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by Aditya

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21

The only other factor you need is the height of the object in real life (otherwise you could be photographing a model which is much closer to the camera). The maths isn't actually that complex, the ratio of the size of the object on the sensor and the size of the object in real life is the same as the ratio between the focal length and distance to the ...


20

The flattening or compression effect is not caused by a particular kind of lens, it applies to all lens in the same way. Actually, this property of lenses applies to our own eyes as well. The factor that affects flattening is the distance from the camera to the subjects. Consider the following exercise: Place two friends 1 meter away from each other. ...


15

I believe the effect has to do with the RATIO of distances from the camera to various parts of the subject / scene. For example, if you take a wide-angle shot of a person's face, their features are exaggerated because the camera-to-nose distance might be half of the camera-to-ear distance. On the other hand, consider the same shot taken with a telephoto ...


13

In this sort of situation, the normal approach is to use a focusing rail which allows fine and controlled adjustments to be made. There are several available on the market, with some under £50, but it would be possible to make one yourself, if you have the right tools.


11

Many (possibly most) modern SLR lens systems return focus setting data to the camera. Potentially the precision of data returned could be high - something better than 1% of range would be possible and meaningful with modern systems. However, it appears that most if not all systems use a simple gray-coded* system with perhaps 16 steps. Number of steps ...


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

As noted @matt-grum, the most simple formula to estimate distance to the object is pinhole projection formula: where x is the size of the object on the sensor, f is focal length of the lens, X is the size of the object, and d is distance from nodal point to the object. x and f, and X and d are measured in the same units, e.g. mm and m respetively (to ...


6

A non-obvious answer is to increase the light. Because macro photography has such a small depth of field we're often forced to balance the aperture setting with avoiding camera vibration. Adding some off-camera strobe will nicely fix both problems. The burst itself will act as the shutter if you want it to, by getting about 5-6 stops between the flash ...


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 ...


5

TLDR: No because you need an additional variable, either the height of what the subject is that fills the frame or the distance at which you are focused (infinity doesn't work though). Long answer... To do this you can use simple trigonometry, but you would need to know either the current or desired distance to subject or the size of your subject ...


5

the only relevant thing is that the same light which falls on your main subject falls on your card. The field of view, distance and so on don't enter in the equation. but you have to be sure that by standing near the subject you don't influence in any way the light, which is not always easy since our eyes are very quick to adapt to variation and they ...


4

Yes for most camera systems: For Canon EOS, select EF and EF-S lenses transmit distance information through the EF mount. For Nikon, D- and G-type Nikkor lenses transmit distance information through the F mount; this is what the D designation means. G lenses are the same, only without an aperture ring. For Sony, all current lenses transmit distance ...


4

It's actually far simpler than any of the answers posted so far! You don't need trigonometry, or field of view calculators at all, all you need is multiplication and division! Firstly (all else being equal) the size of your object in the image is directly proportional to the focal length (if you double the focal length you double the size). So if you know ...


4

According to the FAQ for the particular calculator you're using, the calculations are performed for a thin lens, and hence the front nodal point of the lens should be used. The author recommends using the front surface of the lens, on the assumption that the front nodal point is somewhere inside the lens, and this assumption will yield a conservative ...


4

You are misunderstanding the compression. The face appears flat because you are observing it from a distance and thus the angles of it ARE flat. If you move in closer, you are viewing it more from the side/off-axis and you get more depth. The focal length is simply used to crop the image for you. You could take a portrait with an 135mm lens at 3 meters ...


4

My answer will be the same as @MattGrum's, but I would like to add a teeny bit of explanation. Say the diameter of the sun/moon is h, the distance is d and your lens' focal length is f. Assume that the image has come to a focus at a distance l behind the lens and has a diameter of i on the sensor. If you quickly sketch out things assuming a thin lens you ...


3

Inversely linear is a good approximation. Imagine a 1,7m tall girl at 1 m distance b. Her head is at point B. How does the size/length of an object vary with distance? Let the girl walk away from you. Her size stays the same. She appears smaller, because she is appearing under a smaller angle. Her angular size changes. Using arctangent to calculate ...


3

I'm sure this is a duplicate, but I can't find a good answer to the question in the archives so here goes. The relationship between object size and distance is an inverse linear relationship, i.e. size is 1 / distance. This makes sense when you think about it as if you double the distance the size halves. This is why you appear to be observing an ...


3

The relationship is a simple inverse, i.e. object size in image = Object size * focal length / object distance from camera If you keep the same object and the same focal length you get: size = 1/ distance (the =-sign should be proportional-sign).


3

If your lens has a distance scale it will indicate the approximate distance to the point of focus as measured from the film/sensor plane. In the image below, the point of focus is 1.5 meters (5') in front of the film/sensor plane. As the point of focus approaches infinity, the distance measurement becomes less precise. In general, the longer the focal length ...


3

You can go through the lens magnification. M = f / ( f - d) You know the projected size and want the real size, so you divide the projected size by M H_r = H_p / M In this case you get: M = 38 / ( 38-14.000.000), H_r = 24 / -0.0000027143 = -8.818.105mm ~ -8,818km Note the inverted projection by the negative M. The measurement is most accurate if your ...


3

It's probably not the place to ask, but the answer is fairly easy to arrive at using similar triangles. Given that the feature fills your shot, you have two ratios that should be the same: focal length : sensor height distance to feature : height of feature Since you have a 38mm focal length (at 35mm equivalent and a 35mm frame height is 24mm) then you ...


3

Theoretically the only factor you should need to adjust for in the following equation when zooming is the focal length: distance to object (mm) = focal length (mm) * real height of the object (mm) * image height (pixels) --------------------------------------------------------------------------- object ...


3

This is in the FAQ for the calculator site you refer to. In short, the equations used assume an abstract mathematical lens, not a real-world one. Except for macro or other extreme-closeup photography, this is perfectly fine — it doesn't really make much difference where you measure from. Assuming the measurement is from the front of the lens gives a ...


3

In addition to @jrista's detailed answer, you may also wish to experiment with focus bracketing. This technique involves taking many photos of the same macro scene with many different focus points, and then composing a final, fully-in-focus-image from the images. I haven't done it myself, but I imagine that it could be a very useful technique for your ...


3

Macro focusing can be a difficult task, even with the proper tools, as DOF can be so thin (sometimes just millimeters thick, or with extension tubes, even thinner.) There are a couple techniques you can use to focus at extreme macro scales. The cheapest, and obviously the simplest, is to move the object if it is mobile. You can usually get a very fine degree ...


3

I think, whenever you reverse a lens by itself, your working distance becomes tiny - of the order of the flange to film distance, since that is how the optics work out. Your best bet for getting good magnification AND good working distance is simply to go for longer and longer focal lengths with short extension tubes. Here are some things you could try re: ...


3

The minimum and maximum focus distances have no direct relation to the focal length of the lens. All lenses have a minimum focus distance. For Macro lenses this is closer than for normal lenses (is the main part of what makes them Macro). The majority of lenses have an infinite maximum focus distance (focus at infinity). Under certain conditions ...


3

It really depends on the kind of shot and what conditions allow. That's why I answered in terms of focal lengths.) Shooting around the 65-85mm effective range is generally considered the most natural and most common, but if you want to flatten the image more, you can push it out to the 105-155 range. There are also some shots that work well in the 24-50 ...



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