Points in Focus Photography

Depth of Field (DoF), Angle of View, and Equivalent Lens Calculator

Camera and Lens Settings


Enter a crop factor (eg. "6.7" or "0.25") or dimensions in mm as width x height (eg. 36x24 or 15.2 x 9.5). See notes section for details.
mm
f/

Advanced Options (Click to Show)

See notes on extended functionality.

A resolution in MP (eg. "24MP" or "24 MP"), or in pixels (eg. 6000x4000 or 6000 x 4000).

Shutter speed in fractional (eg. 1/500) or decimal notation.

Equivalent Lenses

Format Focal Length Av Setting* F-number
Results go here.

* The Av setting is the camera aperture setting, in 1/3rd-stops, that will closely approximate source
lens's depth of field. Available values range from f/1 to f/64 in 1/3rd stops.

Note: The experimental values and extended functions are part of an ongoing experimental concept that I’m currently developing. Please don’t contact me and ask me what it is or how to interpret it. It is only included and displayed so I can move forward on developing an idea.

Troubleshooting: I’ve tested this script in most desktop browsers as well as the current version of iOS. If the script is not loading or working properly, and you’ve used it the past, it’s likely that your browser is caching an old copy of the JavaScript. Before reporting a problem, please try force reloading the page. To do this hold shift and click reload in most desktop browsers, or tap and hold the reload icon in iOS and select “Reload without content blockers”.

Help

The following subsections cover various bits of specific functionality.

Custom Frame/Sensor Sizes

All of the supported calculations can be preformed for any arbitrary custom frame/sensor size; not just the standard photographic ones that are provided in the sensor size menu.

To enter a custom frame, select “Custom (NaNx)” from the sensor size drop down (at the bottom). Then enter the custom size in the “Custom Sensor” box that appears below it.

Custom sizes can be entered as either a crop factor relative to the 135-format (36 x 24 mm) film frame. In which case the custom size will retain the 3:2 aspect ratio. Additionally values larger than 1 will produce a smaller frame and a value smaller than 1 will produce a larger one (e.g. entering 0.5 will produce a frame larger than 36 x 24 mm).

The other option is to enter the dimensions of the frame in millimeters, formatted as “width x height” (e.g “36 x 24” or “17.3×10.8”). Spaces will be ignored. Decimal values can be used.

Custom Apertures

Support for entering non-1/3 stop apertures such as may be used by scientific or specialty lenses.

Click the “Custom AV” button and enter the aperture value in the adjacent box.

To return to the normal aperture drop down mode, click “Custom AV” again.

Custom Circle of Confusion

By default, the calculator uses the standard circle of confusion used by most current lens manufacturers (diagonal / 1500). However, the circle of confused can be changed under advanced options. Four preset methods are:

Modern Standard Method (Default) Frame’s diagonal / 1500
Zeiss Formula Frame’s diagonal / 1730
Kodak Formula Focal length / 1720
Archaic Standard Frame’s diagonal / 1000

Additionally a custom circle of confusion can be specified in the custom box under advanced options.

Click the “[reset to def.]” link next to the displayed “Circ. of Conf.” to reset the circle of confusion to the default method.

Depth of Field Graph

The depth of field graph presents depth of field curves for the full-stop aperture settings from f/1.4 to f/16. This is done using the current settings, except aperture, specified in the input section.

For improved readability, there are limits imposed on the scale of the graph.

  1. The maximum depth of field that the graph will show is 10 meters (33 feet); depth of field approaches infinity at the hyperfocal distance for each aperture.
  2. The minimum distance is 4 times the focal length.
  3. The maximum distance is given by:  dmax ~= focalLength * ( 750 * focallength / log(focalLength) ).

Each curve’s display can be toggled on or off in the graph by clicking the name or color in the key.

Rendered charts can be saved as a PNG image by right clicking on it and selecting “Save image as…” or the equivalent in your browser’s context menu. The depth of field graphs are provided under the Creative Commons Attribution (CC-BY) license. Feel free to save them for use in your own content, so long as attribution to this page is provided.

Pano Shift Angles

Pano shift angles are intended to make it easier to determine useful increments for shooting stitched panorama images.

Pano shift increments target a 35% overlap of the previous frame rounded to the nearest 2.5° increment (first column), and then shows the actual overlap that will occur with that setting (second column). To be used to calculate the rotation of a panning base divided in 2.5° increments, such as the base on a Really Right Stuff tripod head or PC-LR/PRO panning clamp.

The pano shift angles are clickable, and will take you to my Pano Tripod Head Angle calculator with the calculated shift angle already filled in.

Methodology

Depth of Field

Depth of field and hyperfocal distances are calculated using the following formulas.

Hyperfocal distance equation.

Near limit of depth field equation.

Depth of field far limit equation.

  • H is the hyperfocal distance
  • f is the focal length
  • s is the focus distance
  • Dn is the near DoF limit
  • Df is the far DoF limit
  • N is the f-number
  • c is the circle of confusion

Equivalent Lenses

Equivalent lenses are calculated so that the focal length produces the same approximate angle of view and the f-stop results in an aperture diameter equal to the base lens’s aperture diameter which will produce the same depth of field. Some variation will occur when the format aspect ratios do not match.

Angle and Field of View

Angle of View Method 1: Angle of view equation 1, simple geometry method. θ the angle of view
s is the focus distance
h is the frame dimension
f is the focal length
Angle of View Method 2: Angle of view equation 2, adjusted effective focal length for overall extension lenses method.
Field of View: Field of view equation.

The calculator calculates angle of view using two different methods. Both arrive at the same result at infinity focus, but differ at closer focusing distances. Since fields of view are calculated using the angle of view, there are also two calculated values for the field of view at the specified distance; one for each method. Further, neither method is completely accurate at very close focusing distances. For more information on this, see Calculating the Angle of View: When Theory Meets Practice.

Method 1 is a “naive” method that assumes that the angle of view for the format, which is the angle of view at infinity focus, does not change when the lens is focused closer.

Method 2 compensates for lens extension when focusing closer than infinity.

Which method should you use?

  Use Method
Does your lens extend when focusing? 2
Does your lens use an inner focusing system or is a video or cine lens? 1

Update History

Update History
Date Update
2022-04-28 Added all video formats for the Canon R5C, and Nikon Z9.
2022-02-08 Added standard 8mm film format.
2019-03-16 Added 16:9 aspect ratio 2/3″/B4 mount lens, uses 11mm diagonal for a sensor size of 9.6 x 5.4 mm. Matches published angles of view for many ENG lenses.
2017-12-26 Major fix to angle and filed of view calculations to compensate for the shift needed to focus closer than infinity.
2017-12-25 Revised page content to be more reflective of current capabilities and features. Revised update history to make it more terse.
2017-10-20
Added 4 preset circle of confusion calculations. Circles of confusion are calculated on the fly instead of using ratios of 0.029.
2017-07-18 Added panorama shift increments to the calculator.
2017-05-18
Added DoF at aperture chart.
2017-05-17
Added ability to specify custom circle of confusion under the extended functions area (click where it says click to show extended functions).
2017-05-16
Major engine upgrade, changes include:

  • Crop factors are calculated using the format’s diagonal, not standard approximations.
  • Circles of confusion are calculated as the crop factor ratio of 0.029.
  • Changed precision of displayed output values (metric 3 sig. figs., imperial <1% error)
  • Changed URI hash fragment format (bookmarks using the old format will still work).
  • Grouped format list by typical use.
  • Added new format(s): Canon’s 5D mark IV and 1DX mark II 4K crop factors, Red’s Epic/Weapon, and Arri’s 65mm format.
2017-05-08 Fixed bug with custom apertures. Improved mobile support. Extended experimental functions.
2017-03-13 Added custom aperture value. Reduced precision of displayed values.
2016-04-01
Added custom frame and crop factor support.
2016-10-11
Bug fixes for distance calculations as reported by Frank.
2016-02-25 Added new format(s): 5×7 (requested by Nameless Person), Super8, and 16mm movie.
2014-04-12 Added new format(s): 645D/IQ250/H5D-50c.
2014-05-13 Updated script engine, Added support to store current settings in the URI hash to allow linking to a specific settings.
2014-02-05 Added new format(s): 1/1.7″ CCD used in many compact cameras and the Pentax Q mirrorless cameras.
2014-02-03 Clicking on the format in the equivalent lens chart will update format and lens settings to match.
2014-01-28 Added angle and field of view calculations.
2014-01-12
Added support for metric distances. Added preset focus distances roughly corresponding to various types of portraits.
2011-01-17 Added features for AF calibration uses: AF Test Dist button, sets focus distance 50 times the focal length; 1/8 DoF to outputs.

