Hello all.
I’ve taken it upon myself to do a little write-up on what the lenses in our projectors do. I’ve posted this information in various threads, this is basically a collection all in one place that can maybe be stickied.
Essentially, lenses bend light. When light enters the lens at an angle, it is bent at a refraction angle determined by the optical properties of the lens material, as compared to the material that they left (Typically air.) I’d suggest taking a look at HowStuffWorks.com “How fresnel lenses work” as some basic information, as well. There are also links there to a lot of other interesting things having to do with lenses and projection.
First off, the general drawing from the Guide gives a good idea of light transmission though the projector, but it’s often misinterpreted as being the light path which gets projected. This is simply not the case. Brain’s drawing is not one of the projection, but simply of the usable and needed light path through the projector in order to hit the projection lens. (Fig. 1)
Full Version: What lenses do
We’ll start with the collimator fresnel. What this lens does is in the diagram, it takes the diverging light rays and bends them to make a set of parallel rays. It does NOT focus the light rays, and in fact, as we’ll see later, at this distance, it is impossible for it to focus the light from the lamp on anything. In order for it to do so, the lamp would have to be at a distance greater than the focal length of the lens.
The field fresnel could, in fact focus the light, and if you were to remove the triplet and place a “screen” at an appropriate distance from the field fresnel, you would get a focused image of the lamp. It would not be a high quality focused image of the lamp, but it would be there. It is important to distinguish that neither of the fresnels actually focus on the LCD, they are merely there to direct the light from the lamp through the triplet. It is entirely possible to make a projector without these fresnels at all, however, it would far less efficient, requiring much greater light input and delivering much lower output.
So, let’s get to the interesting lens… The triplet.
This is the lens that actually focuses on the LCD. In theory for a monochrome image, there would be no advantage to a triplet over a single lens. This is because lenses treat different colours differently. Red light has less energy, and is deflected at a different angle than the higher energy blue light. This is why crystals produce rainbows – the same thing is actually happening in your projection lens, but the triplets are aligned in such a way as to re-integrate the light at the output, while still having the desired focusing effect.
For purposes of this write-up, we’ll treat the triplet as a “perfect” single convex lens.
The focal distance of a convex lens is defined as the point where parallel light entering the lens will converge to a point. While this exact distance is useful for starting fires with a magnifying lens, it’s actually not much good in a projector. The following figure represents an LCD at the focal distance of the lens. You can see that what you get projected onto the screen isn’t the 3 distinct points on the LCD, but a big mess. In fact, what you may notice is that the individual points are each the same size as the lens. This is “infinite focus” Where a screen which is far enough away that this limited resolution is not a problem. (Fig. 2)
The field fresnel could, in fact focus the light, and if you were to remove the triplet and place a “screen” at an appropriate distance from the field fresnel, you would get a focused image of the lamp. It would not be a high quality focused image of the lamp, but it would be there. It is important to distinguish that neither of the fresnels actually focus on the LCD, they are merely there to direct the light from the lamp through the triplet. It is entirely possible to make a projector without these fresnels at all, however, it would far less efficient, requiring much greater light input and delivering much lower output.
So, let’s get to the interesting lens… The triplet.
This is the lens that actually focuses on the LCD. In theory for a monochrome image, there would be no advantage to a triplet over a single lens. This is because lenses treat different colours differently. Red light has less energy, and is deflected at a different angle than the higher energy blue light. This is why crystals produce rainbows – the same thing is actually happening in your projection lens, but the triplets are aligned in such a way as to re-integrate the light at the output, while still having the desired focusing effect.
For purposes of this write-up, we’ll treat the triplet as a “perfect” single convex lens.
The focal distance of a convex lens is defined as the point where parallel light entering the lens will converge to a point. While this exact distance is useful for starting fires with a magnifying lens, it’s actually not much good in a projector. The following figure represents an LCD at the focal distance of the lens. You can see that what you get projected onto the screen isn’t the 3 distinct points on the LCD, but a big mess. In fact, what you may notice is that the individual points are each the same size as the lens. This is “infinite focus” Where a screen which is far enough away that this limited resolution is not a problem. (Fig. 2)
So if we look at an example of the same image focused, the image source (LCD in our case) is further than the focal distance of the lens. You will see that we have distinct points back on our screen, in a scaled reversal of the points on the LCD. The light represented on the left side of the lens is only the light that we’re interested in. There is generally going to be more of it than what I’ve shown, but only the light that hits the lens gets projected. On the right side of the lens, however, is the most light that there can be. (Fig. 3)
Now we can step back to the field fresnel. This lens alters the light path somewhat, but does not actually produce a focus. The reason for this is that the lens is far too close to the LCD to do what the projection triplet does. Remember that we’re only interested in the light that hits the triplet. Due to the nature of what the collimator fresnel does, the majority of the light leaving the LCD is not headed for the triplet, but is instead going straight forward, in parallel beams. Naturally the LCD scatters this a bit. The field fresnel redirects that light so that it hits the LCD. I’ll say it again, it does not focus it. The following figure shows what generally happens to the light leaving the LCD, passing through the fresnel (Not to be confused with the previous marker for focal distance) and hitting the projection lens. You will see that at the projection lens what you get is the sum of ALL of the light passing through the LCD. You sill also see that from the point of view of the projection lens, the LCD screen would be magnified, appearing close than it actually is. This modifies the effective focal length of the projection triplet, requiring a greater distance from the combination to generate an effective focus. This is also how the field fresnel assist with keystone, by being closer to the projection lens, the magnification effect becomes greater. You will see that as the light progresses, it is still diverging from it's point source, and as such it will never be able to truely come to a focus. As mentioned before this is because the LCD is too close to the lens relative to its focal distance.(Fig 4.)
I hope this information is useful in understanding what the individual lenses do, and how it all comes together for our projectors.
I hope this information is useful in understanding what the individual lenses do, and how it all comes together for our projectors.
O.K. I am a super newbie here.
As long as we are on the discussion of the fresnels, I have a few questions.
Keystoning. I am going to let gravity put the collector freznel in place. I would assume that the fresnel should stay leve, thus putting the hinge on top. But in a few pictures, I've seen the hinges on bottom. And in the guide it shows it swaying left to right.
Point me in the correct direction.
Second question. The fresnels should be sandwitched in between 2 panes of glass to keep them perfectly straight and clean. But, the panes of glass makes for a loss of lumens. Has anyone come up with a way to adhere the fresnel to 1 pane of glass?
Thank for you input.
As long as we are on the discussion of the fresnels, I have a few questions.
Keystoning. I am going to let gravity put the collector freznel in place. I would assume that the fresnel should stay leve, thus putting the hinge on top. But in a few pictures, I've seen the hinges on bottom. And in the guide it shows it swaying left to right.
Point me in the correct direction.
Second question. The fresnels should be sandwitched in between 2 panes of glass to keep them perfectly straight and clean. But, the panes of glass makes for a loss of lumens. Has anyone come up with a way to adhere the fresnel to 1 pane of glass?
Thank for you input.
Very good SupraGuy, I think that this will significantly help everyone understand whats happening inside thier PJ. Just wondering what your backround is? Do you work in the optics field? This thread should be pinned and be made required reading.
You're my hero Supra.
QUOTE (DKaeser @ Mar 30 2005, 01:50 PM)
Keystoning. I am going to let gravity put the collector freznel in place. I would assume that the fresnel should stay leve, thus putting the hinge on top. But in a few pictures, I've seen the hinges on bottom. And in the guide it shows it swaying left to right.
Actually, you want it vertical, or parallel to the screen. Only a certain amount of tilt is acceptable, more than that will cause you problems with focus. You can also tilt it horizontally, if your projector won't be perfectly pependicular to the screen. Though personally, I think the picture in the Guide is a side view, not a top view. Note that it doesn't matter where the hinges are, if you're not using gravity to correct the angle, hinges at the bottom ensure that the bottom of the lens is as close to the LCD as practical, ensuring the best quality image. You can adjust the top hinge location to get the same effect. It matters where the lens is, not how it's held in place.
QUOTE (DKaeser @ Mar 30 2005, 01:50 PM)
Second question. The fresnels should be sandwitched in between 2 panes of glass to keep them perfectly straight and clean. But, the panes of glass makes for a loss of lumens. Has anyone come up with a way to adhere the fresnel to 1 pane of glass?
