Almost finished my first pinhole camera

[Existing Light]

Well, I've almost finished my first pinhole camera. It's a 7.5 inch focal length camera made out of a birch plywood that I found at hobby Lobby. I just have one more step: making the pinhole. Since my math skills are comparable to the math skills of a drunk monkey, can someone check my numbers to make sure I havent missed something or screwed up some simple calculation?

I'm using a formula that Ralph Lambrecht posted on a pinhole thread here, which is PinholeDiameter=sqrt( (2.44)(wavelength)(Focal Length) )

I found on a seperate website (cant remember where) that 5500Angstroms is the wavelength of average daylight, so I guess I'll go with that. That same website said 1Angstrom = 1nm, which is 0.00055mm, right?

To convert 7.5inches to millimeters, I multiplied 7.5 by 2.54 to get 19.05cm. I multiplied that by 10 to get 190.5mm.

i plugged the wavelength and focal length in millimeters in to the formula. Assuming I did my math right, the diameter should be about 0.5mm. Is that right? If not, could someone point me in the right direction?




BTW, I have some 5x7 HP5+ coming from Freestyle. Ihope to be taking pictures with it by this weekend :

 

[grahamp]

1 Angstrom = 0.1 nanometres

I did all my x-ray diffraction in the old units, and had to adapt when the publication standard changed to SI units.

(Apologies to the Scandianavian readers for not accenting the A correctly)

 

[RalphLambrecht]

...I'm using a formula that Ralph Lambrecht posted on a pinhole thread here, which is PinholeDiameter=sqrt( (2.44)(wavelength)(Focal Length) )

I found on a seperate website (cant remember where) that 5500Angstroms is the wavelength of average daylight, so I guess I'll go with that. That same website said 1Angstrom = 1nm, which is 0.00055mm, right?

To convert 7.5inches to millimeters, I multiplied 7.5 by 2.54 to get 19.05cm. I multiplied that by 10 to get 190.5mm.

i plugged the wavelength and focal length in millimeters in to the formula. Assuming I did my math right, the diameter should be about 0.5mm. Is that right?...

Looks all right to me!

 

[Existing Light]

1 Angstrom = 0.1 nanometres

I did all my x-ray diffraction in the old units, and had to adapt when the publication standard changed to SI units.

(Apologies to the Scandianavian readers for not accenting the A correctly)

oops. That's what I meant to type

 

[Existing Light]

Looks all right to me!

Thanks for the confirmation, Ralph.You telling me it looks right makes me feel better.

I do have a question, though. I was looking on the website where I found the wavelength measurement (http://bizarrelabs.com/pin3.htm). It says "1Å = .1 nm or 0.0000001 mm." Is that right? Isnt 0.1Å equal to 0.00000000001? If so, wouldnt that make 0.1Å equal to 0.00000001mm? Perhaps it's just late and I'm not thinking straight. I'm not used to dealing with numbers that small. Millimeters is usually as small as I deal with on a regular basis, so I'm probably adding too many zeros or something

 

[Romary]

Do not worry to much about the size of the hole. There are plenty of formula.

The article is in French but you will understand the formula : http://www.photo-argentique.com/modules/wordbook/entry.php?entryID=26

Be careful, I notice this week end I made a mistake in the Combe formula. I put five of them. According to Eric Renner in his (very intersting) book « Pinhole Photography, rediscovering a historic technique » there are more than 50.

Personally, I still have very big difficulty to understand the purpose of the light wavelength. I strongly think thta they other parameters such the thickness of the material, the shape of the hole which are first order when the wavelentgh is far far behind. Not counting the fact that the light on earth is not monochromatic (is it English?).

 

[RalphLambrecht]

Thanks for the confirmation, Ralph.You telling me it looks right makes me feel better.

I do have a question, though. I was looking on the website where I found the wavelength measurement (http://bizarrelabs.com/pin3.htm). It says "1Å = .1 nm or 0.0000001 mm." Is that right? Isnt 0.1Å equal to 0.00000000001? If so, wouldnt that make 0.1Å equal to 0.00000001mm? Perhaps it's just late and I'm not thinking straight. I'm not used to dealing with numbers that small. Millimeters is usually as small as I deal with on a regular basis, so I'm probably adding too many zeros or something

Just forget the Angstrom business. The visible spectrum ranges from roughly 400-700 nm. A good medium is 550 nm wich are 0.000550 mm.

 

[RalphLambrecht]

Do not worry to much about the size of the hole. There are plenty of formula.

The article is in French but you will understand the formula : http://www.photo-argentique.com/modules/wordbook/entry.php?entryID=26

Be careful, I notice this week end I made a mistake in the Combe formula. I put five of them. According to Eric Renner in his (very intersting) book « Pinhole Photography, rediscovering a historic technique » there are more than 50.

