I have a quick question regarding a focal length equation that I've seen taht looks like this
fl = Throw/(M + 1)
the geometry that makes the most sense to me is like this. If M just defines the object to image ratio, where does the + 1 come from?? I believe this is related to why the lens to lcd distance does not match focal length, I just don't understand why!
fl = Throw/(M + 1)
the geometry that makes the most sense to me is like this. If M just defines the object to image ratio, where does the + 1 come from?? I believe this is related to why the lens to lcd distance does not match focal length, I just don't understand why!
your formula is ok as well.
is just some operations with those formulas, all off them are true;
1/F=1/D + 1/T F=focal D=lcd-triplet dist T=throw
T/D=M=I/S= magnification
I=projected image diag
S=source image diag
play with those formulas and you would get the one you post.
refering to your question about why the focal of the triplet is always shorter than the D (lcd-triplet dist) is because if you place the lcd on the focal of your lens, the proection image would be infinite size and the throw would be infinite as well. Have you got such a big room? 😀
is just some operations with those formulas, all off them are true;
1/F=1/D + 1/T F=focal D=lcd-triplet dist T=throw
T/D=M=I/S= magnification
I=projected image diag
S=source image diag
play with those formulas and you would get the one you post.
refering to your question about why the focal of the triplet is always shorter than the D (lcd-triplet dist) is because if you place the lcd on the focal of your lens, the proection image would be infinite size and the throw would be infinite as well. Have you got such a big room? 😀
Everytime I think I understand it, I realize how much I don't understand about it!
I agree that all those equations match and are consistent with each other. I also agree, that given those equations, placing the lcd on the focal point would yield an infinite throw distance. I think what I'd really like to see is the derivation of those formulas. Perhaps I'd better find an optics book!
The part that has me confused is references I've seen regarding cameras where the focal length is the same as the lens to (in this case) object distance. There, a whole different set of formulas hold and since both sets contradict, only (atmost) one can be right.
I could experiment and verify each set, but I want to be able to prove it to myself. Not as much fun building something if you don't know exactly how it works!
I agree that all those equations match and are consistent with each other. I also agree, that given those equations, placing the lcd on the focal point would yield an infinite throw distance. I think what I'd really like to see is the derivation of those formulas. Perhaps I'd better find an optics book!
The part that has me confused is references I've seen regarding cameras where the focal length is the same as the lens to (in this case) object distance. There, a whole different set of formulas hold and since both sets contradict, only (atmost) one can be right.
I could experiment and verify each set, but I want to be able to prove it to myself. Not as much fun building something if you don't know exactly how it works!
OOooohhh!!! That clicks...
ok, I think I had a poor idea of exactly what focal length is. I think I have it a lot better now, thanks for the help! 🙂
ok, I think I had a poor idea of exactly what focal length is. I think I have it a lot better now, thanks for the help! 🙂
Tried to edit that last post...
Been thinking about this over lunch and tried to edit my last post but my time expired to do that. My understanding as it stands now...
So with the fresnels, we assume one side is parallel light or in other wirds, an infinite throw, so we use the focal length as the actual lens to object distance. Have I got it?
(assuming non-split fresnel setup)
This means that the 2nd fresnel is placed exactly the FL distance from the optical center of the lens. It's size then should be big enough to cover the light cone from the lens center through the LCD.
The LCD is placed at the calculated LCD to Lens distance based on the lens FL and throw.
The first fresnel would similarly be placed it's FL from the light source.
I REALLY hope I have it this time. Now how does the use of a condenser effect the placement of that first fresnel? Thanks guys! I hope to also be getting a digital SLR soon and hope to document this well so others can understand it faster than I.
Been thinking about this over lunch and tried to edit my last post but my time expired to do that. My understanding as it stands now...
So with the fresnels, we assume one side is parallel light or in other wirds, an infinite throw, so we use the focal length as the actual lens to object distance. Have I got it?
(assuming non-split fresnel setup)
This means that the 2nd fresnel is placed exactly the FL distance from the optical center of the lens. It's size then should be big enough to cover the light cone from the lens center through the LCD.
The LCD is placed at the calculated LCD to Lens distance based on the lens FL and throw.
The first fresnel would similarly be placed it's FL from the light source.
I REALLY hope I have it this time. Now how does the use of a condenser effect the placement of that first fresnel? Thanks guys! I hope to also be getting a digital SLR soon and hope to document this well so others can understand it faster than I.
you have got it!
Yes, all of those statements are correct. You can fudge things a bit by changing the lamp arc to condensor fresnel distance, but your statements reflect the ideal design.
If you have the right focal length fresnels to match the projection lens, then the condensor fresnel has to "see" the light as if it is coming from the focal point. If you insert a pre-condensor lens in that path, then you have to place it so the same size and shape cone of light strikes the condensor fresnel. (That will still get you parallel rays coming out the other side of the fresnel.)
To get that shape of light cone coming out of the pre-condensor lens, you have to put the lamp closer to the fresnel than it would go without the pre-condensor lens. In fact, the whole point of using a pre-condensor lens is to put the lamp very close to it in order to capture as much light as possible.
There is a thread on this forum you can search for that has a graphical procedure for finding the optimum spacing, but you can also do it experimentally: Put the pre-condensor lens across the path of light from the lamp arc at the focal point to the fresnel, so the full diameter of the lens would just intersect all the light. Then move the lamp arc toward the lens until the cone of light it sends to the fresnel just fills the fresnel from corner to corner.
Yes, all of those statements are correct. You can fudge things a bit by changing the lamp arc to condensor fresnel distance, but your statements reflect the ideal design.
If you have the right focal length fresnels to match the projection lens, then the condensor fresnel has to "see" the light as if it is coming from the focal point. If you insert a pre-condensor lens in that path, then you have to place it so the same size and shape cone of light strikes the condensor fresnel. (That will still get you parallel rays coming out the other side of the fresnel.)
To get that shape of light cone coming out of the pre-condensor lens, you have to put the lamp closer to the fresnel than it would go without the pre-condensor lens. In fact, the whole point of using a pre-condensor lens is to put the lamp very close to it in order to capture as much light as possible.
There is a thread on this forum you can search for that has a graphical procedure for finding the optimum spacing, but you can also do it experimentally: Put the pre-condensor lens across the path of light from the lamp arc at the focal point to the fresnel, so the full diameter of the lens would just intersect all the light. Then move the lamp arc toward the lens until the cone of light it sends to the fresnel just fills the fresnel from corner to corner.
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