the ideal long-throw design
All of the pieces to make the "holy grail" projector will soon be available:
15.8" 1280 by 1024 LCD (can support 720P HD format)
550 mm fl fresnel
500 mm wide-FOV triplet (coming soon)
The two things we are missing are:
A good cheap hot mirror large enough in diameter (ie. 5")
A good cheap dichroic spherical reflector.
Combine all of these with a standard 220 mm fl fresnel, a piece of Lexan XL10 as a UV filter, and a cooling fan or two, and you will have an HD-compatible projector that blows the sox off anything a DIY builder has ever made.
All of the pieces to make the "holy grail" projector will soon be available:
15.8" 1280 by 1024 LCD (can support 720P HD format)
550 mm fl fresnel
500 mm wide-FOV triplet (coming soon)
The two things we are missing are:
A good cheap hot mirror large enough in diameter (ie. 5")
A good cheap dichroic spherical reflector.
Combine all of these with a standard 220 mm fl fresnel, a piece of Lexan XL10 as a UV filter, and a cooling fan or two, and you will have an HD-compatible projector that blows the sox off anything a DIY builder has ever made.
https://secure.lumenlab.com/shop/index.php?ref=&
it isn´t available yet, but I believe it will be soon.
it isn´t available yet, but I believe it will be soon.
I wonder if I could have my math checked:
Projection Lens: 11.6" - 19.4" VFL; 11" barrel, 4" diameter
Screen Size: 81" diagonal
Throw: 110"
LCD: 17" Diagonal
So, using:
Screen Size = (Throw - FL) x (LCD Size) / FL
81 = (110 - FL) x (17) / FL
FL = 19.1" or 485mm
Now to find the distance from LCD to projection lens:
1 / (Focal Length) = 1 / (LCD to Lens Distance) + 1 / (Throw)
1 / 19.1 = 1 / (LCD to Lens Distance) + 1 / 110
Distance = 23.3" or 591mm
Now this is where I am having a problem. I have read you want to mark the distance from the LCD to the MIDDLE of the projection lens. Is this correct or should I go from the LCD to the back end of the projection lens?
And... For the condensing fresnel lens, does its focal length have to equal or exceed the projection lens' focal length (485mm) or does it have to equal or exceed the LCD to projection lens distance (590mm)? Do I measure from the fresnel lens to the back end of the projection lens? Or do I measure from the middle of the projection lens?
This info makes a difference. There is a 550mm fresnel lens available, but if I have to go 590mm... Thanks for the help!
Projection Lens: 11.6" - 19.4" VFL; 11" barrel, 4" diameter
Screen Size: 81" diagonal
Throw: 110"
LCD: 17" Diagonal
So, using:
Screen Size = (Throw - FL) x (LCD Size) / FL
81 = (110 - FL) x (17) / FL
FL = 19.1" or 485mm
Now to find the distance from LCD to projection lens:
1 / (Focal Length) = 1 / (LCD to Lens Distance) + 1 / (Throw)
1 / 19.1 = 1 / (LCD to Lens Distance) + 1 / 110
Distance = 23.3" or 591mm
Now this is where I am having a problem. I have read you want to mark the distance from the LCD to the MIDDLE of the projection lens. Is this correct or should I go from the LCD to the back end of the projection lens?
And... For the condensing fresnel lens, does its focal length have to equal or exceed the projection lens' focal length (485mm) or does it have to equal or exceed the LCD to projection lens distance (590mm)? Do I measure from the fresnel lens to the back end of the projection lens? Or do I measure from the middle of the projection lens?
This info makes a difference. There is a 550mm fresnel lens available, but if I have to go 590mm... Thanks for the help!
yes, you need to consider that 19" focal lengh triplet as a thin lens. That is the equivalent thing lens that would be placed on the optical center of your varifocal lens (well, we use to say fisical center=optical center, this is not true exactly but it is a close enough aproximation).
now you need the field fresnell to condense light on the optical center as well (591mm from lcd so if you are going for unsplit, add 20mm to your ideal field fresnell; 611mm focal lengh, or 571mm if you go for split design. Now you wont find any of those fresnells, you need the closest 550 or 650mm that are available for us.
you can always play with lamp positioning to fine tune the focusing point on the optical center of the triplet. Good luck
now you need the field fresnell to condense light on the optical center as well (591mm from lcd so if you are going for unsplit, add 20mm to your ideal field fresnell; 611mm focal lengh, or 571mm if you go for split design. Now you wont find any of those fresnells, you need the closest 550 or 650mm that are available for us.
you can always play with lamp positioning to fine tune the focusing point on the optical center of the triplet. Good luck
Thanks Rox! So the middle of the lens is where I want to measure my distances from. Both the fresnel distance and the LCD distance is measured from the middle of the projection lens.