My field fresnel is in a frame, no glass at all. I believe that Brain mentions in the Guide that he used binder clips to hold the collimator fresnel to a piece of tempered glass. Any kind of clip should do the trick, provided that there's enough (but not too much) clamping force.
Pagercam: No, I don't actually work with optics, but I have done considerable reading on the subject. I'm interested in just about everything science related, and my father was a land surveyor for a while, until he became a general contractor. (Long story...
) So I had lots of opportunity to play with extremely high quality optics when I was a kid, and when I asked my Dad how it all worked, he handed me one of his University texts. A 3" thick textbook titled Fundamentals of Optics and told me to go read it. (It's still in my bookshelf, actually.) Small print and thin pages, too. A bit heavy reading for an 8 year old If the information that I've presented here is useful, I'm sure that Brain (or a mod) will pin the thread, to make sure that it's avaialable. In the meantime, now that it's here, I can just post a link to it when I see someone asking the same questions.
any possible explanations of how a triplet could cast a second image with all colors. I understand how it could with a specific wavelength, but see no reason how it could with a complete image.
Our triplets are messed up in a sense that you have your regular image and then you have another image cast over it that is blurry. The blurry image that is cast can be focus either in the center or at the edges, if you focus at the edges then you lose focus on the first projected image and the entire thing appears to be out of focus.
Our triplets are messed up in a sense that you have your regular image and then you have another image cast over it that is blurry. The blurry image that is cast can be focus either in the center or at the edges, if you focus at the edges then you lose focus on the first projected image and the entire thing appears to be out of focus.
Thank you Supra. You are most helpfull.
QUOTE (foamcows @ Mar 30 2005, 09:15 PM)
any possible explanations of how a triplet could cast a second image with all colors. I understand how it could with a specific wavelength, but see no reason how it could with a complete image.
Our triplets are messed up in a sense that you have your regular image and then you have another image cast over it that is blurry. The blurry image that is cast can be focus either in the center or at the edges, if you focus at the edges then you lose focus on the first projected image and the entire thing appears to be out of focus.
Our triplets are messed up in a sense that you have your regular image and then you have another image cast over it that is blurry. The blurry image that is cast can be focus either in the center or at the edges, if you focus at the edges then you lose focus on the first projected image and the entire thing appears to be out of focus.
How out of focus do you think the edges would be if you were using a regular double convex lense? I'm sure that you could construct a higher quality triplet...or even a more effective arrangement of lenses...but how much do you want to spend on a DIY projector?
ive already dropped about a grand on everything for this one including my screen materials. whats weird though, is my old projector with the old standard lenses worked great
Quick lense question. Do different wavelengths of light bend at diffrent angles in the fresnel? IE woul having the LCD between the fresnels introduce more chance for a color seperation campared to having the LCD in front of both fresnels?
QUOTE (MadCraiZ @ Mar 31 2005, 04:20 PM)
Quick lense question. Do different wavelengths of light bend at diffrent angles in the fresnel? IE woul having the LCD between the fresnels introduce more chance for a color seperation campared to having the LCD in front of both fresnels?
lambda won't come into play until you're dissecting the IR band, or working with precision lasers.
w.r.t. Thin Lens Physics
The only geometry that changes the light path is the refractive index between glass and air, or in our case plastic and air. The value for N is constant for all visible wavelengths.
N Glass to Air should be ~1.5 if I remember right from Physics 3 back in the day.
The optical poly-carbonates will probably ballpark glass but will definitely be > 1.5.
In any case.. N is a constant. So you won’t have to worry about other factors. Just think in Rays and you’ll be coolio.
Yes, all lenses bend different wavelengths of light differently, however because the fresnel is so close to the LCD, this is unlikely to cause a problem. There will be a small amount of colour distortion due to the fresnel, getting greater proportional to the distance from the center of the image. Fortunately, the amount of distortion is also proportional to the distance of the lens from the image.
You can experiment for yourself. Take the fresnel, and hold it at arms length. Look at (for example) a flourescent light fixture, which should be more than arms length away from the lens. With the light fixture at the center of the fresnel, you will not see much colour distortion. (There may be a slight orangeish rim around the light.) With the light fixture toward the edges, you'll see a blue rim to the inside, and an orange rim to the outside, which gets wider as it gets closer to the edge of the lens. Now, use a closer light source (I'd suggest a small flashlight) at a distance of less than an inch (20mm or so from the lens) Keep the lens at arm's length. It will be difficult to notice any of the colour distortion effect (though it is still there)
Foamcows: Without seeing and measuring the lenses for myself, I'd suggest that it's possible that the lenses are out of round, with part of the lens having a slightly different focal length than the main part. This can also happen if one of the lenses in the triplet "bounces" light off of the outside of the lens casing at a very shallow angle, or if there's internal reflections happening inside the lens triplet. Remember that every part of the lens is handling every part of the image, so a missed focus in any area of the lens will project a proportionally dimmer (By the percentage of the area affected) whole image as at the different focal length.
The most likely place for a lens to have this kind of error is in either the very center (flat spot) or around the very edges. You could try masking off the outside edge of the lens, or a small disc covering the very center of the lens, and see if the problem improves or goes away. If so, that would tell you the problem, at least. Either method would affect overal image brightness. Any masking would have to be done on the projection side of the lens, as any light entering the lens may still hit the affected areas. You only want to block light EXITING the affected areas.
This is, of course pure speculation.
EDIT: Just read your post in the "Pro Lenses" thread, and it seems that this may be in fact the problem. The ring blocking the outside of the lens will have less effectiveless if the error is with lenses other than the very outer one, but will still help. I think that gives a good indication as to a problem with the pro lenses....
You can experiment for yourself. Take the fresnel, and hold it at arms length. Look at (for example) a flourescent light fixture, which should be more than arms length away from the lens. With the light fixture at the center of the fresnel, you will not see much colour distortion. (There may be a slight orangeish rim around the light.) With the light fixture toward the edges, you'll see a blue rim to the inside, and an orange rim to the outside, which gets wider as it gets closer to the edge of the lens. Now, use a closer light source (I'd suggest a small flashlight) at a distance of less than an inch (20mm or so from the lens) Keep the lens at arm's length. It will be difficult to notice any of the colour distortion effect (though it is still there)
Foamcows: Without seeing and measuring the lenses for myself, I'd suggest that it's possible that the lenses are out of round, with part of the lens having a slightly different focal length than the main part. This can also happen if one of the lenses in the triplet "bounces" light off of the outside of the lens casing at a very shallow angle, or if there's internal reflections happening inside the lens triplet. Remember that every part of the lens is handling every part of the image, so a missed focus in any area of the lens will project a proportionally dimmer (By the percentage of the area affected) whole image as at the different focal length.
The most likely place for a lens to have this kind of error is in either the very center (flat spot) or around the very edges. You could try masking off the outside edge of the lens, or a small disc covering the very center of the lens, and see if the problem improves or goes away. If so, that would tell you the problem, at least. Either method would affect overal image brightness. Any masking would have to be done on the projection side of the lens, as any light entering the lens may still hit the affected areas. You only want to block light EXITING the affected areas.
This is, of course pure speculation.
EDIT: Just read your post in the "Pro Lenses" thread, and it seems that this may be in fact the problem. The ring blocking the outside of the lens will have less effectiveless if the error is with lenses other than the very outer one, but will still help. I think that gives a good indication as to a problem with the pro lenses....
QUOTE (fore1337 @ Mar 31 2005, 04:50 PM)
QUOTE (MadCraiZ @ Mar 31 2005, 04:20 PM)
Quick lense question. Do different wavelengths of light bend at diffrent angles in the fresnel?
lambda won't come into play until you're dissecting the IR band, or working with precision lasers.
I have a hard time with this statement. If this were true, there'd be no such thing as chromatic aberration in lenses, and chromatic aberration most definitely IS an issue in visible-light optics. You see it yourself when you see a rainbow, or light shone through a prism, or a glass of icewater.
This is a simple page, but it gives a bit of information: chromatic aberration page
I'll yank the ol' Physics book out from beneath the couch when I get home from work.
I didn't think about prismatic effects.
Like I said. My optics physics is a bit rusty.
stay tuned.
1-Apr
Edit:
Yeah, as far as I can tell, all visual optics employ Achromatic Lens builds to avoid Chromatic Aberration. I asked some of my co-nerds @ work, and we all chalked it up to the non-uniform index of refraction (impurities in the lens). If I ever see one of my best pals from college again (j/k), I'll have to ask his wife about this. She's working on a Phd in optics & lasers.
syscrush even provided a link describing this in limited detail.
cheers m8s.
