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
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>> 
>> 
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