I am trying to get a grasp on the relationships between all of the lenses, the light, and the LCD. Kinda complex.
I will have to toy with the lenses and the throw and the screen size to make it fit a 650mm fresnel lens.
I am trying to get a grasp on the relationships between all of the lenses, the light, and the LCD. Kinda complex.
I will have to toy with the lenses and the throw and the screen size to make it fit a 650mm fresnel lens.
95% energy of LCD pixels
I'm just wondering, I was asked by the optics shop with this question and
they don't quite understand this, the lens must transfer 95% energy of LCD pixels to the screen.
They asked me instead with the resolution of this lens or MTF data.
Could you answer this Guy or Rox?
Thanks!
I'm just wondering, I was asked by the optics shop with this question and
they don't quite understand this, the lens must transfer 95% energy of LCD pixels to the screen.
They asked me instead with the resolution of this lens or MTF data.
Could you answer this Guy or Rox?
Thanks!
Another Question,
Using this formula
1/focal length = 1/LCD to lens + 1/lens to screen
I came up with the figures below, does it mean I cannot use the 550mm Fresnel or
I have to increase the distance of "Field Fresnel to LCD" to (550mm-503.2mm=46.8mm)
This always confuses me because I understand that we have to put the arc image of the bulb to the
center of the triplet so the distance will be 550mm from the Field Fresnel to Center of the triplet(bulb image at the center).
But if we use the above equation, we have to move the "Objlens to LCD" distance so the arc image is not in the
center now?
LCD Size= 17"
FL of Obj= 17.72(450mm)
Throw Dist.= 168"
Screen Sz(Diagonal) = 144.2
Lcd 2 LCD = 19.81 (503.2mm)
Ideal Fresnel = 503.2mm+20mm=523.2mm
Using this formula
1/focal length = 1/LCD to lens + 1/lens to screen
I came up with the figures below, does it mean I cannot use the 550mm Fresnel or
I have to increase the distance of "Field Fresnel to LCD" to (550mm-503.2mm=46.8mm)
This always confuses me because I understand that we have to put the arc image of the bulb to the
center of the triplet so the distance will be 550mm from the Field Fresnel to Center of the triplet(bulb image at the center).
But if we use the above equation, we have to move the "Objlens to LCD" distance so the arc image is not in the
center now?
LCD Size= 17"
FL of Obj= 17.72(450mm)
Throw Dist.= 168"
Screen Sz(Diagonal) = 144.2
Lcd 2 LCD = 19.81 (503.2mm)
Ideal Fresnel = 503.2mm+20mm=523.2mm
I haven't been able to find a 600mm or a 650mm fresnel lens in a size big enough for a 17" LCD. Soooooooooo.... I might do the split fresnel design. The only problem with this is I have been reading that you want to put the field fresnel about 20mm after the LCD (in a split design). With a 550mm FL fresnel and the projection lens placed 590mm from the LCD, I would need to make the space 40mm from the LCD. Is this too far?
95% of energy
>the resolution of this lens or MTF data.
>Could you answer this Guy or Rox?
Resolution of a particular spot size does not mean that all the light from a particular spot on the LCD ends up at a particular spot on the screen. If 95% of the light from everywhere in an LCD pixel ends up within a one pixel-sized square on the screen, then the image will look fine. 15" LCDs have 0.293 mm square pixels, so with magnification around 6.7, the screen pixels end up being about 1.95 mm.
To see if a particular spot size would meet that goal could be solved with the right calculus equation, but I would just write a program to try different spot sizes until the sum of all the light coming from different spots in the LCD pixel was 95% inside the screen pixel boundaries. Have at it!
>the resolution of this lens or MTF data.
>Could you answer this Guy or Rox?
Resolution of a particular spot size does not mean that all the light from a particular spot on the LCD ends up at a particular spot on the screen. If 95% of the light from everywhere in an LCD pixel ends up within a one pixel-sized square on the screen, then the image will look fine. 15" LCDs have 0.293 mm square pixels, so with magnification around 6.7, the screen pixels end up being about 1.95 mm.
