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Physics help (again) please - energy given up at reflection - Page 3

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Posted by Morris Dovey on June 20, 2009, 4:56 am
Morris Dovey wrote:

Going with the 1/9 absorption of the incident energy at each reflection,
the numbers look fairly good. The first column is the reflection number
and the others are percentages of the original input energy...

R#   Energy    Absorbed  Reflected
-- ---------- ---------- ----------
  1 11.1111111 11.1111111 88.8888889
  2  9.8765432 20.9876543 79.0123457
  3  8.7791495 29.7668038 70.2331962
  4  7.8036885 37.5704923 62.4295077
  5  6.9366120 44.5071043 55.4928957
  6  6.1658773 50.6729816 49.3270184
  7  5.4807798 56.1537614 43.8462386
  8  4.8718043 61.0255657 38.9744343
  9  4.3304927 65.3560584 34.6439416
10  3.8493268 69.2053852 30.7946148
11  3.4216239 72.6270091 27.3729909
12  3.0414434 75.6684525 24.3315475
13  2.7035053 78.3719578 21.6280422
14  2.4031158 80.7750736 19.2249264
15  2.1361029 82.9111765 17.0888235
16  1.8987582 84.8099347 15.1900653

At that rate, after just 12 reflections a little more than 3/4 of the
input would have been absorbed - and that's even without any selective
coatings (such as anodization, plating, etc).

Here's the code in case anyone would like to play "what if":

#include <stdio.h>
#include <stdlib.h>

int main(int argc,char **argv)
{  unsigned i;
    double fraction        = 1.0 / 9.0;
    double absorbed_energy = 0;
    double impact_energy   = 0;
    double input_energy    = 100;

    if (argc > 1) fraction = atof(*++argv) / 100;
    printf("\nR#   Energy    Absorbed  Reflected\n"
           "-- ---------- ---------- ----------\n");
    for (i=0; i<16; i++)
    {  impact_energy = input_energy * fraction;
       absorbed_energy += impact_energy;
       input_energy -= impact_energy;
       printf("%2u %10.7f %10.7f %10.7f\n",
    return 0;

Morris Dovey
DeSoto Solar
DeSoto, Iowa USA

Posted by daestrom on June 20, 2009, 6:36 pm


I didn't mean to say that 1/9 of all light is absorbed each reflection, my
numbers above were just to illustrate the point that the reduced energy is
from fewer photons, not lower frequency.

I guess I don't understand your use of reflective collector.  In most
reflector designs, it is best to have as little absorption as possible so
that as much energy as possible is reflected from the collector surface and
sent on to the target.

Your design seems to be *wanting* to absorb as much of the energy as
possible?  So you are using the collector surface to absorb the energy and
have some sort of heat collector behind the surface?


Posted by Morris Dovey on June 20, 2009, 6:56 pm
 daestrom wrote:

Your example was actually 1/10 - and I suspected that you'd pulled a
number out of the air, which was fine because the idea you were trying
to get across was clearly illustrated.

What I'm after is the ability to make an accurate prediction of the
actual fraction(s), which is pretty difficult right now because I
haven't identified all the variables - nor their relationships.

I want the reflector to do its normal job, what I'm messing with is the
surface of the /target/.

Morris Dovey
DeSoto Solar
DeSoto, Iowa USA

Posted by Jean Marc on June 26, 2009, 12:40 pm

Sun light is polarised in every direction. Or see it as a couple of
perpendicular vectors. By applying a polarising filter, you filter out the
"perpendicular to" component of the light. The result is one of the 2

Then every reflection somewhat polarises light, depending on the reflection
"old time" polariser consisted of several angled pieces of glass in an open
wood box. Each piece of glass would transmit only xx% of one pol dir, and
almost 100% of the perpend one.
Mr Land system consists of very tiny lines somewhat printed on the lens.

Hope it is clear enough. There must be some info on Wikipedia.


Posted by Jean Marc on June 26, 2009, 12:41 pm
"Jean Marc" <jean-marc.brun> a écrit dans le message de news:

Should be "somehow"

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