Personally, I still have very big difficulty to understand the purpose of the light wavelength. I strongly think thta they other parameters such the thickness of the material, the shape of the hole which are first order when the wavelentgh is far far behind. Not counting the fact that the light on earth is not monochromatic (is it English?).

Yes, there are many very old (1850s) and very arbitrary formulas. Eric Renner's book lists a few and gives some historic background. I like his book for its creative content and the inspiration it provides, but it does not go too deep into the mathematics of pinhole imaging.

Recent research and MTF studies at the Royal Institute of Technology in Stockholm, Sweden have shown that there are only two equations with a solid mathematical foundation. One for max resolution (based on Rayleigh criterion), and one for max sharpness or contrast (based on Airy disc). You can safely ignore all others as they all lie somewhere between the two equations or are just in a different format. Since most people value sharpness over resolution, I based the above table on the optimum pinhole diameter for max sharpness, which is based on the Airy disc:

d = SQRT (2.44 * wavelength * focal length)

or sometimes shown as:

d = 1.56 * SQRT (wavelength * focal length)

As to the wavelength, the visible spectrum ranges from roughly 400-700 nm and film is sensitive to the most part of it. Plugging these values into the equation gives quite a spread (almost 200%). Unfortunately, you have to pick a value, because pinholes cannot be corrected for chromatic aberration. It makes sense to pick a medium value at which the eye is most sensitive and push the threshold of sharpness to the boundaries of the visible spectrum. Hence the suggestion of 550 nm.

Pinhole shape has a big influence. You can use that creatively (see Renner's book) or you can go for a laser-cut pinhole to optimize image quality. I can offer a contact at the University of Munich, Germany who can laser-cut pinholes into thin brass plates for you at a reasonable price. he made all of mine.

 

[Romary]

Recent research and MTF studies at the Royal Institute of Technology in Stockholm, Sweden have shown that there are only two equations with a solid mathematical foundation.

Do you the reference of the paper (or papers)? I would be interested.


If you are interested, Stenocamera in France manufacture very good hole also : http://www.stenocamera.fr/ (shop : http://s192257538.onlinehome.fr/stenocamera/Boutique du site.html )

 

[RalphLambrecht]

Do you the reference of the paper (or papers)? I would be interested.


If you are interested, Stenocamera in France manufacture very good hole also : http://www.stenocamera.fr/ (shop : http://s192257538.onlinehome.fr/stenocamera/Boutique du site.html )

I have it somewhere and will take a look.

 

[wildbillbugman]

Pinhole shape has a big influence. You can use that creatively (see Renner's book) or you can go for a laser-cut pinhole to optimize image quality. I can offer a contact at the University of Munich, Germany who can laser-cut pinholes into thin brass plates for you at a reasonable price. he made all of mine.[/QUOTE]

Hi,
I am "new" to piholes,as I have not done it in several years. And even when I was trying to do it I was nowhere near the leval of accuracy that people on this Forum are about. But it seams to me that one cannot speak of pinhole diameter and ignore thickness of the material used.
One gentleman,on another thread, talks about hammering,filing,over and and over,until microscopic inspection shows perfect roundness and a very thin wall.
Laser drilling will leave a wall the same thickness as the thickness of the original foil.
Regards to all,
Bill

 

[Existing Light]

Thanks for the advice guys. I'm really not too worried about the wavelength and pinhole diameter stuff. I'm just kinda curious about the physics and reasoning for doing what I'm doing, even though I'm not the best mathmatician in the world

 

[Romary]

I have it somewhere and will take a look.

Thanks Ralph.

Just in case you do not know this one : http://inside.mines.edu/~mmyoung/PHCamera.pdf

 

[Romary]

But it seams to me that one cannot speak of pinhole diameter and ignore thickness of the material used.

Bill

I totally agree, I strongly believe (= I have not done any trial or calculation, which is not scientific) the thickness is one of first order parameter.

You can find here a work (sorry in French) made by a gentleman call Claude Boutet. He concludes, what intiutively looks logical, that the thiner the material is the better it is. The only proble that a infitively thin material does not exist. And from what I can see mathematical approach usually assume an infinitively fine material.

 

[Steve Smith]

Laser drilling will leave a wall the same thickness as the thickness of the original foil.

Depending on the laser used, it may be possible to laser etch the area around the hole so that the area local to the hole is much thinner than the rest of the material - like a lasered counter-bore.

My laser cutter at work cannot cut metal but I do this with plastic materials.


Steve.

 

[Steve Smith]

Something else has just come into my mind (always dangerous!).

