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#### custard

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Is it possible, in theory, to get a screen that reflects MORE energy in total compared to the ref?
theoretically a material could reflect all light 100% that is emitted onto it without any absorption.
but the total reflection cannot increase above 100%.

i think the reference comes very close to the ideal as does titanium dioxide that you have mentioned previously.

the total light reflected would be proportional to the area under the gain curves. which in turn is reduced if the material is darker in color (absorbs more light)

#### custard

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robert,
first of all well done with the calculations :clap:

i've given this abit of thought and i think there is abit of a problem with just using the integral to find the area under the curve and then relating that to total light/energy reflected.

when we are talking about total light reflected we need to think in 3 dimensions rather than 2. we cannot simply use the area under the curve.

the area under the curve, or the integral of the gain curve will provide values which are 2-dimensional,
while the total energy reflected should be a 3-dimensional value.

or to put it differently, we are only taking the 'x' and 'y' axis into account and not the 'z' axis.

i'll try and draw a diagram to show why i think that there is alot more light energy at larger angles than is shown on the gain curves.

i will also try and explain this abit further when my brain starts remembering what it was taught all those years ago.:dizzy:

#### custard

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my first thought is that we need to take into account all the angles beginning at on axis at 0 degrees to 90 degrees where no light is reflected.

now the hard part of putting the next thought into words.

as we move from 0 to 90 degrees we need to apply an extra value to multiply to the integral. this may result in an exponential realationship between gain and degrees to take into account that at larger angles more light is being projected.

at 10 degres the circle formed by the reflected light is smaller than the circle formed at say 50 degrees.

e.g.
if gain was 1.5 at 10 degrees
and gain was 1.5 at 50 degrees

then there is still more total light being reflected at 50 degrees as the circle of light is larger at this angle.

i will try and explain this abit better with some diagrams at some point.

the formula may need to apply 'tan' to the degrees values as a starting point. i 'll try and calculate a formula but i may need your help as my maths brain has not worked at this sort of thing since my school days.

#### custard

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hopefully this diagram shows as the angle increase the total light at the mentioned angle increases aswell.
the perimeter of the circles would denote the amount of points where light has reflected to at one angle.
the larger the perimeter the more points are present.

the total reflected light/energy at a particular angle would be the sum of all those points.

so there are more points at 50 degrees than 10 degrees.

the area under the curve only takes into account 2 points for all angles.

#### custard

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Interesting. I use math in my work as Submarine Engineer so I might be able to help you in that area.
wow, that is a job i would find really interesting!

#### custard

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But I am beginning to understand what you are after, and I do believe that you are heading in the correct direction. But I believe that the path you are heading is pointing towards absolute figures. You are correct that if you regard a "cone of light" that is narrow, the "amplitude" i.e. the height of the cone can be very high, even though it actually does not hold as much volume as a more wider but lower cone.

But what I am trying to wrap my brain around is a way to present to the user a set of relative figures that all relates to the famous Magnesium Block. I totally agree that for an absolute quantification of the reflected light you would have to do a so called curve integral. But the normal viewrws do not sit in a circle *wink*. We sit in a row in fron of the screen. And measurements show that there are screens reflecting substantially more light energy "in the relevant line of sight of the observer".
yep, the total light figure is not relevant to the viewers at all.

my only reasoning for looking at these calculations was to show that a screen cannot reflect more than 100% of the light.

