you can start, i give you the word

Yes we have been fooling you all for many MONTHS

We are a psychotic bipolar with multiple personalties.

ahahaha

I didn´t know that funny facet of scuba we have discussed several times but become friends finally

well, I will start with this last discussion:

I hold that introducing a 17" where there was a 15" before does inmediatelly increase the projector output in term of lumens.
I call it the 17" projector is BRGIGHTER than the 15".

please scuba, tell us what is your initial position to that discussion before we can start arguing...

this is funny

So do we call you scubarox?

DJ

I dont have time to make a long scientific post the defines the lunar and solar alignment with a galaxy 2trillion light years away, so I will start with a simple question.

In your calculations in the 1000W thread you said

120,000 LUMENS = 120,000 LUX

You then factored some loss that took it down to 108kLux (not important right now)

then you take 108kLux x .069 (Meters^2)

You took that value, ran it throught the LCD, and called that the LUMENS OUTPUT.

My question is.....

What if the surface area in Meters^2 was greater than 1.0

Example:

108kLux x 1.2 (Meters^2) = 129.6kLux

That is more light than what you started with, HOW can you gain light? Does it become somesort of LIGHT AMPLIFIER?

Please answer this question, and I will then respond to one of yours when I get home tonight!

EDIT

I dont think we can actively TRANSPOSE

lm/m^2 with lm

There may be 100 lumens within 5 square meters, but it will be LESS intense than 100 lumans spread over 2 square meters.

So while a 17" panel may have the light spread over a larger area, it will be LESS INTENSE.

EDIT #2

a 15" LCD (E-machines for example) that has 250cd/m^2

a 17" LCD (E-machine from the same line) has 250cd/m^2

that is candels per meter^2

In order for them to achieve that SAME light, over the LARGER 17" panel, they have a bigger BACKLIGHT.

If they left the same backlight in, then the cd/m^2 number would be lower.

If you put a 40,000 Luman lamp behind a 15" LCD (Let just say we collect all 40,000 lumans)

Then the same 40,000 lumans will be spread across the 17' LCD. Since this is a larger area, the light will be more spread out, and thus LESS INTENSE, and LESS concentrated.

A PJ that produces a 100" screen at 100 ANSI lumens, will be LESS INTENSE than a PJ that produces 200 ANSI lumens at 100"

Is it not the LIGHT INTENSITY that we care about, Dark scenes being dim because the light is less intense

mmm, if the square area is more than 1 meter square... then no problem, just multiply it...

lux*meters square=lumens.

example;
you know there is a known trees density of 100 trees/mile^2
if there is 0.5mile^2 there must be 50 trees.

now if there is 2.7 mile^2 then there are 270 trees.

no creation of trees by GOD or any kind of miracle...

So in my example above

108,000 Lumens = 108,000 LUX

108,000 Lux x 1.2m^2 = 129,600 Lumens

129,600 Lumens = 129,600 LUX

The math doesnt work here. According to what you are saying

a 108,000 Lumen light shinning on a surface area of 1.2m^2 will have 129,600 Lumens to it.

Something is mathematically wrong! Where are the extra 20,000 Lumens coming from?

1

hold on, we are going to fast...

let's keep it simple.

We have a 17" projector and we have that powerstrip program for streching the image to exaclty a 15" size.

The 15" size will have an outer black square up to 17"... this ASUMING a costant lux (this is not true but think of the acerage lux value) at the lcd will directly mean that the AREAS relation is the lumens relation on the two setups.

The 17" is 30% larger area than 15", so ideally it is 30% more lumens there...

I think more lumens projector=brighter projector final conclusion is evident.

how could X lumens equal to Y lux?

unit mismatching there... where did i confuse the units?

Yes that is correct, because the FRESNEL will ALSO be 17" therefor WASTED light will go into the DARK area.

However, if we have a 15" fresnel that fills up the 15" LCD then all the collected light will go through the image.

It would be an UNFAIR comparison to say

17" fresnel with 17" LCD > 17" fresnel with 15" LCD

a 15" fresnel at proper FL should collect about the same amount of light as a 17" fresnel at the proper FL, correct?

If its just a lamp shining (No fresnel lens) then its common sense that the larger object will collect more light, because its larger

But since BOTH fresnels irregardless of size, focus on a POINT source, then they should collect approximatly the same amount of light.

lets pretend we have a perfect point source of light.

