Archive for January, 2008


What will my pinhole camera see?

January 28, 2008

OK, last post I ran through (or rode roughshod) what it takes to figure exposures for a pinhole converted camera.  But what about the angle of view?  What will be in the picture?

Without going into a lot of derivations on the physics of light and how it bends and how the pinhole’s diffraction effect is doing all our heavy lifting, the equation is:

angle = 2 * arctan(d / 2f)

Here the value “d” refers to the distance across the film.  This can be the horizontal or vertical measurement of the film or more typically the measure of the diagonal.  And f is the focal length of the camera.

Great, now how do I get the diagonal of the film?  Well, you could measure it directly or tap into the spirit of a dead Greek, Pythagoras.  I’ll do that one for you.

So, example time:

Lets say I have an APS-C sized sensor dSLR.  Google tells me the sensor measures 23.7mm x 15.7mm.  This varies slightly from camera maker to camera maker but it may be listed in your manual.  Pythagoras tells me that the diagonal measurement will be the square root of the sum of the squares of height and width.  Huh you say?

d = sqrt(h^2 + w^2) = sqrt(23.7^2 + 15.7^2) = 28.4mm

In the other post I suggested we might have a focal length of 45mm with a body cap on the dSLR.  So f=45mm in this example too.

Quick word of warning about cheap calculators, and Excel spreadsheets.  Often times they like to use radians as the unit of measure for angles, not degrees.  What is a radian you ask?  Well there are “pi” radians in a circle.  There, did that help?  No?  OK, just make sure you have your calculator in DEGREES mode or read carefully in the Excel help files how to convert from radians to degrees.  We want the answer in degrees because the little protractor we used in 7th grade is marked in degrees, not radians. 🙂

Also, “arctan” may be given as “tan-1” or similar text on the calculator buttons  YMMV!

2*arctan(28.4 / 2*45)  =  35 degrees (dropping the fraction).  What does this mean to you and me?  Well, depending on who you ask and how you measure things, we have a field of view about 55 degrees wide.  So 35 degrees is narrower.  So the camera is seeing a bit less than you.  It will be a mildly telephoto depiction of the scene.  By the way, the pinhole angle of view (and a lens for that matter) is really a cone, not square or rectangular like your film or sensor.

So now that I know the angle, I still want to know how to estimate the view for my picture before I take it!.  OK, get out the little plastic protractor you had in Jr. High.  Or go buy one at the dollar store.  On some cardboard, we are going to trace out the angle above.  Do this by making three marks.  One is the center point of the straight edge of the protractor, remember which point this is (the apex), it will be important later.  The other two will be at 1/2 the desired angle, left and right from the center point of the curved part of the protractor.  We said 35 degrees, so that would be 17 1/2 degrees each side.  Don’t fret the accuracy.  Now connect the dots, all three ways and you have a triangle.  Cut it out.

Hold the triangle such that the important apex point is over the pinhole of your camera (or close to it) and the flat side (base) faces you.  Sight along the sides of the triangle and you have the field of view left to right.  Move the triangle to the side of the camera and you can sight the view up and down.

If you are using a dSLR, you can check your work immediately.  Don’t be surprised if things aren’t EXACTLY the field of view.  But with your little triangle you can “frame loose” and crop a little later if needed.


Pinhole exposures – quick how to, OK, not so quick

January 28, 2008

I’ve seen more than a few posts where people are trying to figure out how to take a pinhole photograph with their dSLR. A common problem seems to be that they are still expecting the dSLR to figure the exposure or at least give them a preview.

I don’t have a dSLR so I can’t diagnose that problem but I’m betting that the dSLR CAN’T figure the exposure and can’t give you a preview for the same reasons a film SLR can’t. There is no interconnect between the f-stop (your pinhole), and the camera’s meter any longer and it is just too dim to see through the finder. So quit trying to use it that way.

This will require some thinking on the operator’s part so get ready (FYI, you can pre-compute most of what you need and just carry a cheat-sheet). This stuff works for any pinhole camera. I’m making some simplifications here so if possible, always bracket exposures.

