of a Camera and Lens

Computes Field of View seen by camera and lens, for any sensor size or aspect shape (film or digital). Computes both the dimensional size of the Field Of View seen at a specified distance, and also Angle Of View (degrees, which is independent of distance). And even a Milky Way star trail blur calculator is one option. Usage descriptions are below calculator.

Uses of course include computing dimensional field of view size at the subject distance. We don't often care about precise field size, but suppose you plan a portrait to include a 2x3 foot subject area. You know you need to stand back six or eight feet for proper portrait perspective. What focal length is that field size and distance going to require? (Option 5, and it depends on your sensor size). And the background may be six feet farther back yet, then how large is it? This calculator can plan or verify your choice. It can also compare expected fields seen by different lenses or sensors or distances.

There's a large table of Field of View (angular, in degrees) for many lens focal lengths and a few popular sensors on the next page (or Option 1 to 4 below shows more specific cases). Perhaps somewhat related, another calculator can compute distance or size of an object in a photo.

This calculation requires accurate sensor size and focal length. These values may be difficult to determine for compact cameras and phones and camcorders, but DSLR values should be more easily known. You can specify crop factor as a way to compute actual sensor size. The Exif data often reports focal length. Do NOT specify 35 mm Equivalent focal length (at least Not unless you also specify 35 mm film size, 36x24 mm, or crop factor 1, which is 3:2 aspect ratio). Use the actual lens focal length with the actual sensor size.

**Options 1-4 are four ways to specify sensor size**. Enter Focal Length and Distance, select a sensor size in Option 1-4, and click Compute. Then FOV is computed from focal length, distance, and sensor size. Or Options 5-8 are more special purpose.

**Numbers only** (any NaN result means input is Not A Number). Periods as decimal points are OK.

Please report ( Here ) any problems with the calculator, or with any aspect of this or any page. It will be appreciated, thank you.

Distance units can be feet or meters, miles or cubits, any units, your choice. FOV dimension results will be those same units (BUT, Magnification applies only to feet or meters).

Clicking the little **green button** (by Distance, marked **In** for Inch) will assume feet, and will repeat Dimensions in feet ' and inches " format. You can click the green button again to toggle this option off or on.

The **blue Flip button** (near Option 5) will toggle to swap finding either Focal Length or Distance in option 5 and 8, to compute either one from the other. Options 6 and 7 are not affected by it.

Options 5 and 6 are applicable to the sensor size specified in options 1-4.

The calculator is NOT for fisheye lenses (their view will be wider), and it is not for close macro distances (their view will be more narrow, at end below). The lens focal length number increases with closer distance, so field accuracy will be better at distances of at least a couple of feet (magnification 0.1 or less). The marked value of focal length applies to Infinity focus, but at normal distances, the focal length difference is small for the focus range of regular lenses.

Select an Option, and click the Compute button (for all Option numbers).

Options 1-4 are four ways to specify sensor size.

**Option 1**- Best accuracy is when entering actual exact sensor dimensions (width and height in mm, from camera specifications, any aspect ratio).**Option 2**- If for a 16:9 frame in a 3:2 or 4:3 sensor, this option assumes the 16:9 is full width of the sensor. That may not be quite true in every case, your camera manual might specify.Or 4:3 in a 16:9 camcorder is available in Option 3. This case may take some attention to determine what you actually have.

**Option 3**- Or, if you can determine the lens crop factor (from the lens specifications), sensor size can be computed. See crop factor notes and 16:9 notes below. For mixed cases (of still and movie formats in same sensor), if the two crop factors are available, the crop factor to use is the native case, still for still cameras, or movie for camcorders (the slightly smaller crop factor number of the two).**Option 4**- You may be able to select one of the general sensor descriptions. Film sizes should be accurate, and the larger sensors with actual WxH dimensions too, but the "1/x inch" sensor numbers are Not meaningful. The "1/x inch" description is NOT the sensor dimension, not even related to the sensor. It's a false specification (like fraud), it compares the sensor to the picture size of an old glass vidicon tube (1950s, before CCD). But the "1/x inch" dimension was that outer glass tube diameter, and there is absolutely nothing about the digital sensor that is that dimension (its diagonal is probably no more than 70% of that dimension). It is an inflated false number. Instead, we need to know the sensors actual real W x H size in mm. Or an accurate Crop factor can compute it here. The list here tries to provide some known sensor sizes for the 1/x numbers, but there are usually a few different sensor sizes claiming the same 1/x number, so it can be wrong. The correct calculation really needs the correct sensor size, WxH in mm.See more at Determine Crop Factor.

