Building a 5x7 P&S and have some newbie questions
I've got it in my head to build a wide to ultra wide 5x7 fixed focus P&S. I've seen similar cams built over on LFF and I have a cunning plan but I've never built a camera before and I have some questions I've been trying to figure out on my own but my math doesn't seem to be making sense.
The design is basically a back with fixed wide lens that will have a set focus and DOF will get me something like 3m to infinity in focus.
My questions revolve around image circle and setting focus:
I've been looking at charts and see lenses with 216mm, 214mm, 236mm, etc listed as good for 5x7. Now calculating it myself I came up with 228mm for coverage with no movements. Yet Ebony states on their website that their coverage estimates are very conservative and include movements. So what am I doing wrong in my figuring? What would be the minimum image circle I could get away with considering that there won't be any movements?
To set focus to work as described above I'm guessing I need to look at a DOF chart, figure the distance and aperture I need to achieve the circle of focus I want, measure the distance out and figure the distance between lensboard and back that I need to have focus at that distance, correct?
Thanks in advance for any advice!
I don't use anything above 4x5 but I think your estimate of focal length for w/a is a little off - someone here will know for sure but I think you're looking for something closer to 100mm.+ - I've no idea what's available except maybe an XL?
Wide-angles for 4x5 are available quite cheaply but I suspect there may be a big jump in price to cover 5x7.
A helical focus mount (using scale focus) would be worth the extra effort.
Everything but the viewfinder should be easy enough to find. I did notice just today a "director's viewfinder" - needing repair to the aspect ratio mechanism so may go cheap - on eVilbay (UK or France I think). Newly listed. Didn't say how wide it goes.
Originally Posted by unclemack
For focal length I was thinking either 72, 75, or 90mm. The numbers I listed above were circle of coverage not focal length.
OOPS! Not newly listed, ends in 3 1/2 hours...sorry, failing memory. I keep forgetting it's failing though.
Item number is 220577610896.
for minimum image circle on a 5x7, I get...
25.4 * sqrt( 5*5 + 7*7) = 218.5mm
The actual opening in a 5x7 film holder is slightly smaller than 5x7 however, so one could get away with even less coverage. Assuming the opening is really only 4.75 x 6.75 (for example, I do not have a film holder to measure) you could get away with around 210mm of image circle.
Sponsored Ad. (Subscribers to APUG have the option to remove this ad.)
Sorry to have completely missed the point! Missed my afternoon nap today too...
This free program may help you in your project:
It is geared more towards digital cameras, but it does include LF film sizes in its dropdown lists.
To find the length of any side of a right triangle, when the length of two of the sides are known, use the formula a squared plus b squared equals c squared. C is the diagonal. A and b are the two sides that are at right angles to each other.
You are looking for the diagonal, because that is how much coverage you need. You know the lengths of the two sides that are at right angles to each other, so you can figure it out:
5 inches squared plus 7 inches squared equals the diagonal squared.
Therefore 74 square inches = the diagonal squared.
You are trying to find c, but you have c squared at the present time. To eliminate the exponent on c squared, to just make it plain-ol' c, you can find the square root of c squared. You know this because the square root of something that is squared is obviously just the base number. After doing this, to make the equation continue to hold true, you must do the same to that which is the other side of the equals sign. Therefore, the square root of 74 square inches equals the diagonal (c).
So, the answer boils down to, what is the square root of 74? If you know your basic multiplication tables, you know it is not a whole number. So, it is slow if not impossible to get it exact in your head, but you can get close enough for this application. 8 inches x 8 inches = 64 square inches and 9 inches x 9 inches = 81 square inches, so the square root is somewhere in between the two. Let's just call it 8.5 inches (finding the square route also turns the units of square inches into plain-ol' inches). Seems close enough. Multiply that by approximately 25 to convert it to millimeters (25.4 to be exact, but I am trying to show how to do it without a calculator), and you get about 210 mm.
So, you need a lens that will cover about 210mm at the f/stop at which you will be setting it.
Last edited by 2F/2F; 03-30-2010 at 08:19 PM. Click to view previous post history.
"Truth and love are my law and worship. Form and conscience are my manifestation and guide. Nature and peace are my shelter and companions. Order is my attitude. Beauty and perfection are my attack."
- Rob Tyner (1944 - 1991)
"Assuming the opening is really only 4.75 x 6.75 (for example, I do not have a film holder to measure) you could get away with around 210mm of image circle."
Keep in mind that you will never try to focus the lens at infinity, you are aiming for the hyperfocal distance, so you don't need infinity coverage. This should give you adequate coverage from lenses which don't cover 5x7 at infinity. I don't know how to calculate how much a given lens covers at the hyperfocal distance though.