Hello
I am working out a projector project based on a Sony Laptop.
I plan on using the entire laptop within the design of the projector.
My question is on the size of the lense.
The LCD size is 16.1" UXGA = 13" x 9.75" or 330mm x 247.65mm.
1. What fresnel size should I look for?
2. If they are larger than the LCD is there any problems with cutting them down?
3. I am trying to stay under 100" around 10 feet away. I would be happy with 80".
Once I determine the fresnel...what are your thoughts on the triplet? What would work best for this LCD.
Sorry for all the questions - -> Thanks in advance!!!
Greg
I am working out a projector project based on a Sony Laptop.
I plan on using the entire laptop within the design of the projector.
My question is on the size of the lense.
The LCD size is 16.1" UXGA = 13" x 9.75" or 330mm x 247.65mm.
1. What fresnel size should I look for?
2. If they are larger than the LCD is there any problems with cutting them down?
3. I am trying to stay under 100" around 10 feet away. I would be happy with 80".
Once I determine the fresnel...what are your thoughts on the triplet? What would work best for this LCD.
Sorry for all the questions - -> Thanks in advance!!!
Greg
lucky U!
That is a great LCD. You are in for a wonderful image. I'm a little shocked you would tear apart a laptop with that kind of screen. It must be fairly new. Most do 1024x768. uxga is 1280x1024.
There is a chinese company posted here with big fresnels awi industries.
If you only have a fresnel between the lamp and the lcd it should be bigger and pass light through the lcd at a slight converging angle. This design has a clearer image.
If you go with 2nd frenel between lcd and objective lens both fresnels need to be about the same size as the lcd. A little bigger with covered edges.
post pics when your done!
Good luck!
That is a great LCD. You are in for a wonderful image. I'm a little shocked you would tear apart a laptop with that kind of screen. It must be fairly new. Most do 1024x768. uxga is 1280x1024.
There is a chinese company posted here with big fresnels awi industries.
If you only have a fresnel between the lamp and the lcd it should be bigger and pass light through the lcd at a slight converging angle. This design has a clearer image.
If you go with 2nd frenel between lcd and objective lens both fresnels need to be about the same size as the lcd. A little bigger with covered edges.
post pics when your done!
Good luck!
The laptop was given to me in a non-booting situation. I got it working, added a HD and now I am about to strip the LCD. I plan on using the entire laptop within the design. That way I can plug it in...add a DVD and fire up a movie - all within 1 movable box.
I still am unsure about cutting the lenses.
I see that diyprojectorcompany.com has a 17 inch fresnel.
# Maximum dimensions: 15.5x15.5 inches (~390x390 mm)
# Largest Square Aperture: 15x15 inches (~376x376 mm)
They are about bigger than the Sony LCD, is there any danger in cutting them (degrade the picture quality?).
If I do end up buying them, not many other places have larger size fresnels...what should I use for the triplet?
Thanks!
Greg
I still am unsure about cutting the lenses.
I see that diyprojectorcompany.com has a 17 inch fresnel.
# Maximum dimensions: 15.5x15.5 inches (~390x390 mm)
# Largest Square Aperture: 15x15 inches (~376x376 mm)
They are about bigger than the Sony LCD, is there any danger in cutting them (degrade the picture quality?).
If I do end up buying them, not many other places have larger size fresnels...what should I use for the triplet?
Thanks!
Greg
Hezz or Guy are probably the best to speak on the lens.
Regarding cutting the fresnels, yes there is danger here. Whatever you use to hold the lens steady while cutting has to be very delicate. Hold - do not squeeze. You want nothing to deform the rings. This will ruin your image.
Use a very sharp shear or a dremmel cutting disk. Be wary of a dulled paper shear that could fracture the lens.
Regarding cutting the fresnels, yes there is danger here. Whatever you use to hold the lens steady while cutting has to be very delicate. Hold - do not squeeze. You want nothing to deform the rings. This will ruin your image.
Use a very sharp shear or a dremmel cutting disk. Be wary of a dulled paper shear that could fracture the lens.
projection lens
For a 99.6" image, you need a focal length of 424 mm.
For a 80.5" image, you need a focal length of 508 mm.
You can buy a big 450 mm fl triplet from some of the DIY projector stores for about $120. That would probably give you the best image.
You could try making a 510 mm fl symmetric duplet, if you are interested. Cost is around $20, but the image is only good enough for watching video. If you want to use that high-res windows display, then the triplet would be worth it.
You can cut PVC fresnels with a shear, but you should be using acrylic fresnels anyway! Use a mototool thin cutting disk to cut the correct amount off the left and right sides, then cut the correct amount off the top and bottom. (You need to keep the center of the uncut lens, in the center of the lens once it is cut.)