I didn't think about prismatic effects.
Like I said. My optics physics is a bit rusty.
stay tuned.
1-Apr
Edit:
Yeah, as far as I can tell, all visual optics employ Achromatic Lens builds to avoid Chromatic Aberration. I asked some of my co-nerds @ work, and we all chalked it up to the non-uniform index of refraction (impurities in the lens). If I ever see one of my best pals from college again (j/k), I'll have to ask his wife about this. She's working on a Phd in optics & lasers.
syscrush even provided a link describing this in limited detail.
cheers m8s.
It's not impurities that cause the aberration, it's a property of any lensing material that the amount of refraction is, at least in part, a function of the amount of energy (ie wavelength) in the wave.
great post, the information is very helpful. Thanks 
I think this would be the correct place to post this question. I've been building my PJ, and have all the components and parts in place. When I fire it up, I can see my image (white square) in the triplet, but there is also a soft light "glow" around the triplet on the inside of the box. Is this normal? I plan on finishing the PJ this weekend, and sending an actual image to the triplet (i.e movie), but was just curious if this is something normal or correctable. Thx for any input.
about the light that does not enter on the triplet, there is a point i would like to discuss if there is any interested on;
if you check the first picture supraguy posted, you will see that the field fresnell focuses "idealy" the light at a farther point than the triplet (so full triplet aperture intersects with the light cone), now do you think this is a good starting point?
if you check the first picture supraguy posted, you will see that the field fresnell focuses "idealy" the light at a farther point than the triplet (so full triplet aperture intersects with the light cone), now do you think this is a good starting point?
Rox: No. But it would be a good place to start if we had a true point source light, which we know that we don't.
I think this to be the place to post this... If not, please direct me. I have bought a old pj for 200 and the quality is not so great. 
What I was curious about is the lense on it. Dont have numbers or anything, but I do know that it is a zoom lense and has a focus at the end. It happens to have a few scratches on it 
, but I dont see them on the projected image....but then as I said, the quality sucks. My question is if I can use it for the projector that I am building... and if so, will there be a way to know how to mount it right? 
Just a newbe with alot of Qs. Thanks for the help
What I was curious about is the lense on it. Dont have numbers or anything, but I do know that it is a zoom lense and has a focus at the end. It happens to have a few scratches on it , but I dont see them on the projected image....but then as I said, the quality sucks. My question is if I can use it for the projector that I am building... and if so, will there be a way to know how to mount it right? Just a newbe with alot of Qs. Thanks for the help
The "Help" forum would be more appropriate.
In short, though, the lens from another projector is probably not appropriate, as it's focal length and field of view are likely unsuited for the relatively large LCD panels that we're using.
In short, though, the lens from another projector is probably not appropriate, as it's focal length and field of view are likely unsuited for the relatively large LCD panels that we're using.
First, thanks for this explanation. It's clear, but still I have two questions .
1. When the light goes parallel to the LCD (split setup), this explanation shows the LCD works the same as a diverging lens (like this: <). But at wich angle? You must know the angel in order to calculate the distances... (as far as I know there isn't an angle, but just different intensities at all angles)
2. So if I'm right, the light that comes to the triplet is not converged light, but diverged light cones that come to one point?
.1. When the light goes parallel to the LCD (split setup), this explanation shows the LCD works the same as a diverging lens (like this: <). But at wich angle? You must know the angel in order to calculate the distances... (as far as I know there isn't an angle, but just different intensities at all angles)
2. So if I'm right, the light that comes to the triplet is not converged light, but diverged light cones that come to one point?
QUOTE (Tommy The Cat @ Apr 9 2007, 12:48 PM)
First, thanks for this explanation. It's clear, but still I have two questions .
1. When the light goes parallel to the LCD (split setup), this explanation shows the LCD works the same as a diverging lens (like this: <). But at wich angle? You must know the angel in order to calculate the distances... (as far as I know there isn't an angle, but just different intensities at all angles)
2. So if I'm right, the light that comes to the triplet is not converged light, but diverged light cones that come to one point?
.1. When the light goes parallel to the LCD (split setup), this explanation shows the LCD works the same as a diverging lens (like this: <). But at wich angle? You must know the angel in order to calculate the distances... (as far as I know there isn't an angle, but just different intensities at all angles)
2. So if I'm right, the light that comes to the triplet is not converged light, but diverged light cones that come to one point?
If I understand your question correctly, I think you've misunderstood something. Picture the LCD as a simple transparency such as would be used on an overhead PJ, and not like a lens. The light is collimated by the lamp side fresnel, so it's traveling approximately perpendicular to the plane of the LCD. It passes through the LCD, and the field fresnel takes the parallel light and converges it to a point. The triplet should be placed so that the rearmost lens intersects the converging cone of light. The bigger the "image" at the rearmost triplet element, the better. Then adjust for focus.
I'm sure I screwed that up someplace, but that's how I picture it.
QUOTE (tenzip @ Apr 10 2007, 07:41 PM)
If I understand your question correctly, I think you've misunderstood something. Picture the LCD as a simple transparency such as would be used on an overhead PJ, and not like a lens. The light is collimated by the lamp side fresnel, so it's traveling approximately perpendicular to the plane of the LCD. It passes through the LCD, and the field fresnel takes the parallel light and converges it to a point. The triplet should be placed so that the rearmost lens intersects the converging cone of light. The bigger the "image" at the rearmost triplet element, the better. Then adjust for focus.
I'm sure I screwed that up someplace, but that's how I picture it.
I'm sure I screwed that up someplace, but that's how I picture it.
Right, so if I understand right, you say the light that leaves the LCD (by a split setup) is still parallel light and not diverged light.
In that case, the whole explanation of supraguy is worthless...
What do I miss? I thought I had found the explanation for the fact that the distance of the LCD - triplet is important by a split setup, but now I doubt again
.(I always thought, if the light goes parallel after the LCD, only the place of the second fresnel is important... )
QUOTE (Tommy The Cat @ Apr 10 2007, 03:26 PM)
Right, so if I understand right, you say the light that leaves the LCD (by a split setup) is still parallel light and not diverged light.
In that case, the whole explanation of supraguy is worthless...
What do I miss? I thought I had found the explanation for the fact that the distance of the LCD - triplet is important by a split setup, but now I doubt again.
(I always thought, if the light goes parallel after the LCD, only the place of the second fresnel is important... )
In that case, the whole explanation of supraguy is worthless...
What do I miss? I thought I had found the explanation for the fact that the distance of the LCD - triplet is important by a split setup, but now I doubt again
.(I always thought, if the light goes parallel after the LCD, only the place of the second fresnel is important... )
In the split setup, the triplet is not focusing on the actual LCD or on the fresnel, but the "virtual" LCD as seen through the field fresnel. The virtual LCD's distance is not the same as the actual LCD.
In the unsplit setup, the triplet focuses on the actual LCD. That's why it can have better image quality, there's nothing between the triplet and the LCD.
Maybe that's what's confusing you? Think of one of those trick boxes with the mirror in it, where it looks like there's a penny floating in the hole in the top of the box. The image is here, but the object is there.
QUOTE (tenzip @ Apr 10 2007, 09:56 PM)
In the split setup, the triplet is not focusing on the actual LCD or on the fresnel, but the "virtual" LCD as seen through the field fresnel. The virtual LCD's distance is not the same as the actual LCD.
In the unsplit setup, the triplet focuses on the actual LCD. That's why it can have better image quality, there's nothing between the triplet and the LCD.
Maybe that's what's confusing you? Think of one of those trick boxes with the mirror in it, where it looks like there's a penny floating in the hole in the top of the box. The image is here, but the object is there.
In the unsplit setup, the triplet focuses on the actual LCD. That's why it can have better image quality, there's nothing between the triplet and the LCD.
Maybe that's what's confusing you? Think of one of those trick boxes with the mirror in it, where it looks like there's a penny floating in the hole in the top of the box. The image is here, but the object is there.
It's too vague. I focus at the exact happenings with the light in combination with, lcd, fresnel and triplet. I must have exact images and explanations to understand
. Supraguy gave me that for 75%, but my questions remain. Also your conclusion conflicts with supra's guide.So it remains unclear.
