Im new here, and a noob to DIY Projectors. I've toyed with the idea for a couple years, but didn't have the resources or time.
Anyways, I recently pulled 3 projector lenses from an old Zenith rear-projection TV. They arent marked in any way except for a Zenith sticker with the part number "A-8718" and the name DELTA IV. Being that these are from such an old TV (circa early 1980's), I cant find any documentation on them anywhere.
Is there any way to tell the focal length on them? Also, do you think this is a viable lens for a projector using a 17" NEC LCD?
Here's a photo. Sorry about the quality, the only camera I have at the moment is my camera phone.
Anyways, I recently pulled 3 projector lenses from an old Zenith rear-projection TV. They arent marked in any way except for a Zenith sticker with the part number "A-8718" and the name DELTA IV. Being that these are from such an old TV (circa early 1980's), I cant find any documentation on them anywhere.
Is there any way to tell the focal length on them? Also, do you think this is a viable lens for a projector using a 17" NEC LCD?
Here's a photo. Sorry about the quality, the only camera I have at the moment is my camera phone.
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I pulled one of them apart this morning and found 3 lenses in the body. The one closest to the image source is flat on the back, deeply concave on the front. The middle is convex on both faces, shallow on the rear, more agressive on the front. The last is also convex on both faces, but shallow, and the lens is thin overall. I'll try to build a rig tonight to test it.
Found the formula for focal length in another thread. I've been looking for days, but when I originally posted, I hadnt found it yet. PERSEVERANCE!!!
Anyways, using the formula:
1/f = 1/o + 1/i (where i is the projected image, o is the object being projected (lcd), and f is the calculated focal length)
1/f = 1/1500mm + 1/180mm
1/f = ~6.6666 + ~0.0055
1/f = ~0.00622
f = ~160mm
That's an awfully short FL... According to the calculator on diyprojectorcompany.com for a 64" diagonal, I would need a throw of 30" ... not an option.
So, I either need to find a way to lengthen my FL, or find a new lens.
Anyone want 3 rear-projection lenses??
🙄
Anyways, using the formula:
1/f = 1/o + 1/i (where i is the projected image, o is the object being projected (lcd), and f is the calculated focal length)
1/f = 1/1500mm + 1/180mm
1/f = ~6.6666 + ~0.0055
1/f = ~0.00622
f = ~160mm
That's an awfully short FL... According to the calculator on diyprojectorcompany.com for a 64" diagonal, I would need a throw of 30" ... not an option.
So, I either need to find a way to lengthen my FL, or find a new lens.
Anyone want 3 rear-projection lenses??
🙄
your maths are wrong... or some data is missing....
you have 7" LCD and need 1500mm diagonal Image, this is not enough for determining focal... any focal can give you that projection at the propper trhow...
you have 7" LCD and need 1500mm diagonal Image, this is not enough for determining focal... any focal can give you that projection at the propper trhow...
17" LCD, triplet objective lens with a FL of 160mm, and a goal of 64" diagonal projection on the wall, gives me a required throw of only 30". Thats according to that online calculator. But just from playing around with the lenses and experimenting, it seems reasonably accurate. Either way, I dont think these lenses will work for me... not for what I want them for anyways.
yes, 17"--64" image 16cm lens... 30" would be the trhow. This is correct.
"1/f = 1/o + 1/i (where i is the projected image, o is the object being projected (lcd), and f is the calculated focal length)
1/f = 1/1500mm + 1/180mm
1/f = ~6.6666 + ~0.0055
1/f = ~0.00622
f = ~160mm
"
but this is wrong... what does "o" mean? what does "i" mean...?
180mm I thought that was LCD 7" size...
as you sugest, i and o are SIZES, that equation is false. It only works with distances being o= object to lens distance and i=image to lens distance.
"1/f = 1/o + 1/i (where i is the projected image, o is the object being projected (lcd), and f is the calculated focal length)
1/f = 1/1500mm + 1/180mm
1/f = ~6.6666 + ~0.0055
1/f = ~0.00622
f = ~160mm
"
but this is wrong... what does "o" mean? what does "i" mean...?
180mm I thought that was LCD 7" size...
as you sugest, i and o are SIZES, that equation is false. It only works with distances being o= object to lens distance and i=image to lens distance.
Using my laptop screen facing the wall, I slowly moved the lens backwards and forwards until I could see a focused image on the wall. I measured the distances out to be 18cm from lens to wall, and 150cm from lens to laptop screen surface. Thats where those numbers are coming from.
Is this wrong still?
If I hold the lens 1 cm from the surface of the screen, I get a focused image on the wall from about 3-4 feet. And it's huge. Unuseably huge.
Is this wrong still?
If I hold the lens 1 cm from the surface of the screen, I get a focused image on the wall from about 3-4 feet. And it's huge. Unuseably huge.
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