USE FULL SEALED EYE PROTECTION! and a dustmask is good too. Cutting acrylic this way does not crack it, but it throws tiny sharp shards everywhere. You do not want to breathe it or get it in your eyes!
For a 99.6" image, you need a focal length of 424 mm.
For a 80.5" image, you need a focal length of 508 mm.
You can buy a big 450 mm fl triplet from some of the DIY projector stores for about $120. That would probably give you the best image.
You could try making a 510 mm fl symmetric duplet, if you are interested. Cost is around $20, but the image is only good enough for watching video. If you want to use that high-res windows display, then the triplet would be worth it.
You can cut PVC fresnels with a shear, but you should be using acrylic fresnels anyway! Use a mototool thin cutting disk to cut the correct amount off the left and right sides, then cut the correct amount off the top and bottom. (You need to keep the center of the uncut lens, in the center of the lens once it is cut.)
USE FULL SEALED EYE PROTECTION! and a dustmask is good too. Cutting acrylic this way does not crack it, but it throws tiny sharp shards everywhere. You do not want to breathe it or get it in your eyes!
Some Math - More questions
Thanks for the reply.
NOTE:
I am trying to make this simple so future projects can find a easy way to determine this information (...sorry if it seems basic)
OK, to review this is what I know.
1. DIY seems to be the only one with a fresnel that I can use with my LCD.
=========================================================================
2. Lets Calculate the lense:
*Big thanks to Guy for the formulas from a previous post*
Known information:
I want 100" screen
Throw (lense to wall distance) = 10'
LCD size = 16"
Soo....
The magnification would equal = 100/16 = 6.25
Plugging this information into this eq:
fl = throw / (M+1) = > => (10 feet * 304.8 mm/foot) / (6.25 + 1)
works out to:
3048/7.25 = 420.41 mm focal length lens
=========================================================================
So far soo good...
Now that is a pretty big lense...and DIY also seems to be the only one that can handle something that big. Now this lense is not without controversy. ROX has been debating it's usefulness in a number of posts...but...one person got it to work:
(Diyeitor17) - http://diyprojectorcompany.com/phpBB2/viewtopic.php?t=381)
Although he took it apart...and it now seems to work OK = lots of extra work.
I guess that is my only option right now...roll the dice and see what happens.
It looks like I can be the test subject on this issue.
UNKNOWN:
1. I can't seem to figure out the formula to determine the LCD to lense.
I have seen it in action...but still cannot figure it out.
2. The spacing of the [fresnel] to the [LCD] - > Formula? Trial and error?
Note: I plan on using only one to simplify things
3. The [light] to the [fresnel] - > Formula? Trial and error?
How do you make the $20 version?
The quest continues!
Thanks for the reply.
NOTE:
I am trying to make this simple so future projects can find a easy way to determine this information (...sorry if it seems basic)
OK, to review this is what I know.
1. DIY seems to be the only one with a fresnel that I can use with my LCD.
=========================================================================
2. Lets Calculate the lense:
*Big thanks to Guy for the formulas from a previous post*
Known information:
I want 100" screen
Throw (lense to wall distance) = 10'
LCD size = 16"
Soo....
The magnification would equal = 100/16 = 6.25
Plugging this information into this eq:
fl = throw / (M+1) = > => (10 feet * 304.8 mm/foot) / (6.25 + 1)
works out to:
3048/7.25 = 420.41 mm focal length lens
=========================================================================
So far soo good...
Now that is a pretty big lense...and DIY also seems to be the only one that can handle something that big. Now this lense is not without controversy. ROX has been debating it's usefulness in a number of posts...but...one person got it to work:
(Diyeitor17) - http://diyprojectorcompany.com/phpBB2/viewtopic.php?t=381)
Although he took it apart...and it now seems to work OK = lots of extra work.
I guess that is my only option right now...roll the dice and see what happens.
It looks like I can be the test subject on this issue.
UNKNOWN:
1. I can't seem to figure out the formula to determine the LCD to lense.
I have seen it in action...but still cannot figure it out.
2. The spacing of the [fresnel] to the [LCD] - > Formula? Trial and error?
Note: I plan on using only one to simplify things
3. The [light] to the [fresnel] - > Formula? Trial and error?
How do you make the $20 version?
The quest continues!
More questions & answers
1. I can't seem to figure out the formula to determine the LCD to lense. I have seen it in action...but still cannot figure it out.
1/(lens fl in mm) = 1/(throw in mm) + 1/(lcd to lens in mm)
Usually you already know the lens fl and throw distance, so on the windows calculator you do:
<lens fl> 1/x - <throw> 1/x = 1/x
to get the lcd to lens distance. But this is an approximation, since it is the equation for a thin lens (which a triplet is NOT) and the distances are measured to two different principle planes of the triplet that may not even be within the lens itself.
