i need a lens that is 3 to 4 inches away from my 5" lcd and from 15 feet it will project about 67"
no
You want a 67" image from a 5" LCD, so your magnification will be 67/5 = 13.4. You want to do that with a throw distance of 15 feet, so by:
fl = throw / (M+1) = (15 feet * 304.8 mm/foot) / (13.4 + 1) = 317.5 mm focal length lens
It is very easy to find such a projection lens, since this is almost exactly the focal length of overhead projector lenses.
But the distance from the lens to the LCD will be:
1/fl = 1/LCD to lens + 1/lens to screen
1/317.4 - 1/(15 * 304.8) = 1/341
LCD to lens = 341 mm or 13.4 inches
You want a 67" image from a 5" LCD, so your magnification will be 67/5 = 13.4. You want to do that with a throw distance of 15 feet, so by:
fl = throw / (M+1) = (15 feet * 304.8 mm/foot) / (13.4 + 1) = 317.5 mm focal length lens
It is very easy to find such a projection lens, since this is almost exactly the focal length of overhead projector lenses.
But the distance from the lens to the LCD will be:
1/fl = 1/LCD to lens + 1/lens to screen
1/317.4 - 1/(15 * 304.8) = 1/341
LCD to lens = 341 mm or 13.4 inches
throw distance
The only effect the throw distance has is to increase the size of the image, unless you are running your projector in a really dusty environment (ie. coal mine).
But increasing the size of the image DOES spread the same light over more area (and increases the size of the pixel screedoor pattern). You would get a dimmer image from 15 feet than from 10 feet, using the same projector lamp & lenses, etc. Since the light gets spread over an area, the brightness will go down by the square of the diagonal image size difference. That is a linear function of the throw distance, so the difference would be:
(15/10) ^2 = 2.25 times as bright with the 10 foot throw distance
The only effect the throw distance has is to increase the size of the image, unless you are running your projector in a really dusty environment (ie. coal mine).
But increasing the size of the image DOES spread the same light over more area (and increases the size of the pixel screedoor pattern). You would get a dimmer image from 15 feet than from 10 feet, using the same projector lamp & lenses, etc. Since the light gets spread over an area, the brightness will go down by the square of the diagonal image size difference. That is a linear function of the throw distance, so the difference would be:
(15/10) ^2 = 2.25 times as bright with the 10 foot throw distance
Guy,
If you have a projected image in focus at 10 feet then you move it at 7 feet to get a smaller image. Do you have to open your projector case again then adjust all the lenses, field lens or projector lens or all of them?
If you have a projected image in focus at 10 feet then you move it at 7 feet to get a smaller image. Do you have to open your projector case again then adjust all the lenses, field lens or projector lens or all of them?
adjustment after moving
If you are an obsessive-compulsive type, then you would adjust everything so the condensor system put the projected arc image at exactly the right place in the new position of the projection lens.
Normal people would only adjust the projection lens position relative to the LCD, and ignore the condensor adjustments. Changing from 10' to 7' throw distance would only move the projection lens position about 60 mm with a 450 mm fl lens.
But if you do move from 10' to 7' and the image does not get brighter, then you might be able to improve it by readjusting the condensor system fresnels, lamp position, etc.
I have seen OHPs that had a knob for adjusting the lamp position to handle different throw distances. You could design that into a DIY projector too.
If you are an obsessive-compulsive type, then you would adjust everything so the condensor system put the projected arc image at exactly the right place in the new position of the projection lens.
Normal people would only adjust the projection lens position relative to the LCD, and ignore the condensor adjustments. Changing from 10' to 7' throw distance would only move the projection lens position about 60 mm with a 450 mm fl lens.
But if you do move from 10' to 7' and the image does not get brighter, then you might be able to improve it by readjusting the condensor system fresnels, lamp position, etc.
I have seen OHPs that had a knob for adjusting the lamp position to handle different throw distances. You could design that into a DIY projector too.
Thanks guy! I asked because i want to make a portable projector that I can carry to different places. Does a varifocal would solve the problem about adjusting the fixed triplet lens?
Based on specs, what do u think is the best varifocal for our application.
http://www.diyprojectorcompany.com/...id=81&osCsid=bb44b5ef2d13fcd284746e3e5bf21c71
http://www.exclusiv-online.com/shop/index.php?main=product&art=OSET1
In terms of projected image quality, which is better varifocal or fixed triplet .
