Sometimes, they ask me what distance will be to the subject, if you photograph on a particular lens. In this article, I derived a simple calculation formula.

For calculations, I used a full-format camera with a physical sensor size of 36 X 24 mm.

I recommend reading the text under the images.

The viewing angle can be found in brochures, instructions or on the official websites of the lens manufacturer. But there is one small nuance, which for some reason few people take into account - the angle of view of the lens is indicated for the diagonal of the frame.

I work as a photographer and do not shoot “diagonal shots” at all (to take a shot with a diagonal fill of the frame), and therefore this data gives me only an approximate idea of the angle of view when shooting in normal portrait (vertical camera orientation) or landscape (horizontal camera orientation) mode ...

**Output:** the physical size of the matrix w * h and the focal length of the lens f.

**Search for:** formula for calculating the viewing angle diagonally, vertically, horizontally. Check the found Beta angle for f = 50mm.

Thus, the data taken from the official website (47 °) and verification (46,79 °) are the same.

Now let's find the angle of view horizontally (Xi) and vertically (Tau):

It turns out that if we shoot a portrait at a 50 mm focal length (vertical position of the camera), then the viewing angle at which we will need to enter the model will be only 40 degrees.

Now find **formula for calculating the distance L**, with which we will need to shoot so that an object with the given dimensions fits in the frame **H**.

Thus, if we shoot a model with a height of 180 cm on a full-frame camera with a lens that has a 50 mm focal length, then in order for the heels and the crown of the head to get into the frame with the vertical orientation of the camera, we will need to move back 2.5 meters, horizontally, to fit the entire model into the frame, you will need to step back 3.75 meters.

To be more precise, 5 cm of the focal length (or any other number of the focal length) from the focus plane to the plane of the matrix should be added to these numbers, because the distance is calculated from the object to the focal plane. And you also need to take into account the effect of changing the viewing angle of the lens at different focusing distances, because the same fifty dollars has the declared 47 ° only when focusing on infinity, more about this here.

If we shoot the same model for the same fifty dollars with the horizontal orientation of the camera, but already on the Nikon DX camera (Kf = 1.5), then we will need to move 5,6 meters. And if you take into account that in addition to the model itself, you still need to capture a bit of space from below and from above, then by fifty dollars it will be necessary to retreat by 7 meters.

To use the count for cropped cameras, use the formulas to specify the width w and height h for your camera. For Nikon DX cameras: w = 23.5 mm, h = 15.6 mm. The focal length f should be taken as indicated on the lens without any conversion. Basic formulas are highlighted. If you cannot find the value of w and h in the instruction, then usually w = 36 / Kf, h = 24 / Kf, where Kf is the value crop factor cameras.

It is very simple to find out the focusing distance to the subject from the already taken photo. To do this, just check EXIF photo using http://regex.info/exif.cgi (The site supports any photo format)

*Thank you for attention. Arkady Shapoval.*

Arkady thanks for the useful formulas!

I saw with them 0_o !!

I just wanted to know which blurred background is 200mm F / 2.8 or 135mm F / 2 for a full-length portrait. Applying your formulas, I realized that blurring the background (depth of field) is not related to the focal length))

And then everything fell into place F hole is responsible for the depth of field and it is the same for any focal length. Those. and 28mm F2 and 135m F2 have the same depth of field relative to one object (naturally, with 28mm it will be necessary to approach the object much closer)

And the focal length only affects the perspective (the degree of compression of space).

By the way, having the vertical and horizontal viewing angles of the lens, we can multiply these values by 17 and get the viewing angles in “thousandths” (1 degree = 17 thousandths). For example, for fifty dollars on FF horizontally 39.59 degreesX17 = 673 thousandths, vertical 26.99X17 = 459 thousandths. Thus, we can determine the width of the field of view horizontally (673m / 1000m, 67.3m / 100m, 6.73m / 10m, 3.36m / 5m, 1.7m / 2.5m, 0.85m / 1.25m 0.67m / 1m, etc.). ) and vertically (459m / 1000m, 45.9m / 100m, 4.59m / 10m, 2.3m / 5m, 1.15m / 2.5m, 0.58m / 1.25m 0.46m / 1m, etc.)

http://www.vision-doctor.co.uk/optical-calculations/calculation-object-size.html

And where to find what would be clear? what is 2 * arctg, etc. for those who are zero in mathematics like me. A clear example in numbers with clear words is what to divide and multiply by.

This is the arctangent.

2 * arctg is "2 times the arctangent"

The same thing turns out for the Canon 600D for 50 mm it’s 4000/6000! And for 135 mm it’s 97200/16200! Something too much.

Or am I counting incorrectly?

Hello. At your leisure, according to your technique, I calculated viewing angles at standard focal lengths, from 8mm to 200mm, for FX DX matrices. The goal was to calculate the number of frames (clicks on the rotator of the panoramic head) for each focal with a 360 degree coverage. To quickly shoot panoramas later, you already know how many clicks there are on the head. Counting overdue personnel 25%. And with great surprise I realized that the angles / clicks on the rotary heads partially do not fall on the required angles. On wide-angle lenses, the head values fall into the required with minor errors, and with focal lengths of about 50 mm, a very strong deviation. For example, the matrix FX vertical orientation of the frame F 20mm, according to the calculation of 7,8 frames per head 8, we can say the perfect hit. If you take the focal length of 200mm, according to calculations, you need 70 frames, there are 72 or 36 on the head. There are even more deviations on DX matrices. The question is, why are such angles chosen on rotators?

It was nevertheless necessary to indicate for those who know little mathematics that arctg is the arctangent. Despite the suitability of this formula, unfortunately, it is not suitable for calculating fish. I wanted to calculate Sigma 15mm, which on the diagonal should be 180 degrees - the test formula for determining the degrees on the diagonal was in the region of 110, which does not correspond to reality. Maybe there is not 180, but clearly more than 110 per eye. But thanks for the work anyway.

Duc, for fish and will not do. But for the wide-angle (i.e., with corrected distortion) just around 110 and will be

Surprised statements about so unaware of mathematics that they do not understand these formulas. The same geometry and trigonometry in the volume of high school.

Many thanks to the author. The work is very useful.

“Not knowing” with a dependent word (not knowing mathematics) is written separately. This is the Russian language at the level of secondary school. Unlike arctangents, it is used a little more often, sometimes for combat purposes.

Arkady, thanks for the formulas! I would only like to add that the equation for determining the distance to the object L is easily derived without any trigonometry, for this, 2 formulas are used, the thin lens formula and the lens magnification formula. In the final version, to determine the distance to the object, we only need to know 3 values, the focal length of the lens in mm, the height of the object in mm, the height of the matrix in mm. Actually, the formula is very similar to yours, but with a slight addition. Calculations according to it show that in your first example you got 2500mm, and according to this formula you get 2550mm, in the second example you got 3750mm, and according to this formula you get 3800mm. Of course, a difference of 1-2% is not critical for a rough estimate.