2. The spacing of the [fresnel] to the [LCD] - > Formula? Trial and error? Note: I plan on using only one to simplify things
Most people think a pair of fresnels works a lot better. (You'll see!) You can use the same formula above for the fresnel(s). It works best if you focus the image of the lamp arc to the center of the triplet. A pair of fresnels in close contact have an effective focal length of:
1/EFL = 1/fl1 + 1/fl2
3. The [light] to the [fresnel] - > Formula? Trial and error?
Fresnels work a lot better when the rays on the ringed side are parallel and on the smooth side they all meet at the focal length. That is why a pair of fresnels work very well with the lamp arc at the focal length of the condensor fresnel, and the arc image in the projector lens at the focal length of the field fresnel.
4. How do you make the $20 version?
You get two #15.0320 +1.0 Diopter acrylic positive meniscus lenses from Rolyn Optics. Then you mount them about 80 mm apart (convex curves facing out) in a piece of black plastic sewer pipe. Sand the inside of the pipe to get rid of the shine before you mount the lenses. I use rings cut from a cardboard mailing tube for the mounting. It doesn't do any aberration corrections at all, and you can't focus the edges and center at the same setting, but it works pretty well, considering the price.
1. I can't seem to figure out the formula to determine the LCD to lense. I have seen it in action...but still cannot figure it out.
1/(lens fl in mm) = 1/(throw in mm) + 1/(lcd to lens in mm)
Usually you already know the lens fl and throw distance, so on the windows calculator you do:
<lens fl> 1/x - <throw> 1/x = 1/x
to get the lcd to lens distance. But this is an approximation, since it is the equation for a thin lens (which a triplet is NOT) and the distances are measured to two different principle planes of the triplet that may not even be within the lens itself.
2. The spacing of the [fresnel] to the [LCD] - > Formula? Trial and error? Note: I plan on using only one to simplify things
Most people think a pair of fresnels works a lot better. (You'll see!) You can use the same formula above for the fresnel(s). It works best if you focus the image of the lamp arc to the center of the triplet. A pair of fresnels in close contact have an effective focal length of:
1/EFL = 1/fl1 + 1/fl2
3. The [light] to the [fresnel] - > Formula? Trial and error?
Fresnels work a lot better when the rays on the ringed side are parallel and on the smooth side they all meet at the focal length. That is why a pair of fresnels work very well with the lamp arc at the focal length of the condensor fresnel, and the arc image in the projector lens at the focal length of the field fresnel.
4. How do you make the $20 version?
You get two #15.0320 +1.0 Diopter acrylic positive meniscus lenses from Rolyn Optics. Then you mount them about 80 mm apart (convex curves facing out) in a piece of black plastic sewer pipe. Sand the inside of the pipe to get rid of the shine before you mount the lenses. I use rings cut from a cardboard mailing tube for the mounting. It doesn't do any aberration corrections at all, and you can't focus the edges and center at the same setting, but it works pretty well, considering the price.
Guy grotke wrote;
"to get the lcd to lens distance. But this is an approximation, since it is the equation for a thin lens (which a triplet is NOT) and the distances are measured to two different principle planes of the triplet that may not even be within the lens itself"
i found this equation really lose to the reallity. The problem is determining the optical center on the lens. If you meassure the distancies to the center on the projection lens, then this won´t work as accurate as expected. But since you know the exact optical center, its close to ideal.
When we talk about the 450 efl focal, we are talking about the thin flat lens placed on the optical center. Forget you have 15cm long triplet, and think you have 450 focal at somewhere.
Do you know where the optical center is exactly? it is easy to check. (it is 1,5cm closer to the rear lens from the fisical center of the lens)
"to get the lcd to lens distance. But this is an approximation, since it is the equation for a thin lens (which a triplet is NOT) and the distances are measured to two different principle planes of the triplet that may not even be within the lens itself"
i found this equation really lose to the reallity. The problem is determining the optical center on the lens. If you meassure the distancies to the center on the projection lens, then this won´t work as accurate as expected. But since you know the exact optical center, its close to ideal.
When we talk about the 450 efl focal, we are talking about the thin flat lens placed on the optical center. Forget you have 15cm long triplet, and think you have 450 focal at somewhere.
Do you know where the optical center is exactly? it is easy to check. (it is 1,5cm closer to the rear lens from the fisical center of the lens)
Hello Again...
OK, recap reply...
==============================================
Lets Calculate the lense:
Known information:
I want 100" screen
Throw (lense to wall distance) = 10'
LCD size = 16"
Soo....