Based on specs, what do u think is the best varifocal for our application.
http://www.diyprojectorcompany.com/...id=81&osCsid=bb44b5ef2d13fcd284746e3e5bf21c71
http://www.exclusiv-online.com/shop/index.php?main=product&art=OSET1
In terms of projected image quality, which is better varifocal or fixed triplet .
What does a varifocal do?
For instance you design your projector with the throw distance of 10 feet. You set everything, the fresnel the projection lens, etc.
Now you have a perfect in focus projected image with a throw distance of 10 feet and a 67" image on the screen. After an hour of watching DVD, you feel like the screen is too big for your small apartment... If you change the throw distance to 7 feet for a small projected image, do you have to open your box again and do some fiddling with the lenses, etc. ?
Or your image will be out of focus with other throw distance except for 10 feet? Is that the varifocal do? Adjust the throw distance then by just rotating and changing the focal length of the Projection Lens to focus the image?
Many thanks Guy!
For instance you design your projector with the throw distance of 10 feet. You set everything, the fresnel the projection lens, etc.
Now you have a perfect in focus projected image with a throw distance of 10 feet and a 67" image on the screen. After an hour of watching DVD, you feel like the screen is too big for your small apartment... If you change the throw distance to 7 feet for a small projected image, do you have to open your box again and do some fiddling with the lenses, etc. ?
Or your image will be out of focus with other throw distance except for 10 feet? Is that the varifocal do? Adjust the throw distance then by just rotating and changing the focal length of the Projection Lens to focus the image?
Many thanks Guy!
all u have to do is move the objective lens, and focus it.
hey guy here is my question. will the 67" image look the same with 15 feet as it is with 10 feet.
so same size image but different lenses
hey guy here is my question. will the 67" image look the same with 15 feet as it is with 10 feet.
so same size image but different lenses
varifocals
buddy123 & onie: Those two varifocals have very different focal length ranges. I would pick one based on the LCD size, the throw distance I need, and the image size I want. A fixed triplet has been optimized for a particular lens spacing. A varifocal has to work okay at the extreme adjustments, so it will not be quite as good.
If you move a projector with a varifocal lens and the change in throw distance is small, you might be able to re-focus it just by using the lens fl adjustment. But I think you still need to be able to adjust the position of the whole lens. Then the varifocal feature lets you adjust the size of the image at a particular throw distance.
part2wanksta: There might be a slight difference, but I don't know if you would notice it. The difference would be because a shorter focal length projection lens sits closer to the LCD, which makes the viewing angle more extreme around the edges of the LCD. So the version where you have a shorter throw distance might have dimmer edges.
buddy123 & onie: Those two varifocals have very different focal length ranges. I would pick one based on the LCD size, the throw distance I need, and the image size I want. A fixed triplet has been optimized for a particular lens spacing. A varifocal has to work okay at the extreme adjustments, so it will not be quite as good.
If you move a projector with a varifocal lens and the change in throw distance is small, you might be able to re-focus it just by using the lens fl adjustment. But I think you still need to be able to adjust the position of the whole lens. Then the varifocal feature lets you adjust the size of the image at a particular throw distance.
part2wanksta: There might be a slight difference, but I don't know if you would notice it. The difference would be because a shorter focal length projection lens sits closer to the LCD, which makes the viewing angle more extreme around the edges of the LCD. So the version where you have a shorter throw distance might have dimmer edges.
Re: no
so than is there a lens that i can put close to my lcd but from 15 feet away get from 50" to 67" projected image
Guy Grotke said:
so than is there a lens that i can put close to my lcd but from 15 feet away get from 50" to 67" projected image
probably not
For that throw distance, with that magnification, you have to use a lens with that focal length. But the LCD to lens distance refers to the optical center of the lens. If you could find a very long lens (ie. say 20 inches long) then the distance between the LCD and the first lens surface would be more like 13.4" - (20"/2) = 3.4" (This is the idea behind Back Focal Length.)
Most of the lenses like this are for CRT projection and they have shorter focal lengths. Like between 6" and 8". You could use one of these very close to your LCD, but it would have to be a very large diameter CRT lens (At least 5" diameter.) Then you would also have to place the projector much closer to the screen, to get the size image you want.
For that throw distance, with that magnification, you have to use a lens with that focal length. But the LCD to lens distance refers to the optical center of the lens. If you could find a very long lens (ie. say 20 inches long) then the distance between the LCD and the first lens surface would be more like 13.4" - (20"/2) = 3.4" (This is the idea behind Back Focal Length.)