Special Feature: In Option 4, the

*All Sensors*button will instead show a summary of all sensors in the Option 4 list.**Option 6**- Angular field is independent of distance, so you can enter a**known angular goal**, like a relative number to frame the 0.5 degrees of Moon diameter, and just ignore any distance. If distance is blank, Option 6 (and 7) will default to 10 for distance (so the field math doesn't blow up), but just ignore it. The calculator still computes FOV angle for the sensor dimension, and field dimensions for the distance.**Option 7**-**Milky Way**. The Earth rotates. A reasonably long star exposure on a fixed mount will show an elongated trail (blur) instead of a pin point dot. A longer lens seriously magnifies this movement, so usually a 12 to 24 mm lens is used for the Milky Way (also to show a lot of sky). There's a rule of thumb for Milky Way photography called the**500 Rule**. This idea from 35 mm film days says 500 / focal length = seconds is the maximum exposure time retaining sharp stars (if using a fixed mount). But sensor size also matters, and for a cropped sensor, this rule is 500 / (focal length x crop factor). However, this does not take megapixels into account, and more megapixels will show more pixels of trail. These calculations are NOT about proper exposure or ISO or aperture. It is only about focal length and the Earths motion causing star trails. You'll still need proper exposure, and for a dark sky, I'd say start at ISO 3200, f/2.8, and about 25 seconds (you'd like it longer). Since the earth rotates, some trail is of course inevitable without a tracking mount, and this will be a conflict. If serious about it, you may want to investigate an easy-to-build hinge tracker for the camera (not all of these articles mention that the distance from hinge to drive screw is a precise calculation to match the screw thread pitch). This calculation is NOT about intended long circular star trails (those are circular and rotate 15 degrees per hour). This goal is to compute movement of the Earth's rotation on a fixed mount. This FOV calculator already has the sensor information, so stars were added here.Option 7 computes the fixed mount star trail blur based on your focal length, sensor size, image pixel dimensions and shutter exposure. You can choose to enter a shutter time, or to limit the star blur trail to X pixels long, or to limit it's size on the sensor to X times the CoC diameter (relative to the normal DOF limit). The mm length of the elongated pixel trail is compared to standard DOF CoC diameter, to judge how much it matters. The DOF definition is that 1x CoC diameter is where our eyes decide fuzzy instead of sharp enough, which is much more evident at larger enlargement.

For comparison, the calculator also shows the value of the Rules of 500 and 600 adjusted for crop factor. A few pixels of blur may not always be quite as bad as it may sound, except of course in a 100% crop. The adjusted 500 Rule seems a decent compromise, and nearly agrees with 1.21x CoC (if computed with zero added pixels, and computes about 6 pixels of trail if 24 megapixels). Rule of 600 is about 1.45x CoC, a bit more lenient of motion. FWIW, a Rule of 413 (which I just made up) would compute 1x CoC, but is a shorter exposure.

My reason for the

**"added pixels"**. You can set this to zero if desired, and then the math is straight-forward and precise. The calculation is how many pixels and mm a star "moved" over time. The math is exactly as presented, but if starting at one pixel and moving to the next, math calls it a movement of one, but that's two pixels exposed, not one. If it's a fraction of a pixel, that's another pixel. And if the star also starts on a fraction of a pixel, that's another too. It seems reasonable to see about three more pixels, NOT counting any misfocus blur. Meaning that any blur width seen on the width of the line is surely blur also added to the length, at each end.Also it may matter if the camera times the shutter or if you time it manually. Because the camera's nominal shutter speed of 30 seconds is actually implemented as 32 seconds (2

^{5}= 32, and nominal 15 seconds is actually 16). You can correct the time calculation if you expect it. The blur trail lines are likely more like at 45 degrees, so you may want to rotate it to count pixels. The calculator only calculates Width and horizontal dimensions, but magnification is the same in all directions.**Option 8**-**Magnification**. You can assume either feet or meters units if consistent, and any choice of units for distance will work for everything except magnification. This calculator simply calculates magnification for both feet or meters, and one should apply to your use. But Option 8 itself needs to know which way to compute the match.The two units (ft and m) are different numbers (representing different distances), but for example, dividing the distance in feet by 3.28 converts feet to meters, which will then see the corresponding magnification number for meters. The magnification number can drift slightly out past a few decimal places, because all the other values likely only have a couple of significant digits.

**Option 9**- Probably not practical to use, but technically, if you can measure the width and height field of view dimensions at a specified distance, and know the focal length accurately, sensor size can be calculated. The distance ought to be at least a few feet for best accuracy.The apparently excessive significant digit precision used here may not have practical meaning, but the purpose is so Option 9 can precisely recompute sensor size from previous FOV results. For example, to prevent a 36 mm sensor using Option 1 FOV size with fewer digits from coming back as 36.02 mm dimension.

