A question about optics...

If you want conventional imaging, why not start with any 35mm camera lens (eg from a single use camera) - gives you a very wide range of possible focal lengths/aperture ? The size of the whole array is less than a 35mm image

OTOH, perhaps you were thinking of a lens-per-sensor type of arrangement (insect eye) ? In this case you may find that starting with a compact array was not such a good idea - difficult to squeeze in the optics. You can buy small (~ 4mm dia, but machineable) moulded lenses, or you could use pieces cut from a fresnel lens. You could even think about light pipes. Presumably in this case you would be looking to cover a fairly large solid angle anyway ...

A big problem is going to be getting enough sensitivity into any system like this.

Presumably you are trying to do something that cannot be done by hacking an opto mouse ?

Dave

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Maybe try some cheap (plastic) magnifying glasses? A suggestion also might be to put the transistors in a vertical row and then by rotating them (or the robot) you can get more resolution? You might be able to read large home made bar codes for example or detect vertical edges.

John Casey

Hi Dennis,

Let's cover some abbreviations first. ( I use these at work, though optics texts use different. )

EFL=effictive focal length FOV=field of view DOF=depth of field f#

The formula to determine the FOV is 2*ATAN(1/2 sensor surface length/EFL)

So imagine you have a 35mm EFL single element ( no need for anything more complicated ) SQRT(1.5^2+2.5^)=2.9=29 mm ( the diagonal of the sensor array)

2*ATAN((29/2)/53)=45 degree field of view.

blueeyedpop wrote: Hi Mike,

You gave me every bit of the information that I needed. Now to see if I an make the phototransistors sensitive enough to do any thing useful with it! I have LOTS of polaroid adjustable lens setups lying around from hacking the SONARs out of them, now I've got a good use for them.

have fun, DLC

: Hi Dennis,

: Let's cover some abbreviations first. ( I use these at work, though optics : texts use different. )

: EFL=effictive focal length : FOV=field of view : DOF=depth of field : f#

: The formula to determine the FOV is 2*ATAN(1/2 sensor surface length/EFL)

: So imagine you have a 35mm EFL single element ( no need for anything more : complicated ) : SQRT(1.5^2+2.5^)=2.9=29 mm ( the diagonal of the sensor array)

: 2*ATAN((29/2)/53)=45 degree field of view.

: ---- : f# is used to determine DOF

: Since your sensing elements are so large, we can almost ignore this, except : to say that

: f#=EFL/diameter. : Let us assume a 25mm diameter lens element. It will have an f# of 35/25 or : 1.4 : The larger the f#, the more depth of field we will have, and the less light : gathering power we will have.

: The typical progression of f# is : 1.4 : 2 : 2.8 : 4 : 5.6 : 8 : 11 : 16 : 22 : 32 : 45 : 64

: This is what you will see on the side of an SLR camera.

: Every step is halfing the intensity of the light, but increasing your depth : of field. : A 10 mm diameter lens will give you a value of f3.5, or less than 1/4 the : light gathering capability : (this would also apply to pinhole lenses)

: ----

: A lens will focus at infinity when it is 1 focal length from the imager's : surface. Moving it further out will focus on closer objects. Take 5 leds : into a darkened room, arrange them in a recognizable pattern. make a frame : with scotch tape(frosted) and hold the lens at some distance away from the : scotch tape. At some point, you will be able to observe the leds' image on : the scotch tape. I would even sugest making a mask that looked like your : sensor. print clear dots on frosted laser transparency film, and observe.

: If you have a digital camera, you can take pictures of this image to : simulate what is going on with your sensors.

: --- : I would use the cheapest optics you can get your hands on. depending on the : sensitivity, I would even consider a pinhole lens. Take some brass shim : stock, and drill a 1/16th inch diameter hole in it. You can use the above : calculations to determine the EFL of the lens based on its distance from the : subject, and its diameter.

: Mike

: > I've built up a "vision" system that is pretty low cost. A series of : > opto-transistors in a 5x6 array that I'm trying to get to work as a : > cheap vision board. I've run up hard against trying to match a lens : > to the thing. The array is about 2.5cm by 1.5cm. Not small nor is it : > of great resolution, but I'm hoping that it'll work as a motion sensor : > when all is said and done. : > Now my problem. I need to put a lens on it to give me a reasonable : > field of view, but I've no idea how to figure out the proper lens : > type or focal length to mount a lens. Heck, I'm not even sure what : > diameter or strength to make the lens. : >

: > If anyone has any ideas, suggestions or solutions I'd love to hear : > them. : >

: > thanks, : > DLC : >

: > -- : >

: ============================================================================ : > * Dennis Clark snipped-for-privacy@frii.com

: * : > * "Programming and Customizing the OOPic Microcontroller" Mcgraw-Hill 2003 : * : >

: ============================================================================

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Hey Dennis, good to see you again. Um, that's a fairly large-sized array to focus onto. However, I would think that about any "convex" lens should work that is 1-2" (2.5-5 cm) diameter and has a focal length of at least 1-2". With smaller lens diam and/or FL, you may not get good coverage of the edges of the array.