Most of the lenses like this are for CRT projection and they have shorter focal lengths. Like between 6" and 8". You could use one of these very close to your LCD, but it would have to be a very large diameter CRT lens (At least 5" diameter.) Then you would also have to place the projector much closer to the screen, to get the size image you want.
Hi Guy,
I have a friend who can design a fresnel and an objective lens for me.
He's working in a manufacturing of OHP.
He gave me this information of their current triplet
EFL: 190-220mm
Resolution:
Center = Minimum 10LP/mm
Above and Below = Minimum 6LP/mm
Corners=Minimum 4LP/mm
Focusing Distance: About 10mm
Joggle Angle: About 180 degrees
The internals would look like this:
() )( () - aperture about 70
According to him, He could make me an objective and fresnel for free during his free time.
Now my question, what is the best configuration for my 8" LCD?
What is the specs for the fresnel and what is the specs for the objective?
What is the best diameter and FL for the objective?
What is the best FL and grooves for the fresnel?
Best means Ideal, I dont want to have an 8 inches diameter Objective Lens.
If you have all the equipment and tools, what would you design?
Many Thanks Guy! Your the Best!
I have a friend who can design a fresnel and an objective lens for me.
He's working in a manufacturing of OHP.
He gave me this information of their current triplet
EFL: 190-220mm
Resolution:
Center = Minimum 10LP/mm
Above and Below = Minimum 6LP/mm
Corners=Minimum 4LP/mm
Focusing Distance: About 10mm
Joggle Angle: About 180 degrees
The internals would look like this:
() )( () - aperture about 70
According to him, He could make me an objective and fresnel for free during his free time.
Now my question, what is the best configuration for my 8" LCD?
What is the specs for the fresnel and what is the specs for the objective?
What is the best diameter and FL for the objective?
What is the best FL and grooves for the fresnel?
Best means Ideal, I dont want to have an 8 inches diameter Objective Lens.
If you have all the equipment and tools, what would you design?
Many Thanks Guy! Your the Best!
ideal lenses
As long as you can get a set of fresnels that focus the lamp arc image right to the center of your projection lens, there is no reason to go above around 100 mm diameter for the projection lens.
For your 8" LCD, I think you would be very happy with their existing triplet and matching fresnel set! With a 10 foot throw distance, that triplet would give you a 103" to 120" screen image. If you want a different throw distance and/or image size then post it and we can work from there.
Personally, I prefer a long-throw setup so I can put the projector at the back of a longer room. I also prefer a smaller & brighter image, so my ideal lens for an 8" LCD would be a 100 mm diameter achromatic triplet or tessar (4 element) lens with a focal length around 350 mm.
For an 8 inch LCD, the condensor fresnel could be anywhere from about 120 to 220 mm fl. (Shorter means a smaller projector, longer means more even lighting.) An aspheric fresnel design is better for making a clean focussed image. The perfect focal length field fresnel depends on the projection lens fl and the throw distance. You can calculate it like this:
1/LCD to proj lens = 1/proj lens fl - 1/throw distance
Then ideal field fresnel fl = LCD to proj lens + 20 mm
You would then put the two fresnels together (smooth sides out), 20 mm from the LCD.
As long as you can get a set of fresnels that focus the lamp arc image right to the center of your projection lens, there is no reason to go above around 100 mm diameter for the projection lens.
For your 8" LCD, I think you would be very happy with their existing triplet and matching fresnel set! With a 10 foot throw distance, that triplet would give you a 103" to 120" screen image. If you want a different throw distance and/or image size then post it and we can work from there.
Personally, I prefer a long-throw setup so I can put the projector at the back of a longer room. I also prefer a smaller & brighter image, so my ideal lens for an 8" LCD would be a 100 mm diameter achromatic triplet or tessar (4 element) lens with a focal length around 350 mm.
For an 8 inch LCD, the condensor fresnel could be anywhere from about 120 to 220 mm fl. (Shorter means a smaller projector, longer means more even lighting.) An aspheric fresnel design is better for making a clean focussed image. The perfect focal length field fresnel depends on the projection lens fl and the throw distance. You can calculate it like this:
1/LCD to proj lens = 1/proj lens fl - 1/throw distance
Then ideal field fresnel fl = LCD to proj lens + 20 mm
You would then put the two fresnels together (smooth sides out), 20 mm from the LCD.
