answers: 54

  1. Artyom
    02.06.2015

    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).

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  2. Serhii
    31.08.2015

    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.)

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  3. Georg
    08.12.2015
  4. Andrei
    21.07.2017

    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.

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  5. anonym
    10.09.2017

    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.

    Reply

  6. Alexey
    10.09.2017

    Or am I counting incorrectly?

    Reply

  7. Alexander
    14.12.2017

    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?

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  8. Fedor
    06.03.2019

    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.

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    • anonym
      06.03.2019

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

      Reply

  9. Andrei
    04.04.2019

    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.

    Reply

  10. Vlad
    20.09.2020

    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.

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