QUOTE (Tommy The Cat @ Apr 10 2007, 05:23 PM)
It's too vague. I focus at the exact happenings with the light in combination with, lcd, fresnel and triplet. I must have exact images and explanations to understand . Supraguy gave me that for 75%, but my questions remain. Also your conclusion conflicts with supra's guide.
So it remains unclear.
. Supraguy gave me that for 75%, but my questions remain. Also your conclusion conflicts with supra's guide.So it remains unclear.
Hmm. Well, I'll let someone with more optics expertise give it a whirl.
But I don't see where I'm in conflict with what Supra said?
QUOTE (tenzip @ Apr 10 2007, 07:41 PM)
Picture the LCD as a simple transparency such as would be used on an overhead PJ, and not like a lens. The light is collimated by the lamp side fresnel, so it's traveling approximately perpendicular to the plane of the LCD. It passes through the LCD, and the field fresnel takes the parallel light and converges it to a point.
This passage is. Supraguy most essential part is that the light cones to the triplet are actualy diverged cones wich come to one point. This is possible because the light from the LCD (in supra's explanation) is not parallel!
And then I finally understood why the position of the lcd is important, it does affect the breaking angle towards the triplet!
But my most essential question was/is: at wich angel does the light leave the LCD?? Because as far as I know there is no particular angle but much more angles. And the greater the angle is, the less intens the light is.
But in this case, you cannot calculate focal distances!!!
You see??
QUOTE (Tommy The Cat @ Apr 10 2007, 05:48 PM)
This passage is. Supraguy most essential part is that the light cones to the triplet are actualy diverged cones wich come to one point. This is possible because the light from the LCD (in supra's explanation) is not parallel!
That's why I don't like that fourth post's diagram. It's misleading, I think.QUOTE
And then I finally understood why the position of the lcd is important, it does affect the breaking angle towards the triplet!
I not sure I understand what you're saying here, but here's my answer to what I think you're saying. Any light that passes through the LCD that is not parallel to, or very close to parallel to, the Z axis of the LCD won't go through the triplet, and can be ignored.QUOTE
But my most essential question was/is: at wich angel does the light leave the LCD?? Because as far as I know there is no particular angle but much more angles. And the greater the angle is, the less intens the light is.
But in this case, you cannot calculate focal distances!!!
You see??
But in this case, you cannot calculate focal distances!!!
You see??
The light leaving the LCD that will go through the triplet and create the image on the screen will be at 0 degrees relative to the Z axis of the LCD, (plus or minus a few, but you can safely ignore that for calculating focal distances.)
I believe you're overthinking it. Relax.
Just pretend all the light entering and leaving the LCD is parallel to the projection axis, and base any calculations off of that.
I drew up a crappy MS Paint diagram that I think shows a better picture of what happens.
Click to view attachment
Pink is the LCD, light blue is fresnel and triplet, and black is the Z axis of the LCD, or the projection axis.
Notice that the red pixel's parallel light is directed to near the edge of the triplet, the green pixel's light is directed to near the center, and the blue pixel's light is directed about halfway between edge and center.
Now, if you want to, you can draw a diverging cone around those lines, because that's what actually happens, but that light does not matter.
If you extend those lines, they'll all meet at the same point on the Z axis, in theory.
I hope that clears it up for you. If not, why don't you PM Supra, and I imagine he'd be willing to clear up your questions.
Click to view attachment
Pink is the LCD, light blue is fresnel and triplet, and black is the Z axis of the LCD, or the projection axis.
Notice that the red pixel's parallel light is directed to near the edge of the triplet, the green pixel's light is directed to near the center, and the blue pixel's light is directed about halfway between edge and center.
Now, if you want to, you can draw a diverging cone around those lines, because that's what actually happens, but that light does not matter.
If you extend those lines, they'll all meet at the same point on the Z axis, in theory.
I hope that clears it up for you. If not, why don't you PM Supra, and I imagine he'd be willing to clear up your questions.
There’s nothing wrong with Supra’s explanation.
The rays after the rear fresnel won’t be collimated (parallel). You can only get collimated light if you used a theoretical point source. Once an arc is used as a source the rays will be diverging. The larger the arc, the more the rays diverge.
This trace shows the rays from the source to the triplet.
DJ
The rays after the rear fresnel won’t be collimated (parallel). You can only get collimated light if you used a theoretical point source. Once an arc is used as a source the rays will be diverging. The larger the arc, the more the rays diverge.
This trace shows the rays from the source to the triplet.
DJ
QUOTE (DAZZZLA @ Apr 11 2007, 05:03 AM)
There’s nothing wrong with Supra’s explanation.
The rays after the rear fresnel won’t be collimated (parallel). You can only get collimated light if you used a theoretical point source. Once an arc is used as a source the rays will be diverging. The larger the arc, the more the rays diverge.
The rays after the rear fresnel won’t be collimated (parallel). You can only get collimated light if you used a theoretical point source. Once an arc is used as a source the rays will be diverging. The larger the arc, the more the rays diverge.
But we can ignore that for the purposes of calculating the placement of the elements in the PJ, right? I guess that's what I've been trying (badly) to say all along. Ignore practical reality, and use the theoretical model as a base for calculating.
QUOTE
This trace shows the rays from the source to the triplet.
DJ
DJ
Now rotate your trace 90 degrees around the projection axis, so we're seeing the arc from the end. The light gets damn near parallel from that point of view. I'm not saying his explanation is wrong, but in this particular case, I think it's leading Tommy off track. To quote from his first post in this thread:
QUOTE
So if I'm right, the light that comes to the triplet is not converged light, but diverged light cones that come to one point?
and later, QUOTE
But my most essential question was/is: at wich angel does the light leave the LCD?? Because as far as I know there is no particular angle but much more angles. And the greater the angle is, the less intens the light is.
But in this case, you cannot calculate focal distances!!!
But you can calculate them, if you ignore the diverging light.But in this case, you cannot calculate focal distances!!!
To do the calculations for determining where the elements in the projector should be placed, you can use that perfect point source, and not worry about the fact that we're using an imperfect source. To borrow Supra's sig: "In theory, there's no difference between theory and practice. In practice, there is."
Or, as I'd heard it before: "The difference between theory and practice is smaller in theory than in practice."
But you can use the theory to get started!
I'm going to shut up now. (At least in this thread!
)
Whoa. Blast from the past here.
Okay, the problem seems to be with Fig. 4, right? The "cones" of light drawn in that figure represent only light that gets projected.
If we had perfectly collimated light from a perfect point source going through the LCD, then we would not even need a triplet. We wouldn't even need the fresnels, we could simply put our light source behind an LCD and project that on the wall. We don't have a light source like that, so we use lenses to create a focus.
The light DOES diverge from the LCD. In a 'perfect' model it would not, however because 1. The lamp arc is of signifigant size when compared to the level of detail that we want and 2. the LCD has a scattering effect on light passing through it.
In an exceptional build, with a small lamp and very precise alignment, the light from an individual pixel may not hit the entire surface of the lens, like I pictured.
The point is this: Any light that comes from the same point on the LCD (Or virtual LCD as in Fig. 4) and strikes the lens will be focused to the same point on the screen, regardless of where it strikes the objective lens.
Okay, the problem seems to be with Fig. 4, right? The "cones" of light drawn in that figure represent only light that gets projected.
If we had perfectly collimated light from a perfect point source going through the LCD, then we would not even need a triplet. We wouldn't even need the fresnels, we could simply put our light source behind an LCD and project that on the wall. We don't have a light source like that, so we use lenses to create a focus.
The light DOES diverge from the LCD. In a 'perfect' model it would not, however because 1. The lamp arc is of signifigant size when compared to the level of detail that we want and 2. the LCD has a scattering effect on light passing through it.
In an exceptional build, with a small lamp and very precise alignment, the light from an individual pixel may not hit the entire surface of the lens, like I pictured.
The point is this: Any light that comes from the same point on the LCD (Or virtual LCD as in Fig. 4) and strikes the lens will be focused to the same point on the screen, regardless of where it strikes the objective lens.
QUOTE (tenzip @ Apr 12 2007, 12:01 AM)
But we can ignore that for the purposes of calculating the placement of the elements in the PJ, right? I guess that's what I've been trying (badly) to say all along. Ignore practical reality, and use the theoretical model as a base for calculating.