To see if a particular spot size would meet that goal could be solved with the right calculus equation, but I would just write a program to try different spot sizes until the sum of all the light coming from different spots in the LCD pixel was 95% inside the screen pixel boundaries. Have at it!
adjusting the arc image location
If the 550 fl mm fresnel gets parallel rays, then it will focus those to a point 550 mm from the fresnel. But you don't have to send it perfectly parallel rays: If you send it slightly diverging rays, then it will focus the arc image further than 550 mm away.
Likewise, if you put the lamp arc exactly 220 mm from a 220 mm fl fresnel, then you get parallel rays out the other side. But if you move the lamp arc just a bit closer, you get a diverging cone of rays out the other side.
So you can use the lamp arc to condensor fresnel distance to adjust where the field fresnel will focus the arc image.
If you need to do keystone adjustment, then you can try putting the field fresnel about 30 mm from the LCD with a pivet mounting. If not, then you can put it 20 mm from the LCD. Closer is better in terms of less distortion by the fresnel. Too close and you will see the fresnel rings in your image. If you use the new 550 mm fl fresnel, then you can put it 10 mm away and still not see rings. That's where mine is right now.
If the 550 fl mm fresnel gets parallel rays, then it will focus those to a point 550 mm from the fresnel. But you don't have to send it perfectly parallel rays: If you send it slightly diverging rays, then it will focus the arc image further than 550 mm away.
Likewise, if you put the lamp arc exactly 220 mm from a 220 mm fl fresnel, then you get parallel rays out the other side. But if you move the lamp arc just a bit closer, you get a diverging cone of rays out the other side.
So you can use the lamp arc to condensor fresnel distance to adjust where the field fresnel will focus the arc image.
If you need to do keystone adjustment, then you can try putting the field fresnel about 30 mm from the LCD with a pivet mounting. If not, then you can put it 20 mm from the LCD. Closer is better in terms of less distortion by the fresnel. Too close and you will see the fresnel rings in your image. If you use the new 550 mm fl fresnel, then you can put it 10 mm away and still not see rings. That's where mine is right now.
adjusting the arc image location
Did you mean to say, put the lamp arc a bit further away to get slightly converging rays out of the 220mm fresnel, so that the 550mm fresnel will focus a bit closer?
So the design would go;
lamp
220mm FL fresnel (placed at slightly further away then 220mm)
550mm FL fresnel (right after the 220mm fresnel, focusing closer than 550mm, @ 503mm for buddy123)
lcd (10mm from the 550mm fresnel, or as close as possible without seeing rings)
triplet lens (for buddy123 @ 503mm away, focusing on the lcd)
Gunvald
Did you mean to say, put the lamp arc a bit further away to get slightly converging rays out of the 220mm fresnel, so that the 550mm fresnel will focus a bit closer?
So the design would go;
lamp
220mm FL fresnel (placed at slightly further away then 220mm)
550mm FL fresnel (right after the 220mm fresnel, focusing closer than 550mm, @ 503mm for buddy123)
lcd (10mm from the 550mm fresnel, or as close as possible without seeing rings)
triplet lens (for buddy123 @ 503mm away, focusing on the lcd)
Gunvald
I just got my projection lens from ebay in the mail!!! It's a Buhl VFL projection lens. At least 4 lenses in this thing. Heavy. 100mm back lens and 90mm lens in the front.
So, I was trying the lens out with the 17" LCD I am using. No fresnel lenses yet and the LCD is using its own backlight (no MH bulb yet). In a dark room I set up the lens and LCD and found that I couldn't see the whole LCD on the wall. Now, is this because I don't have a field fresnel to condense the picture yet? Or a better question would be, would the field fresnel help the projection lens to "see" the whole LCD?
I also have a thought. The mounting bracket came with the lens. It looks like it mounts right to an LCD (the lens was originally from a projector with a 3.1" diag LCD). So I measured how far it was from the back of the lens to the back of the mount; 193mm. I was thinking if the image from my 17" LCD shrinks to 3.1" diag at 193mm from the lens, the lens would "see" the 17" LCD as a 3.1" LCD. Is this just wishful thinking or is there any truth to my blabber?
This is the projection lens in the mounting bracket
This is the lens out of the bracket. The back end of the lens slides into the bracket. So I measured from the back of the lens to the end of the mounting bracket.
