is not as quite as easy but
many tables, almanacs, programs and websites can give it to you. Google
solar declination.
What if it is not noon? The altitude and azimuth are still relatively easy
to calculate using the classical formulae of spherical trigonometry used by
navigators with sextants. Sin Sin Sin Cos Cos Cos is the first equation to
know. Sin H = Sin Dec x Sin Lat + Cos Dec x Cos Lat x Cos t. Input your
latitude, declination and time as an angle from noon to calculate H, the
altitude angle that determines the shadow length. These intimidating trig
expressions are just numbers, simple numbers that you can add, subtract,
multiply and divide.
But have you considered the Sine effect of the incident light? Light
straight down on a surface such as a flowers leaves is fully effective. As
the angle tilts from straight down to a lower angle, the effective incident
light is diminishes. How much? By the Sine of the altitude. Straight on
the altitude is 90° and Sin 90° = 1. At altitude = 45°, Sin 45 = 0.707, so
the light is 70% as intense. At 30° altitude, the intensity is halved as Sin
30 = 0.5.
Many on this mailing list have found the a little geometry, trigonometry and
even spherical tri can be very useful in solving problems like yours.
Regards, Roger Bailey
--
From: Marcelo mmanil...@gmail.com
Sent: Thursday, January 02, 2014 11:37 AM
To: Sundial List sundial@uni-koeln.de
Subject: Garden planning problem
Hello fellow dialists, how are you?
I'm with a problem here which doesn't concern exactly to sundials, but
since it deals with sun's position and his shadows, I couldn't think
of anyone better than you to help me.
I have a little garden here at home, a walled area where I grow some
plants in pots. I've found that, depending on the place, teher's a
difference greater than 2.5 hours in the sunlight a plant receives,
and that affects greatly its development.
I've measured the shadows cast before and after true noon during
summer solstice (I live slightly south to the Tropic of Capricorn). I
could repeat the process during equinox and winter solstice, but
that's a boring task, and above all, if the weather is cloudy I'll
miss the chance.
So, can you tell me some trigonometrical method for calculating the
shadows, using sun's altitude and azimuth? I couldn't devise one by
myself.
Thanks in advance, and Happy New Year!
Marcial
---
https://lists.uni-koeln.de/mailman/listinfo/sundial
-
No virus found in this message.
Checked by AVG - www.avg.com
Version: 2014.0.4259 / Virus Database: 3658/6969 - Release Date: 01/02/14
-
No virus found in this message.
Checked by AVG - www.avg.com
Version: 2014.0.4259 / Virus Database: 3658/6971 - Release Date: 01/02/14
---
https://lists.uni-koeln.de/mailman/listinfo/sundial
