wich bulb? HQI-TS 250W+spherical or HQI-T 150W+par30?

[blackout]

i'm planning my diy projector, and the first thing i have to build is the lightbox, so i've to choose a bulb.
I've read in this forum that a classic spherical cold reflector with a 250W bulb and a big condenser collect about 65% of lights, so 20.000*0,65=13.000 lumens
i thing this is the only "cheap and easy" setup for a double ended lamp.
With a 150W single ended (g12) lamp i can recycle a par-30 reflector (from some spotlight) and build a setup that collect around 80-90% of the light: 13000*0.8-0.9 = 11.000 lumens.
so i have only a 2000 lumens gap with a 100W difference, so 100W less heat to dissipate.
I'm going to build a 7" lilliput projector, probably with 2 mirrors to save space.
Which bulb have i to choose? is 2000 lumens difference enought to visibly decrease brightness? are my calcolous correct?

thank you all, this forum is the best place to make something yourself!!

[Guy Grotke]

Most people use a lamp, spherical reflector, pre-condensor lens, and then a condensor fresnel to get parallel light through the LCD.

If you want to use a parabolic reflector, then you omit the pre-condensor lens and the condensor fresnel. But you need a really big parabolic reflector so the parallel light that comes out of it lights the whole LCD. You also need to do something to block the direct light from the lamp, or you will get a very bright hot spot in the center of your image. (ie. 20 times brighter than the rest of the image!) The best way to do that is to use a small spherical reflector directly past the lamp arc. This makes the parabolic reflector design too complex and difficult to get working well.

An elliptical reflector is a much better choice. Then the light diverging from the second focal point of the ellipse goes to a condensor fresnel, as if that focal point was the light source. You can even increase the efficiency more by adding a big spherical reflector (with a hole in the middle) to send the light that missed the first time back into the elliptical reflector.

[charlie10]

Do MH bulbs have even lumen distribution? Or (taking the lamp base to be the "south pole") do they produce higher flux near their "equator" than near their "poles"?

I guess a more specific form of my question is, what percentage of a HQI-TS 250W bulb's output lumens is concentrated into the region between the "45th parallels", north and south?

Apologies for the clumsy globe analogy

[Me2!]

The short answer is unsatisfying:

Its hard to say and it doesnt matter.

I'll explain. MH bulbs have a plasma chamber that may be 5-20mm x 5-10mm. Different areas have different intensities (temps) and your optics focus on a small area throwing the rest of the light away. You hope its thrown away because it would fuzz your image otherwise. So you take steps to shut the other light out. Do you know the properties of the arc point you use? No. It depends on the bulb orientation, cooling method, etc.

Osram has a document that goes over the arc in detail. It varies quite a bit! goto their website.

So you might only use 25% of the arc. Each optic may take up 10%. The polarisation on the back of the LCD eats at least 50%. Light doesnt shine through the entire panel due to the electronic control bits ~60%? All this stuff eats that percent of all light that made it to that point so its cummulative.

[Guy Grotke]

If I remove my LCD, and put a piece of white paper in place of the projection lens, then I can see a focussed image of the lamp's arc on the paper. In my projector, that arc image is about 60 mm long which agrees very well with the size of the lamp arc times the ratio of the two fresnels.

If you replaced the LCD with a large piece of sheet aluminum with a single hole in the center, then rays from the entire volume of the lamp arc would pass through that hole. The fresnels would still direct those rays to converge into a focussed image of the arc at the projection lens. If the hole was small, then the arc image would be dim. If the hole was larger, the image would be brighter.

When the different rays passing through such a hole are refracted by the projection lens, they all meet at one point on the screen. (Assuming you have a good lens!)

Each LCD cell acts like one of these holes. So I don't think it is possible to "use only 25% of the arc".

[Me2!]

Sounds like you moved your fresnel closer to get the "extra light" from the rest of the arc. This is wrong. I mean no disrespect, you seem a very bright guy. Your reflector will use the light from the focus and the rest of the arc but you don't want it from the rest of the arc.
See the picture. The 2/3 of the light from the focus that hits the reflector is what you want (in black). The red light from the whole rest of the arc will make a bright circle around your focus but you cant control the vector of these rays.