So, I was trying the lens out with the 17" LCD I am using. No fresnel lenses yet and the LCD is using its own backlight (no MH bulb yet). In a dark room I set up the lens and LCD and found that I couldn't see the whole LCD on the wall. Now, is this because I don't have a field fresnel to condense the picture yet? Or a better question would be, would the field fresnel help the projection lens to "see" the whole LCD?
I also have a thought. The mounting bracket came with the lens. It looks like it mounts right to an LCD (the lens was originally from a projector with a 3.1" diag LCD). So I measured how far it was from the back of the lens to the back of the mount; 193mm. I was thinking if the image from my 17" LCD shrinks to 3.1" diag at 193mm from the lens, the lens would "see" the 17" LCD as a 3.1" LCD. Is this just wishful thinking or is there any truth to my blabber?
An externally hosted image should be here but it was not working when we last tested it.
This is the projection lens in the mounting bracket
An externally hosted image should be here but it was not working when we last tested it.
This is the lens out of the bracket. The back end of the lens slides into the bracket. So I measured from the back of the lens to the end of the mounting bracket.
no cigar!
Sorry, superdaveumo! You are correct: That is wishful thinking. Your lens is designed for a 3.1" FOV when the LCD is the right distance for focussing an image on the screen. That LCD-to-lens distance will change a bit as the throw distance changes, but it will never show more than 3-4" of the 17" LCD.
One thing that should have been a dead give-away: The lens is so long and narrow that the rays from the corners of a 17" LCD could not possible get through that lens to the screen, unless the LCD was 6 feet away! A good lens for a 17" LCD will be larger in diameter than it is long.
Gunvald: "put the lamp arc a bit further away to get slightly converging rays out of the 220mm fresnel, so that the 550mm fresnel will focus a bit closer"
Yes, either way, depending on where you need it to focus. Design with the fresnels that are closest to the ideal distances, and then use this trick to adjust the actual arc image focus location.
Sorry, superdaveumo! You are correct: That is wishful thinking. Your lens is designed for a 3.1" FOV when the LCD is the right distance for focussing an image on the screen. That LCD-to-lens distance will change a bit as the throw distance changes, but it will never show more than 3-4" of the 17" LCD.
One thing that should have been a dead give-away: The lens is so long and narrow that the rays from the corners of a 17" LCD could not possible get through that lens to the screen, unless the LCD was 6 feet away! A good lens for a 17" LCD will be larger in diameter than it is long.
Gunvald: "put the lamp arc a bit further away to get slightly converging rays out of the 220mm fresnel, so that the 550mm fresnel will focus a bit closer"
Yes, either way, depending on where you need it to focus. Design with the fresnels that are closest to the ideal distances, and then use this trick to adjust the actual arc image focus location.
Shoot. That's what I was afraid of. I was able to get a 17" widescreen in focus on the wall (from my laptop) but, as you said, the lens was around 4 feet from the LCD. And, the screen image was smaller than the LCD! DOH! Back to the drawing board.
So, you mentioned a process lens in a previous post. Something that would do 14x11 at 18" to 25" would work? 20x16 even better? Should I watch out for diameter of the process lens? Bigger is better?
Thanks for the info Guy.
Eureka!
>the screen image was smaller than the LCD
Congrats! You have just reinvented the camera! 😀
The thin lens equation says:
1/focal length = 1/object to lens + 1/lens to image
So for every particular lens and object to image distance, there are two points you can put the lens to get a focussed image. Since you put the lens at the point closer to the image, you made a camera. If you put the lens at the point closer to the object, then you get a projector.
Process lenses have terrific performance, in terms of focussing a flat screen image from a flat LCD right out to the edges. But there are two problems:
1) They are hardly ever wide enough for our purpose, so they don't get much light to the screen.
2) They cost $250 to $500, since they are used by large format photographers.
On the other hand, you can buy an opaque projector lens on eBay for around $50. (I have seen them go for as low as $12!) They have a focal length of 18" (perfect for long throw projection), and they are about 5" in diameter (so lots of light on the screen). These lenses are optimized for around 15" diagonal, so they won't focus your LCD subpixels from the center to the corners, but they will give you a very nice bright image.
The main difference in focussing performance between a process lens and an opaque projector lens, is that the process lens will give you very sharp black screendoor all across the image. The opaque projector lens will give you lighter gray screendoor at the sharpest focussed region, and the screendoor will dissappear far from that region. Is that so bad?
