Project 8mm film magnified on a piece of glass a few inches away

Thread Starter

jerujanssen

Joined Mar 27, 2021
6
For an artproject I would like to project 8mm film magnified on a piece of glass a few inches away.
Basically an analog micro filmprojector.

The setup I have thusfar is a lend-ended E10 bulb that lights the 8mm film. This does project a very small image
on the glass.

But this image needs to be enlarged. I've been trying several lenses I have laying around in my studio,
but no luck. I've been looking at lenses online, but have no clue what to specifically look for, I am pretty sure
there is math involved with focal distances and such. Do I need a double convex lens, or plano or any other type?
I am a visual artist and have no experience whatsoever with electronics, and math was my worse subject at school.
As usual I am way over my head with this project...

Any feedback would be greatly appreciated!
 

LesJones

Joined Jan 8, 2017
3,078
You probably require a lens with a short focal length. How large do you want the image to be and how many inches is "A few" ?
The ratio of the size of the image on the glass to the size of the image on the 8mm film will be the same ratio as the distance between the lens and the glass and the lens and the film. (So the lens will be closer to the film than it is to the glass.)
The formula for lens calculations is 1/F = 1/U + 1/V F is the focal length of the lens. U and V are the object distance (Film) and the image distance. (Glass) I can't remember if U is the image distance or the object distance as it is about 60 years since I left school.
It does not not matter for your calculation. Also the units do not matter so long as they are ALL THE SAME. An easy way to measure the focal length of a lens is to use it to project the image of a distant object (Such as the sun) onto a piece of paper.and measure the distance between the lens and the paper.
If you post the size of the image on the 8mm film. the size of the image on the glass and the spacing between the film and the glass many of the forum members will be able to calculate the required focal length for the lens for you. Have you considered scanning the film on a high resolution scanner and printing the image at the required size on transparent film.

Les.
 

Thread Starter

jerujanssen

Joined Mar 27, 2021
6
Thank you so much for your helpful reply Les!

About the film: I am planning to use a loop cartridge to loop the 8mm film. I know there are probably easier (digital) solutions, but I really want to use analog 8mm film, as that’s the premisse of the project.

The measurements are not set yet. It also depends if it’s even possible to have such a small focal length. Otherwise I might have to resort to using mirrors?

Let’s say the glass is about 6cm away from the film and needs to project an image of about 10cm tall.
 

MrSalts

Joined Apr 2, 2020
193
If you use an LED to illuminate it, you a put the light source very close to the film and achieve your effect (possibly) though a lens. Assume a point source of light that projects at 120° arc, to illuminate your whole 8mm width, you’ll need to be just under 8mm away.

you are looking for about 12x magnification so you’ll need to put the light about 5mm from the film and the glass about 60mm from the glass. Just draw it out carefully to estimate the projection with Ray tracing or use some trig. Either way, no lens needed. The biggest challenge will be soldering a high brightness flat SMD LED to a circuitboard if you are not familiar. I can send you one or two if needed - I have circuit boards remade for something similar.

note, you’ll need to use a colored LED (white LEDs have phosphor that makes a board area of light and, without a point source of light, you’ll have “penumbra” shadows (gray) covering your whole image and much of your contrast will be lost). A point source of light will give all “umbra” shadows with the best possible contrast.
 
Last edited:

LesJones

Joined Jan 8, 2017
3,078
I have confirmed that "U" is the image distance so we use the terms correctly.
I agree that the image needs to be about 12 times larger than the image on the film.
So to get 12 times magnification V needs to be 12 times U.
As the distance between the film and the glass is 60 mm U + V needs to be 60 mm. As V = 12 times U then U = 60/13 mm = 4.6.
So V = 4.6 mm * 12 = 55.2 (Note I am only working to one decimal place.)
As 1/U + 1/V= 1/F
1/4.6 + 1/55.2 = 1/F
0.217 + 0.018 =1/F
1/F = 0.235
F = 1/0.235 = 4.26 mm This is the required focal length of the lens.
These dimensions will result in quite a wide field of view.
The tangent of half the field of view will be 50mm/55,2mm = 0.906
The angle with this tangent is 42.2 degrees (Used scientific calculator to get tan ^-1)
So the field of view = 84.4 degrees
Possible sources of lens might be from a webcam or similar or possibly from the optics in a CD or DVD drive.
There will be some distortion of the image with this field of view.