Now rotate your trace 90 degrees around the projection axis, so we're seeing the arc from the end. The light gets damn near parallel from that point of view. I'm not saying his explanation is wrong, but in this particular case, I think it's leading Tommy off track. To quote from his first post in this thread: and later, But you can calculate them, if you ignore the diverging light.
To do the calculations for determining where the elements in the projector should be placed, you can use that perfect point source, and not worry about the fact that we're using an imperfect source. To borrow Supra's sig: "In theory, there's no difference between theory and practice. In practice, there is."
Or, as I'd heard it before: "The difference between theory and practice is smaller in theory than in practice."
But you can use the theory to get started!
I'm going to shut up now. (At least in this thread! )
Now rotate your trace 90 degrees around the projection axis, so we're seeing the arc from the end. The light gets damn near parallel from that point of view. I'm not saying his explanation is wrong, but in this particular case, I think it's leading Tommy off track. To quote from his first post in this thread: and later, But you can calculate them, if you ignore the diverging light.
To do the calculations for determining where the elements in the projector should be placed, you can use that perfect point source, and not worry about the fact that we're using an imperfect source. To borrow Supra's sig: "In theory, there's no difference between theory and practice. In practice, there is."
Or, as I'd heard it before: "The difference between theory and practice is smaller in theory than in practice."
But you can use the theory to get started!
I'm going to shut up now. (At least in this thread!
)The way I understood his question was that the light from the LCD in split fresnels shown on Supra’s forth drawing was diverging and not collimated as he expected it to be for split.
Most of the misunderstandings I see regarding optics are from using a point source for describing our projectors. Although it is easier to understand for beginners it also creates misconceptions further along in learning that are hard to dis-learn. Using a point source for finding angles is possible and convenient, sometimes, but you should have an understanding of what really is happening so that you don’t jump to the wrong conclusions. It’s akin to using a calculator for math but not knowing how to add or subtract.
DJ
QUOTE (Tommy The Cat @ Apr 9 2007, 10:48 AM)
First, thanks for this explanation. It's clear, but still I have two questions .
1. When the light goes parallel to the LCD (split setup), this explanation shows the LCD works the same as a diverging lens (like this: <). But at wich angle? You must know the angel in order to calculate the distances... (as far as I know there isn't an angle, but just different intensities at all angles)
2. So if I'm right, the light that comes to the triplet is not converged light, but diverged light cones that come to one point?
.1. When the light goes parallel to the LCD (split setup), this explanation shows the LCD works the same as a diverging lens (like this: <). But at wich angle? You must know the angel in order to calculate the distances... (as far as I know there isn't an angle, but just different intensities at all angles)
2. So if I'm right, the light that comes to the triplet is not converged light, but diverged light cones that come to one point?
Okay, I just re-read this whole thing over...
1. You've misunderstood. The LCD does not behave like a lens at all. It is merely an image source, and passes light through it. For the purposes of this article, it does not matter what "direction" the light is going when it leaves the image source, as we're only interested in the light that os going to strike the objective lens. Fig. 2-4 all make this assumption. In fact, you can take the points in any/all of these figures as leaving an LCD, or an opague image, whatever. You can also assume that they radiate light in a full 180 degree arc, with only the light that goes to the objective lens (Which is all that the lens can operate on) as pictured. Your final conclusion in parenthesis is actually correct, in that there isn't an angle to be concerned with, but different intensities, with the most intense being along that theoretical centerline.
2. This is ultimately true, though what is more important here is the relationship of the fresnel lenses to the arc length, and what it really determines is the optimum diameter for the objective lens to be. This is beyond the scope of what I was originally writing about, but is relevant to why we're using lenses instead of a pinhole camera. The arc length of most of the 400W MH lamps that we use is about 35mm. (I don't remember the exact length, but it's ABOUT that. We'll use this as an example figure anyway.) Since the 220mm and 317mm fresnels are used, they project a reasonably focused image of that arc, magnified by approximately the ratio of the lenses. 317/220 = 1.44 so the projected arc image (Where the most intense light is) is about 51mm. Since the clear diameter of the standard lens is 63mm, this works well. (Hmmm... I remember this as being a tighter fit, which means that I'm probably on the low end for the arc length.
)
First thanks to you three guys for the replies.
At last I understand. The angle of which I was looking for is the angle created by the ARC which is bigger than a focal point. The scattering of the light caused by the LCD is something we can forget here (that would be your insignificant practical info for the theory, Tenzip 
).
1. Further, if I understand correct the triplet would not be necessary if the ARC was as big as the focal point of the first fresnel if you consider 'chromatic aberration' wouldn't exist. But is does exist, so regardless of the ARC, the triplet would be needed at any time.
2. Secondly in this case the position of the LCD wouldn’t matter also. The position only matters with an ARC bigger than the focal point (at split).
3. But considering the perfect arc, Supra, we would need the second fresnel to create a bigger projection than the size of the LCD.
If these three conclusions are correct, I will be able to make a clarification at a Dutch forum (in my own language).
Edit:
Is the focal calculator of dazla also based at a particular ARC size?
Edit2:
It seems by an unsplit setup all gets messed up! I can't picture the unsplit setup in combination with the diameter of the ARC. This because the light entering the LCD is not a point where the light cones cross!
Edit3:
Considering that the triplet 'solves' the problem of the large ARC, is it accurate to conclude a smaller ARC wouldn't give any better results?
At last I understand
. The angle of which I was looking for is the angle created by the ARC which is bigger than a focal point. The scattering of the light caused by the LCD is something we can forget here (that would be your insignificant practical info for the theory, Tenzip ).1. Further, if I understand correct the triplet would not be necessary if the ARC was as big as the focal point of the first fresnel if you consider 'chromatic aberration' wouldn't exist. But is does exist, so regardless of the ARC, the triplet would be needed at any time.
2. Secondly in this case the position of the LCD wouldn’t matter also. The position only matters with an ARC bigger than the focal point (at split).
3. But considering the perfect arc, Supra, we would need the second fresnel to create a bigger projection than the size of the LCD.
If these three conclusions are correct, I will be able to make a clarification at a Dutch forum (in my own language).
Edit:
Is the focal calculator of dazla also based at a particular ARC size?
Edit2:
It seems by an unsplit setup all gets messed up! I can't picture the unsplit setup in combination with the diameter of the ARC. This because the light entering the LCD is not a point where the light cones cross!
Edit3:
Considering that the triplet 'solves' the problem of the large ARC, is it accurate to conclude a smaller ARC wouldn't give any better results?
Sigh.
1. No. Ignore ALL of that. If you had a perfect point light source you wouldn't need any lenses at all. Your projector would be a point light source and the LCD would be the front opening of it. you'd get a projection like they use in the movies and TV of the image of the LCD, getting larger as you get further away from it.
2. Also no. The size of the arc is absolutely unimportant at the LCD site. The triplet is not focusing on the arc at all, it is focusing on the LCD, and whatever light happens to be passing through the LCD at any given point. This is what the whole tutorial is about.
It does not matter what angle the light leaves the LCD, as long as it will strike the projection lens.
We use the fresnels to ensure that the majority of the light passing through the LCD does in fact hit the projection lens. That's their job -- to make sure that the light from the arc falls within that cone.
3. Yet again No. See point 1. If we had a perfect light source, you could simply project the "shadow" of the LCD onto the wall, and be done with it. It is only because such a light source does not exist that we bother with projection lenses at all.
Re edit: No. The focal calculator works on the theoretically perfect light source model, and also assumes perfect lenses. This is why you don't actually need to consider ANY of this tutorial in order to construct a projector. Arc size (Again beyond the scope of this) is irellevant to the placement of components, it only affects the ideal clear diameter of the projection lens.
Re Edit2: No, unsplit is exactly the same, but use Fig.3 as your reference, since the fresnel in Fig.4 is no longer present. None of these diagrams represent the 'cone of light' from the lamp, they all represent the potential 'cone of light' from the LCD that the objective lens can focus!
Re Edit3: since there will always be some scattering, the more accurate your light source, the more efficient that it will be. If you can find a light source of the same lumen value, but smaller size, you will end up with a better projection. The problem being that smaller = more expensive and shorter lamp life, all else being equal. This is why we don't use the lamps from commercial projectors, because though they're small and bright, they're expensive, and don't last long.
1. No. Ignore ALL of that. If you had a perfect point light source you wouldn't need any lenses at all. Your projector would be a point light source and the LCD would be the front opening of it. you'd get a projection like they use in the movies and TV of the image of the LCD, getting larger as you get further away from it.