 

Comments

tex andrews

Please add Pentax 645D/Z

    Jason Franke  | admin

    @tex andrews, I’ve added the format under 645D/IQ250/H5D-50c.

Yannick Ciancanelli

thx for this :)

Gabe

A custom sensor size or multiplication factor would save you from having to add more formats. This is a great calculator, but I came here hoping to find common CCD calculations. 1/3.2″ for iphone S5, 1/4″, 1/6″, 1/8″, 1/10″ CCDs common on lesser cell phones as well as security cameras.
Regardless, thanks for the nice page.

    Jason Franke  | admin

    Gabe, that’s actually a good idea, I’ll have to look into implementing that.

Nameless Person

Could you add 5×7 as you got 4×5 & 8×10…?

    Jason Franke  | admin

    That’s easy enough, I’ve also added super 8mm and super 16mm film since there seems to be some movement in those formats (Black magic’s pocket cine camera and the new Kodak film camera).

    Nameless Person

    Nice, thanks. I use this a lot!

Andrew Batson

Question. I have a 60D with an APS-C (1.62x) sensor. So I choose that in the drop down above. On the 60D you can either attach a lens that is EF (full frame – used commonly with a 5D), in which the 1.62 crop is inherently applied to the focal length. OR you can attach an EF-S lens (same mount but opening tailored for the smaller sensor) which holds the focal length true for the APS-C sensor.
When inputing the focal length above for a EF Full frame lens attached to the APS-C sensor, I have do the 1.62x calculation first, correct? i.e. an EF 30mm would be effectively a 48.6mm on the APS-C sensor. I should input 48.6mm in the focal length section above? Correct?

    Jason Franke  | admin

    Hi Andrew,

    Thanks for the question.

    The short answer is that you always enter the actual focal length that’s printed on the lens.

    The long answer is a bit more detailed but it’s worth going over at least quickly.

    Focal length is a measure of how strongly a lens bends or focuses light. A lens’ focal length doesn’t care about the sensor behind it at all, it’s a property of the lens and only the lens. The same focal length will have different angles of view if you put a different sensor or frame size behind it, but the lens focuses/bends light the same way (it’s a bit more complicated than this in reality, but that gets into a lot more optical engineering than I want to go into here).

    For example, a 50mm lens on your APS-C camera has a an angle of view that would commonly be described as being “short telephoto”. A lens with the same focal length on a full frame camera, would be described as having a “normal” field of view. And finally a 50mm lens on a medium format camera would best be described as having a “wide angle” view. All of these lenses fundamentally “bend light” the same way, but because the frame gets bigger the angle of view increases too (the edges are further and further away so the same amount of bend results in light coming form wider and wider angle to match).

    Photographers complicate the matter because they often say focal length when what they’re really talking about is angle of view (and given that it’s much easier to say 35mm than 74 degrees 10 minutes it’s no wonder why). This is what is being talked about when you do the “equivalent focal length” dance where you multiply or divide by the crop factor.

    However, the focal length, being a property solely of the lens does not change when you change the size of the sensor; the angle of view does, but not the focal length.

    This calculator uses the actual focal length of the lens, not the “equivalent angle of view focal length” that photographers talk about when comparing things. So there’s no need to do anything to focal length, just use the number that you had the lens set to.

    Also, to clarify a point. EF-S lenses do not “hold the focal length true” for APS-C cameras. The focal lengths of those lenses (as printed on them) are the real focal lengths of the lens, and would be the same as the focal length as on an EF lens. That’s why they have different focal lengths than full frame counter parts for the same approximate angles of view.

    EF-S lenses specifically do 3 things:

    • They provide angle of view ranges that match traditionally defined and useful ranges (i.e. wide to short-telephoto, or ultra-wide to wide angle), by using different focal lengths as required by the smaller sensor. (This can be generalized to all crop lenses, like Nikon’s DX or 3rd party crop lenses.
    • They have a shorter back focus distance, which allows the rear element to be placed closer to the sensor; the practical implication of this is that they can’t be mounted on a full frame camera (3rd party, e.g. Sigma, crop lenses are not EF-S, they use EF mounts), but this offers Canon benefits in designing the lenses (they don’t have to be as strongly of a retro-focal design).
    • They have a smaller image circle that won’t fully cover, or do so with good quality in the corners, a full frame sensor (again this is an engineering trade off as it allows some aspects of the lens to be made smaller.

    Also the equivalence dance applies the same to them as it does to EF lenses. An EF-S 17-85mm zoom is the crop body equivalent to the EF 28-135mm zoom for full frame cameras. If you were going to jump to full frame, you’d want to multiply all your lenses’ focal lengths by the crop factor to figure out what focal lengths you want for full frame lenses to give you the same coverage.

    I know that’s probably a lot more than you were bargaining for, but I hope that answers your question and clears up some possible misconceptions.

Frank

Thank you for posting this valuable tool – I have one question:
How do you calculate the distances for the different test situations (Headshot, Head & Shoulders, etc.) – I naively assumed that here the corresponding distance is based on an equal Field of View (e.g. Diagonal), however, I find different values for Field of View with different focal lengths (equal sensor size). Where is my mistake?
Best regards, Frank

    Jason Franke  | admin

    @Frank,

    Nope, that’s not your error at all. I was improperly simplifying the trig that handled calculating the distances on the assumption that the error wouldn’t matter that much for the intended use. I’ve corrected that to use the proper formulas.

    For the record, I use the horizontal (long axis) of the for the distance approximations, since that’s how I expect those kinds of images would be shot (i.e. as portraits). The AF test distance is just 50 times the focal length which is what was recommended by Canon and Nikon in their literature in the past. Other than the AF test distance preset, the intent of the distance buttons is to create an approximate distance number, not to be an exact standard.

    Thanks for taking the time to check things.

    Frank

    Dear Jason,
    thank you for the very rapid answer and the correction. However, I am still somewhat puzzled:

    E.g., in the tool the “Head&Shoulder” preset for Canon APS-C gives out a horizontal FOV of 30 cm @ 20 cm distance for 15 mm FL, but 68 cm HFOV @ 307 cm distance for 100 mm FL. Is that big variation in HFOV (30 cm to 68 cm) intended ?

    If you would calculate the WD (working distance in mm) from the angle of FOV (AFOV(°)) as

    WD (mm) = HFOV (mm) / (2x tan(AFOV(°)/2))

    and AFOV(°) from the height (h) of the sensor in mm and the focal length (fl) in mm as

    AFOV(°) = 2 x arctan (h/2fl)

    you could preset HFOV as e.g. 500 mm (for head&shoulders) and would receive the following combinations:

    34 cm distance @ 15 mm FL
    227 cm distance @ 100 mm FL

    Is that correct?

