(sorry my english is poor...)
Osram HQI BT 400w
15" cmv 529A
S15 Projection Lens Triplet Focal length: 320mm
220mm focal length fresnel
317 mm focal length fresnel
I don`t know what is the distance between LCD and Fresnel Lenses...and the other distance , just as the pic below
If I want to project a 100"~120" image on the screen , what is the distance with A , B, C,D ?
THANKS
Osram HQI BT 400w
15" cmv 529A
S15 Projection Lens Triplet Focal length: 320mm
220mm focal length fresnel
317 mm focal length fresnel
I don`t know what is the distance between LCD and Fresnel Lenses...and the other distance , just as the pic below
An externally hosted image should be here but it was not working when we last tested it.
If I want to project a 100"~120" image on the screen , what is the distance with A , B, C,D ?
THANKS
A = focal lenght of the first fresnel 220mm
B and C - it is best to make some experiments with the length. If you put the fresnel to close you will see 'wrinkles' in the projected image. If it is to far the image won't be sharp. B and C should be about 10 mm to 20mm.
D = depends on the distance between the projector and the screen but it sholud be about 317mm (focal length of the secnod fresnel) +/- 20mm...
B and C - it is best to make some experiments with the length. If you put the fresnel to close you will see 'wrinkles' in the projected image. If it is to far the image won't be sharp. B and C should be about 10 mm to 20mm.
D = depends on the distance between the projector and the screen but it sholud be about 317mm (focal length of the secnod fresnel) +/- 20mm...
1 / (Projection Lens Focal Length) = 1 / (Distance from LCD to projection Lens) + 1 / (Throw)
first all values must be in the same unit. inches for example(I'm rounding these numbers a bit):
1/12.5=1/X + 1/96(8' throw)
gives us:
.08 = 1/X + .010416 or 1/X = .08 - .010416
so we have,
.069584 = .08 - .010416
to convert it back to inches take 1 divided by each of the numbers and you get:
1/14.37 = 1/12.5 - 1/96
so the answer is 14.37 inches or 365mm from the lcd to the projection lens.
I'm certainly no math genius so hopefully this is correct and there may be an easier way, I don't know. This is what I was shown and it worked for me.
With a 15" LCD at the those numbers(8' Throw) you should project a 100" diagonal screen.
lamp arc center should be at the FL of the rear fresnel so 220mm to start out. then tweak it.
If I'm wrong about any of this someone please say so. I know I answered this on the other forum but who knows how long it will be there and not get deleted by the mods.
first all values must be in the same unit. inches for example(I'm rounding these numbers a bit):
1/12.5=1/X + 1/96(8' throw)
gives us:
.08 = 1/X + .010416 or 1/X = .08 - .010416
so we have,
.069584 = .08 - .010416
to convert it back to inches take 1 divided by each of the numbers and you get:
1/14.37 = 1/12.5 - 1/96
so the answer is 14.37 inches or 365mm from the lcd to the projection lens.
I'm certainly no math genius so hopefully this is correct and there may be an easier way, I don't know. This is what I was shown and it worked for me.
With a 15" LCD at the those numbers(8' Throw) you should project a 100" diagonal screen.
lamp arc center should be at the FL of the rear fresnel so 220mm to start out. then tweak it.
If I'm wrong about any of this someone please say so. I know I answered this on the other forum but who knows how long it will be there and not get deleted by the mods.
math
gabel87: Your calculations are perfect. The thing you did not document is:
Magnification = throw distance / LCD to lens distance
image size = Magnification * LCD size
Interesting note for anybody calculating the distance & focal lengths with: 1/fl = 1/throw + 1/LCD-to-Lens
This can be done in just a few keystrokes on the windows calculator in scientific mode. For example, say you have:
1/450 = 1/x + 1/4000
This is the key sequence to solve for x:
450 1/x - 4000 1/x = 1/x
gabel87: Your calculations are perfect. The thing you did not document is:
Magnification = throw distance / LCD to lens distance
image size = Magnification * LCD size
Interesting note for anybody calculating the distance & focal lengths with: 1/fl = 1/throw + 1/LCD-to-Lens
This can be done in just a few keystrokes on the windows calculator in scientific mode. For example, say you have:
1/450 = 1/x + 1/4000
This is the key sequence to solve for x:
450 1/x - 4000 1/x = 1/x
Is it from the LCD to the projection lens or from the second fresnel (collimator) to the projection lens?gabel87 said:
This is the distance from the LCD to the projection lens. It is different than the field (collimator) fresnel lens. The field fresnel lens wants to be its focal length away from the projection lens.