2. Also no. The size of the arc is absolutely unimportant at the LCD site. The triplet is not focusing on the arc at all, it is focusing on the LCD, and whatever light happens to be passing through the LCD at any given point. This is what the whole tutorial is about.
It does not matter what angle the light leaves the LCD, as long as it will strike the projection lens.
We use the fresnels to ensure that the majority of the light passing through the LCD does in fact hit the projection lens. That's their job -- to make sure that the light from the arc falls within that cone.
3. Yet again No. See point 1. If we had a perfect light source, you could simply project the "shadow" of the LCD onto the wall, and be done with it. It is only because such a light source does not exist that we bother with projection lenses at all.
Re edit: No. The focal calculator works on the theoretically perfect light source model, and also assumes perfect lenses. This is why you don't actually need to consider ANY of this tutorial in order to construct a projector. Arc size (Again beyond the scope of this) is irellevant to the placement of components, it only affects the ideal clear diameter of the projection lens.
Re Edit2: No, unsplit is exactly the same, but use Fig.3 as your reference, since the fresnel in Fig.4 is no longer present. None of these diagrams represent the 'cone of light' from the lamp, they all represent the potential 'cone of light' from the LCD that the objective lens can focus!
Re Edit3: since there will always be some scattering, the more accurate your light source, the more efficient that it will be. If you can find a light source of the same lumen value, but smaller size, you will end up with a better projection. The problem being that smaller = more expensive and shorter lamp life, all else being equal. This is why we don't use the lamps from commercial projectors, because though they're small and bright, they're expensive, and don't last long.
Considering the fact I want to write a clarification of all optical processes I am so pinned to understand all details. But do not feel obligated to respond at my questions. You all done enough already, thanks so much.
I am beginning to understand almost everything. I'll come there. It's also harder for me because english is not my own language.
I understand. There is no use in using just the fresnels and no triplet if the ARC would be perfect. In that case it would be better not to use any lenses at all.
Although a perfect ARC with only the fresnels would create a focused image right (just for the understanding)?
I understand. I’ve drawn a larger ARC in the image presented by Dazla a few posts before. And I understand what you say. When the ARC becomes bigger, nothing happens to the position of the LCD in order to get a focused image. The only thing that happens is that the greater the ARC, the less light will actually fall at the triplet.
And if I understand correct, even the distance between the ARC and the collimated fresnel would not affect the LCD-triplet distance. Only here there is not a perfect light cone to the triplet anymore. Is this accurate?
And here we come to the part I can most hardly understand. In my thinking wheatear or not there will be a focused image is determined by the angles of which the light cones reach the triplet. Considering the perfect ARC in a split setup with all lenses (yes I know, lenses are not necessary here) there will only be collimated (parallel) light leaving the LCD. So the angle of the breaking of the light in this case will only be determind by the FL of the field fresnel. In this case you could put the lcd 1 cm or 10 cm behind the field fresnel, it doesn’t matter because this distance LCD-fresnel does not affect the breaking angle!!!
In other words. In my thinking it is not important where the light gets the information of the LCD (movie) but at wich angle the light is being ‘projected’ at the triplet.
I cannot yet visualize this. What I see at unsplit is that all the ARC light leaving the field fresnel is not coming to one point anymore (see Dazla's draw a few posts earlier). And at split you see it does happen exactly where the LCD is positioned!
So in my thinking it’s important that the LCD is placed at the point all light of the ARC comes to one point, right?
In that case, the position of the collimated Fresnel – LCD is also relevant, is that correct?
So what you’re saying here is that there will be less scattering with a smaller ARC. The scattering means: less light which will actually fall into the triplet, or it will disturb other projected light rays.
In that case, a bigger ARC, but also with sufficient more lumen, will result in the same projection-quality as a smaller ARC with less lumen. Is that correct?
I am beginning to understand almost everything. I'll come there. It's also harder for me because english is not my own language.
QUOTE (SupraGuy @ Apr 12 2007, 10:39 PM)
Sigh.
1. No. Ignore ALL of that. If you had a perfect point light source you wouldn't need any lenses at all. Your projector would be a point light source and the LCD would be the front opening of it. you'd get a projection like they use in the movies and TV of the image of the LCD, getting larger as you get further away from it.
1. No. Ignore ALL of that. If you had a perfect point light source you wouldn't need any lenses at all. Your projector would be a point light source and the LCD would be the front opening of it. you'd get a projection like they use in the movies and TV of the image of the LCD, getting larger as you get further away from it.
I understand. There is no use in using just the fresnels and no triplet if the ARC would be perfect. In that case it would be better not to use any lenses at all.
Although a perfect ARC with only the fresnels would create a focused image right (just for the understanding)?
QUOTE (SupraGuy @ Apr 12 2007, 10:39 PM)
2. Also no. The size of the arc is absolutely unimportant at the LCD site. The triplet is not focusing on the arc at all, it is focusing on the LCD, and whatever light happens to be passing through the LCD at any given point. This is what the whole tutorial is about.
It does not matter what angle the light leaves the LCD, as long as it will strike the projection lens.
We use the fresnels to ensure that the majority of the light passing through the LCD does in fact hit the projection lens. That's their job -- to make sure that the light from the arc falls within that cone.
It does not matter what angle the light leaves the LCD, as long as it will strike the projection lens.
We use the fresnels to ensure that the majority of the light passing through the LCD does in fact hit the projection lens. That's their job -- to make sure that the light from the arc falls within that cone.
I understand. I’ve drawn a larger ARC in the image presented by Dazla a few posts before. And I understand what you say. When the ARC becomes bigger, nothing happens to the position of the LCD in order to get a focused image. The only thing that happens is that the greater the ARC, the less light will actually fall at the triplet.
And if I understand correct, even the distance between the ARC and the collimated fresnel would not affect the LCD-triplet distance. Only here there is not a perfect light cone to the triplet anymore. Is this accurate?
QUOTE (SupraGuy @ Apr 12 2007, 10:39 PM)
Re edit: No. The focal calculator works on the theoretically perfect light source model, and also assumes perfect lenses. This is why you don't actually need to consider ANY of this tutorial in order to construct a projector. Arc size (Again beyond the scope of this) is irellevant to the placement of components, it only affects the ideal clear diameter of the projection lens.
And here we come to the part I can most hardly understand. In my thinking wheatear or not there will be a focused image is determined by the angles of which the light cones reach the triplet. Considering the perfect ARC in a split setup with all lenses (yes I know, lenses are not necessary here) there will only be collimated (parallel) light leaving the LCD. So the angle of the breaking of the light in this case will only be determind by the FL of the field fresnel. In this case you could put the lcd 1 cm or 10 cm behind the field fresnel, it doesn’t matter because this distance LCD-fresnel does not affect the breaking angle!!!
In other words. In my thinking it is not important where the light gets the information of the LCD (movie) but at wich angle the light is being ‘projected’ at the triplet.
QUOTE (SupraGuy @ Apr 12 2007, 10:39 PM)
Re Edit2: No, unsplit is exactly the same, but use Fig.3 as your reference, since the fresnel in Fig.4 is no longer present. None of these diagrams represent the 'cone of light' from the lamp, they all represent the potential 'cone of light' from the LCD that the objective lens can focus!
I cannot yet visualize this. What I see at unsplit is that all the ARC light leaving the field fresnel is not coming to one point anymore (see Dazla's draw a few posts earlier). And at split you see it does happen exactly where the LCD is positioned!
So in my thinking it’s important that the LCD is placed at the point all light of the ARC comes to one point, right?
In that case, the position of the collimated Fresnel – LCD is also relevant, is that correct?
QUOTE (SupraGuy @ Apr 12 2007, 10:39 PM)
Re Edit3: since there will always be some scattering, the more accurate your light source, the more efficient that it will be. If you can find a light source of the same lumen value, but smaller size, you will end up with a better projection. The problem being that smaller = more expensive and shorter lamp life, all else being equal. This is why we don't use the lamps from commercial projectors, because though they're small and bright, they're expensive, and don't last long.
So what you’re saying here is that there will be less scattering with a smaller ARC. The scattering means: less light which will actually fall into the triplet, or it will disturb other projected light rays.
In that case, a bigger ARC, but also with sufficient more lumen, will result in the same projection-quality as a smaller ARC with less lumen. Is that correct?