#### custard

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I apologize for my poor English if I gave anyone the impression that I was promoting the idea of a screen that contradicts the laws of nature. Custard is of course very correct that a screen can never reflect more than the PJ's output. My figures above that indicate something more than 100% is a relative figure when the SUT (Screen Under Test) is compared to the Magnesium block.
as i have spent over a day trying to calculate and quantify how much light is reflected by MgCo in relation to the dalite hp and other screen samples i would like to show some of my formula.

everyone/anyone - feel free to correct me/help me with this as i probably havn't calculated everything right.

to calculate all gain at one angle

gain=G
angle=A

perimeter of circle = 2 x pi x R

R = sin A x G

where R = radius of circle made by angle A at given gain G

therefore perimeter = 2 x pi x(sin A x G) = total gain at that angle A

to calculate total gain at all angles
total gain at all angles = the sum of 'total gain at angle A' from 0 to 90.

total gain at all angles = the area under curve (INTEGRAL) of x,y graph
where x= angle, y = total gain at angle A

trapeziod rule to calculate area under curve in excel gain spreadsheet.

results abit later:bigsmile:

#### custard

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sorry robert :hide:

to give you an idea mgco3 is coming up tops with the carada for 90 to -90 degrees with an an assumption that the value at 90 degrees is equal to 75 degrees. HP is showing less from what i remember.

i'll stop now. i wont pursue this anymore.

#### custard

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Happy that you enjoy our artithmetics :bigsmile:

Basically, what you could do is 10 deg incremental measurements 0-60 deg if that is enough in your opinion.

The easy way out is to do this for every 10 deg interval:
Gain 0 minus gain 10. This gives us the delta.
Divide delta with 2.
Add the half delta to Gain 10. This gives us the average Gain between 0 and 10.
Multiply this average with 10, as the interval between measurments is 10.

Repeat this between all measurements and incrementally add the results.
Then you get the area under the gain curve which could represent an acumulated persieved light reflection of the screen from 0-60 deg.

Custard, I thought excel had some tools for integrals but I cant find any. Can you suggest better, more accurate ways of estimate the integral of the curve based om measurements from mech?
there are some add ons for excel which caculate the area under the curve but i have not used them.
i have simply created the formula myself for one reading and then applied it to the rest.

with the data mech has provided and because we donot have equal intervals between measurements, the closest approximation that i can think of is that trapezoid rule again.

mech has provided readings at 30 and 60 degrees so calculating the values for total perceived light at these is fairly simple.

the 10 degree intervals are going to need more approximations and more calculations..
mech has provided readings for 15, 20, 30, 45 and 60.
so for the 10 degree intervals we are missing 10, 40 and 50.

if we assume a linear change between each interval:

then for 10 degrees we can approximate total gain as 2/3 of the total gain at 15.
the calculation would be:

[(gain at 0 - gain at 15) x 1/3] + gain value at 15 = gain value at 10 degrees

total gain at 10 = (gain value at 0 + gain value at 10)/2 x 10 (trapezoid rule)

for 40 degrees, total gain can be appproximated as total gain at 30 degrees plus 2/3 of the total gain difference between 30 and 45 degrees.
for 50 degrees, total gain can be approximated as total gain at 45 degrees plus 1/3 of the difference between 45 degrees and 60 degrees.

robert - what set of values are you using to calculate your total gain values for -75 to 75. i think the 250 watt readings are better to use than the 1000 watt readings in this instance.

#### custard

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Cust, any ideas on this? I do think it would be nice to have the 50% in there somehow...but how...:scratch:
if i understand correctly you want the angle when total perceived gain is 50%

we need to calculate the total gain from 0 to 90 so an assumption is needed for the 90 degree value (or should i say 89.99999999).

we can either assume it stays the same as 75 degrees or it follows a linear relationship from 60 degrees to 75 degrees to 90 degrees.

next we need to formulate a table which shows the cumulative totals from 0 degrees to 90 degrees. this would be a simple extension of the area under the curve table we have already.

at 90 degrees we will have total cumulative gain 'T'

----------------------

T= total cumulative gain
H= 50% cumulative gain

therefore H=T/2

looking at the table we can also see which two cumulative figures H falls between, and we will also be able to see what angles these two figures are related to.

A= lower cumulative value
B= higher cumulative value
C= angle at lower cumulative value
D= angle at higher cumulative value
E= angle at H, the value we are seeking

then:

E = [(H-A)/(B-A) x (D-C)] + C

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