Would a 5" fresnel collect just as much light as a 15" fresnel, since both fresnels are focused on the SAME point source of light? (speaking in theoretical terms, and assuming fresnels are at proper FL)

mmmm, aren´t we just saiyng the improveents introduced by 17" panel while manteining the rest of the setup?

this includes same lamp, same reflector, same rear fresnell, same field frwnslel, same triplet....

from the 1000W thread



you converted the 120,000 lumen bulb to be 108Klux because of the distance to LCD

correct? You say because of average Candelas Value

If we took this 108Klux * 1.2 (Meters^2)

then we would have 129.6Klumens

This is more light than we started with.

You say that:

Klux * meters^2 = Lumens

My problem is that if you simple plug in 1.2m^2 instead of .069m^2 into your calculations above, you end up with more light than you started with.

ANY number greater than 1.0 is an AMPLIFICATION FACTOR.

If I start off with $10 * 1.2, then I have $12

How are you ending up with MORE LIGHT THAN YOU STARTED WITH?

No we arent assuming SAME FRESNEL.

That would be UNFAIR

someone such as myself using a 15" LCD is more than likely using the Standard Lenses

While someone using a 17" LCD is has to using LARGER fresnel lenses

I agree if everything is the same, then the 17" LCD would be brighter. But it would just be stupid to use 17" fresnels with a 15" LCD and not cut them down. Youd be wasting alot of collected light!

Please tell me our whole debate hasnt been over the ASSUMPTION of using fresnels that are too large for your LCD.

yes, I hold there are 108klux at 33cm from the lamp.

but your LCD are of 1.2m^2 is ULTRAEXAGERATED.

this area is more than the area of the full shperical surphace at 33cm radius.

I said the flat to spherical error was my only mistake, but at 0.09m^2 value, this error is not so important yet.

hold on when i mean same rear fresnell, i mean same focal fresnell, obiously you can cut down the fresnell for a 15" smaller box


edit;

"Please tell me our whole debate hasnt been over the ASSUMPTION of using fresnels that are too large for your LCD"

mmm, what does it matter a 15" proector with a larger fresnell (but same focal) is there any diference on the image?

well, felling aslep here

17" should be brighter ideally... don´t know what more to add

Mind if I jump in for a second so you can stop talking with yourself scubarox.

If we are using the same lamp, the same fresnel (large enough for a 17 inch panel), the same front fresnal, the same triplet and the same distance from the projector to the screen then my guess would be that the projector using the 15" panel would have more intense lighting.
My reasoning is that when you move from a 15" panel to a 17" panel your projection grows in size linearly to the increased panel size, but the increased light collected by the rear fresnal does not increase linearly, the distance from the lamp to the corners is now increased, and the energy of the light hitting the fresnal decreases exponientaly with the distance from the lamp.

Now, if you took the projector with the 17" panel, and moved it closer so the screen size is the same as the previous 15" projecter screen size, then the lighting would be more intense, because you are collecting more light, and projecting it onto the same area.

Is this what you guys are talking about?

I think scubarox has exhausted himself and gone to bed.
I can see that both of you are arguing about two different things and each separate augment won’t agree with the other. So I’m going to try and moderate this in the true sense of the word. I’ll define two questions and I need both ofyou, hmm this is a bit confusing do I use both or just you? , to answer yes or no.
I’ll define the model first.

Apparatus:
A point source that outputs 20,000 lumens.
An optical devise that enables the collection of 100% of the lumens and redirect it at a desired angle.
15” fresnel with a 220mm FL.
15” LCD white screen.
17” fresnel with 220mm FL.
17” LCD white screen.
A 17” collector fresnel and matching triplet (the fresnel has an aperture that can be stopped it down to 15”, FLs are irrelevant).

Method:
Test1 Place the point source at 220mm from the 15” fresnel.
Place the stopped down collector fresnel in front of the condenser fresnel.
Place the 15” LCD in front of the collector fresnel.
Adjust the projector and triplet position so that a screen image of 100” is created.
Now measure the ANSI lumens or average lumens of this image and call this value 15LumenResult.

