135mm long-throw lense kit from diylabs

[jcbklyny]

there are no wrong lens, just wrong setups.

Trust me, there IS such a thing as having 'the wrong lenses' for certain setups and no matter how much you move and adjust it will never work the way it SHOULD!

and Rox, enough with the 135... really. You talk more about that lens then any 15 people on any one of the DIY projector boards and you dont even have one! BUY ONE... and then talk. Or dont and drop it.


-JCB
www.diybuildergroup.com

[Rox]

its glad to me to tell you that i received information from awi.

They aren´t good news for you. But i am still working on it. You will have news soon.

[Rox]

Ok here it is;

THE MAX FIELD ANGLE THIS LENS WAS DESIGNED FOR IS: 24 DEGREES

What does it mean? it was designed for 24 degrees, and you can use a larger field angle but the lens was not designed for more than 24.

So 45 degrees need for 17" are very far and the 41 degrees for 15" are closer but something far as well.

Now this valious information has been reveled. Each one can do what hell the wants.

Good luck.

[Guy Grotke]

I wonder if that was the half angle they quoted you? (The maximum field angle from the central axis.) I calculated a full field angle of around 49 degrees for this lens, (using limited size data), which is very close to twice the 24 degrees you claim. This is also a much more typical full field angle for a triplet design.

If they wanted a projection lens with a full field angle of just 24 degrees, then they could have done it much cheaper.

[Hezz]

ROX,

The FOV angle is usually quoted for the angle which is the angle between the axial center of the lens and the angle for which there has been correction made. This is because most lens systems are symetrical. Therefore the usable total angle is 48 degrees. Now the longer the FL is the farther the lens is away from the image source so the bigger image it can focus.

If you take a 48 degree angle and move it out from a flat surface about 15 to 19 inches and lengthen the angle lines out until they intersect the image plane. That is about the image size that the lens can focus.

If you don't have CAD software you can easily do this with a large piece of cardboard or paper, a protractor and a straight edge.

I will sketch a CAD drawing for you so you will know the total theoretical size that the lens can image. Be back in a few minutes.

Hezz

[Hezz]

ROX,

If you can interpret this sketch one thing should become obvious.

This lens should in theory work with up to a 21 inch LCD. But you would probably need to project at least a 200 - 250 inch screen to get it to work. This is just estimate.

This is going to give some guys with a 600 watt metal halide bulb some
dangerous ideas. If your room was big enough and you made a monster projector you could use a 21 inch LCD and have a really large image.


Hezz

[Rox]

well, i didn´t know that the field angle was the half angle so twice would be the correct one we can use to our setups.

i´ll check it with the manufacturer.
24 degrees (12+12?)

OR

48 degrees (24+24?)

sorry about confusing you.

[Rox]

Originally posted by Guy Grotke
I calculated a full field angle of around 49 degrees for this lens, (using limited size data),

Please could you tell us how?

Those are first hand specs;

450mm EFL ,

F=3,6

Clear aperture= 125mm

barrel lenght=150mm

midle element F stop aperture=109mm

[Guy Grotke]

The clear aperature and the middle element aperature would be important if we were using this triplet for photography. But we are not!

The fresnels are used to create a cone of light that could be focussed at the same plane as the optical center of the lens. The edges of that cone form a virtual aperature, since there is no light outside the cone. Assuming you use a 24 mm arc, 220 and 550 mm fl fresnels, and the LCD to lens distance is 528 mm, then the focussed arc image would be 60 mm long.

Draw a full-scale picture of the LCD and lens cross-section with the LCD on the left side. Draw the 60 mm arc image as a vertical line through the center of the triplet. Then draw a line from the upper edge of the arc image to the upper edge of the lens barrel. Measure the angle of that line from the central axis. That is the half angle. You can also extend the line to the left until it crosses the plane of the LCD. If it passes above the vertical line that represents the LCD, then all of the light passing through corner pixels can get to the arc image in the lens, and thus out the other side to the screen.

Of course, this is just a simple estimate. It assumes that the optical center of the lens is at the physical center, which may not be true. It also assumes the triplet is symmetrical, so light that gets in along the arc image path will also clear the opposite barrel. But it should be pretty accurate, since the lens could focus a clear image of a pixel even if some of the light from that pixel was blocked.

[Rox]

i would say that if we consider the refraction by the first and third lens the angle would be even smaller. forget by now the arc lengh and try to imagine a ligh ray from first lens upside passing throw midle lens elemnt and get outs from botton third elemnt. I think this is the critical ligh stop last ray. then if you consider the refranction by those two lenses (midle lens is not working) you would see the angle of the ray coming from lcd is smaller than the angle inside the triplet. Also we can say that the light would be dimer because of the arc lengh, but i don´t want to complicate more.

here is what i mean;