By the way, Google is your friend (and so is Wikipedia). If you don’t recognize a term, go look it up. Also, Google is very good at converting units so if you don’t want to know that 1 inch = 25.4mm and learn how to apply that, no problem. Tell Google “3 inches to mm” and it will tell you the answer.


  1. Know your equipment. You need to know several things. The film speed (or equivalent film speed if you are using photo paper as your “film”). The pinhole diameter. And the focal length, that is the distance from the pinhole to the film (or sensor if you are using a dSLR). Come to think of it, you need to know one more thing, how to run your camera in 100% MANUAL MODE. 🙂
  2. For the pinhole and focal length from #1, you can now compute the equivalent f-stop. How? The f-stop is the ratio of the focal length to the diameter of the pinhole (or iris opening for a lens). Example: let’s say we measured the depth of the camera from the front of the mounting ring to the sensor (or the filter over the sensor, close enough) to be 35mm. And we measured the thickness of our pinhole-bodycap to be 10mm. This puts our pinhole at 45mm (close enough for our purposes) from the “film”. And we measured the diameter of the pinhole to be 0.012″. Notice I have different units, they must be the same units for the calculation. 0.012″ = 0.3048mm, call it 0.3mm f-stop = 45mm / 0.3mm = 150. f-stop=f-150!
  3. Figure the exposure correction factor versus f-16. Why f-16? Because next step we will use the Sunny-16 rule so just trust me on this one. What we want is a factor to multiply a later exposure time by to correct for our new pinhole fstop. This is done by: (fpinhole / f16)^2. “^” is raise to the power so we are squaring things. (150 / 16)^2 = 87.89. Let’s be simple at call it 88. We don’t need a jillion decimal places for these things.
  4. Sunny-16 exposure rule (Alert – Google for more info) says if we have a sunlit subject, strong shadows (sunlight over our shoulder) the base exposure is 1/ISO for an fstop of f16. ISO? That is the film speed (or equivalent for photo paper, but that is another post). So, what speed? If we were using an SLR I’d suggest 200 or 400 speed film for outdoors. So set your dSLR on 400 or even 800 if you can. Now back to the Sunny-16 rule. 1/400 at f16 (assuming you picked ISO400). Great, but our pinhole is f150, now what?
  5. Take the factor from step 3 and multiply that by the speed from#4. 88 x 1/400 = 88/400 = 0.22. You probably don’t have 22/100 setting on your camera but you will have something close. Maybe a quarter second? Again, each camera is different, but a “Real” camera would have the quarter second marked with just a 4 (1/4). Get as close as you can.

So, now take your picture. 1/4 second exposure is tough to hand-hold so use a tripod or brace the camera. That is unless you are looking for the blurred exposure effect. You can’t frame the shot in your finder but you can probably see the result on your LCD afterward.

The simple calculations above are starting points. You may find that you need a little more or a little less exposure. This is where a cheat sheet helps because once you get the baseline you can write down the base exposure (Sunny-16) and then write down what you need for other common conditions like weak shadow, hazy, full shade, indoors etc in terms of + or – stops of light. What, dSLR shooters aren’t familar with the concept of adding or subtracting a stop? Google and Wikipedia. Go learn it… 🙂

And now a word for film shooters. Schwarzschild. Film doesn’t continue to respond equally to longer and longer exposures, and this may also affect a dSLR sensor but perhaps not in exactly the same way. Referred to as reciprocity failure or the Schwarzschild effect. So for long exposures with a pinhole camera, usually those greater than 1 or 2 seconds using film, you have a second compensation factor to apply.