    Best regards, Frank

    Jason Franke  | admin

    @Frank,

    The distance presets should be working correctly now. More buggy code on my end. When I updated the script to support a custom frame size, I changed everything but the distance preset function to use that code, so it wasn’t properly determining the frame size it should have been using (in addition to the goof on the trig).

    Again, thanks for spotting the bug.

sfc

Hi Jason,

I’ve been scouring the internet and speaking with people, but I’ve not found anyone who knows the answer to my problem. I am working with old (1960) aerial photography and in the software I’m using, I am asked for the 35mm focal Equivalent. I know this typically applies to digital photography, but does it also refer to analog film too? If it does, then how do I calculate it? I am not a photographer, so I’m not certain of all of the terms involved with the camera and film. The camera used was a Fairchild with a Bouche & Lomb lens. The calibrated focal length is ~153mm and the film size is 9×9 inches. The aperture (is that also the ‘f/stop’?) is possibly f/8. The altitude is 20,000 feet. In your calculator above, you list the “Sensor size”? Will that be the same size as the film or something else entirely because this is information that I do not currently have (and trying to find a camera manual hasn’t been the easiest).

I would very much appreciate any time an consideration you have to address this.

Regards,
S

    Jason Franke  | admin

    Hi sfc,

    Thanks for the questions, I hope these answers help.

    in the software I’m using, I am asked for the 35mm focal Equivalent. I know this typically applies to digital photography, but does it also refer to analog film too?

    Yes, the “35mm focal equivalent” applies as a conversion to any size frame that’s not a 24 x 36 mm (“35mm” or full frame digital).

    If it does, then how do I calculate it?

    You need to determine the ratio between the frame you’re working with and the 35mm frame, then divide the focal length by that ratio. Normally this would be the diagonal of one divided by the diagonal of the other.

    However, you may want to use the short edge of the frame (24mm) instead since aspect ratio is different between your square frame and the 3:2-aspect ratio film frame.

    My calculator here uses the diagonal of the frame to compute the equivalent focal lengths.

    The calibrated focal length is ~153mm and the film size is 9×9 inches.

    Since your frame is larger than a 35mm frame, you should expect the 35mm equivelent focal length to be much shorter than the 153 mm specified.

    In your calculator above, you list the “Sensor size”? Will that be the same size as the film or something else entirely because this is information that I do not currently have (and trying to find a camera manual hasn’t been the easiest).

    Yes, sensor size is the same as film size. More precisely I should probably call it “frame” size as the size as that’s what is really being sought.

    Ben M

    Hi Jason,

    This is a great resource – thanks for this.

    Question – regarding the calculations, how can I modify to work with anamorphic lenses/sensor settings?

    Thanks.

    Jason Franke  | admin

    Hi Ben,

    That’s a good question, in part because I don’t really know the answer to that. Worse, I have precisely 0 experience with anamorphic lenses so I’m not even sure where to start in trying to answer that. Sorry that’s probably not the answer you were hoping for.

    Ben M

    No problem, thanks for the reply.

    I’ll keep digging.

Chris

Hi Andrew, I have a question about FOV calculations.

Thank you for this post, it’s been very helpful! I’m writing some code to automate this process, taking sensor size, focal length and object distance as the inputs.

I’ve got to the point where I can calculate the horizontal angle of view from:
2*atan(sensor width/(2*focal length)), that works fine.

I can’t work out how you’re getting from angle of view to the metric field of view (although I know your calculations are correct!)

I’ve been using:

2*object distance*tan(angle/2), but this doesn’t work for some reason.

I might be doing something dumb, but wanted to check first what equation you’re using, if you don’t mind.

Again, thanks for this post!
Best wishes,

Chris

    Jason Franke  | admin

    Hi Chris,

    I’m not Andrew, not sure who you’re addressing there, but since this is my site and my code, I’ll be answering your question today. :-)

    The equation I use is 2 * object distance * tan(angle of view/2) for the appropriate angle of view (horizontal, vertical, or diagonal).

    If you’re not getting the correct results from that, make sure that you’re using the right units for the angle. The trig functions for most, if not all, languages use radians not degrees. Internally my script does all calculations in mm and radians and only converts to degrees and meters or inches to display the value.

    Hope that helps.

Artur

Hello Jason,

first of all thank you for that great tool. I tried many DOF-Calculators but yours is nearly perfect. And now my question to make. Is it much of a trouble to include a custom F-Stop function? Most of the lenses I use or try to calculate with have unusual values. I tried to mess with the URL and change the values, but they are obviously not connected to the code.

Best regards,
Artur

    Jason Franke  | admin

    Artur,

    It certainly would be possible, but I can’t make any promises as to how quickly I’d be able to get to it. I will add it to my to do list. I am working on adding some new features to the tool currently, so I should be able to fit it in reasonably soon.

    Thanks.

terry stahly

I just spent two hours making charts or cheat sheets for my iPhone in Photos a folder I created that is DOF charts for each of my lense on another calcualtor before I found yours and wasted all my time. Before I go through all of that again with yours does the fact that a Sony a7rII is mirrorless affect any of these formulas with EF full frame Sony lenses? Thanks this looks like a real valuable tool I am going to try and load the bookmark for it and see if can to that on my phone or ipad if necessary.

    Jason Franke  | admin

    @Terry Stahly,

    Being mirrorless or not is not a factor. Since your A7R II is full frame, you can pick “Full Frame/Nikon FX” from the list and go from there.

Jeff Armstrong

Why would the depth of field of a 6×4.5 with a 105mm f/2.5 framing for the upper body be less than the depth of field of a 6×7 with a 105mm f/2.5 framing for the upper body?

Wouldn’t the 6×7 have shallower depth of field due to the need to be closer to the subject to achieve the same crop as the 6×4.5?

I have tested both, and have found that I must be further from my subject with the 6×4.5 to achieve the same crop as the 6×7. However, the calculator says that I must be equal distance from the subject with both formats to achieve the same upper body crop.

How can this be?

    Jason Franke  | admin

    Hi Jeff,

    To answer your last question, “how can this be?” It can be, because I’m an imperfect programmer that reversed width and height values for the 6×7 frame in the the script.

    I calculate my “preset subject distances” by using the width of the frame. Since in most cases, the width is the long edge, and since the subjects are “portrait” kinds of images, this usually works fine. The oops is that I seem to have specified 6×7 as 7×6 in the script. So the script was using the same width as 645 and 6×6 when calculating the subject distances. Same width = same distance, and yep, things don’t work right.

    Thanks for catching this, I’ve updated the script to put the right dimensions in the right fields.

Paul

Thought you might like to know – I haven’t been able to access your site for a few months. Every time I tried it asked for my Apple ID keychain info! that was on Safari. I’ve just tried on Chrome and it worked. Weird.

    Jason Franke  | admin

    Paul,

    I haven’t seen that problem but, I’ll look into seeing if I can figure out what might be causing that.

    Update: I think I’ve tracked this down and fixed it. I think this was a side effect of a server configuration change I made on my end. Sorry about the problems.

Tomas Lind

Hi Jason,

Accuracy of field of view for taking pictures at close distances, say less than a few meters.

Thanks for a nice webpage. A great resource. I have a Nikon DX camera with a standard 18-300 mm superzoom lens so I specify “Nikon DX/APS-C (1,53×)” on the webpage.

When calculating field of view the figures for 1 m distance for focal lengths between 18 mm upp to 100 mm the figures given on the webpage are close to what I get when I´m taking pictures of a ruler. Now, When using 300 mm (actual focal length “labeled on the lens”) at 1 m distance the result tells me that 8 cm would fit in horisontally on the Picture. In reality, when taking a picture of a ruler 16 cm fits in! Maybe there are some approximations in the formulas used to calculate the field of view. It would be great if they were closer to reality.

A second question. Does the depth of field, angle of field and field of view results from the webpage also work for macro lenses? (I don´t have any macro but it would be nice to know.)