Les.
 

Danko

Joined Nov 22, 2017
1,060
Let’s say the glass is about 6cm away from the film and needs to project an image of about 10cm tall.
Graphical solution.
Solved in Microsoft Visio:
1616898833466.png
--------------------------------------------------------------------------------------
http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/image.html
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ADDED: US $29.99 IVEC CCTV Box Camera Lens 2.5mm WIDE ANGLE NEW "open box"
https://www.ebay.com.au/itm/IVEC-CCTV-Box-Camera-Lens-2-5mm-WIDE-ANGLE-NEW-open-box-/361613229303
 
Last edited:

MrSalts

Joined Apr 2, 2020
193
Li lh loou book

You are right! All of those cameras and projectors that have we’ve been sold are a scam! Perhaps we should initiate a class action suit and get our money back.

Bob
Or, you can do the math and realize that meeting the OPs requirement of thin stack up, short range and wide angle doesn’t require any lenses. But, feel free to add extra components just because, ummm.., because, a... because... they exist?
 

Thread Starter

jerujanssen

Joined Mar 27, 2021
6
Thank you all for your indepth replies, much appreciated!

I will order a few things and see if I can get a proof of concept working...
 

LesJones

Joined Jan 8, 2017
3,078
When I wrote post #5 I did not have time to check on the actual image size on 8mm film. Danko in post #6 has checked the image size so his result is more likely to be correct than mine. (As an exercise you could do the calculation for the correct image size using the formula that I showed you.) I think you will find it difficult to get this to work well as your size limitations force the use of a lens with a very short focal length.

Les.
 

MrChips

Joined Oct 2, 2009
23,069
You need to increase the optical distance between film and screen. You can do this with a pair of mirrors or prisms as they do in monoculars and binoculars.
 

BobTPH

Joined Jun 5, 2013
3,000
Or, you can do the math and realize that meeting the OPs requirement of thin stack up, short range and wide angle doesn’t require any lenses. But, feel free to add extra components just because, ummm.., because, a... because... they exist?
OP's requirements, as I understand them:

Size of film frame 4.8 x 3.5 mm.
Size of projected image 10cm (I'll assume best case, this is the width, not the height)
Distance from screen: 6cm.
Calculated image magnification required: 100 mm / 4.8 mm = 20.88 X

I will add, and the TS can correct me if the requirements are less:

Bright enough to see in a normally lighted room.
Sharp enough to say, recognize head shots of a famous persons.

If you can use an LED and no lens to project an image according to those specifications, you are a far better engineer than I.

Bob
 

Thread Starter

jerujanssen

Joined Mar 27, 2021
6
I will try both options, with and without a lens. I do think I might have to use mirrors to enlarge the distance though.

Thusfar with a setup with a lens ended lightbulb it’s impossible to enlarge the image without a lens. That might change with a high powered LED, but we’ll see.

It will take quite a while for the bought items to arrive, but I’ll show pictures if I have something to show!
 

BobTPH

Joined Jun 5, 2013
3,000
The only way works without a lens is when the light is a point source. There is no such thing. But, if the light source is small compared to the frame you are projecting it will approximate a point source. It is going to be hard, if not impossible to find an LED with a small enough emission area and enough power to do this. It would be easier it it were a 35mm frame.

Bob
 

ApacheKid

Joined Jan 12, 2015
377
For an artproject I would like to project 8mm film magnified on a piece of glass a few inches away.
Basically an analog micro filmprojector.

The setup I have thusfar is a lend-ended E10 bulb that lights the 8mm film. This does project a very small image
on the glass.

But this image needs to be enlarged. I've been trying several lenses I have laying around in my studio,
but no luck. I've been looking at lenses online, but have no clue what to specifically look for, I am pretty sure
there is math involved with focal distances and such. Do I need a double convex lens, or plano or any other type?
I am a visual artist and have no experience whatsoever with electronics, and math was my worse subject at school.
As usual I am way over my head with this project...

Any feedback would be greatly appreciated!
I think you'd be wise to use some form of efficient LED light source, one of the problems old cine projectors faced is that if the film ever stopped for any length of time it would melt, the film had to be kept running, if you paused a projector the frames being displayed would melt in a few seconds.

If you have a stationary piece of true 8mm film you run the risk of destroying the film itself.
 
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