The magnification would equal = 100/16 = 6.25
Plugging this information into this eq:
How to:
fl = throw / (M+1) = > => (10 feet * 304.8 mm/foot) / (6.25 + 1)
works out to:
3048/7.25 = 420.41 mm focal length lens
==============================================
Lets Calculate the LCD to the lense:
KNOWN INFORMATION:
lens fl in mm = 450mm
throw in mm = 3084mm
lcd to lens in mm = ?
Formula:
1/fl=1/throw+1/lcd
How to:
1/450=1/3084-1/lcd
0.002=0.00032-1/lcd
0.002-0.00032=1/lcd
0.00168=1/lcd
lcd=1/0.00168
lcd=595mm or 23.4251969 inches
================================================
Question:
1. Does the LCD to lense answer look right?
2. If I really wanted a "true" measurement...do I now subtract 1.5cm (previous Rox reply) from the 595mm value?
Note:
Oh, from a previous post I said I wanted to use 1 fresnel...I actually meant 2...non-split design - 330 mm focal length fresnel lenses
I am trying to keep thing easy so other people can look to one note and get what they need formula wise...
🙂
OK, recap reply...
==============================================
Lets Calculate the lense:
Known information:
I want 100" screen
Throw (lense to wall distance) = 10'
LCD size = 16"
Soo....
The magnification would equal = 100/16 = 6.25
Plugging this information into this eq:
How to:
fl = throw / (M+1) = > => (10 feet * 304.8 mm/foot) / (6.25 + 1)
works out to:
3048/7.25 = 420.41 mm focal length lens
==============================================
Lets Calculate the LCD to the lense:
KNOWN INFORMATION:
lens fl in mm = 450mm
throw in mm = 3084mm
lcd to lens in mm = ?
Formula:
1/fl=1/throw+1/lcd
How to:
1/450=1/3084-1/lcd
0.002=0.00032-1/lcd
0.002-0.00032=1/lcd
0.00168=1/lcd
lcd=1/0.00168
lcd=595mm or 23.4251969 inches
================================================
Question:
1. Does the LCD to lense answer look right?
2. If I really wanted a "true" measurement...do I now subtract 1.5cm (previous Rox reply) from the 595mm value?
Note:
Oh, from a previous post I said I wanted to use 1 fresnel...I actually meant 2...non-split design - 330 mm focal length fresnel lenses
I am trying to keep thing easy so other people can look to one note and get what they need formula wise...
🙂
math errors
3048, not 3084 for the throw distance.
Then I think you introduce a lot of error by rounding. The equation subtracts two very small numbers so the error is huge without high precision.
Put the Windows calculator in scientific mode. It has a "1/X" key. Then use this sequence:
450 1/X 3048 1/X - 1/X
That gives me 528 mm. And that would be very close to the distance from the LCD to the center of a thin lens. With a thick compound lens like the big triplet, the 528 mm has to be measured to the correct principle plane of the lens (and the throw distance to the other principle plane). So this is just a starting point! You will adjust your LCD to projection lens distance to focus your screen image, once you have it running.
Assuming a thin projection lens with 528 mm between the LCD and the lens, a non-split design would work best with a 548 mm field fresnel 20 mm before the LCD. With a thick lens like the triplet, that "528 mm point" will not be right in the middle of the lens, but it doesn't matter! When the lens is positioned to give you a focussed screen image, the fresnel will be focussing the arc image AT THAT SAME POINT 528 mm from the LCD.
So I don't think you need to bother figuring out exactly where the principle planes are.
3048, not 3084 for the throw distance.
Then I think you introduce a lot of error by rounding. The equation subtracts two very small numbers so the error is huge without high precision.
Put the Windows calculator in scientific mode. It has a "1/X" key. Then use this sequence:
450 1/X 3048 1/X - 1/X
That gives me 528 mm. And that would be very close to the distance from the LCD to the center of a thin lens. With a thick compound lens like the big triplet, the 528 mm has to be measured to the correct principle plane of the lens (and the throw distance to the other principle plane). So this is just a starting point! You will adjust your LCD to projection lens distance to focus your screen image, once you have it running.
Assuming a thin projection lens with 528 mm between the LCD and the lens, a non-split design would work best with a 548 mm field fresnel 20 mm before the LCD. With a thick lens like the triplet, that "528 mm point" will not be right in the middle of the lens, but it doesn't matter! When the lens is positioned to give you a focussed screen image, the fresnel will be focussing the arc image AT THAT SAME POINT 528 mm from the LCD.
So I don't think you need to bother figuring out exactly where the principle planes are.
i just tould this 1,5cm distance from the fisic center to the optical center just to have more precision. Thats because i was told by JCB that his setup does not follow the formulla (he has to much error, something like 6 cm) then i thought it would be more accurate with we aplly most of imformation as posible.
Actually my tests are exactly as expected. Never got more error than 5mm.
Actually my tests are exactly as expected. Never got more error than 5mm.
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