25

I got a little carried away with formatting my answer... This drawing is adaptable and can automatically calculate different scenarios, I'll give LaTeX/Python source to anyone who wants it. Edit: I've put the source code here. I must warn potential viewers that it's difficult to read and badly formatted because of nesting python inside LaTeX.


21

I'm going to sort of disagree with all the other answers that talk about DPI or PPI rules of thumb, and suggest two different 'rules' (based on PPD, from another answer of mine) Rule 1 — The 'Retina' rule (aka the Pixels-Per-Degree (PPD) / 'better than your eye can see' rule) This comes pretty much straight from Apple's Retina display designs, the idea ...


20

I was tempted to mark this to be closed as "primarily opinion based" but then realized that I can prove that the "rule" of thirds is not a matter of opinion. Well, sort of. In one specific way. Maybe. First, accept that it's not a rule. Appropriated from Pirates of the Carribbean: Curse of the Black Pearl, Barbosa says "...more what you'd call 'guidelines' ...


17

Depth of field depends on two factors, magnification and f-number. Focal length, subject distance, size and circle of confusion (the radius at which blur becomes visible) jointly determine the magnification. Depth of field does not depend on lens or camera design other than the variables in the formula so there are indeed general formulas to calculate ...


15

You wanted the math, so here it goes: You need to know the CoC of your camera, Canon APS-C sized sensors this number is 0.018, for Nikon APS-C 0.019, for full frame sensors and 35mm film the number is 0.029. The formula is for completeness: CoC (mm) = viewing distance (cm) / desired final-image resolution (lp/mm) for a 25 cm viewing distance / enlargement ...


14

The key is to find areas of the image with a lot of parallax, such as a foreground building and a background tree. Try to pick a point as close to one edge of frame as possible. Now walk left/right (green) to find the correct point of intersection from the old photograph. Now that you've done that, you've established a straight line to move along (red). ...


12

The calculators you posted are for fairly standard, rectilinear lenses. This means you can use the Pinhole camera model to calculate the information. This graphic fairly well shows what is going on: On the horizontal axis you see f. This is the focal length of the lens. Then, the arrow labeled Y1 is the image plane (where the sensor sits). If the sensor ...


12

It's pretty much strictly linear unless you're talking about very close focusing distances or macro distances. For everything else, what little you may be off is probably less than the rounding error between the actual focal length and the marketed focal length of the lenses in question. For example, a lens with a focal length of 192mm will probably be sold ...


11

The equation assumes a simple single element lens that is bilaterally symmetrical. The camera lens, to mitigate the 7 major aberrations (shortcomings that degrade) is constructed using several individual glass lens elements. Some are positive in power, some with negative power. Some are air-spaced apart and some are cemented together. Because this array ...


10

Here's what you're missing: that larger formats have less depth of field for the same framing, not at the same focal length. A 100mm lens is much wider on medium format than it is on 35mm film. If you keep that and the aperture constant, DoF will be identical assuming you print with the same enlargement (that is, the medium format print will be much larger). ...


10

The quote from the book does not state that the numbers are intended for any specific lighting situation, so it appears to be indeed only concerned with showing how aperture and shutter speed can vary while maintaining the same exposure. And anyway, the "Sunny 16" rule is only a broad rule of thumb and one f-stop difference is not that big. Maybe the author ...


10

Doubling the number of lights at the same power doubles the output. Assuming both speedlights are the same to begin with, and if they're both set to ¹⁄₆₄th power, that'll be like setting just one of them to ¹⁄₃₂nd. That's in terms of just the output, though. Since both flashes can't be in the same place, and therefore their relation to your subject not the ...


10

Your intuition is right. To validate it, we can dig into basic high-school geometry. Although a camera lens is actually a complex lens made from many elements, conceptually and mathematically for most practical purposes, this reduces to an ideal, where you can imagine a pinhole exactly a distance from the sensor equal to the focal length. Light might fall ...


9

Image rotation is a lossy operation, but rotating an image once then rotating it back likely loses very little detail, especially compared to typical JPEG compression. Image rotation works like this mathematically: A grey level image consists of luminance values L_(x,y) at integer pixel positions x, y. First a real-argument function f(x,y) is constructed ...