Test2
Place the point source at 220mm from the 17” fresnel.
Place the collector fresnel in front of the condenser fresnel.
Place the 17” LCD in front of the collector fresnel
Adjust the projector and triplet position so that a screen image of 100” is created.
Now measure the ANSI lumens or average lumens of this image and call this value 17LumenResult

Test3
Place the point source at 220mm from the 15” fresnel.
Apply the optical device to the point source and adjust its angle so it illuminates the 15” fresnel.
Place the stopped down collector fresnel in front of the condenser fresnel.
Place the 15” LCD in front of the collector fresnel.
Adjust the projector and triplet position so that a screen image of 100” is created.
Now measure the ANSI lumens or average lumens of this image and call this value Compensated15LumenResult.

Test4
Place the point source at 220mm from the 17” fresnel.
Apply the optical device to the point source and adjust its angle so it illuminates the 17” fresnel.
Place the stopped down collector fresnel in front of the condenser fresnel.
Place the 17” LCD in front of the collector fresnel.
Adjust the projector and triplet position so that a screen image of 100” is created.
Now measure the ANSI lumens or average lumens of this image and call this value Compensated17LumenResult.


Question 1: Will the value of 17LumenResult be greater than 15LumenResult.

Question 2: Will the value of Compensated17LumenResult be equal to Compensated15LumenResult.

Now remember just a yes or no answer from both of you… I mean you… I mean… you know what I mean.
If you can’t answer yes or no for some reason then ask me to clarify. I can’t stand you fighting with yourself, it just breaks my heart.

DJ

There is a lot of strangeness here. I blame math .

2 of the things that increase projected brightness (that I am willing on defending ):

1. Increase the average wave amplitude of light that impacts the effective area of the first fresnel.

The average amplitude does not change with a change in fresnel size, so that (#1) is out.

2. Impact the effective area of the first fresnel with more (tranverse) waves of light.

If the waves can be pre-focused such that the same number of waves impact any sized fresnel then #2 will not have applied as well.


It seems when fresnel size does matter is when a fresnel gets so small as to no longer collect as much light as a larger size (#2). Seems this is the case with all setups without pre-focused light.

For example: as I see it increasing the fresnel size (and panel to go with) in the standard LL setup will increase brightness. How am I wrong?Yes.No. (Were you trying to make that so sneaky?).

I think he's looking for yes, yes

There are no trick questions. I’ve redefined the questions to make it clearer.
Question 1: Will the value of 17LumenResult be greater than 15LumenResult by more than 5%?.


Question 2: Will the value of Compensated17LumenResult be more than 5% different than Compensated15LumenResult?

Yes.
Yes.

And as you can see by my answer I think that last one is a sneaky question. To be less sneaky you could have left the stopped down part out of Test4. To add to the sneakiness your original question switched from asking is greater than to is equal to. And now the new question goes from 5% more to 5% different. GAH! . Then to make things sneakier still you added the 5% bit so even if you hadn't stopped down Test4 I would need to have worked out the exact losses to be sure. I have just taken my chances with this answer but it may be a change of less than 5% (but surely greater than 1%) as I have not checked. And then to really finish me off you said this:Sneaky (maybe even tricky), no ?

Mark.

Question 1: Will the value of 17LumenResult be greater than 15LumenResult by more than 5%?.
NO

Question 2: Will the value of Compensated17LumenResult be more than 5% different than Compensated15LumenResult?
NO


The question is sneaky though, as the typical diy projector does not use "An optical devise that enables the collection of 100% of the lumens and redirect it at a desired angle"

As I read it question #1 (Test1 and Test2) is the case of a standard LL projector and question #2 (Test3 and Test4) is if you could apply all of the light from the bulb to iether sized fresnel. If this is not the case and I've read this wrong then all bets are off.

The major tricky part as I read it is that Test4 masks the fresnel edges by being stopped down, even though all of the light has been focused to a 17" size. That means that some light will be blocked by the mask. I don't think that DAZZZLA meant to do this, but my answers reflect that he did.

Put differently here is question 1 and 2 without the tests as I read it:

Question 1: In the standard LL setup, will using larger fresnels and a panel to match result in a brighter image?

My answer to this is: Yes.

Question 2: In the standard LL setup, if you always focused all of the light from the lamp onto any sized fresnel, would you achieve a different brightness with larger fresnels and a panel to match?