If you move your fresnel closer so they are in it's imaging window you will see the whole arc chamber as you said. This reduces your contrast and softens the image. Read unfocused.

[Guy Grotke]

Please draw optical diagrams with the light moving left to right. That is pretty much a universal standard.

You seem to have the idea that fresnels have some kind of "imaging window" that you can use to mask unwanted rays from the ends of the lamp arc. They don't. Rays from all parts of the lamp arc will go through the fresnel. If you put the lamp arc center exactly on the central axis at the focal length of the fresnel, then light from the center of the arc will exit mostly perpendicular to the fresnel (see the blue rays in the diagram below). Light from the ends of the arc will not be be perpendicular to the fresnel, but it will still get refracted to parallel paths by the fresnel (see the red and green rays below).

Of course, rays come from every part of the arc to every point on the fresnel surface. So the drawing just shows the extremes. The light coming through every point on the fresnel forms a diverging cone. There's no way to stop it!

Every individual LCD pixel will "see" a large spot on the fresnel that consists of rays from all parts of the arc. No way to stop that either.

If you move the lamp away from the fresnel focal distance, then the cone coming through each point on the fresnel will be centered around a line that converges toward the central axis, but it will still be a cone. Move the lamp closer to the fresnel, and the cone will tilt outward, again still a cone.

[Me2!]

This seems to be changing the subject.

Of course you can stop it. How are you stopping the rays expanding directly from your lamp that dont hit your reflector? You make a wall with a hole that limits the light to your desired cone.

I dont want the fresnel to see only the desired light, it only gets the desired light! lol And this is from a small part of the arc. Im switching to a smaller 150w G12 so a greater % of light created is used with the same system. I dont expect to lose brightness but i will see when it arrives. People with BIG arcs have missed the point of imaging optics.

When I first started I thought the structure of the lcd itself would image the light passing through so more light was better. I was wrong. You have to be precise with the light source.

So back to the original question - you use a bit of the arc. See the lamp manufacturer for the details of it. They assume an open area around it not a box with cooling design and sound damping design so your arc will be different.

[Guy Grotke]

Oh, I see now. You are talking about changing to a lamp with a smaller arc length, or using a restrictive aperature at the second focal point of an elliptical reflector. I agree that either one of those will give you a smaller light source.

When you wrote: "If you move your fresnel closer so they are in it's imaging window you will see the whole arc chamber...", I thought you were implying that you could limit the part of the arc of an existing lamp that supplied rays to the fresnel, just by adjusting the distance. Since your drawing of the elliptical reflector light engine showed the whole virtual arc image at the second focal point, I did not think you were proposing a restrictive aperature.

But I don't understand what you mean by "the structure of the lcd itself would image the light passing through"?

And I also think that lamp arcs are not all that much smaller until you get down to <5 mm, but those lamps run for so few hours and cost so much money that you might as well just buy a used commercial projector with a used-up lamp, and then put a new lamp in it. The DIY concept is to use the inexpensive 10000+ hour lamps that cost < $70 each. I use a Ushio retrofit lamp with a 24 mm arc, and I get an image so sharp I can see the individual red, green, and blue stripes within each pixel, from the center of the screen all the way to the corners. So arc length doesn't have to limit your image quality, if you have a good projection lens.

Maybe you could give us a URL for that OSRAM paper? I have never been able to find much good info on any of the many OSRAM websites.

[Me2!]

Sure, i'll try to find it.

The title is "Technology and application
Metal halide lamps Photo Optics"

Poor wording on my part. Originally I imagined the LCD would collimate the light to some extent so that you only had to provide a strong source directly behind it.

Yes the large arc is cheap but you design the system to create a small point like an expensive bulb. You throw a lot of light away and it bugs you. oh well.

So you get that good an image without anything limiting the source to your focus. thats cool. Do you find you lose contrast or reduce the vivid colours?