9

It helps to start with exactly what a "stop" means. See What is one "stop"?, but, fundamentally, each stop is a doubling or halving of the exposure. So, given two shutter durations, you can find the number of stops between them by calculating the binary logorithm (log₂) of each, and subtracting. (If you don't remember your elementary school math ...


8

If you want to see a practical implementation of the depth of field formulas you can check out this Online Depth of Field Calculator. The source of the linked HTML page has all the formulas implemented in Javascript.


8

Why work so hard? I think you are making this much harder than it has to be. Getting something to be exactly 640 pixels in the image will be difficult, require careful measurements and accurate calculations and is very error prone (and for something really close the internal construction of the lens makes a huge difference and this will be impossible to ...


8

You're missing a calibration step. You're calculation is probably correct, the focal length is indeed 478.634, it's just not 478.634 millimeters. The vanishing points you've calculated in image space have no units unless you know the size of the camera sensor. So from a formula with no units you can't apply mm to the answer and expect it to make sense! ...


8

The size of the sun or moon in mm in the sensor plane will be approximately f / 110 where f is your focal length. A typical APS-C sensor is 16mm tall (or 15mm for Canon), hence a 1760mm lens would be required to fill the frame (vertically). 800mm would get you about half the frame, 400mm one quarter etc. A "full frame" sensor is 24mm tall, so you'd need ...


8

I can't summarize the whole optical physics theory behind pinhole (mostly because I don't have the proper knowledge!), but I try to explain why there are different values for constant C. One reason that there are different values for C is the fact that one parameter in the calculation of optimum diameter of hole is missing! Let us refer to the wikipedia ...


8

The aperture size is a property of the lens only and does not depend on the crop factor. It does depend, however, on the actual focal length of the lens (not the "equivalent" focal length). So you need to obtain the actual focal length by dividing by the crop factor. actual-focal-length = equiv-focal-length / crop-factor You can then calculate the size (...


8

How do I calculate the aperture size and area You divide the focal length by the aperture/F-stop value. Infact, that's what the F-stop/aperture value is. It's a divider. Sometimes written as ƒ2.8 (as an example) but a lot of people leave out the vinculum and should be written as ƒ/2.8. Replace the ƒ with the focal length and that's the diameter of the ...


8

Well, the main thing is that crop factor doesn't really affect focal length. It just affects the field of view by making it narrower. So, what you really have is a 400x1.4x => 560mm lens combination on a crop body, which has the same FoV that an 896mm lens would have on a full frame body. So, unless you shoot full frame enough to translate focal lengths to ...


7

(1) By inspection it is "clear" [tm] that a relatively wide angle lens is in use. In a 35mm full frame system a guesstimate far far far closer to 18mm than 480 mm would be arrived at. (2) Without having got my brain fully around the triple vanishing point method described in a reference, I would think that for vanishing points to be relevant you would need ...


7

First, there is no way to get the focal length from the aperture. Your camera's max aperture is f/3.1 at 6.3mm and f/5.9 at 18.9mm - but there isn't anything that say the camera must use the max aperture, it's completely possible to use f/5.9 at 6.3mm - so there no relationship what so ever between the aperture used and the focal length. Even if you force ...


7

Like Michael says, the top chart is just hypothetical, and it only uses some random exposure, and is NOT about any specific exposure condition. Instead, it tries to show two concepts. It is an Equivalent Exposure guide, indicating that f/2.8 at 1/1600 sec, or f/4 at 1/800 sec, or f/5.6 at 1/400 sec, etc, etc, are Equivalent Exposures. You can choose any ...


7

What am I missing? You're missing that Depth of Field is subjective. In actuality, there's only one plane that's in focus — everything else can't be. Then, there's an area around that plane that is indistinguishable from out of focus because it's beyond the technical limitations of your camera to distinguish from perfectly in-focus. But, that's often quite ...


7

First, a word about what depth-of-field is and is not: In a way, depth-of-field is an illusion. There is only one plane of focus. Everything in front of or behind the point of focus is out of focus to one degree or another. What we call DoF is the area where things look, to our eyes, like they are in focus. This is based on the ability of the human ...


7

Short answer: if you halve the width, you have (effectively) doubled the focal length.


7

first of all, kudos on your effort to break a photography problem down to first principles. The discrepancy you've observed stems from a common oversimplification. Your 100mm Lens is actually what optical engineers refer to as a "lens assembly" As you likely know, it is comprised of multiple lens elements in groups working in tandem to form, refine, and ...


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