My answer to this first part is: No.

How about if you masked off the edges of the first fresnel but did not adjust the cone of light to match (such that some light hits the mask)? The second part here is the question as I read it.

My answer to this second part is: Yes.

I have got to get some work done .

Doh, I didn’t mean to have test4 stopped down, that’ll teach to use cut and paste and not read it. Good catch, I’ve edited it here.



Apparatus:
A point source that outputs 20,000 lumens.
An optical devise that enables the collection of 100% of the lumens and redirect it at a desired angle.
15” fresnel with a 220mm FL.
15” LCD white screen.
17” fresnel with 220mm FL.
17” LCD white screen.
A 17” collector fresnel and matching triplet (the fresnel has an aperture that can be stopped it down to 15”, FLs are irrelevant).

Method:
Test1
Place the point source at 220mm from the 15” fresnel.
Place the stopped down collector fresnel in front of the condenser fresnel.
Place the 15” LCD in front of the collector fresnel.
Adjust the projector and triplet position so that a screen image of 100” is created.
Now measure the ANSI lumens or average lumens of this image and call this value 15LumenResult.

Test2
Place the point source at 220mm from the 17” fresnel.
Place the un-stopped down collector fresnel in front of the condenser fresnel.
Place the 17” LCD in front of the collector fresnel
Adjust the projector and triplet position so that a screen image of 100” is created.
Now measure the ANSI lumens or average lumens of this image and call this value 17LumenResult

Test3
Place the point source at 220mm from the 15” fresnel.
Apply the optical device to the point source and adjust its angle so it illuminates the 15” fresnel.
Place the stopped down collector fresnel in front of the condenser fresnel.
Place the 15” LCD in front of the collector fresnel.
Adjust the projector and triplet position so that a screen image of 100” is created.
Now measure the ANSI lumens or average lumens of this image and call this value Compensated15LumenResult.

Test4
Place the point source at 220mm from the 17” fresnel.
Apply the optical device to the point source and adjust its angle so it illuminates the 17” fresnel.
Place the un-stopped down collector fresnel in front of the condenser fresnel.
Place the 17” LCD in front of the collector fresnel.
Adjust the projector and triplet position so that a screen image of 100” is created.
Now measure the ANSI lumens or average lumens of this image and call this value Compensated17LumenResult.


Question 1: Will the value of 17LumenResult be greater than 15LumenResult by more than 5%?.


Question 2: Will the value of Compensated17LumenResult be more than 5% different than Compensated15LumenResult?

hi people,

I see new people here; you are welcome

Answer; question 1)YES question 2)YES

I see how you noticed scuba's arguing direction and mine... so you cleverly asked about those two questions

Now my question;

projector A is 1000 lumens
projector B is 1200 lumens

is B projector brighter than A?

I see the questions have changed... Dynamic answers are needed to address dynamic questions

Question 1: YES... by about 28% (theoretically)
Question 2: NO

Absolutley.... as long as A and B project images of equal size.

I havnt had the time to draw any pictures, or make long post here.

If larger images are brighter with the same amount of light behind them, then why do they become dimmer?

This is not a question of light shining on a larger ojbect.

This is a question of FRESNEL THEORY, and LIGHT COLLECTION

If we were talking about a 17" peice of wood, versus a 15" peice of wood, then I would say Yes, the 17" peice of wood would collect ~30% more light. Thats common sense.

However, A fresnel does not have light just shined on it.

The fresnel lense will collect light, and if it collects light from a FOCAL point source (arc of bulb), then how could it collect anymore light?

Does a larger fresnel focus to a larger point?

Another illustration, some with a PJ and a lightmeter. Sadly I dont have a lightmeter or I would do this

Place Lightmeter dead center of screen. (4:3 ratio)

Display a pure white image in a 16x9 aspect ratio. Collect reading. Collect reading only from the center of the screen. (black bars top and bottom)

Now, without changing anything at all, display a white image that fills the whole screen. Keep the light meter at the center ONLY. Would the reading change? I would believe not. How could light intensity (WHATS IMPORTANT TO US) change?

According to what is being said here, the 4:3 image will magically increase light intensity in the middle somehow, because its a bigger image.