Basically all you need is a magnifying glass. You might just try experimenting with whatever MG's you have around the house, including any reading glasses, bifocals, etc [which are also just magnifying lenses]. I have a cheapo plastic MG, 2" diam and prolly 2X, and it focuses long-distance objects to about 4" - which is approx the FL of the lens. The smaller insert on the MG is prolly 6X-8X and focusses at

1.5". Would nicely cover your array.

I also have a 5X loupe which focuses at 2". You won't be needing anything special like an achromatic lens for this app. BTW, the formula for a convex lens FL is exactly the same one as for resistors in parallel - sound familiar :).

1/Fo + 1/Fi = 1/FL, where Fo = object distance, Fi = image distance, and FL = focal length.

The image size vs object size is directly related to the 2 distances Fo and Fi.

Xi : Fi :: Xo : Fo [I believe this is right]

For the same FL lens, if you make Fo smaller, then Fi will be larger. For an object at infinity, Fi = FL. If Fo is closer, as the case for focusing on your moving targets, then Fi will be a bit farther back. So,

Fo = far, then Fi = FL. Fo = 18" and FL = 2", then Fi = 2.18", etc.

Also, if the target size = 4" and Fo = 18", then the image size Xi = (2.18/18)*4" = .48" (hmmm, a little smaller than your array - but prolly about right for tracking from cell to cell).

If you want a lot of light, you'll need a wide field-of-view, and then the depth of field will not be good [you know your 35mm camera basics]. However, for the lens given, Fi will not vary that much for your target distance going 12-36",

Fo = 36" and FL=2", then Fi = 2.12". Fo = 12" and FL=2", then Fi = 2.40".

This prolly won't matter to your 5x6 array. Of course, you get better D-o-F by using a smaller aperture, but then you'll also loose light. The usual tradeoff. I think this is all correct - been a while with the optics.

have fun,

- dan michaels

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Not sure what exactly your application is, but have you considered using an optical mouse sensor like the kind Agilent (HP) makes? 16x16 array, spi interface, 2300 frames per sec...built in motion detection....

The SPI transfer of the mouse is slow, whereas transistors are fast

The optical requirements of the mouse sensor are difficult. A 1 mm lens is like a 53 deg field of view. Getting that wide is difficult, wider is near impossible without a lot of hassle. The sensor is deeply entrenched within the package compared to the focal length of the lens.

Phototransistors can be fast, and interfacing optically is easier in many ways.

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dan michaels wrote: Dan, et. al,

Great info, thanks.

DLC

: Hey Dennis, good to see you again. Um, that's a fairly large-sized : array to focus onto. However, I would think that about any "convex" : lens should work that is 1-2" (2.5-5 cm) diameter and has a focal : length of at least 1-2". With smaller lens diam and/or FL, you may not : get good coverage of the edges of the array.

: Basically all you need is a magnifying glass. You might just try : experimenting with whatever MG's you have around the house, including : any reading glasses, bifocals, etc [which are also just magnifying : lenses]. I have a cheapo plastic MG, 2" diam and prolly 2X, and it : focuses long-distance objects to about 4" - which is approx the FL of : the lens. The smaller insert on the MG is prolly 6X-8X and focusses at : 1.5". Would nicely cover your array.

: I also have a 5X loupe which focuses at 2". You won't be needing : anything special like an achromatic lens for this app. BTW, the : formula for a convex lens FL is exactly the same one as for resistors : in parallel - sound familiar :).

: 1/Fo + 1/Fi = 1/FL, where Fo = object distance, Fi = image distance, : and FL = focal length.

: The image size vs object size is directly related to the 2 distances : Fo and Fi.

: Xi : Fi :: Xo : Fo [I believe this is right]

: For the same FL lens, if you make Fo smaller, then Fi will be larger. : For an object at infinity, Fi = FL. If Fo is closer, as the case for : focusing on your moving targets, then Fi will be a bit farther back. : So,

: Fo = far, then Fi = FL. : Fo = 18" and FL = 2", then Fi = 2.18", etc.

: Also, if the target size = 4" and Fo = 18", then the image size Xi = : (2.18/18)*4" = .48" (hmmm, a little smaller than your array - but : prolly about right for tracking from cell to cell).

: If you want a lot of light, you'll need a wide field-of-view, and then : the depth of field will not be good [you know your 35mm camera : basics]. However, for the lens given, Fi will not vary that much for : your target distance going 12-36",

: Fo = 36" and FL=2", then Fi = 2.12". : Fo = 12" and FL=2", then Fi = 2.40".

: This prolly won't matter to your 5x6 array. Of course, you get better : D-o-F by using a smaller aperture, but then you'll also loose light. : The usual tradeoff. : : I think this is all correct - been a while with the optics.

: have fun, : - dan michaels :

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