What does “angle of view equivalent to that of some lens in 35mm format” mean?

Question

Does "angle of view equivalent to that of 22.5-585 mm lens in 35mm [135] format" for a digital camera mean that the lens is equivalent to a 22.5-585 mm lens for a DSLR camera?

Asked by Jack

Answer

The answers at What is "angle of view" in photography? should help. In short, the "equivalent" gives you a way to compare the angle of view of these lenses, by putting them all in the terms of a common format.

The format usually used as the standard is that of normal 35mm film — called "135" because that's the standard film cartridge format of that size. This format is also used by "full-frame" DSLRs. However, that's (currently and for the forseeable future) the realm of high-end cameras, usually well above $2000 for the body with no lens. Most DSLRs use a smaller format called "APS-C", which is about half the area. There's actually a number of slightly different sizes that go by this name.

For these smaller formats, you can get the "equivalent" field of view by multiplying by the "crop factor". For Sony, Nikon, and Pentax, the value is 1.5x. For Canon, it's 1.6x.

To put that in concrete terms, with the Nikon D5000, a 15-390mm lens would have that same 22.5-585 equivalent angle of view as your example. (Because 15 x 1.5 = 22.5, and 390 x 1.5 = 585.)

The lenses in compact cameras are often specified only in terms of their equivalents, because a) bigger numbers are more impressive and b) there's a dizzying array of sensor sizes in compact cameras, so using some standard makes sense. A typical sensor size is called 1/2.3", and that has a crop factor of about 5.6 — so a compact camera advertising a 22.5-585mm lens may really have a 4-105mm lens.

But don't be too swayed by the impressiveness of that gigantic zoom range. It comes at a cost — one of course being that it's easier to do with the smaller sensor coverage, and those pinky-finger-nail-sized sensors are at a disadvantage, especially in low light. But that's not all, and the lens will certainly also have significant compromises in sharpness, chromatic aberration, distortion, bokeh, and every other aspect of image quality. It still may be very capable of good results, but you should be aware of where the gear you are using makes compromises. See What does 'how much zoom' mean? for more on this.

Finally, remember that zooming is functionally equivalent to cropping — that's why the term is "crop factor" and why this equivalence works out in the first place. That means that if you have an image shot at a "300mm equivalent" focal length (for example, with a 200mm lens on a typical DSLR), and you take 25% off each edge, the resulting cropped image has a field of view like a 600mm lens. Because of the larger sensor and potentially better lens, this cropped image probably will look even better than the "full" image from the point-and-shoot with the 585mm-equivalent lens. (A decade ago, pixelation might have been a concern, but now even entry-level DSLRs have more megapixels than needed for reasonably-sized prints even after such a crop.)

Answered by mattdm

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How can I calculate vertical field of view from horizontal field of view?

Question

I have an Axis M1011 (specs available here), which I need to find the vertical field of view. The online specifications tell me the horizontal field of view is 47 degrees, and the image size is 640x480. Does this mean that the vertical field of view will be 32.25 degrees?

Asked by Joel

Answer

The formula for this calculation is:

a = 2 arctan(d/2f)

Where 'a' is the angle, 'd' is the size of the sensor in the direction of interest, and 'f' is the focal length. The real angle of view of this combination is actually much better than what you get but, because it's video (which is technically off topic here, but the basic question isn't), you're lopping off some info.

The camera's sensor is 6.35mm (1/4 inch according to doc), so if it holds to standard 3:2, then the vertical is about 4.23mm. So...

• True horizontal is: 71.63 degrees
• True vertical is: 51.35 degrees

If I do a little bit of math, it would appear that the sensor is about 167 pixels per mm, so the vertical is using about 2.87mm of the sensor. That leads to the formula returning: 36.13 degrees or thereabouts, call it probably 36. How did I arrive at the 167? Well, the formula to yield an angle of 47 degrees would imply the horizontal is using about 3.83mm of the sensor and if there are 640 pixels, the math works out to about 167 pixels per mm.

Mind you, without real specs on the sensor, this is all approximation. However, the formula I gave you is correct. :)

Answered by John Cavan

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How can I get better results photographing dry erase boards?

Question

As far as photography goes, I'm less than an amateur. I have a camera my friend (the head of the photography dept. at a local college) recommended to me that I really want to learn to use, but haven't had time (Nikon D90 - a gift from a relative). I used to do a little more work years ago with film, but I've forgotten more than I remember.