And NO ONE HERE has still answered my question on dealing with an area greater than 1meter^2. How can you have a light amplifier?

That’s not a yes or no answer.

edit:
Don’t worry scuba I’ll ask another question to satisfy your augments

GOOD!! Now the exact same logic applies to the fresnels. If both fresnels have the same focal length, and you place the same lamp at the focal point of each one, the 17" will collect a bigger angle - There for it will collect MORE LIGHT.


As for intensity... Well lets define intensity (NOT to be confused with lux - which is the measurement unit for Elluminance - exact unit is lumens/m^2)

Intensity = Flux / Solid angle = Lumens / Steradian... also known as candela.
It is a function of collected flux and spread angle. It's used to describe the the amount of light in a given direction.
Eg. A single 5mm LED may have a bigger intensity than a MH lamp within a 5 degree angle of spread - but that does not mean the LED outpus more light, it only means that in a particular direction of 5 degrees spead and ONLY within this the 5 degrees spread the LED is brigher.

Have a look at the setups below. Assuming all things equal exept the fresnels and LCD size... Can you see that setup A will have more lumens at the screen than setup B ?? Intensity will be the same for both setups because the ratio of lumens/steradin is the same. But Elluminance is higher for setup A than setup B - This is because both setups have the same screen size but setup A is throwing more lumens at the screen - therefore you will measure more lumens/m^2 for setup A.

In this case A will appear brigher than B. This was accomplished because a larger fresnel was used in setup A, which allowed you to capture more light at the very start of the optical system.

Click to view attachment Click to view attachment

[EDIT] ... Had to fix incorrect terminology.

Ok, I still believe there is a cross referncing of terms, And I admit part of it is on my part.

When I say the word Intensity, im refering to the physical intensity of the light on the screen. I may have mis-used the word scientifically, but my intent was a more common base expression of the word

for example,

a dark scene on the PJ is dim, a scene with an explosion or something is brighter. The light is more intense in this scene. Its brighter to the human eye. ect.

Something that is larger, will have to collect more light to have the same brightness.

Hypothetical numbers here,

A 5" LCD may have a 25W backlight behind it to achieve 250Cd/m^2

But a 15" LCD may have to have a 100W backlight behind it to achieve 250Cd/m^2

Granted the 15" LCD very well will capture more light, it has to capture more to break even.

If you were to take the 15" LCD, and place the 25W backlight behind it, then it would no longer be 250cd/m^2

So in order for a PJ with a 17" LCD to actually produce a %30 brighter picture than a 15", it would have to collect MORE than 30% more light

Lets say both the 15" LCD and 17" LCD collected the SAME LIGHT. Lets say they both collect 20,000 of the 40,000 lumens.

Then 20,000 lumens spread over 15" would be BRIGHTER than 20,000 lumens spread over 17" Simple physical concept of dilution, Equal matter is more dilluted over a larger surface area

Now, realistically due to the angle, the 17" will capture more of the light, but it has too, in order to just MAINTAIN the brightness the 15" had.

30% is a huge gain!

One lux is equal to one lumen per square metre.

so

100 Lux = 100lm/m^2

100 Lux = 500lm/5m^2

the 5meter area wouldnt be brighter

If you have an object that is illuminated 10lm/1m^2, and another oject that is 100lm/10m^2 then they would appear equally bright, would they not? In both cases you have 10 lumens of light per 1 meter of area.

Of course in one case, you have more ENERGY and have lit up a larger area, but that does not mean it is brighter.

100lm/.069m^2 = 130lm/.09m^2

meaning that for a 17" panel it would take 130 lumens behind it to light the panel with the same intensity that 100 lumens would light a 15" panel.

Its just like a factory warehouse. Put a 1500W bulb in your living room and its gonna make your paint peel, but put the same 1500W bulb in a large warehouse and it will barely be noticeable. The warehouse wouldnt be brighter because it has more surface area, it would be DIMMER. This is all I have been saying the WHOLE time.

Focus actually has little to do with collection. All focus tells us is to what angle the wave will be bent to on the opposite side of the fresnel. So focus is independant of collection. A fresnel will collect some of the light from nearly any direction and pass it through to the other side regardless of wether that light was at the focal point or not. And light that has made it from one side of the first fresnel to the other can be thought of as passing the first stage in being useful later on and could contribute to overall brightness. Ultimately, all a fresnel is is a sheet of clear plastic.