QUOTE (Tommy The Cat @ Apr 13 2007, 08:47 PM)
Considering the fact I want to write a clarification of all optical processes I am so pinned to understand all details. But do not feel obligated to respond at my questions. You all done enough already, thanks so much.
I am beginning to understand almost everything. I'll come there. It's also harder for me because english is not my own language.
I understand. There is no use in using just the fresnels and no triplet if the ARC would be perfect. In that case it would be better not to use any lenses at all.
Although a perfect ARC with only the fresnels would create a focused image right (just for the understanding)?
I understand. I’ve drawn a larger ARC in the image presented by Dazla a few posts before. And I understand what you say. When the ARC becomes bigger, nothing happens to the position of the LCD in order to get a focused image. The only thing that happens is that the greater the ARC, the less light will actually fall at the triplet.
And if I understand correct, even the distance between the ARC and the collimated fresnel would not affect the LCD-triplet distance. Only here there is not a perfect light cone to the triplet anymore. Is this accurate?
And here we come to the part I can most hardly understand. In my thinking wheatear or not there will be a focused image is determined by the angles of which the light cones reach the triplet. Considering the perfect ARC in a split setup with all lenses (yes I know, lenses are not necessary here) there will only be collimated (parallel) light leaving the LCD. So the angle of the breaking of the light in this case will only be determind by the FL of the field fresnel. In this case you could put the lcd 1 cm or 10 cm behind the field fresnel, it doesn’t matter because this distance LCD-fresnel does not affect the breaking angle!!!
In other words. In my thinking it is not important where the light gets the information of the LCD (movie) but at wich angle the light is being ‘projected’ at the triplet.
I cannot yet visualize this. What I see at unsplit is that all the ARC light leaving the field fresnel is not coming to one point anymore (see Dazla's draw a few posts earlier). And at split you see it does happen exactly where the LCD is positioned!
So in my thinking it’s important that the LCD is placed at the point all light of the ARC comes to one point, right?
In that case, the position of the collimated Fresnel – LCD is also relevant, is that correct?
So what you’re saying here is that there will be less scattering with a smaller ARC. The scattering means: less light which will actually fall into the triplet.
In that case, a bigger ARC, but also with sufficient more lumen, will result in the same projection-quality as a smaller ARC with less lumen. Is that correct?
I am beginning to understand almost everything. I'll come there. It's also harder for me because english is not my own language.
I understand. There is no use in using just the fresnels and no triplet if the ARC would be perfect. In that case it would be better not to use any lenses at all.
Although a perfect ARC with only the fresnels would create a focused image right (just for the understanding)?
I understand. I’ve drawn a larger ARC in the image presented by Dazla a few posts before. And I understand what you say. When the ARC becomes bigger, nothing happens to the position of the LCD in order to get a focused image. The only thing that happens is that the greater the ARC, the less light will actually fall at the triplet.
And if I understand correct, even the distance between the ARC and the collimated fresnel would not affect the LCD-triplet distance. Only here there is not a perfect light cone to the triplet anymore. Is this accurate?
And here we come to the part I can most hardly understand. In my thinking wheatear or not there will be a focused image is determined by the angles of which the light cones reach the triplet. Considering the perfect ARC in a split setup with all lenses (yes I know, lenses are not necessary here) there will only be collimated (parallel) light leaving the LCD. So the angle of the breaking of the light in this case will only be determind by the FL of the field fresnel. In this case you could put the lcd 1 cm or 10 cm behind the field fresnel, it doesn’t matter because this distance LCD-fresnel does not affect the breaking angle!!!
In other words. In my thinking it is not important where the light gets the information of the LCD (movie) but at wich angle the light is being ‘projected’ at the triplet.
I cannot yet visualize this. What I see at unsplit is that all the ARC light leaving the field fresnel is not coming to one point anymore (see Dazla's draw a few posts earlier). And at split you see it does happen exactly where the LCD is positioned!
So in my thinking it’s important that the LCD is placed at the point all light of the ARC comes to one point, right?
In that case, the position of the collimated Fresnel – LCD is also relevant, is that correct?
So what you’re saying here is that there will be less scattering with a smaller ARC. The scattering means: less light which will actually fall into the triplet.
In that case, a bigger ARC, but also with sufficient more lumen, will result in the same projection-quality as a smaller ARC with less lumen. Is that correct?
Try reading this. It may help answer a few of your questions.
QUOTE (Tommy The Cat @ Apr 13 2007, 03:47 AM)
Although a perfect ARC with only the fresnels would create a focused image right (just for the understanding)?
I understand. I’ve drawn a larger ARC in the image presented by Dazla a few posts before. And I understand what you say. When the ARC becomes bigger, nothing happens to the position of the LCD in order to get a focused image. The only thing that happens is that the greater the ARC, the less light will actually fall at the triplet.
I understand. I’ve drawn a larger ARC in the image presented by Dazla a few posts before. And I understand what you say. When the ARC becomes bigger, nothing happens to the position of the LCD in order to get a focused image. The only thing that happens is that the greater the ARC, the less light will actually fall at the triplet.
This would be true if the fresnels were also perfect, but then fresnels are by definition imperfect lenses.
QUOTE (Tommy The Cat @ Apr 13 2007, 03:47 AM)
And if I understand correct, even the distance between the ARC and the collimated fresnel would not affect the LCD-triplet distance. Only here there is not a perfect light cone to the triplet anymore. Is this accurate?
Abolutely correct.
QUOTE (Tommy The Cat @ Apr 13 2007, 03:47 AM)
And here we come to the part I can most hardly understand. In my thinking wheatear or not there will be a focused image is determined by the angles of which the light cones reach the triplet. Considering the perfect ARC in a split setup with all lenses (yes I know, lenses are not necessary here) there will only be collimated (parallel) light leaving the LCD. So the angle of the breaking of the light in this case will only be determind by the FL of the field fresnel. In this case you could put the lcd 1 cm or 10 cm behind the field fresnel, it doesn’t matter because this distance LCD-fresnel does not affect the breaking angle!!!
You are reading too much into this.
In a split setup the collimated light passes through the LCD. In an unsplit setup, light that is already converging on the triplet passes through the LCD. In both cases, what is important is that light which has passed through the LCD reaches the triplet.
With the triplet at the appropriate distance from both the LCD and the screen, it will reassemble that information by making sure that all the light that leaves a given point of the LCD will arrive at the corresponding point of the screen, regardless of where it actually hits the lens.
QUOTE (Tommy The Cat @ Apr 13 2007, 03:47 AM)
In other words. In my thinking it is not important where the light gets the information of the LCD (movie) but at wich angle the light is being ‘projected’ at the triplet.
You have it backwards. The angle of the light hitting the triplet is unimportant (As long as it ddoes actually hit the triplet.) What is critical is where the light leaves the LCD. That is what the lens focuses.
QUOTE (Tommy The Cat @ Apr 13 2007, 03:47 AM)
I cannot yet visualize this. What I see at unsplit is that all the ARC light leaving the field fresnel is not coming to one point anymore (see Dazla's draw a few posts earlier). And at split you see it does happen exactly where the LCD is positioned!
This is really just a convention with Dazzla's drawing. Since the LCD is where the information comes from, it's convenient to think of it in this manner, but absolutely unnecessary. Split or unsplit is the exact same as far as the projection lens is concerned, the only difference is where the triplet is positioned in order to focus on the LCD. This moves somewhat for a split setup, because the collector fresnel makes a 'virtual' LCD, and the objective lens needs to focus on that.
QUOTE (Tommy The Cat @ Apr 13 2007, 03:47 AM)
So in my thinking it’s important that the LCD is placed at the point all light of the ARC comes to one point, right?
No. Not at all. Though it is convenient to think of it that way, there is no actual 'point' that the light comes to. What is being drawn here are ficticious boundaries to the "useful" light -- that is light that passes through the LCD. No matter where the LCD is positioned, since the arc has a nonzero size, there will be this 'point' at it's edges simply because light that is outside of that arc is useless to us, so we disregard it. Light outside of that point still exists, but since it does not pass through the LCD to the triplet, it is irrelevant.
QUOTE (Tommy The Cat @ Apr 13 2007, 03:47 AM)
In that case, the position of the collimated Fresnel – LCD is also relevant, is that correct?
So what you’re saying here is that there will be less scattering with a smaller ARC. The scattering means: less light which will actually fall into the triplet, or it will disturb other projected light rays.