I'm a writer and when I'm working on a story, I use white dry erase boards of different sizes (from about 9x11 inches up to about 2'x3', but one or two are a little larger than that). This works fantastically well for me when I'm diagramming plots or doing scatter notes, before I sit down and do the writing. With the dry erase boards, it's easy to change things around by just wiping a section out and restarting.

I'd really like to be able to take pictures of these whiteboards and save the image when I'm done (and perhaps, later, print it out).

The problem is, with the larger boards, if I'm back far enough to get a photo that fits all of it into the frame, it's too hard to read what I've written on it. (Also, the surfaces are reflective.) I've been told there is software I can use so I could take a number of photos of sections of one board and have the software put them together in one big image. I don't know how well that works (and I don't know what that process is called), so I'm worried about the cost of the software as well as being able to keep the scale the same from shot to shot.

I don't mind building something to do this if it's not something huge, or if I can set it up quickly once I'm done. I just don't want to have to worry about lots of details when I'm in "work mode."

Is there a fairly simple way I can use my camera to take images of these boards that include the whole board and still let me see enough detail so I can read what I've written on them?

Answer

I think you want simple, so I'd recommend using Program mode, which gives you the greatest control without making you learn about exposure. Program mode (the 'P' on the Auto-P-S-A-M dial) gives you these advantages:

• it sets exposure correctly for the lighting conditions, so you won't get a too-dark or too-light image, and you don't have to think about shutter speeds or aperture values or what they mean.
• it won't try to raise the flash, which would reflect off the whiteboard
• you can set exposure compensation, which helps if the camera isn't guessing the exposure correctly.
• you can set the ISO.
• you can set the white balance.
• if you really want to, you can change the shutter speed / aperture combination that your camera chose. For your particular use case, you want to spin the dial all the way to the right (I think) to get the fastest shutter speed.

So, with all that, I'd follow this checklist:

• set your image quality to the highest possible. Don't use RAW.
• set into Program mode.
• unless you're using a very small whiteboard or can't get closer to it, zoom all the way out (to give the camera the widest range of apertures to choose from).
• set the ISO to 100.
• set exposure compensation to 0.
• set white balance to Auto.
• take a sample picture (aim the camera, fill the viewfinder with the whiteboard, half-press the button so it will auto-focus, then while calmly breathing out, slowly and steadily press the rest of the way down).

Now look at the results. Apply any of the following corrections, as needed, and take another sample picture:

• Zoom in to a sample of the text: if the picture is too blurry, raise the ISO and try to hold the camera steadier.
• If the picture is too dark, increase exposure compensation (+1). If the picture is too light, decrease exposure compensation (-1).
• If the colors don't look right, change the white balance. I know very little about white balance, so you'll have to get someone else to help with this.

When the "sample picture" doesn't have any problems, promote it to "finished picture" and you're done.

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How to calculate viewing distance for a print size?

Question

I'm working on a photomontage (35mm landscape). The client is asking for what size they should print it at and what is the viewing distance.

I'm planning to print the final image on either A1 or A2 sized paper.

I have read numerous guides on how to work out the viewing distance. But none of them make much sense. The advice note from the Landscape Institute suggests it is not guess work by I'm more confused by it.

The Diagonal x 1.5 rule seems to produce a large viewing distance. I thought a value of around 400mm for an A1 print would be more suitable, but looking for a way calculate it rather than guessing it. Any help is appreciated.

Answer

The viewing distance of an image is based on two factors; first is the diagonal image size and second are the pixels per inch required at that distance to give a sharp image.

Firstly the rough rule of thumb is that the viewing distance should be 1.5 to 2 times the diagonal length. This will give you an optimal viewing distance for the overall printed size based on the human eye's ideal viewing angle. You have to understand, however, that for a landscape this may not be optimal as you may actually want the viewer to pan around the image, and you may want the size of features within the image to be the basis of this calculation. This is an artistic decision though, based on the composition of your image.

Secondly for the image to look good at the distance you choose, there need to be sufficient pixels per inch (ppi) to fool the eye into seeing a smooth image that isn't pixelated. The minimum ppi needed for a print with acceptable quality is calculated by dividing the value 3438 by the viewing distance. Anything above this ppi will look good at the distance chosen.

So: minimum ppi = 3438/Viewing Distance

where 3438, a constant for human vision, was derived as follows:

1/ppi = 2 x Viewing Distance x tan(0.000290888/2)

1/ppi = Viewing Distance x tan(0.000290888)

ppi = 3438/Viewing Distance

where 0.000290888 radians (1 arc minute) is known as the 'visual acuity angle' and represents how much resolution a human can see.