To focus on our light source really only means that all of our light that hits the fresnel will be bent to the same place (the triplet). Only light that makes it to the triplet is useful. This is the only reason why we keep all the light at the focal point. Light outside that point will still pass, but will miss the triplet.

So in our setup, simply having more light impact the fresnel will contribute to a brighter image. The bigger the fresnel, the more light that will impact the fresnel. Therefore, the bigger the fresnel the brighter the image.

Hope that helps ,
Mark.

from http://en.wikipedia.org/wiki/Lux

Lux versus lumen

The difference between the lux and the lumen is that the lux takes into account the area over which the luminous flux is spread. 1000 lumens, concentrated into an area of one square metre, lights up that square metre with an illuminance of 1000 lux. The same 1000 lumens, spread out over ten square metres, produces a dimmer illuminance of only 100 lux.

Achieving an illuminance of 500 lux might be possible in a home kitchen with a single fluorescent light fixture with an output of 12000 lumens. To light a factory floor with dozens of times the area of the kitchen would require dozens of such fixtures. Thus, lighting a larger area with the same number of lux requires a larger number of lumens

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

the link is saying the same thing I am.

So therefor

Thus, lighting a larger area(17" LCD) with the same number of lux requires a larger number of lumens(more light).

I still fail to see how a 17" LCD would be 30% brighter than a 15" LCD, even if it collects 30% more light, I dont believe it would collect 30% more, but I could be wrong.

I think having fresnel lens will differ the physics over having a 400W MH shining on a 17" peice of wood. Meaning that a 400W lamp shing on a 17" fresnel will behave different than a 400W shining on a peice of 17" driftwood. If so, focal lengths wouldnt be so important to us for the rear fresnel

To illustrate

a 17" fresnel with FL 220mm, and a 17" fresnel with FL 300mm.

If both fresnels are placed 260mm from the bulb (neither at focal point), the light each of the two collect will be different. Which simply means the physics introduced with the lens are more advanced than all the diagrams in this thread of illustrate.

If it were as simple as this light shines on this lens and collects this much light, then any fresnel the same exact dimensions regardless of FL would capture the same amount of light, alas they do NOT however.

Scenario A - a 17" 220mm FL fresnel placed 260mm from the lamp



Scenario B - a 17" 300mm FL fresnel placed 260mm from the lamp



Now, according to everyones explanation. According to RymGB's illustations, and according to ROX's mathematical formulas the only VARIABLES included are: Light, and SIZE

both scenarios according to ROXs calculations would be expressed

40,000Lm = 40Klx

Scenario A = 40Klx * .09 = 3.6Klm

Scenario B = 40Klx * .09 = 3.6Klm

Now, Im not an expert on what happens when fresnel lenses are placed at INCORRECT distance, but Im pretty certain that the light collect in Scenario A will NOT equal the light collect in scenario B. What happens if I introduce Scenario C, which is a 220mm FL fresnel placed at the CORRECT FL of 220mm. Would the light collect in all 3 scenarios be the same?

Final thoughts.

I have linked wikipedia, that shows that it takes MORE light to make a larger area have just as much brightness as a smaller area with a proportional smaller light. I could have been wrong in my assumptions in believe that both a 15" fresnel and a 17" fresnel would collect ~similar light, however irregardless the 17" LCD panel will need more LIGHT dispearsed accross its larger surface area to remain just as bright as a 15"

Ive also introduced reasons why you just cant ASSUME that a fresnel lens will collect light in the same manner as a peice of driftwood.

This is where theory butts heads with practicality!

at the end of a day, a 15" PJ and a 17" PJ with all other things created equal will have ~the same lm/m^2 ratio displayed on the wall.

The only way to prove my statement wrong, would be to DISPELL the wikipedia link, and DISPELL "Thus, lighting a larger area with the same number of lux requires a larger number of lumens".

Oh yeah, I meant to post this drawing:

Click to view attachment

It is light waves (lines) coming uniformly out of an arc tube and impacting a fresnel. The average amplitude of each of those waves, and the number of those waves that impact the fresnel are the contributing factors in getting a brighter image. You can see that by extending the fresnel I would have drawn in more lines at the edges because they too would now apply.