The most usable general understanding to compare magnification of focal lengths (for same sensor and same distance) is that the resulting image size is the simple **ratio of the two focal lengths**. Compared to a 50 mm lens, a 400 mm lens will show an enlarged view 8x the subject size and 1/8 the field of view (400/50 = 8). This example 1/8 is true of frame dimensional field of view, or 8x for subject size, however the numeric angle of view number (in degrees) is Not linear for short focal lengths.

The biggest risks to FOV accuracy are in not actually knowing the specific accurate sensor size, or not actually knowing the specific accurate focal length, or your guess about distance. Compact and phone cameras may be difficult to know (and Crop Factor might be your best tool). DSLR cameras do better describing those specifications.

Entering exact sensor dimensions above would be the most precise. The camera manufacturers specify the equivalent 35 mm crop factor from the diagonal ratio to 35 mm film (because many of us are very familiar with 35 mm film, and crop factor tells us what view to expect now). We may not know sensor size or focal length on compacts, except at either end of the zoom range, but then we can determine crop factor, for example, if they specify their 6.1 mm lens is equivalent of 24 mm lens on a 35 mm camera, then obviously their crop factor is 24/6.1 = 3.93. And after the fact, the Exif (Manufacturers Data section) often shows the zoom focal length used for the picture (see an Exif viewer that will show this).

**Magnification** is a related property here. Greater magnification reduces FOV, and also, FOV calculation becomes inaccurate if magnification exceeds about 0.1 (if focus is too close, because focal length increases with close distance). However, other than macro lenses, normal lenses typically don't focus quite that close. This FOV calculation is not accurate for macro distances. Macro purposes find it much easier to use Magnification for calculations instead of FOV, that being: Magnification = object size on sensor / object size in real life.

Magnification (for cameras) can be computed in two very standard ways. The obvious way was just mentioned:

Magnification = (Sensor dimension / FOV dimension), horizontal dimension for example.

Magnification = (focal length / subject distance) computes same number (similar triangles).

Magnification = (Sensor dimension / FOV dimension), horizontal dimension for example.

Magnification = (focal length / subject distance) computes same number (similar triangles).

When these dimensions or distances are equal (when image size on sensor is equal to subject real life size, or, when subject distance is equal to sensor distance), this is 1x magnification, called 1:1 reproduction. But camera "magnification" is normally a reduced size on the sensor, < 1.

Note: I'm saying "focused distance to sensor" is "focal length", which it is when focused there. The focal length number marked on the lens applies ONLY to focus at infinity. Focal length necessarily becomes longer when focused closer. This affects f/stop numbers too, and it becomes significant in math when magnification grows to approach 0.1 (slightly closer than most lenses will focus, except macro).

Binocular and telescope magnification numbers are a different system, being "viewing devices", and their magnification number is relative to the size our naked eye sees at 1x. If a magnifying eye piece is used (like binoculars use), then their magnification is (main lens focal length / eye piece focal length).

If no eye piece is used (if telescope is attached like a camera lens, called prime focus photography), then the normal camera M = f/d applies. If our Moon is 3474 km diameter, and if its image is 0.5 mm, that's an extreme size reduction, and not likely a meaningful number. So some astronomers do attempt to compare it as if a 50 mm lens gives 1x magnification (so a 2000 mm telescope directly attached as prime focus lens might be said to give 2000/50 = 40x). That's an approximation based on a 50 mm lens being the "normal lens" if on 35 mm film. However of course, in a different sensor size situation, 50 mm and its field of view may not be useful to your sensor. A different sensor size would be a different situation, but still in this 2000 mm case, 2000/(your comparison lens focal length) would give a meaningful comparison size number of those two lenses.

But cameras are a "reproduction device" and the magnification number is relative to the actual real life size of the subject being reproduced. So obviously, sensor dimension / FOV dimension is the actual magnification. For example, reproduction size of 1/100 is 0.01x or 1:100.

Other camera uses: In contrast, the DSLR **viewfinder** magnification specification is compared to the eye's view (called 1x), regardless of the lens attached (its only about how well we see the viewfinder). But magnification of the lens image is compared to the size of the distant field of view (affected by focal length and distance).

Magnification is Not affected by a cropped sensor or smaller sensor size. Both sensor and field of view may be cropped proportionally smaller, but an object size in the lens image is unchanged (if focal length and distance are unchanged). Magnification is f/d, and is directly proportional to focal length or inversely to distance. 2x focal length is 2x more magnification of subject size. 2x distance is 1/2 the magnification of subject size. Therefore 2x focal length AND 2x distance together combine to remain the same magnification and same subject size. See the Depth of Field page for more about using that.