So I think you should try an opaque projector lens. The next step up from that will probably be one of LumenLab's new 100 mm diameter 500 mm fl triplets, when they get them in their store.
>the screen image was smaller than the LCD
Congrats! You have just reinvented the camera! 😀
The thin lens equation says:
1/focal length = 1/object to lens + 1/lens to image
So for every particular lens and object to image distance, there are two points you can put the lens to get a focussed image. Since you put the lens at the point closer to the image, you made a camera. If you put the lens at the point closer to the object, then you get a projector.
Process lenses have terrific performance, in terms of focussing a flat screen image from a flat LCD right out to the edges. But there are two problems:
1) They are hardly ever wide enough for our purpose, so they don't get much light to the screen.
2) They cost $250 to $500, since they are used by large format photographers.
On the other hand, you can buy an opaque projector lens on eBay for around $50. (I have seen them go for as low as $12!) They have a focal length of 18" (perfect for long throw projection), and they are about 5" in diameter (so lots of light on the screen). These lenses are optimized for around 15" diagonal, so they won't focus your LCD subpixels from the center to the corners, but they will give you a very nice bright image.
The main difference in focussing performance between a process lens and an opaque projector lens, is that the process lens will give you very sharp black screendoor all across the image. The opaque projector lens will give you lighter gray screendoor at the sharpest focussed region, and the screendoor will dissappear far from that region. Is that so bad?
So I think you should try an opaque projector lens. The next step up from that will probably be one of LumenLab's new 100 mm diameter 500 mm fl triplets, when they get them in their store.
I almost got a process lens. Rodenstock APO-Ronar 1:9 FL=485mm. Had a field angle of 48 degrees and a viewing circle of 430mm. Someone outbid me in the last 5 seconds! ARGH!
Are opaque projection lenses called something else? I did a search on ebay and nothing showed. Are they rare to find there? How wide are process lenses usually? They seem like a perfect candidate.
Are opaque projection lenses called something else? I did a search on ebay and nothing showed. Are they rare to find there? How wide are process lenses usually? They seem like a perfect candidate.
lenses
That Rodenstock lens is very nice, but I think it is f9. That means it is 485 / 9 = 54 mm in diameter. It's very difficult to get most of your light through a lens so small. Even without the LCD in the way, the fresnels form an image of the lamp arc that is about arc length * field fresnel fl / condensor fresnel fl.
If you have a 24 mm arc length, 220 mm fl condensor fresnel, and 330 mm fl field fresnel, that gives you:
24 * 550/220 = 60 mm arc image
Most of the light comes from the two ends of the arc, and the LCD diffuses a lot of the light. Ergo, not much light gets through a lens smaller than 80 mm diameter. 100 mm is much better.
If you watch the process lenses (also in large format photography section), you will see the final prices usually run $250 to $500. That often happens in the last few seconds.
I just search for "opaque projector lens", and I see them now and again. Keep looking.
That Rodenstock lens is very nice, but I think it is f9. That means it is 485 / 9 = 54 mm in diameter. It's very difficult to get most of your light through a lens so small. Even without the LCD in the way, the fresnels form an image of the lamp arc that is about arc length * field fresnel fl / condensor fresnel fl.
If you have a 24 mm arc length, 220 mm fl condensor fresnel, and 330 mm fl field fresnel, that gives you:
24 * 550/220 = 60 mm arc image
Most of the light comes from the two ends of the arc, and the LCD diffuses a lot of the light. Ergo, not much light gets through a lens smaller than 80 mm diameter. 100 mm is much better.
If you watch the process lenses (also in large format photography section), you will see the final prices usually run $250 to $500. That often happens in the last few seconds.
I just search for "opaque projector lens", and I see them now and again. Keep looking.
Ahhh. Very good point. So that's what all this f# is about. Take the focal length and divide the smallest f# to get the lens size. The key points for this lens: picture angle needs to be around 48 degrees (for a 480mm FL) and the lens diameter needs to be big enough to "see" the complete arc length. Man, this is getting more difficult by the message! hehe. The other problem I have run into is fresnel lenses with a focal length longer than 550mm (for under $50). I heard people using a 760mm or a 790mm? I can't find that anywhere! 600mm - 650mm ish would be nice too. Although if I can find a good 480mm or so, a 550mm would just fine.
I really appreciate this info. I WILL find a suitable lens.
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