This image also demonstrates how the square law applies to our fresnels in visual terms. If you divide the panel up into quarters, you can see there are less waves of light impacting the outer quarters than the center. So we will get dimmer edges. By extending the fresnel we will have increased the total number of lines impacting the fresnel. And the total is all that matters. By having more lines impact the fresnel you will have more lines hit the screen. The more lines that hit the screen, the more likely it is that a line will get redirected to your eye and excite the chemicals in your retina at the point of impact. The more excited those chemicals: the brighter the image both percieved and measured.

Another way of looking at this:



The blue rectangle represents the collector fresnel
The surface area of the sphere represents the maximum amount of light (in lumens) that could be collected at that distance from the light source.

Therefore dividing the area of the fresnel by the area of the sphere gives us the maximum percentage of light that can be captured by the fresnel

Larger fresnel at same distance from light source = more light collected (ie lumens)
As long as the projected image is the same size, then more lumens delivered = brighter image

I find this approach more intuitive than converting to mCd



Not sure, but I think that Rox's calculation (and this one) is only valid (ie it is a crude and approximate model) if the size of the fresnel is somewhat smaller than the distance between it and the lamp

edit:

agreed Mark, I misread the question
Q1 - Yes
Q2 - No

So, if that is the case, then simply have 1000W bulb can very well EASILY contribute to a much brighter picture, which again is the root cause of this discussion.

The bigger the fresnel the more light that fresnel can collect I agree, However the BIGGER the fresnel the MORE AREA that light has to cover.

a 100" PJ displaying an image 100cd/m^2 is dimmer than a 3" IPOD displaying the same image at 250cd/m^2

Who cares if a larger fresnel collects more light, if at the end that light has to be spread out over a larger area and thus comes out with similar brightness on the wall.

100 LUX = 6.9lm/.069m^2 = 9lm/.09m^2

To get a 30% brightness increase then gain would have to go to 130 LUX

130 LUX = 11.7lm/.09m^2

Therefor in order for this scenario to be 30% brighter than you would need a 69.5% increase in LUMANS( 11.7 - 6.9 / 6.9)

The first 30% gain in lumans is used to MAINTAIN 100 LUX, then the next 39.5% gain is used to acheive the additional 30 LUX.

I think I see the problem with this discussion.

Do we only look at the lumen value AFTER the LCD??

My understanding is that the front fresnel basically magnifies the image as the triplet projects it. Kinda crude example but.

Lets say we have our beloved scenario with our two PJ, one 15" and the other 17".

Lets say both LCD have the same LUX, which I will use

500lm/.069m^2 = 7.2kLux (15")

648lm/.09m^2 - 7.2kLux (17")

If you were to stare at your LCD direct (triplet removed) then both of these LCDs would appear about equally bright to the human eye.

Now, what happens when projecting both of these scenarios.

I guess most of you are saying that the (m^2) component at the LCD is NOW irrelavant? I however dont believe that it is.

I believe if both PJs throw a 100" image, and that since the LUX is approximatly the same then approximately the same brightness would appear on screen. I dont think you can just disregard the m^2 component of LUX .

Using a larger fresnel without adjusting anything else will result in a larger image at the wall. But there will be more waves of light hitting the wall. Agreed?

You can now move the projector closer to the wall or make focal length adjustments to make the image the same size as before. You now have the same increased number of waves as the larger projection, but with the original projection size. This is the scenario that DAZZZLA was painting. It seems to me your image will then be indisputably brighter due to a larger fresnel.

But don't forget you now have a projection that is bigger in area than the 15". Now you adjust down the size of that projection to match.

When dealing with theory, you cannot create any bounds.

So lets take what you said to the extreme. Since you said FL ect can be adjusted, basically implying that the m^2 component of the LCD lux was moot.

Purely theoretical here, but lets say we have a 50" LCD in our PJ.

and lets just say that it since the M^2 component isnt important, lets compare the 50" LCD to a 5" LCD both with 500 Lumens of light AFTER THE LCD.

500 Lumens/.774m^2 = 646 LUX (50" LCD)

500 Lumens/.00774m^2 = 64600 LUX (5" LCD)

With only 500 Lumens of light behing a 50" LCD probably wouldnt be bright enough to barely light the pixels up.