I very much appreciate your webpage.

Thanks

Tomas

    Jason Franke  | admin

    Hi Tomas,

    There area number of factors in real lens designs that can cause the actual behavior of a lens to deviate from the predicted idealized behavior that I’m calculating here. Some of these can become more significant at close focusing distances, and I don’t consider in these calculations (nor do any other DOF calculators that I know of for that matter), largely because the extent of their effect is dependant on the specifics of the design of the lens and I would need to have specific profile information for every possible lens. It’s important to understand that any of these kinds of calculations are idealized and not necessarily going to be reflective of the reality of your specific lens under every specific condition.

    For example, internal and rear focusing still photography lenses (your Nikon 18-300 is an internal focusing lens) tend to change focal length, and thus angle of view, when focusing closer than infinity. The listed focal length is only the actual focal length of the lens when the lens is focused at infinity (in my experience they get wider in terms of angle of view).

    For example, when I was testing filter holders on ultra-wide angle zooms, a holder might not vignette at 16mm and infinity focus, but would if I focused the lens at the minimum focus distance.

    At normal working distances, the changes in focal length from the internal/rear focusing design aren’t that significant. But at the very close end of the focusing range scale, and again depending on the lens, they can be.

    Since this error in focal length is specific to each individual lens design, I don’t even attempt to compensate for it in the calculations. Like all DoF calculators, these numbers are more of an ideal than a completely accurate prediction for any specific lens.

    As far as macro goes, the calculations should provide a reasonable estimate, but they almost certainly won’t be accurate; again because of the complexities imposed by real lens designs instead of idealized calculations.

    Hope that helps explain some of the differences you’re seeing.

May WANG

Hi Jason,

A nice webpage!

I want to know how to get the size of circle of confusion. I googled that the circle of confusion is not a constant value, which is related to focal length,object distance,etc. However in your webpage as the sensor is chosen,the circle of confusion is fixed.

“Distance” in your tools is the distance between object and image plane?

Thanks.

May WANG

    Jason Franke  | admin

    May,

    Yes, distance is the distance between the object and image planes.

    As for the circle of confusion… To put it simply this is complicated and highly depends on the assumptions you’re going to use and what those assumptions are.

    The most commonly used formula for the circle of confusion as it relates to depth of field calculations is d/1500 where d is the diagonal of the frame. This is what the vast majority of camera makers use when they provide depth of field information for their lenses.

    That said, there are other formulas, such as “Zeiss formula” which is d/1730 or the “Kodak formula” which is f/1720 where f is the focal length.

    The circle of confusion, is fundamentally a simplification for a much more complex set of assumptions. In very broad strokes, is that the CoC we use is a transformation of the acuity of human vision looking at a print of the image at a specified distance and scaling that down to the size of the film/image sensor. If you’ve cropped your image relative to the full frame size of your format, then the “actual” CoC changes because the magnification ratio from the image to the final print changes. If you change the viewing distance of the final print, then the CoC changes. If you’re viewer has better or worse than average vision, then the CoC changes. Change the viewing distance of the final print, and the CoC changes.

    Calculating and dealing with all those variables is not practical to try and figure out every detail, so for depth of field calculations we distill the myriad of variables into something that provides a reasonable approximation across many cases, and use that to calculate the “depth of field”. These are the “standard” circle of confusion values.

    Perhaps what may be most important to understand is that depth of field calculations are not precise values with hard physical cut offs and you can’t treat them that way. Depth of field values are at best reasonable estimates of what should be acceptably in focus for a reasonable set of assumptions.

    Hope that helps.

Tom

Wouldn’t the correct technical term be “Out of Focus Potential” not “Blur Potential”? Blur is caused by movement. Maybe Bokeh potential is more apt?

    Jason Franke  | admin

    Tom,

    In thinking about it, I think the most precise technical term would probably be “background defocus potential” given what and how I’m calculating the number. Bokeh potential could work as well too. That said, I’m not settled on the language.

    Thanks for the comment though, I hadn’t really thought about the terminology as I’ve been more focused on the math and what I was trying to represent.

Al Naqba

This is a great web page and very useful.
Thank you for making it.

Jon Williams

I’m shooting video using a Panasonic LUMIX GH5 with Metabones Speedbooster and EF lenses for an effective crop factor of 1.42 (2x.71). What should I do to calculate the distance from subject necessary to achieve a certain shot (i.e. Full Body, Head and Shoulders, etc.) in 16:9 format rather than 3:2?

    Jason Franke  | admin

    That’s a good question.

    The calculator isn’t set up, per se, to do what I think you’re trying to do, but it could be made to work. Being a still photographer, I calculate the subject distances for the portrait like shots based on the long edge of the frame, under the presumption that they would be shot holding the camera in portrait orientation. That’s probably not the case for you shooting video. Aspect ratios are dictated by the sensor/frame that’s specififed, so for example, the Canon 1Dx mark II 4K mode (1.42x) option uses a 1.9:1 aspect ratio frame.

    As a workaround, until I can figure out a better solution, you could trying using a custom sensor size with the width and height reversed. (Pick custom under the sensor size drop down, and enter the dimensions in the box that appears below it.) For 16:9 in native 4/3rds format, you want 9.7×17.3. For your Speed Boostered configuration you could try 13.7×24.4. The width and height values in the AoV table will be backwards, but the calculated distances should be “correct” for a landscape framing.

    I’ve not completely thought through how to deal with a Speed Booster, since it’s neither something I use or, until today, have been asked about.

Kevin

Please consider adding 6×9 as a medium format option. Thanks!

    Jason Franke  | admin

    Added.

    If it doesn’t show up, you may need to clear your cache and refresh the page (typically hold ctrl or shift and click the reload button.

Pablo Yamamoto

The best calculator ever. Great piece of work! It would be nice to have it as an AppStore and PlayStore app. I would definitely buy it.

    Jason Franke  | admin

    Hi Pablo, thanks for the kind words.

    I have looked into putting together a standalone iOS app for this, but unfortunately it’s currently not feasible for me to do so.

Marc

Hello Jason, your calculator was recommended by the manufacturer of a high-speed camera I purchased. The camera is to be embedded in a machine. I want to use your calculator to help me estimate the camera layout to satisfy a HFOV requirement given limited available working depth in the machine. Using the camera fitted with a 50mm fixed focus lens I have on hand, I made some images of a checkerboard test pattern at two known distances, ≈546 and ≈846mm. I compared the HFOV measured in the images with those given by your calculator, so that I could characterize its estimates. I was encouraged by the manufacturer to share the results with you, as both predicted values for HFOV were greater than expected. At 546mm the measured HFOV is ≈147mm, the calculator had 191mm. At 846mm, a ≈252mm HFOV was calculated to be 296mm. The calculated values are 30% and 17% over respectively and interestingly the same amount in both cases, +44mm. I am uncomfortable with the 30% ‘error’, given the relative simplicity of the 50mm lens. Perhaps you can have a look. Any help will be greatly appreciated. The sensor is custom, 17.5×11.8. Thanks.

    Jason Franke  | admin

    Hi Mark,

    At close focus distances, which your 550 and 850 mm distances would generally be, there are more complexities involved than my calculator currently deals with — many of which are specific to individual lens designs and not something that can be calculated for in general.

    That said, I can certainly run some empirical measurements with my own 50mm lens (a Canon EF 50mm f/1.8 STM), and see what kinds of results I get for those kinds of focusing distances. I’ll also see if there’s anything in my optical engineering texts for a generalized AoV formula that is more accurate at close focusing distances.

    If you’d like, you can drop me a line using the contact form and I’ll let you know directly what I find directly as soon as I have a chance to do some testing. If not, I’ll just reply back here.

Neil Thomson

Hi,

Your calculator is brilliant and invaluable, thanks for doing the work and making it free to use!