**Aspect Ratio**: Aspect ratio is the SHAPE of the sensor. Numerically it is the ratio of width to height. If Aspect Ratio is unknown, look at the size of your images (pixels) straight from the camera. For example, maybe the size is 4320x3240 pixels. Divide Width by Height. For example, 4320 / 3240 is 1.3333. Common camera Aspect Ratio values are:

1.3333 is 4:3 (also = 4/3) - Compacts and phones

1.5 is 3:2 (also = 3/2) - DSLR

1.7778 is 16:9 (also = 16/9) - Camcorders

**HDTV 16:9 movies:** Special considerations. The differences in the Aspect menu in Option 3 above is about the difference in camcorders from DSLR, compacts and phones. Someone might find some exception about some camcorder, but generally:

Still cameras (including DSLR, compacts, phones) today are usually more or less around 10 to 24 megapixels. They have 3:2 or 4:3 sensors, and take 3:2 or 4:3 photos of that size, and their diagonal fits the diameter of the lens circular view. Their HDTV movies possibly might be slightly smaller than sensor width, but can't be larger - and of course **will necessarily be resampled to 1920x1080 or 1280x720 pixels, up to about 2 megapixels** - which is the maximum that HDTV can use. The 16:9 movies in still cameras are constrained within that still camera sensor size, *limited to the same width*.

Whereas **camcorders are typically only about the necessary HDTV 2 megapixels** (a 4K camcorder might be 12 to 18 megapixels). They have 16:9 sensors, which *fits the full diagonal*, not constrained by any 3:2 or 4:3 sensor width. Generally, if camcorders also take still pictures, those are 16:9 too. But if they provide 4:3 still photos (Sony does), and if less than 2 megapixels, those are necessarily horizontally cropped to be constrained within the height of the 16:9 sensor, *limited to the same height*. These situations will require some attention to know what you have.

This is a complication about the field of view of movie recordings in still cameras, and in still pictures from camcorders. We're not always too sure about the sensor area used. Option 3 above can differentiate these differences (aspect can match the yellow drawings).

In this image at right, the blue circle represents the diagonal of the image that the lens projects. 16:9 in camcorders is wider, but not if constrained within the more narrow frame sizes. I suppose there could be some exception, but camcorders should have a 16:9 sensor, and DSLR, compacts and phones typically have 4:3 or 2:3 camera sensors. Their 16:9 width will be the same as the 4:3 width, but then the height is less.

Determine aspect ratio by dividing image width by the height (both in pixels), which is its aspect ratio. 3:2 divides as 1.5, 4:3 divides as 1.3333, and 16:9 divides as 1.7778. So if your camera takes still pictures of aspect 1.7778, then it is probably a camcorder with a 16:9 chip.

Some photo cameras will use their full sensor width for their HD movie width (Nikon D7100, D600, D750 do specify their movies are full sensor width) and that is assumed here. But for example, the Nikon D800/D810 use slightly less width (these D8xx manuals specify the sensor image area for FX HD movies is 32.8 x 18.4 mm FX, and DX HD is 23.4 x 13.2 mm). That is 35.9/32.8 = 91% of full width. An iPhone 5S (on a tripod) measures the movie field of view to be 91% less too. This field of view reduction could be computed by using effective Focal Length = 1/0.91 times longer (10% longer), OR a crop factor 1/0.91 times larger (10% larger).

But actually knowing the actual sensor size is the key to FOV accuracy.

There are approximations in calculations. The math is precise, but the data is less so. Focal length is a little vague, as might be the precise sensor size. However, the results certainly are close enough to be very useful in any practical case. My experience is that the field is fairly accurate (at distances of at least a meter or so), assuming you actually know your parameters. Some problems are:

The Marked focal length of any lens is a *rounded nominal number*, like 50 or 60 mm. The actual can be a few percent different. Furthermore, the Marked focal length is *only applicable to focus at infinity*. Focal length necessarily increases when lens is extended forward to focus closer. Also zoom lenses can do other internal tricks with actual focal length (some zooms can be shorter when up close, instead of longer). Focal length will be less accurate at very close distances, and field of view becomes a little smaller. So this calculation does not include macro distances (if magnification > 0.1). You also have to measure your distance and field dimensions accurately too. And of course, we are only seeking a ballpark number anyway, we adjust small differences with our subject framing.

And a fisheye lens is a different animal, wider view than this formula predicts. A regular lens is rectilinear, meaning it shows straight lines as straight lines, not curved. A fisheye is rather unconcerned about this distortion, and can show a wider view, poorly purists might say, but very wide, and very possibly interesting.

