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 :

"Gotta little problem with personal space, and I've been pounding the Jager. My breath and behavior have been driving the patrons away" -"Whipped Cream" by Ludo

...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?...

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

"Gotta little problem with personal space, and I've been pounding the Jager. My breath and behavior have been driving the patrons away" -"Whipped Cream" by Ludo

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

Last edited by Existing Light; 01-03-2010 at 11:38 PM. Click to view previous post history.
Reason: typo

"Gotta little problem with personal space, and I've been pounding the Jager. My breath and behavior have been driving the patrons away" -"Whipped Cream" by Ludo

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?).

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.

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.

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.