I’m re-using the equations in a spreadsheet and went of the arithmetic you have under Methodology, but the field of view equation is not what you actually use? Your previous comment gives fov = 2 * object distance * tan(angle of view/2) which gives me the correct results but the equation image you have is fov = 2 * object distance * arctan (angle of view/2*focal length)

the latex command for the correct equation is:
w = 2s\times\tan\left(\frac{\theta}{2}\right)

thought it might help others.

Thanks.

    Jason Franke  | admin

    Neil,

    Oops, it would be me to fat finger that while updating the documentation at too late at night. Thanks for the catch, the equation is fixed now.

    Sean Dowling

    I still see this in the equations noted in the methodology section…a regression in the page?

    Jason Franke  | admin

    Yea, looks like it. Should be fixed now.

lee r.

For 645 lenses, there should be ZERO change in DOF between crop sensor & full frame, given the *SAME* lens & focal distance. There are *NO* “crop” lenses for 645, so nothing changes except either 1) the crop or 2) the focus distance. That is, ZERO changes in COC, DOF, etc. The lens drives those factors, not the camera body.

    Jason Franke  | admin

    Hi Lee,

    The circle of confusion used to calculate depth fo field is driven by the frame’s size*. Since “crop” 645 cameras have a different sized sensor than the standard 645 frame, they will have a different CoC, and as a result will have a different DoF. Since the sensor/format controls the size of the frame, not the lens, it’s absolutely correct to have “crop” 645 cameras separate from “full frame” ones — ultimately this is no different than having APS-C cameras separated from 135-format ones.

    Also, lens do not “crop”. So called “crop format” lenses are merely lenses with focal lengths chosen to provide the same angles of view that photographers have been accustomed to using on the format they’re designed for. If you mount an 80 mm 645 lens on a 135-format camera, the lens is still just an 80 mm lens, and will behave no differently than an native 135-format 80 mm lens.

    Given that the depth of field calculations are being done correctly (all of the methodology for which, including the calculations for the CoC, are laid out in the text of the article), I will not be changing the behavior of the calculator at this time.

    Additionally, if you wish to specify a specific CoC for your calculations (e.g. such that you want to use the CoC for 645 full frame but the frame size for the Phase One IQ250 sensor), you can specify a custom circle of confusion under the advanced options.

    * This is true for all of the currently accepted and used methods for calculating the circle of confusion. Kodak proposed/used a method that was based on the lens’s focal length. If you wish to use this method instead, you can switch to it by clicking on the Advanced Options header and changing the Circle of Confusion Method to Kodak (f/1720).

    lee r.

    Comparing two sensors, the Circle of Confusion doesn’t change IF:
    1) The pixel pitch doesn’t change
    2) The lens doesn’t change
    3) The focal distance doesn’t change
    4) The crop changes due to the change in sensor size.

    Therefore you are correct in this sense:
    1) The CoC changes if any of those fixed variables changes.
    2) A crop 24MP sensor’s COC is different from the full frame 24MP sensor’s.

    But that’s not what your calculator indicates, so there’s a problem here. You assumptions, like most everyone trying to understand this issue, are based on assumptions that are neither clear nor consistent.

    Look at it this way: Crop lenses, set to achieve the same FOV at the same focal distance, have different bokeh, hence different COC. The higher acutance, or resolving power, (due to finer grain pixels) doesn’t effect the COC.

    AFAICT what’s been passed around as canonical is in fact misleading, much like the nonsense about the magical telephoto effect of crop sensors (only true if both crop & FF sensors are of the same resolution, the crop of greater resolving power and the same lens sufficiently sharp for both…).

    https://petapixel.com/2017/03/29/dispelling-myths-around-depth-field-crop-factor/

    Jason Franke  | admin

    Lee,

    Comparing two sensors, the Circle of Confusion doesn’t change IF:

    1. The pixel pitch doesn’t change
    2. The lens doesn’t change
    3. The focal distance doesn’t change
    4. The crop changes due to the change in sensor size

    Points 1 and 2 are immaterial to the circle of confusion. They are also immaterial to depth of field, but will have an impact on bokeh (i.e. the quality of rendered blur) in the resulting image.

    The focal distance (point 3) is relevant to the depth of field, but it has no bearing on the circle of confusion at all.

    The frame size (point 4) is the only relevant factor here with respect to the circle of confusion. Though whether the frame size is dictated purely by the sensor’s dimensions or by a combination of that and cropping in post doesn’t matter.

    The fundamental driving factors for the circle of confusion is the visual acuity of the human eye and the behavioral relationship between print size and the distances at which people are most comfortable viewing them (which will generally remain similar, with people viewing larger prints from further away).

    The CoC changes with the size of the recorded frame because the enlargement ratio between the frame and the assumed print changes (it wouldn’t change if you only made proportionally smaller prints from smaller sensors; e.g. 8x10s from a full frame camera, and 4x5s from a 4/3rds camera). Smaller formats have to be enlarged more to make the same size print. This enlargement ratio scales the eye’s resolution spot from the print to the sensor where it becomes known as the circle of confusion.

    Also, it’s very important to understand that depth of field is not a physical property of an image, it’s an illusion based on our ability to resolve detail in an image we view. Depth fo field will change in any case where you deviate from the assumptions (and the tolerances built into those assumptions), such as moving closer or further away form the viewed image.

    Look at it this way: Crop lenses, set to achieve the same FOV at the same focal distance, have different bokeh, hence different COC.

    Lenses designed for smaller formats achieve the same AoV[1] by having a shorter focal length. Other factors in what you’re calling “crop lenses”, such as a smaller image circle and a different back focus distance, are immaterial to DoF or the CoC.

    The result of having a shorter focal length is that the diameter of the aperture is smaller for the same f-number (f no. = focal length ÷ aperture diameter) which alters the optical geometry of the light path. This in turn changes the size of the spot created in image space by something in object space, and as a result how the depth of field appears. However, this has nothing to do with the circle of confusion.

    Incidentally, this is the point that your linked article is making. If you hold the angle of view, and the aperture diameter fixed (e.g. they used a 50mm f/4 lens on full frame, and a 25mm f/2 lens on 4/3rds), the depth of field will be the same for the same composition (meaning equal focus distances, aspect ratio, and framing). The CoC for the two cameras in this case will not be, and cannot be equal or the depth of field will change.

    You’ll find that if you plug these settings into this calculator, you’ll get same results. For a FF sensor with a 50mm f/4 lens focused at 10 feet, DoF is 2’10” (865mm). For a 4/3rds sensors with a 25mm f/2 lens focused at 10 feet, DoF is 2’9.1″ (841mm). The difference between the two is a result of 4/3rds sensors not being exactly a 2x crop factor.

    I would point out, too, that I don’t use the popular “crop ratios” in the calucations; all of my calculations are driven based on the dimensions of the frames/sensors. Moreover, my circles of confusion are not derived from popular values either, but calculated for the specific frame based on the formulas used by lens and camera makers to calculate depth of field — as used in things like calculating the position of aperture marks on distance scales.

    The higher acutance, or resolving power, (due to finer grain pixels) doesn’t effect the COC.

    This is correct, neither the acutance or resolving power of the lens or sensor have any bearing on the circle of confusion.

    Fundamentally, depth of field works like this:

    1. You look at a print, your eyes have a resolving power[2] to distinguish details in that print.
    2. You stand a distance from the print to view it, this allows the angular resolution of the eye to be translated to a linear size (spot diameter) on the print.
    3. This gets scaled down by the enlargement factor[3] to a spot on the sensor/film; this is the circle of confusion.
    4. Taking the circle of confusion, combined with the diameter of the aperture stop, and a the focus distance, geometry is used to translate diameter of the spot from being coplanar with the film plane to being an distance axial to the lens that would define that spot; this is the depth of field.