The company I work for does precision industrial screen printing. It would be possible to print an opaque layer of ink onto an optically clear material such as glass or polyester with a small opening. This would effectively make a pinhole in an 8 micron thick material (cured ink) bonded to a clear support. I might even get round to trying it one day!


Steve.

 

[RalphLambrecht]

I totally agree, I strongly believe (= I have not done any trial or calculation, which is not scientific) the thickness is one of first order parameter.

You can find here a work (sorry in French) made by a gentleman call Claude Boutet. He concludes, what intiutively looks logical, that the thiner the material is the better it is. The only proble that a infitively thin material does not exist. And from what I can see mathematical approach usually assume an infinitively fine material.

I'm sure, the thinner the better. My pinholes are laser cut into a 0.04 mm foil, which means the hole/thickness ratio is about 10:1. With this ratio, thickness has little influence. As the material gets thicker, light rays bouncing off the hole inner walls will cause optical disturbances and thickness (and the smoothness of the inner walls) become a factor. But this is not an issue if the hole diameter increases by the same factor, because the ration is important. As far as I know, there is little research on this topic.

 

[RalphLambrecht]

Something else has just come into my mind (always dangerous!).

The company I work for does precision industrial screen printing. It would be possible to print an opaque layer of ink onto an optically clear material such as glass or polyester with a small opening. This would effectively make a pinhole in an 8 micron thick material (cured ink) bonded to a clear support. I might even get round to trying it one day!


Steve.

This is how zone plates and photon sieves are made.

 

[RalphLambrecht]

...Laser drilling will leave a wall the same thickness as the thickness of the original foil...

Looking at the hole under the microscope, I cannot detect that with my pinholes, but I saw it once. I returned it and got a clean replacement. Attached is a picture of my 0.25 mm pinhole.

 

[Existing Light]

My film came in today, along with some powdered fixer and a bulb for my safelight. I'll probably mix the fixer up today and try to take a few shots tomorrow after work.

I bought a 25 pack of 5x7 HP5+, and a bit intimidated by the size of it. The largest film I've shot so far has been 4x5, and that was back in 2007, so it's going to take me a while to get used to this large format thing. I think I can handle it, though. Hopefully I'll have some prints to show in a few days

 

[Ric Johnson]

Don't forget to look at the f295 website. More pinhole information when you think exists.

 

[Existing Light]

I just developed my first pinhole print. Results were less than amazing It's waaaay underesposed. It's my first try, though, so I'm not too disapointed.

The picture was of a couch in the living room. I figured out the exposure for f/16 and then used a little math to figure out that my exposure time of about 70 minutes. There's a bit of an image, but not much.

I have the second one going now. I'm exposing my next sheet outside where it's a bit brighter. I determined that my exposure should be a little over 4 minutes, but I set my timer to 15 since my last sheet was so underexposed. Hopefully this one will turn out better

If anyone wants to know, I'm shooting HP5+ rated at EI 200. I'm developing it in Rodinal 1:50 at 20C. My aperture is 381, but I rounded up to the next fullest aperture. I dont have my notecard with me, so I dont know what that next full aperture is. (My notecard is taped to my camera).

I'll post scans when my film dries

 

[Existing Light]

Don't forget to look at the f295 website. More pinhole information when you think exists.

thanks for the link. I have it bookmarked

 

[RalphLambrecht]

I just developed my first pinhole print. Results were less than amazing It's waaaay underesposed. It's my first try, though, so I'm not too disapointed.

The picture was of a couch in the living room. I figured out the exposure for f/16 and then used a little math to figure out that my exposure time of about 70 minutes. There's a bit of an image, but not much...

Share your math with us.

Also, don't forget to adjust for reciprocity failure.

 

[Existing Light]

Share your math with us.

Also, don't forget to adjust for reciprocity failure.

Uh oh. If you're asking for me to share my math, there must be something wrong with my multiplication (which is very likely).

I finished my exposure and brought my camera and notecard in. I have an aperture scale in one of my photo books that only goes to f/64. I extended that out to f/512, which is the first full stop past my aperture of f/381


I figured out the aperture of my camera by dividing the pinhole diameter by the focal length. My focal length is 7.5 inches, which multiplied by 2.54 to get 19.05 cm. i multiplied that by 10 to get 190.5mm. I divided that by 0.5, which is my pinhole diameter. That gave me an aperture of 381.

I rounded up to f/512 because I knew underexposure was a problem. I didnt adjust for reciprocity the first time because I was like "well, let's see what happens before I start adjusting times." The negative turned out to be more underexposed than I anticipated

My second negative is ready to be developed, so I'll see how that one looks before reporting back here