It's a bird

by Vian Esterhuizen

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30

There are some general rules you can use to determine the "maximum" (I use that term loosely) print size. Keep in mind that the quality of a print is often more dependent on what is being printed than its size in megapixels, and even if your image size is not dense enough to mathematically fit onto a certain page size, you can still blow most images up ...


17

The information above is quite good, so I won't try to compete, but here is a nice infographic: The boxes are the number of megapixels for a print of the size in inches according to the scales on the axes. This is at 300ppi, which is a standard for the print resolution of many images. This great graphic comes from an article at D215.


15

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 ...


13

jrista has the start of the formula, and it covers images viewed at arm's length quite well. But that 'conventional wisdom' devolves into unreasonable numbers as soon as you get to anything "big", say even a 16x20... requiring 5-6000 px. And if you hit poster size, say 30x40... 9000x12000... 108 MPix?! When you're talking about really big prints, it's ...


13

The formula for the percentage of image filled is focal_length x subject_size x 100 _________________________________ distance x sensor size All units are millimeters. Use the width of the subject/sensor to work out the horizontal fill % and the height of the object/sensor to work out the vertical fill %


13

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.


12

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). ...


9

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). ...


9

So, it helps to start with knowing 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 ...


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

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

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 ...


7

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 ...


7

Here's the nice thing — the relationship between focal length and sensor size are directly related in a simple way: If you double the focal length, that's exactly like cropping in half (in each dimension, so a quarter of the area). That means if you put your existing lens at 300mm, and then crop to 75%×75%, you'll see what a 400mm lens will get you, since ...


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

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 ...


6

The focal length and f-number are photographically relevant quantities, so the formula is expressed in terms of those convenient variables. Photographers don't generally know the radius of their lens aperture for every f-stop.


6

You should add more details with respect to the specific special effect. In general, given two function f(x) and g(x) and their Fourier transforms F(k), G(k), the product of the two transform F(k)G(k) when brought back to the coordinate space (that is, when anti-transformed) is equal to the convolution of the original f and g. Suppose that you are ...


6

It seems to be called Seam Carving, and the Gimp has this feature as Liquid Rescale. You're right -- it collapses out parts of the image with less detail where you won't notice it. Liquid Rescale also has options to mask parts of the image you want to preserve and parts you want to eliminate. There's a great demo in Episode 14 of Meet the Gimp


6

I think you got it backwards, EV increases when brightness increases. See the wikipedia article for details


6

You're fundamentally right, because the equation you've given has the sign reversed. The adjustment should go on the left side of the equals sign, not the right side; if you want to move it to the right side, it needs to be subtracted. Or, to answer the title question: it doesn't. But, I think part of the confusion comes from looking at what EV is meant for ...


6

Using the EXIF data, you should be able to use the following formula: (Resolution in pixels/Focal plane resolution in dpi) X 25.4(mm/in)=size in mm For my Canon 5DII, the horizintal and vertical numbers figure out to: (5616p/3849.21ppi) X 25.4mm/in = 37.058mm (3744p/3908.14ppi) X 25.4mm/in = 24.33mm Once you have the horizontal and vertical measurements, ...


6

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 ...


6

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 ...


6

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 eys, like they are in focus. This is based on the ability of the human ...


6

Some do actually. You are right that most do not though. The main problem is that there is no such thing as an exact depth-of-field. DOF actually depends on the viewing size, actually the angle extent of the image as being seen. Most DOF tables assume a fixed 8x10" print seen as from 12" away for someone with 20/20 vision. Today software allow you to enter ...


5

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 ...



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