    [1] We can simplify this point by talking about angles of view as two lens and camera systems will the same angle of view will have the same field of view at the same focal distance.

    [2] Properly put, lens resolution is measured as an angle. The measures commonly quoted by photographers, e.g. lines per picture height, or lines per mm, are translations from this fundamental point, not fundamental points themselves (though they’re also generally far more useful and approachable than talking about a lens’s resolving power in arcseconds in most cases). For the human eye, the resolving power of a person with 20/20 vision this will be about 1 arcminute.

    [3] The enlargement factor must include any cropping done on the image; it’s not just the ratio sensor/frame size to print size.

    lee r.

    >> 2) A crop 24MP sensor’s COC is different from the full frame 24MP sensor’s.

    This is a critical assumption, that the finer grain pixels have an impact on COC. They may actually not, but I’m allowing that higher resolving power & acutance may afford a minor difference. I don’t actually believe it’s a functional difference.

    I love your calculator, BTW, it’s especially helpful w/ planning interior wide angle reprographic fine art photography.

    IAC, I don’t think either one of us are alone in our hopeless quandary of actually getting to the root of this issue, much less how it bears any practical impact on our shooting.

    Now consider this:
    Take a sufficiently sharp APS-C lens (Oh, let’s say 150 lp/mm), drive it with a 1.4x teleconverter (or w/ the Sony A7r2, its Clear Image Zoom feature) & what do you get? With the TC, at least 1 stop less light. With the A7r2’s CIZ, a wholly different camera-lens system. The A7r2 is dumping 42MP files still. OK, with the image circle falloff, maybe 32-36MP of usable image. But *not* 18MP S35/APS-C files. What’s our COC now, eh? LOL. :)

    Jason Franke  | admin

    Lee,

    This is a critical assumption, that the finer grain pixels have an impact on COC. They may actually not, but I’m allowing that higher resolving power & acutance may afford a minor difference. I don’t actually believe it’s a functional difference.

    See my previous comment, no assumption is ever made about the pixel pitch; it doesn’t matter. With respect to the depth fo field. It will matter with respect to the quality of the bokeh that is rendered (along with other factors).

    I love your calculator, BTW, it’s especially helpful w/ planning interior wide angle reprographic fine art photography.

    Glad you find it useful.

    Now consider this:
    Take a sufficiently sharp APS-C lens (Oh, let’s say 150 lp/mm), drive it with a 1.4x teleconverter (or w/ the Sony A7r2, its Clear Image Zoom feature) & what do you get?

    To start with, it being an APS-C lens is immaterial other than the fact that the image circle may not be sufficient to cover a larger 135-format frame.

    The only thing that matters is the focal length. We could just as easily take a Schnieder Kreusnatch 150mm LS f/2.8 IF lens for Phase One’s MF system and results — ignoring effects on the quality of the bokeh by different optical designs — will be the same.

    Adding a 1.4x TC changes the lens that’s on the camera: it physically alters the optical design so a lens and a lens + tc can be considered two different lenses. For DoF calculations, a e.g. 150 mm f/2 lens with a 1.4x TC is equivalent to having a 210 mm f/2.8 lens with no TC.

    CIZ is digital zoom and is therefor identical to cropping the image. That CIZ uses image processing to generate additional information to make a higher pixels count file is immaterial. A CIZ camera providing 50% magnification (equiv to an APS-C sensor) should be treated as an APS-C sensor.

    With the TC, at least 1 stop less light.

    The amount of light passes is immaterial other than how it effects the exposure.

    With the A7r2’s CIZ, a wholly different camera-lens system.

    Yes, with CIZ the A7R2 is just a camera with a smaller sensor behind the same lens.

    The A7r2 is dumping 42MP files still. OK, with the image circle falloff, maybe 32-36MP of usable image. But not 18MP S35/APS-C files. What’s our COC now, eh? LOL.

    You’re conflating image resolution (number of pixels) with the resolution of a camera or lens. Though none of which matter here either way.

    This doesn’t matter to depth of field or the CoC. The number of pixels in a file need not depend on the resolution of the sensor, only that the area they cover is the same. E.g. Canon’s M and S raw formats have fewer pixels, but because they are derived from the same sensor area (the full size of the sensor) the native CoC for the sensor is used when calculating the depth of feild.

    If you’re using CIZ on an A7R2 to match the frame size of an APS-C sensor, then the CoC you use is the CoC for APS-C. It doesn’t matter if the resulting file is 42MP or 18MP, the difference in the case of CIZ is tantamount to having a 42MP APS-C sensor instead of an 18MP one.

Paul

I love this tool but just found that it’s no longer working on iPhone since recent update?

    Jason Franke  | admin

    Hi Paul,

    It’s likely that your phone is caching an old version of the javascript for some reason. There’s nothing in the script that should stop it from working on an iPhone (I too use an iPhone, so I test on that platform).

    You should be able to force Safari to refresh the page (and all the scripts) completely by holding on the refresh icon in the address bar until the reload page options pop up, and then taping on “Reload Without Content Blockers” from the options that come up.

Elizabeth

What a fantastic tool, thank you!!!

Nadrog

Hi,

1/8th is mentioned in DoF: could you please explain how is it supposed to be used?

Thank you!

    Jason Franke  | admin

    Hi Nadrog,

    1/8 of a depth of field is the adjustment increment used by autofocus micro adjustments. It’s included so you can calculate the allowable error in building or using an AF calibration chart.

Gerard

Hi,
a quick question on the calculator, which I can’t see an answer too in the text below it. Is the DISTANCE value entered the distance from the focal plane of the camera or from the front surface of the lens assembly? I suspect it’s the focal plane but just want to confirm.

Thanks

    Jason Franke  | admin

    So long as the subject distance is much larger than the focal length, then yes. For macro distances (where the subject is less than about 10 focal lengths from the camera) the calculations become a lot more complicated (because real lenses aren’t idealized thin lenses) and isn’t fully handled with these equations.

Richard Michalak

Thanks for this fabulous tool.
Obviously a labor of love and fascination and very much appreciated here in Sydney, Australia.
I am a cinematographer but in this case using Sketchup 3d modelling.
Sketchup only gives vertical angles of view (because the horizontal is changeable) so your calculator was very useful finding equivalent real world lenses to match Sketchup views.
Richard Michalak

Kelly Dodds

I have a 2/3″ B4 Mount 9.5×50 broadcast lens. I use your calculator all the time. I think by doing the math though, that height and width measurements for such a sensor are using the 4×3 aspect ratio. If that is indeed the case, I wonder if you can modify the parameters to apply to 16×9 cameras, which is basically all the modern 2/3″ cameras are. Otherwise, what a great resource, thank you for your work.

    Jason Franke  | admin

    Hi Kelly,

    Do you have the dimensions for a 2/3″ 16:9 aspect ratio sensor? I’ve been looking for it, but the only size I can find is 8.8 x 6.6 mm (4:3 aspect), are 16:9 aspect sensors cropped 4:3, so 8.8 x 4.95 mm? Or are they the full 11mm diagonal of 9.6 x 5.4 mm? I have to appologize, B4 mount gear is not something I have any experience with.

    I’ll add a 16:9 aspect 2/3″/B4 mount setting as soon as I can find the the correct sensor dimensions.

    Kelly Dodds

    You rock!

    Jason Franke  | admin

    I’ve updated the script to support 16:9 2/3″ lenses, I’m using a sensor size of 9.6 x 5.4 mm for these lenses as that seems to match what I’m finding for most other sources, including the published angles of view form various lens manufacturers. You may need to hard refresh the page (hold ctrl+f5 [win] or cmd+f5 [mac]) to get the new formats to show up.

