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- Thread starter MustangSVT
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Therefore without knowing your exact circumstances it is impossible to calculate an exact distance and even with everything I'm gussing it would be very difficult to calculate. If you do need to fix it then your best bet is a bit of trial and error before finalising your design.

HTH,

Steve

It is actually very easy to calculate the distances:

1/focal length = 1/LCD to lens + 1/lens to screen

If you know the throw distance and the projector lens focal length, then you can solve for the LCD to lens distance, which is what I think you are asking:

1/LCD to lens = 1/focal length - 1/lens to screen

For example: (300 mm fl lens & 10 foot throw distance)

1/LCD to lens = 1/300 mm - 1/3048 mm (that's 10 feet)

LCD to lens = 332.75 mm (always more than lens focal length)

Those distances are measured from the principle planes of the lens, which are right about the center of a simple lens. For a triplet, they are probably at each end. For your projector, try making the middle of the adjustment range the calculated distance from the LCD. Then you will have some play for focussing.

1/focal length = 1/LCD to lens + 1/lens to screen

based on the focal length from these equations:

An externally hosted image should be here but it was not working when we last tested it.

If so, then you want to figure out the focal length based on how far the projector is from the screen (the throw) and how big the image on the screen is.

If not, how do you take into account the image size on the screen?

Yes, the focal length is all the same in all of those equations. But I suspect you are confusing that with the LCD to lens distance. Focal length is a fixed property of a lens: If you point the lens at the sun, the distance to the smallest bright circle of light on the ground is the focal length.

If your screen was an infinite distance from the lens, then the lcd to lens distance would equal the focal length. Most people use a closer screen.

Other useful equations:

Magnification = (throw distance) / (LCD to lens distance)

Image size = LCD size * Magnification

sorry Guy, what do you mean by that?

the optical center (plane) on a triplet is at F distance from the smallest focusing point when you are focusing the sun (paralell light). It is easier to work with the back focal lengh. the 135 triplet has a 38cm back focal lengh, it means that there is 7cm between the rear lens end and the optical center.

With a simple thin lens, we use the thin lens approximation equations because the two principle planes of that lens are so close tegether that they make no difference.

If you try to do the same with a thick lens, then all of your calculated distances, etc. will be off (maybe by a lot). With a thick lens, the focal length is relative to the principle plane on that side of the lens. There is an equation to find the principle plane, based on the index of refraction, the radius, thickness, and focal length, but the simplest way to think about it is that the principle plane is "where the lens starts to work" to give the focal length you observe.

Here is a simple example: Imagine a duplet projection lens made from two 1000 mm fl PCX lenses. If you put a point source of light 1000 mm from the first lens, then it will send parallel rays out the other side. When they pass through the second lens, they will be focussed to a spot 1000 mm from that lens. If the two lenses were touching each other, then the two focal spots would be 2000 mm apart. But what if the two lenses were 1000 mm apart? Then the two focal spots would be 3000 mm apart! Each lens does its own refraction relative to its own principle plane. If you try to do ray tracing, the ray bends at each principle plane.

the two 1000 focal lenses forming the douplet, lets say 1000 space between the two lenses as you said... then we have 3000 space between the two points (between the object plane and the image plane, those are the lens equation planes).

but then you can consider the douplet as if it was a singlet perfect thin lens of F=1000mm focal (conbiation of thw two and 1000 separation). That imaginary lens would be placed exactly at the mudleway between the two real lenses (so at 1500mm from the light source point and 1500mm from the projected light point as well). So resuming the principal plane (obtical center plane) of the douplet is at center.

When we talk of 450mmfocal lens triplet we are doing exactly the same thing; we are considering it as if it was a perfect thin lens of 450mm focal. The optical center plane or the principal plane does tell us where that imaginary lens is.

I think Guy just showed us that you cannot. Indeed, 1/1500+1/1500>1/1000

That approximation works reasonably for thinner objectives, though. If you want more precision you have to consider separate light rays 'pivot' points. I think they are called entrance and exit pupil. It's possible they have something to do with the weird 2 FOV values you got from the 135mm dia triplet manufacturers.

zzonbi said:I think Guy just showed us that you cannot. Indeed, 1/1500+1/1500>1/1000

finally got it, took me some headacke;

the optical center on a doublet is not at it's center. This is why your 1500mm values do not give you 1 meter focal EFL.

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