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How rapidly does the focal length-equivalent image change by physically moving the camera?

Question

I have recently begun moving from a Nikon APS-C DSLR to the Micro Four Thirds camera while also transitioning from using a viewfinder to using the LCD screen for framing my shots. Because of the size and cost of lenses, I am now shooting with a 14mm (28mm equivalent) lens on the M43 camera instead of the 35mm (52.5mm equivalent) lens on the Nikon. I am getting used to shooting with a wider lens, but I am also curious how my change from viewfinder to screen use effects the actual field of view of the images. Since I no longer hold the camera against my face, the lens is positioned at least 6 inches further forward than it was when using a viewfinder. In other words, rather than going from shooting at 52.5mm to 28mm, I feel like the 28mm lens may behave more like a 35mm lens (relative to my own eyes) due to its different physical position relative to my body.

Do you have any information that can help me better understand the relationship between focal length, field of view, and camera-holding technique?

Answer

I have attempted to illustrate mattdm's answer, and maybe expand on a few points.

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In this first example, two cameras using different focal lengths are photographing the same scene. They have been positioned such that each will capture the full length of the fence behind the house and trees. Notice how because of the different perspective of the wide-angle camera, the second tree appears partly hidden behind the house.

In this second example, the wide-angle camera has been moved to recompose the scene. Now the arrangement of the composition of the house and nearest two trees looks more like what we get from the narrow-angle camera. But notice that by changing our perspective, we are now capturing a wider view of the background than the narrow-angle camera. The fence no longer fills the entire frame.

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EDIT:

For a more mathematical explanation of the relationship between focal length (field of view), and scene width (subject size), you might want to check out the answers the question Estimating focal length range required for shooting scenario.

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Does using a lens adapter change focal length?

Question

Does using a lens adapter change the field of view?

I recently bought a Sony NEX-5, an extra Minolta MD 50mm 1.7/f lens and a MD to E-mount adapter and I was wondering, as the adapter increases the distance between the sensor and the lens by ~10mm, does this effectively make my 50mm lens a ~60mm lens? Or am I just misunderstanding things?

Answer

It won't change the focal length of the lens, that's what a multiplier will do. You'll see these out there, often as a 1.4x or 2x multiplier with optical elements in them. These are designed to multiply the focal length of the lens.

In general, extension tubes, which is what is normally used to move a lens further away, are there to allow closer focus than a lens would normally have, thus giving them macro, or close to macro, ability. However, in doing so, they'll result in loss of infinity focus (which isn't an issue in macro photography). Nevertheless, they do not change the focal length of the lens.

In the case of your camera, however, the adapter is moving the lens further away from the sensor but it has to in order to continue to focus the same. The lens is designed to sit on an SLR and that means it's designed to focus onto a sensor (or film) that is further away from the rear element of the lens than your NEX-5.

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Why do my 100mm EF macro lens and my EF-S zoom set at 100mm appear to give a different field of view?

Question

I have a Canon EF 100mm Macro and a Canon EF-S 18-200mm Lens, and using a Canon EOS 60D.

I believed that when setting the Zoom Lens to 100mm it would show the same field of view as the one from Macro.

Turns that I was wrong. Even putting the Zoom Lens to 200mm, the field of view from the Macro Lens is "closer". But why is that?

I tried to google and just read that there should not be any differences between the EF and the EF-S model regarding the crop factor.

Here the two pictures in question (both taken from tripod):

Image 1 - 100mm Macro Lens - Exif <-- It's more zoomed in than Image 2

Image 2 - 200mm Zoom Lens - Exif

Answer

When both set to 100mm, the field of view is the same - the difference comes where the macro lens can focus much closer; which gives much higher magnification ratios of the image on the sensor.

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What is the formula for percent of frame filled at a specific distance and focal length?

Question

I am thinking about buying a new lens specifically to take photos of distant targets (birds, bears, etc). I currently own the Canon 70-300 IS USM, which is a nice lens, but I find that if I want to take a photo of anything further away than about 10 meters, the target does not fill a large amount of the frame (meaning I have to significantly crop in post production).

I am currently looking at the Canon prime 400mm, however I would like to determine at what distance an object will fill a reasonable proportion of the frame.

Is there a formula or rule of thumb I can apply that will help me in this situation? I realise the size of the target I am shooting will play a role here, so if we need to make an assumption about the size of the target please let me know.

Answer

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 %

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