Most film manufacturers will give you a simple table that suggests how much extra exposure you need for times between 1 and 100 seconds. Some give you a chart and you can interpolate from that. And a few will give you the Schwarzschild coefficient and then you have to do some math. You can also Google for the coefficient or perform your own testing to find it. The testing involves some equipment you probably don’t have on hand so lets skip that for now. Roughly, a coefficient of 0.8 works pretty well for several different films so I’ll use it as the example here. There are astronomy websites where people have done quite a bit of testing on films for long exposures and they come up with other numbers but for all intents and purposes, they tend to center around 0.8. Lacking any other information, it is a decent starting point.

p = 0.8, the Schwarzschild coefficient.

t_new = (t_old + 1) ^ (1/p) -1, the magic equation

t_old = the number we came up with in step 5 above and t_new is what we will really use for our exposure.

Lets say, for example, that our fstop is f500 and we are going to use 100 speed film.

(500 / 16)^2 = 976.56, call it 977 is our first correction factor for exposure. Sunny-16 says 1/100 at f16, so 1/100 * 977 = 9.77 seconds. I’m not going to round this off to 10, keep it at 9.77 for the next step.

Apply the magic equation,

(9.77 + 1) ^ (1/0.8) – 1 (the order of operations is 9.77 + 1, then raise to the power then subtract 1).

(10.77) ^ 1.25 – 1 = 18.5. Now I’ll round this up to 19 seconds. So while step 5 above told me my exposure should have been nearly 10 seconds, it would have been underexposed because of the reciprocity failure of the film. I should try 19 seconds instead. If possible, I’d still bracket and do 19, and maybe 13 and 24 seconds.

One more time, a cheatsheet would help here. Set up a spreadsheet in Excel or OpenOffice and you are done.

One last thing, still haven’t told you how to figure out the angle of view have I? Can’t use your viewfinder so how will you know what is in or out of view of the pinhole? See next entry.


What to do when it is cold outside…

January 27, 2008

Modify cameras!

Good ole’ Ebay coughed up a 6×7 roll film back to fit my Pacemaker Graphic. Then raid the matteboard scrap pile and fabricate a new lensboard for the Graphic. Add a pinhole and shutter. Last thing it to make a focal length ruler and tape it to the camera bed. Now you have a multi-shot, variable focal length, medium format pinhole camera!

I don’t have (yet) a 6×7 negative carrier for my enlarger so I’m cropping to square by using my 6×6 carrier.

Samples – (Fomapan 100 film, developed in Pyrocat-HD, printed on Ilford MGIV RC for scanning)

Magnetic Monkey Capt. Biff Manley, Space Patrol


Capt. Biff Manley, Space Patrol

January 27, 2008


Capt. Biff Manley, Space Patrol, in a rare moment of introspection pauses to consider that perhaps the mission is not proceeding quite as planned…

Frankenroid Pinhole, Fuji FP-100C


Exposure Time

January 17, 2008

Interesting comment made in one of the podcasts at .  The curator of photography threw out the number 400,000 as an estimate of how many photographs the Eastman House has.  Interesting but nothing really special in that number.  But then he suggested you think about that number in terms of the TIME captured.  Early photographs were seconds to minutes.  20th century photography is mostly sub-one second times…

Just for fun, lets say the average is one second.  400,000 photographs = 400,000 seconds

400,000 / 60 / 60 / 24 = 4.63 or just about 4 and 2/3 days of “TIME” captured.

Take this one step further…  I seem to remember a number like $50 Million as the total sales of digital cameras in a recent year.   Lets say the average camera cost is $300 so that is over 160,000 new cameras in a year.  Another guestimate might be that on average there are 5 exposures made per camera per day for a year.  I picked this to cover both the occasional shooter that makes 10 or 20 snapshots per year and the working professional that is making possibly hundreds per day.  And each exposure is on average 1/100 of a second.

So just for those 160,000 “new” digital cameras, in one year:

160,000 x 5 x 365 / 100 = 2,920,000 seconds which is a little less than 34 days captured.  Push the average exposure time up to 1/500 and that time drops to 6.8 days…

This isn’t scientific and it isn’t really what is happening with time but to me it is an interesting way of considering what happens to all the images made (digital or film) over time.



January 12, 2008

Just fooling around with FrankenRoid pinhole camera and FP-100C…



(20 second exposure, 120 second developing time)