Hi Dave,
My initial recollection was of a cosine effect. So I drew a little sketch to
clarify the situation. I specified the angle as the elevation, the altitude
measured from the horizontal. It would be the cosine for the angle measured
from the zenith. The area of the direct projection is Pi R squared. The area
of the ellipse on the projected on the ground is Pi R times the semi-major
axis. This is R/ Sin Alt so the area is Pi R squared/Sin Alt. I considered
the illumination inversely proportional to the area so directly related to
the sine of the elevation, the altitude angle. QED. or is my logic wrong?
At sunrise the elevation is zero, sine =0, the intensity is zero. Directly
above the elevation is 90° and Sine 90 = 1. The intensity is full,
undiminished by spreading over a larger area. Radiation from the sun follows
the inverse square law. Twice the distance from the sun gets 1/4 the
intensity but that is not the effect being discussed in this "gray posting".
Regards, Roger
--------------------------------------------------
From: "David Bell" <[email protected]>
Sent: Friday, January 03, 2014 2:06 PM
To: "Roger Bailey" <[email protected]>
Cc: "Marcelo" <[email protected]>; "Sundial List" <[email protected]>
Subject: Re: Garden planning problem
One thought on that gray posting, Roger:
I may remember incorrectly, but I thought illuminance on a surface was
proportional to the square of the cosine of the incidence angle.
Dave
Sent from my iPhone
On Jan 2, 2014, at 8:07 PM, "Roger Bailey" <[email protected]> wrote:
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" <[email protected]>
Sent: Thursday, January 02, 2014 11:37 AM
To: "Sundial List" <[email protected]>
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
-----
No virus found in this message.
Checked by AVG - www.avg.com
Version: 2014.0.4259 / Virus Database: 3658/6973 - Release Date: 01/03/14
-----
No virus found in this message.
Checked by AVG - www.avg.com
Version: 2014.0.4259 / Virus Database: 3658/6974 - Release Date: 01/03/14
---------------------------------------------------
https://lists.uni-koeln.de/mailman/listinfo/sundial