In that case, a bigger ARC, but also with sufficient more lumen, will result in the same projection-quality as a smaller ARC with less lumen. Is that correct?
So what you’re saying here is that there will be less scattering with a smaller ARC. The scattering means: less light which will actually fall into the triplet, or it will disturb other projected light rays.
In that case, a bigger ARC, but also with sufficient more lumen, will result in the same projection-quality as a smaller ARC with less lumen. Is that correct?
Well, the position of the collimator fresnel is relatively unimportant, provided that is is at least 1cm from the LCD. This is actually so that the rings in the fresnel do not come into the focal range of the triplet.
A larger arc with more lumens is better, unless the arc is too large to then 'fit' within the triplet. The reason is that with a too large arc, you will get much better efficiency near the center of the fresnels than you will near the edges. This will result in a more extreme difference between the lighting at the center vs, the lighting at the corners. In all cases a smaller arc with the same lumens will be better. It is only for more lumens that a larger arc is acceptable.
I hope this addresses your concerns.
Thank you so much for your persistence out of free will.
I understand this completely, I really do, look I will show it to you: In theory I say to myself: Tommy, the fresnel only is there to make sure the light comes to the triplet, but the triplet is focussing at the LCD< and you place the LCD behind it's focal point so the triplet can focus it at the projection screen!!! And I say to myself: Tommy, you fool, it does not matter at which angle light passes the LCD, as long as it reaches the triplet!!! The triplet will create an exact copy of it because if from one pixel light would fall to all edges at the triplet, the triplet would get all those beams to one point at the other side!! So if you would place two lamps right behind the lcd and all light would go everywhere and all light beams would get messed up, the triplet doesn't care!!! He can handle it!
And than (for the part where triplet focuses at LCD at split) I see this what I have drawn below and my head all gets messed up again and it's so frustrating!!!
Considering a perfect ARC (theoratic focal point). What's the bloody difference for the triplet!! The virtual image which it sees of the LCD is in both situations exactly the same!!! I'm so sorry for my stupidity.
I understand this completely, I really do, look I will show it to you: In theory I say to myself: Tommy, the fresnel only is there to make sure the light comes to the triplet, but the triplet is focussing at the LCD< and you place the LCD behind it's focal point so the triplet can focus it at the projection screen!!! And I say to myself: Tommy, you fool, it does not matter at which angle light passes the LCD, as long as it reaches the triplet!!! The triplet will create an exact copy of it because if from one pixel light would fall to all edges at the triplet, the triplet would get all those beams to one point at the other side!! So if you would place two lamps right behind the lcd and all light would go everywhere and all light beams would get messed up, the triplet doesn't care!!! He can handle it!
And than (for the part where triplet focuses at LCD at split) I see this what I have drawn below and my head all gets messed up again and it's so frustrating!!!
Considering a perfect ARC (theoratic focal point). What's the bloody difference for the triplet!! The virtual image which it sees of the LCD is in both situations exactly the same!!! I'm so sorry for my stupidity.
Now wait a minute. Maybe there would be a difference according to my next interpretation (after re-reading dazla's link):
Is this an accurate picture of the difference?
Damn I think I got everything now. Is it right that indeed there would be no difference at split BUT only when we use a theoretic focal point (perfect arc)? If this is a YES, I am absolutely done trying to understand!!!!!!!
Is this an accurate picture of the difference?
Damn I think I got everything now. Is it right that indeed there would be no difference at split BUT only when we use a theoretic focal point (perfect arc)? If this is a YES, I am absolutely done trying to understand
!!!!!!!
I think you pretty much understand this, though I don't understand your diagrams in the last post there. (I don't know what it is that each thing you've drawn is.)
The diagrams that confuse you are basically 'pinhole camera' diagrams, which is exactly how the lens ultimately behaves. This works as a simplified model for a projector, so it is adequate to explain the workings to someone so that they can build one.
The diagrams that confuse you are basically 'pinhole camera' diagrams, which is exactly how the lens ultimately behaves. This works as a simplified model for a projector, so it is adequate to explain the workings to someone so that they can build one.
QUOTE (SupraGuy @ Apr 14 2007, 10:24 PM)
I think you pretty much understand this, though I don't understand your diagrams in the last post there. (I don't know what it is that each thing you've drawn is.)
The diagrams that confuse you are basically 'pinhole camera' diagrams, which is exactly how the lens ultimately behaves. This works as a simplified model for a projector, so it is adequate to explain the workings to someone so that they can build one.
The diagrams that confuse you are basically 'pinhole camera' diagrams, which is exactly how the lens ultimately behaves. This works as a simplified model for a projector, so it is adequate to explain the workings to someone so that they can build one.
Well what I've drawn is two similar situations with the only difference the place of the lcd. Using the link Dazla profided what you see here is what the difference in LCD distance means for the triplet at split. Furthering away the triplet increases the virtual LCD image which is seen by the triplet. I hope this is accurate.
At least I think this is how I must see it to understand why this distance is important. But only when you work with a non perfect ARC, right?
Let me put it this way for maybe the best understanding.
Where ever you place the LCD at split or unsplit, the light that leaves a particular pixel always is diverged (like this: < ) because there is an infinite number of light beams going multiple directions (non perfect ARC) although with a certain maximum angle determind by the ARC size. But considering the light from that individual pixel that get's into the triplet, the angle is determined by the LCD distance to the triplet (or the triplet's point of view). And this distance in combination with the focal length of the triplet will determine if you get a focus. So that's what the objective focusses at, like this: <>
All the light beams drawn in Dazla's drawing a few posts above is only a schematic picture to see if the fresnels make a good cone to the triplet, but that's a different story and only makes sure all light physically passes through it. Don't let these schematic light beams narrow your brain's view. This all doesn't affect the focus story of the LCD and the triplet because of the infinite numbers of light beams (ooh the LCD and triplet really love each other, don't they
).
Wow, did I write this? It seems it just popped up in my head. Let me only hope it's correct.
Where ever you place the LCD at split or unsplit, the light that leaves a particular pixel always is diverged (like this: < ) because there is an infinite number of light beams going multiple directions (non perfect ARC) although with a certain maximum angle determind by the ARC size. But considering the light from that individual pixel that get's into the triplet, the angle is determined by the LCD distance to the triplet (or the triplet's point of view). And this distance in combination with the focal length of the triplet will determine if you get a focus. So that's what the objective focusses at, like this: <>
All the light beams drawn in Dazla's drawing a few posts above is only a schematic picture to see if the fresnels make a good cone to the triplet, but that's a different story and only makes sure all light physically passes through it. Don't let these schematic light beams narrow your brain's view. This all doesn't affect the focus story of the LCD and the triplet because of the infinite numbers of light beams (ooh the LCD and triplet really love each other, don't they
).Wow, did I write this? It seems it just popped up in my head. Let me only hope it's correct
.
Close enough. Except that the LCD is a faithless harlot and those 'infinite number' of beams don't just hit the triplet, they also hit the area around the triplet, the side of the case... Basically everywhere. The difference with these is that they don't get projected, so we really don't care about them. The fresnels are there to ensure that MOST of the beams hit the triplet.
To abuse your lovers analogy a bit more, the triplet has a bit of a wandering eye as well, and will also project anything else it 'sees'. (This is why we tend to paint the inside of the projector black.)
To abuse your lovers analogy a bit more, the triplet has a bit of a wandering eye as well, and will also project anything else it 'sees'. (This is why we tend to paint the inside of the projector black.)
Can you explain the difference in quality while using lenses (triplet) with different focal lengths?
I see that the pro lens in the lumenlab store is 500mm and standard is 320mm.
In photography, higher the focal length lesser the quality, of course keeping the film speed, aperture etc constant. Ofcourse the you focus closer.
I'm not sure how it works in a projector. Wouldnt increasing the distance of the triplet from the LCD increase the focal length?
And would this work if I'm using a 15" LCD?
Cheers!
I see that the pro lens in the lumenlab store is 500mm and standard is 320mm.
In photography, higher the focal length lesser the quality, of course keeping the film speed, aperture etc constant. Ofcourse the you focus closer.
I'm not sure how it works in a projector. Wouldnt increasing the distance of the triplet from the LCD increase the focal length?
And would this work if I'm using a 15" LCD?
Cheers!
This is a "lo-fi" version of our main content. To view the full version with more information, formatting and images, please click here.