Ex.
You use a 330mm field fresnel and use the 365mm LCD to projection lens distance already calculated. Place the LCD 365mm from the projection lens. The fresnel STILL needs to be 330mm from the projection lens. This will place the fresnel 35mm after the LCD (between the LCD and projection lens).
So you have: LCD -> 35mm after is the 330mm fresnel -> 330mm after is the projection lens
field fresnel distance
If you put the field fresnel at it's focal length from the optical center of the projection lens, then it will work with parallel rays coming through the LCD.
But 35 mm is a bit far to space the field fresnel from the LCD. You will see some magnification at that distance, which will give you some distortion in your screen image. Better if you put it about 20 mm from the LCD, and then fiddle a bit with the lamp-to-condensor-fresnel distance. By shortening that distance (just a tiny bit) you make the rays through the LCD diverge. Then the field fresnel will focus them at a longer distance than its focal length.
The best way to do this is by setting up a temporary optical bench where all distances are easily adjusted. You include both fresnels, the LCD, the projection lens, and make the throw distance the same as in your projection room. Move the projection lens until you get a focussed image on the screen. Then remove the LCD and projection lens, and put a piece of white paper where the center of the lens was. (The fresnels must not be moved!) Then you can play with the lamp-to-condensor fresnel distance until you get a tight arc image focussed on the piece of paper. That is the lamp-to-fresnel distance for your projector.
Or you can just make your lamp position adjustable in the final projector design. Then whenever you move the projector, you focus the image by moving the projector lens and adjust the lamp distance for a nice bright even image.
If you put the field fresnel at it's focal length from the optical center of the projection lens, then it will work with parallel rays coming through the LCD.
But 35 mm is a bit far to space the field fresnel from the LCD. You will see some magnification at that distance, which will give you some distortion in your screen image. Better if you put it about 20 mm from the LCD, and then fiddle a bit with the lamp-to-condensor-fresnel distance. By shortening that distance (just a tiny bit) you make the rays through the LCD diverge. Then the field fresnel will focus them at a longer distance than its focal length.
The best way to do this is by setting up a temporary optical bench where all distances are easily adjusted. You include both fresnels, the LCD, the projection lens, and make the throw distance the same as in your projection room. Move the projection lens until you get a focussed image on the screen. Then remove the LCD and projection lens, and put a piece of white paper where the center of the lens was. (The fresnels must not be moved!) Then you can play with the lamp-to-condensor fresnel distance until you get a tight arc image focussed on the piece of paper. That is the lamp-to-fresnel distance for your projector.
Or you can just make your lamp position adjustable in the final projector design. Then whenever you move the projector, you focus the image by moving the projector lens and adjust the lamp distance for a nice bright even image.
bochinlin, How did this all work out for you? How is the lens set you got? (I am thinking of getting the same set). Did the process explained work for you?
In our design, we have provided scope for easy adjustment of all the elements. For details please click the following link of our web page:-
http://www.drtsolutions.com/Projector.htm
http://www.drtsolutions.com/Projector.htm
bochinlin said:
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The theoritical figures for the optical circuit are A - slightly less than 220 mm, D - slightly less than 317 mm, B and C - 10 to 20 mm. In practice, one needs to adjust these distances with reference to the screen. First adjust D for a sharp image followed by A, B and C and then finally again D. Looking to these aspects, we have designed our enclosure in such a manner so that all such adjustments can be carried out easily. You may see the details by clicking http://www.drtsolutions.com/Projector.htm of our web page.
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