Actual focal length can be determined by the Magnification (Wikipedia). The distance from the front nodal point to the object in the subject plane (s1), and the distance from the rear nodal point to the image plane (s2) are related by this Thin Lens equation (Wikipedia). If OK with a little geometry and algebra, you can see the derivation of this classic Thin Lens Equation at the Khan Academy.

In this equation, we can see that if the subject at s1 is at infinity, then 1/s1 is zero, so then s2 = f. This is the marked focal length that applies when focused at infinity.

Also if at 1:1 magnification, then s1 = s2, saying that the working macro distance in front of the lens is equal to the (extended at 1:1) distance to the sensor image plane (at 2x focal length). These are basics.

The focal length measures from lens node to sensor. We compute the right triangle on center line, of half the sensor dimension, so the half lens angle = arctan (sensor dimension / (2 * focal length)). The Subject distance is in front of lens node, with same opposite angle. One way is that field dimension = 2 * distance * tan (center line half angle). The problem is that focal length f becomes longer when focused at close distances. That becomes insignificant in field of view at normal distances, 1 meter or more.

Working Distance = d = S_{1} (distance in front of lens)

Focal Length = f = S_{2} (distance behind lens)

Focus Distance = d + f

Focal Length = f = S

Focus Distance = d + f

Multi-element camera lenses are "thick" and more complex. We are not told where the nodes are designed, normally inside the lens somewhere, but some are outside. For telephoto lenses, the rear node (focal length from sensor plane) is in front of the front lens surface. The designer's term telephoto is about the reposition of the nodal point so that the physical lens is NOT longer than its focal length. Yet, this rear node is generally behind the rear lens surface of a wide angle lens (lens moved well forward to provide room to allow the larger SLR mirror to rise... 12 mm lens, 24 mm mirror, etc). This nodal difference is only a few inches, but it affects where the focal length is measured. And it shifts a bit as the lens is focused closer. Repeating, the focal length marked on the lens is specified for focus at infinity.

The Subject distance S is measured to the sensor focal plane (it is the "focus distance"), where we see a line symbol like Φ marked on the top of the camera (near rear of top LCD). The line across the circle indicates the location of the sensor plane (for focus measurements). However, the Thin Lens Equation uses the distance d in front of lens. This is why we often see in equations: (S - f) used for d.

For **Macro**, computing magnification is more convenient than focal length (since we don't really know focal length at macro extension). Focal length and subject distance determine Magnification, which is the ratio of size of image to size of actual subject. Or the ratio of size of sensor to the size of the remote field. We could compute for magnification here, but we likely don't know new focal length at that close magnification. Just using magnification has more significance up closer (easier for macro), which is where our knowledge of the actual focal length is weakest. We could measure the field to compute the actual magnification. However Magnification is simply:

m = s2/s1. Or m = f/d. Or m = f/(S-f).

So from this, we know macro field of view is simply the sensor dimensions, divided by the magnification. Let's say it this way:

1:1 macro, the focal length f is same as the distance d in front of lens (each with its own node).

1:1 macro (magnification 1), the field of view is exactly the same size as the sensor.

1:2 macro (magnification 0.5), the field of view is twice the size of the sensor.

1:4 macro (magnification 0.25), the field of view is four times the size of the sensor.

This is true of any focal length for any lens (or method) that can achieve the magnification. Focal length and subject distance are obviously the factors determining magnification (it is still about them), but magnification ratio is simply easier work for macro.

The easiest method to determine field of view for macro is to simply put a mm ruler in the field. If a 24 mm sensor width sees 32 mm of ruler, then that is the field of view, and the magnification is 24/32 = 0.75 (this scale of magnification is 1 at 1:1, and is 0 at infinity).

The definition of macro 1:1 magnification is that the focal length and subject distance are equal (distances in front of and behind the lens nodes are necessarily equal, creating 1:1 magnification). In this Thin Lens Equation, if s1 and s2 are equal, the formula is then 2/s1 = 1/f, or 2f = s1. So lens extension to 2f gives 1:1. And since f/stop number = f / diameter, then if 2f, then f/stop number is 2x too, which a double f/stop number is 2 stops change, which is the aperture loss at 1:1. We know those things, this is just why.

But the point here, if f is actually 2f at 1:1 macro, the field of view changes with it. None of the FOV calculators are for macro situations (too close, magnification is instead the rule there). Field of View calculators expect subject distance to be at least a meter or so, reducing the focal length error to be insignificant.

Next page is a table of angular Field of View (degrees) for many lens focal lengths and a few popular sensors.