However, since you say the distance from triplet to wall can be adjusted and the FL ect, then in both cases 500 lumens will end up on the wall, and be equally bright.

This CANT be true, that low a lux at the LCD probably couldnt even be projected.

My point is, is that the m^2 component cant be FULLY disregarded. Perhaps to an extent, I wouldnt be certain. But it cant just be written off as if it didnt exist.

What bounds am I missing?It the same total number of waves headed for the triplet, so the same number of waves will make it to the screen.It does exist. But is just a measure. Just as you have discussed the impossibility of amplifying light, this measure cannot destroy light. It simply allows us to facilitate the measure of how light spreads with distance. It doesn't tell us how light cancels with distance, as that does not happen. If every time we spread the light out we condense it back down again (which we do) then any light we add to the system regardless of how it spreads things out is going to make things brighter.

As a measure it exists any time you want to know the number of waves of light in a given area at a given interface. If the interface sizes stay the same, or balance out to the same size in the end then the only measure that really matters is the total. Increasing fresnel size increases the total.

shit, I closed my eyes once and this thread exploted...

first of all, thanks for your inputs.

second; shit! I answered wrong daz's questions; 1)Yes 2) no.

this was what i wanted to answer form the first moment, but confused the answer by mistake.

third; A projector is brighter because it is more lumens on it's output, no matter what image size you are projecting. This means scuba, that you can have a 50" image (300 lux) next to 100" image (100 lux) and we can say the 100" inch projecting projector is brighter.

the 100 inch projecting PROYECTOR is brighter because it is more lumens there. obviously the 50" IMAGE is brighter than the 100" one.

Check some comertial units specs, they have one called BRIGHTNESS, tipically expresed on MAX ANSI LUMENS.
edit;


thanks michaelJ for your spherical to flat error picture, this is exactly the only error I did on the ideal calcs I have done so far. (remarked 5 times whit this one included).

the area being more than 1 meter square is no problem if you are working on a flat suface scuba. Now your were multiplying the 1.2Meters square area assumed flat, when it was in really spherical (huge error introduced here), also in our example this area was even larger than the total shpere's area for that radius. This is not a valid example, you can't consider an area larger than the total spherical available.

Dude... you post so quick ... Lets talk about one thing at a time.


This is out of context. If you look at the post again, I said all things in the two setups are equal (exept for fresnel and LCD size). That means both fresnels have equal fl, and the lamp is placed AT the focal of each (otherwise both setups are useless, because you're not using the lenses correclty). It also means both screens are displaying an image of equal size.

This is the only way to investigate the effects of fresnel size. You must hold ALL other variabes constant and change only the size.

No one is arguing with this... It is 100% correct. This was really the whole point of my post. If both projectors project an image of the same size - then the 17" setup will be brigher, because MORE lumens were collected and MORE lumens were projected to the screen. More Lumens over an EQUAL image area= brighter image.

Then I retire again!

If all you wanna argue is the number of lumens flying out of a PJ then whats the point.

I thought the discussion was about percieved brightness to us, a human being.

But if you wanna say a 100 Lux 100" image is brighter than a 300 Lux 50" image then whats the point. So what? The 50" image is gonna look better, the 50" is gonna show detail in dark scenes ect.

Its like taking just the magnitude of a vector with no direction, it becomes another pointless statistic, that means very little into the quality of image on the screen.

2 Lumen spread over 10 acres is brighter than 1 lumen spread over 1 acre, Right ROX. Because 2 TOTAL lumens are present.

If you then adjust the size of those fields/warehouses to be equal (which we do very easily), then what you have could be 2 Lumens over 1 acre compared with 1 Lumen in one acre. We do this by moving the projector closer to the wall (the lamp closer to the floor of the warehouse).

Question: Do you not agree that if we adjust our screen size down to be equal, then a bigger fresnel results in an indisputably brighter image?

ok i think we have at last ended that discussion. Just wana say a last coment;

I choosed my post's words with some time, so please take your time as well for reading each word;

The 17" PROJECTOR is brightner than 15" PROJECTOR.

you just swiched PROYECTOR by IMAGE.

As i have remarked to you, the projector's brightness is stated by the amount of MAX ANSI LUMENS, no one talks about screen lux here.