I'm thinking about this setup:
Reflector-Light 120mmFL||220mmFL+(20mm)LCD--------->x(220mmFL)
I really appreciate your time answering our questions.
Many Thanks to you!!!
Reflector-Light 120mmFL||220mmFL+(20mm)LCD--------->x(220mmFL)
I really appreciate your time answering our questions.
Many Thanks to you!!!
too short!
You don't want a 220 mm fl field fresnel with a 220 mm fl projection lens! The projection lens would be 220 mm from the LCD only if your screen was an infinite distance away. If the screen is closer than say 30 feet, you need to calculate the true LCD to lens distance.
Let's say you will use a throw distance of 10 feet. The equation is
1/fl = 1/LCD to lens + 1/lens to screen
so the LCD to lens distance would be 237 mm. Then you need to add the 20 mm from the field fresnel to the LCD, so you get 257 mm focal length for the field fresnel. It is better if it is a bit too long than too short, since you can always increase the fresnel to LCD distance to compensate. But you can't put the fresnel too close to the LCD or you will see rings in the screen image.
You don't want a 220 mm fl field fresnel with a 220 mm fl projection lens! The projection lens would be 220 mm from the LCD only if your screen was an infinite distance away. If the screen is closer than say 30 feet, you need to calculate the true LCD to lens distance.
Let's say you will use a throw distance of 10 feet. The equation is
1/fl = 1/LCD to lens + 1/lens to screen
so the LCD to lens distance would be 237 mm. Then you need to add the 20 mm from the field fresnel to the LCD, so you get 257 mm focal length for the field fresnel. It is better if it is a bit too long than too short, since you can always increase the fresnel to LCD distance to compensate. But you can't put the fresnel too close to the LCD or you will see rings in the screen image.
Guy,
I know the Projection Lens FL should be greater than the size of LCD. Do you know the computation for that?
I have this Zerox Lens w/Mirror 200mm ƒ3.1. Pretty nice specs but
I dont know if this would work with my 8 inches LCD.
200mm = 7.87" pretty close to 8", do you think this would work?
Thanks!
I know the Projection Lens FL should be greater than the size of LCD. Do you know the computation for that?
I have this Zerox Lens w/Mirror 200mm ƒ3.1. Pretty nice specs but
I dont know if this would work with my 8 inches LCD.
200mm = 7.87" pretty close to 8", do you think this would work?
Thanks!
200 mm fl copy lens
A copy lens with that focal length will probably work with an 8 inch LCD. Think about it: Copy lenses do an excellant job making a 1:1 copy of an 8.5" by 11" piece of paper. When you make a 1:1 copy, the distance from the object to the lens and the distance from the lens to the image are equal. There is a formula for these distances:
1/fl = 1/d1 + 1/d2
With a 200 mm fl lens, that works out to 400 mm distance on each side of the lens. You can compute the field angle for the triangle formed by the 400 mm base and half the diagonal of the sheet of paper as the height. I get 23.8 degrees for the half angle, or 47.6 degrees for the full field.
Using the same equation above, let's see what happens with a 10 foot throw distance and your LCD:
1/200 = 1/d1 + 1/(120 inches * 25.4 mm/inch)
so the LCD to lens distance is 214 mm. We can draw another triangle with a 214 mm base and a height of 4 inches, to represent your LCD. That triangle has a field half angle of 25.4 degrees. That's a bit more than the piece of paper, but not by much. So I think it will work okay.
But what does the mirror do?
A copy lens with that focal length will probably work with an 8 inch LCD. Think about it: Copy lenses do an excellant job making a 1:1 copy of an 8.5" by 11" piece of paper. When you make a 1:1 copy, the distance from the object to the lens and the distance from the lens to the image are equal. There is a formula for these distances:
1/fl = 1/d1 + 1/d2
With a 200 mm fl lens, that works out to 400 mm distance on each side of the lens. You can compute the field angle for the triangle formed by the 400 mm base and half the diagonal of the sheet of paper as the height. I get 23.8 degrees for the half angle, or 47.6 degrees for the full field.
Using the same equation above, let's see what happens with a 10 foot throw distance and your LCD:
1/200 = 1/d1 + 1/(120 inches * 25.4 mm/inch)
so the LCD to lens distance is 214 mm. We can draw another triangle with a 214 mm base and a height of 4 inches, to represent your LCD. That triangle has a field half angle of 25.4 degrees. That's a bit more than the piece of paper, but not by much. So I think it will work okay.
But what does the mirror do?
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