Cliff Gawel

Hi,
I do Technical Support for Panasonic Broadcast cameras and this lens calculator is by far the best I’ve ever used. I use it and refer it to customers and colleagues all the time. I hope it never goes away. Thank you very much for making my job easier.

Mitch

It looks like the calculator might be down? It won’t work in Firefox or on my cell phone in Chrome browser…

P.S. – I love this calculator and use it frequently.

    Jason Franke  | admin

    Hi Mitch,

    Sorry about that, it should be working now.

Mitch

Awesome!

Emiliano Arias

Hi Jason,

thanks a lot for this tool! I’ve got a question about object distance. Do you measure distance from the focal plane or the nodal point?
I’m asking because I’m doing copy art and I can’t find what’s wrong with my calculations.
I read your very interesting article “Calculating the Angle of View: When Theory Meets Practice”, but that doesn’t explain my case, at least not entirely. I’ve found inconsistencies, like my laser meter can have a 2mm error, and my Tamron 90mm macro is in fact a 89,4mm, according to these calculators and if we trust the angle of view from the manufacturer. Which I do, since they have two 90mm with different angles of view in the specs. Even though that can explain little deviations but not what I’m getting. So maybe I’m leaving aside much more basic stuff.

Thanks!

I’m doing this for my studies at university, with your permission and following the terms of use, this site and attributions are gonna be in the resources section of my project.

    Jason Franke  | admin

    Hi Emilano,

    Object distance is measured from the fist principal plane (what I think you’re calling the nodal point).

    First, if a 2mm error in your laser distance measurement equipment is a big enough source of error for you, I don’t think you’re going to be able to use standard photographic approaches for depth of field and angle of view. There are too many simplifying assumptions built into the equations for them to be that accurate.

    Second, there’s a huge list of potential sources of error that you could be running into. Such as: Published angles of view are measured at infinity focus, at closer focuses they will likely be different, especially for internal focusing lenses that aren’t specifically designed to hold angle of view constant (e.g. cine lenses), and especially for macro lenses at macro distances. The size of the camera’s sensor may not match what the published AoV was calculated for. E.g. Canon’s APS-C sensors are smaller than Sony’s and Nikon’s; even then, sensor sizes can change subtly across models and over time, usually not a lot but potentially enough that it might matter in your calculations. The lens’s distortion can affect the angle of view, as can manufacturing anomalies such as de-centered and misaligned lenses — again this should be small, but if a 2mm error in distance is a factor, then it could be enough to matter.

    If you’d like to talk about this in more depth, feel free to use the contact page to send me an email.

    Finally, you’re more than welcome to use and cite this article in an academic work.

andres

Camera
Sensor 1/2.3” CMOS
Effective pixels:12 M
Lens FOV 94° 20 mm (35 mm format equivalent) f/2.8
ISO Range
100-3200 (video)
100-1600 (photo)
Electronic Shutter Speed 8 – 1/8000 s
Image Size 4000×3000
Still Photography Modes
Single Shot
Burst Shooting: 3/5/7 frames
Auto Exposure Bracketing (AEB): 3/5 bracketed frames at 0.7 EV Bias
Timelapse
Video Recording Modes
2.7K: 2704 x1520p 24/25/30 (29.97)
FHD: 1920x1080p 24/25/30
HD: 1280x720p 24/25/30/48/50/60

can some one help me i need the measurmet for dji phantom 3 pro 4k camera area coberture.

    Jason Franke  | admin

    According to Wikipedia, the Phantom 3 Pro uses a 6.17 mm by 4.55 mm sensor.

    That said, that doesn’t necessarily tell you the area used when shooting 4K video. Though you could figure it out relatively easily by shooting a picture of a ruler first as a 12 MP still (which should use the entire sensor area) and then using the 4K video mode. The difference in angle of view would tell you the crop factor, which would tell you how much of the sensor is being used in 4K mode.

    Helpful

    If there is no pixel-binning, line-skipping, or other in-camera processing tricks, you could also find the pixel pitch (usually in microns) and multiply that by the number of pixels in either direction to arrive at the sensor’s dimensions (in millimeters)

Lukas

would be nice to have cctv also, check this, sick:) it is 1200mm on ff

€ 32,87 | QILENS 300mm CCTV camera lens 1/3″ Image Format Long Viewing Distance M12 Mount Horizontal View Angle 1.15D Manual Focus
[link removed]

    Jason Franke  | admin

    Hi, 1/3″ sensors are already in the calculator.

Garry George

Jason

This post has been going a while, but I still have a question.

It stems from you saying

“ Method 1 is a “naive” method that assumes that the angle of view for the format, which is the angle of view at infinity focus, does not change when the lens is focused closer.

Method 2 compensates for lens extension when focusing closer than infinity.

Which method should you use?

Use Method
Does your lens extend when focusing? 2
Does your lens use an inner focusing system or is a video or cine lens? 1”

On my Canon 100mm f/4L macro, the FoV clearly changes as I focus between infinity and the minimum focus distance. The lens is, I believe, an inner focusing lens, hence I should use method 2, surely. The entrance pupil diameter also changes radically.

But you say, for inner focusing use method 1, ie ignore the bellows factor (1+m/p).

I’m confused by what you say. Hopefully you can put me right.

    Jason Franke  | admin

    Sorry for the confusion. So to be clear, neither method is likely going to give you the right numbers for your lens (or most other inner/rear focusing lens).

    The reason I suggest using method 1 for inner focusing lenses, is that in my experience, they (at least the ones that don’t maintain angle of view) tend to behave the opposite of an extending lenses, so the answer will be wrong, but less so. For an extending lens, like a EF 50mm f/1.8 STM, the FoV will narrow as you focus closer because the effective focal length gets longer. However, a lens like say the EF 24-70mm f/2.8L II USM, will do the inverse, it gets wider as you focus closer, and method 1 will be less wrong than method 2 (but will still be wrong).

    For your 100mm macro, if it behaves anything like my old Sigma 150mm f/2.8 macro, it’ll get wider as you focus to MFD, probably quite a bit, and I have no way to calculate that accurately. If you really need to know the AoV, the most accurate thing you can do is determine it empirically.

    Hope that helps.

Iqbal Khuraishi

Can we have this as an android app?

    Jason Franke  | admin

    Sorry, but I’m currently not in a position to develop and support apps for Android (or iOS for that matter).

Naif Alsalem

Hi Jason,
Thank you for making this amazing, informative page in your website. As you know, different researchers use different formulas to calculate their iFOV and FOV based on their detector’s details. I came here to verify my iFOV and FOV calculations, and they are extremely close to the output of your calculator here.

Thank you again and wish you a safe 2021.

Best Regards,

    Jason Franke  | admin

    No problem, it’s been added.

    Michl

    :D Thank you very much!!!

Zdeněk

Hello,
I’m a student from Europe and this calculator helped me with writing my paper. I wanted to thank you and at the same time ask you if you have any problmes with me using data gained here (only results from calculator) in it. I would obviously note that I used this site and add link to it.

Once again, thank you.

    Jason Franke  | admin

    Hi Zdeněk,

    I have no problem with you using the results, including the generated generated graphs, in an academic paper. Best of luck in your studies.

Murphy

Does the distance in this calculator refer to distance from image sensor, distance from the end of the lens, or some other distance?

    Jason Franke  | admin

    It’s the distance from the image sensor.

Steven Gutierrez

I use this tool frequently when planning shots so thank you very much for making it, but I’ve run into a couple problems now that I’m doing a lot of macro work.

I recently tested a 90mm fujinon swd on a camera with a bellows extension of 21” by pointing it at a screen of known pixel density. I counted the pixels per inch and calculated magnification to be 8.9:1.

This didn’t match the calculations I had done with M=di/f, which gave 5.9:1. That’s weird, so I tried to calculate the do so I could plug it into your calculator and using 1/f=/do+1/di I got 10.8 cm. That’s weird, the distance is way too big.

I plug it into your calculator and I get weird values. The fixed AoV gives 0.6” wide which is weirdly kinda close but still wrong and it’s the wrong way to get the answer, and the formula that shifts AoV with focus, like my camera, gives negative values.

So I tried to estimate an approximate value for the correct do and do the math to get my bellows draw and check my sanity. But when I do I get negative bellows draw from 1/f=1/di+1/do for any sane input.

And when I plug sane inputs into your calculator I get totally different weird and wrong answers with negative areas of view and stuff.

So now I’m questioning the basic formulas.

Can you help me understand what’s going wrong here?

    Jason Franke  | admin

    Most DoF and AoV calculations use simplifying assumptions that the differences between F and d (i) can be ignored, that d (o) is much larger than F and d (i), and that the lens can be approximated by the thin-lens approximation (d (o) and d (i) are measured from a coincident point). When you get to macro level operations, most of those assumptions fail and the results with them.

    For example, my angle of view calculation ignores the physical lens’s dimensions entirely. It simply uses the focal length and frame dimension (width, height or diagonal) to create a triangle to calculate the AoV. In a real lens, especially at macro distances, the lens’s physical size will change some of those dimensions, which will affect the results.

    In practice, you’d have to account for di and do not being coincident, as well as the actual focal length not being exactly equal to the stated focal length. And that’s before you get the possibility for distortion and spherical aberrations affecting the AoV too.

    Unfortunately, I don’t expect accurate results at macro scales from my calculator due to those limitations (and the inherent limitations baked into things like the DoF equations).

    Hope that helps some.

    Steven Gutierrez

    That makes perfect sense. I think with that information and a bit of research I can try to come up with a formula I can use for these higher magnification situations that will work a little better, or at least not give impossible outputs.

    Thank you, and thank you again for your work. Truly essential to the community.

Nick

Hey Jason,

Thank you so much for this page. I’m working on a so-called “Perkiscope” or “Digital Obscura” camera like the ones featured on DIY Perks and Media Division.

Would it be possible for you add the ability (or post an alternative version of this script with the ability) to see more significant figures? I’d like to see the f-stop down to the hundredths if possible. When working with these very very small effective f-numbers it would be a big help.

Thanks again!

    Jason Franke  | admin

    You can specify any arbitrary f-stop by clicking the “Custom AV” button (next to the f/number drop down) and typing in whatever f-number you need to whatever precision that you want.

    Nick

    Sorry, I wasn’t clear. I’d like to see f-stops calculated down to 0.01 in the “Equivalent Lenses” section. Is that something you could maybe add in the Advanced Settings?

    Nick

    Or alternatively, f-stops below f1.0 could automatically be displayed to the hundredths place, and f-stops below 0.20 could be automatically displayed to the thousandths place.

    Just as you have it automatically add the tenths decimal place below f8. Same concept, just two more splits at 1.0 and 0.2, respectively

    I know this is a rare use case, but I would really appreciate it.

    Jason Franke  | admin

    That’s easy enough to do, so it’s done. That said, I’m not 100% certain I trust the accuracy of the values for apertures less than f/0.20. But it should now display equivalent lenses with a precision of 1 when the f number is >8, 0.1 between f/8 and f/1.2, 0.01 between f/1.2 and f/0.2, and 0.001 for apertures smaller than f/0.2.

    Nick

    THANK YOU! This is very helpful.
    There’s some minor disagreement between your calculator and that found at mmCalc.com – but small enough that I don’t really care. I just want to be able to have numbers down to those sig figs made by the same calculator. Yours previously didn’t go low enough (but mmCalc did), and mmCalc didn’t offer a sensor size as high as I wanted (yours does). Now your calculator does everything I need. So THANK YOU again!

    Nick

    Actually, I may have been wrong about the “minor disagreement” – just depends on which precise crop factor near 1.0 you’re reading on the “Equivalent Lens” chart. I am just reading the 1.00 line, where mmCalc is probably using closer to 1.05.

    Jason Franke  | admin

    In looking at some results v. mmCalc, I think the differences may come down to subtle differences in frame sizes. For example, mmCalc has 4×5 listed as 102 x 127 mm, while I have it as 101.6 x 127 mm. 101.6 mm is exactly 4 inches in mm (4 * 25.4). I have to wonder if even though mmCalc shows 2 decimal points of precision, if they’re not rounding the dimensions for the larger formats. If I change my script to use their frame dimensions I get something closer (6.640106241699867mm and f/0.27).

    Also something to keep in mind, when push comes to shove there’s so much slop in lens designs or at least their markings, that it’s not really useful to calculate things to 2 or 3 decimal points. In a very practical sense, most lenses aren’t exactly the focal length or aperture that’s listed on the barrel. Getting close for comparative purposes is about the best you can hope for, but there’s not a lot of practical value in trying to calculate any of this stuff to more than a decimal place or so.

    Steven Gutierrez

    I just wanted to mention that the imaging area of a 4×5 piece of film is not actually 4”x5”. The film itself is actually slightly smaller than that in order to allow it to fit in the film holder but more importantly the film holder covers a significant portion of the edge of the piece of film. I just measured the image area of a piece of film that I shot and the actual dimensions of the image are 97mm x 120mm. I assume this is true for all standardized sheet film sizes though I only shoot 4×5 so I can’t measure the larger formats. Hope that helps. Thanks for all your hard work.

    Jason Franke  | admin

    Thanks for the heads up. I’ve updated the size in the calculator. You may have to hard refresh (ctrl+refresh) to clear the old cached copy to see the new results.

    That said, from a DoF perspective, I’m not sure that the slight difference matters that much in practice (though it does account for a few degrees in the AoV calculations).

Luca Vanzella

Thanks for the great calculator! I used the calculator to determine the Hor AoV for a Canon EF 8-15mm f/4L USM lens at 10mm FL with an APS-C sensor. The calculated values for Hor AoV are 94.8° and 96°. But when I measured the angular width of an image that I took with the lens at 10mm FL, using horizon landmarks of known azimuths, I get a Hor AoV of about 129° – quite a bit wider! I think I am missing something. (I use https://app.photoephemeris.com/ to determine azimuths of horizon landmarks.)

    Jason Franke  | admin

    So the EF 8-15mm f/4L USM is a fisheye lens, which means its design is not corrected to create a rectilinear projection on the sensor. Unfortunately, that’s kind of a fundamental assumption in the AoV calculations, and they don’t take into account the distortion that is present in a fisheye lens. As a result, the calculations are going to be narrower than that type of lens actually produces. If you compare the full frame AoV’s for the 8-15 and say the 15-35 (e.g. 142° horizontal for the 8-15 @ 15, versus 100°25′ for the 15-35 @ 15) you can see the discrepancy is on order of what you see in your tests.

    I should probably have called this out in the docs for the tool somewhere, but fisheye lenses tend to be a specialized/niche enough thing that I didn’t think about it (or that it would be necessary).

    Hope that helps, and sorry for the confusion.

    Luca Vanzella

    Thanks Jason. I came across Bob Atkin’s site where he discusses methods to calculate AoV for fisheye lenses. For the most common equisolid and equidistance fisheyes, the formulas yield values very close to what I measured.

    http://www.bobatkins.com/photography/technical/field_of_view.html

16mm Film to DVD

Your dedication to creating this tool shines through. The ability to customize sensor sizes beyond standard options sets this apart. I appreciate the clear instructions for inputting custom frame dimensions and apertures. The circle of confusion explanations, especially the preset methods, demystify technicalities. The depth of field graph’s limitations make sense for practical use. However, visual examples could aid those new to these concepts. Overall, a robust tool for photography enthusiasts. Keep up the innovative work!

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