العوامل المؤثرة في المسافة فوق البؤرية ٢-a .. كتاب آلة التصوير

Notes on the supfocal distance calculation table shown on page 201:

First: This table was calculated according to the following formula: suprafocal distance = diameter of the actual diaphragm aperture x 1000

Second: This table is valid for calculating the superfocal distance for any lens in any camera, regardless of its size, but on condition that the following rules are observed:

(a) Since this table has been set on the assumption that the diameter of the permissible mixing circle is 1/1000 of the focal length of the medium lens, so if it is deplorable (for any reason, such as for the sake of accuracy, for example), we consider the diameter of the permissible mixing circle smaller than that (for example 1/1500 of the focal length of the lens) then multiply the hyperfocal distance by 1/2 1, or if he wants to consider the diameter of the mixing circle as 1/2000 of the focal length, for example, then multiply the supfocal distance by 2. We also find that increasing the diameter of the allowed mixing circle leads to the necessity of multiplying the hyperfocal distance x by the amount of this increase.

For example, if the diameter of the allowed mixing circle = 1/500 of the focal length, we must multiply the hyperfocal distance by 1/2, and so on.

(b) Developing this table considering that the lens has an average focal length, which is the case in which the focal length is equal to the chord that connects two opposite corners of the image area (refer to page 102, Figure 69, regarding what came about the law of the chord).

If the lens has a long focal length, or a telephoto one, or it has a short focal length, or an obtuse angle, then we must multiply the suprafocal distance (which we take from this table) by another factor, which is the lens’s magnification power.
It is always equal to the actual focal length of the lens over the suitable medium focal length of the camera.

For example, if the lens has a long focal length, so that the power of its magnification of the image increases by 2/3, or the lens is telephoto and its power
(× 1,5) also and used which one on the same camera, then multiply the hyperfocal distance by 1, 1/2 as well. If the lens has a telephoto power (x 2) or it has a long focal length and its magnification of the image of the object is twice as large as if the lens has a medium focal length, then the focal distance is multiplied by 2 as well.

Thus, we see that decreasing the diameter of the mixing circle or increasing the magnification leads to the necessity of increasing the suprafocal distance and multiplying it by a factor equal to the amount of decrease in the diameter of the permissible circle or equal to the amount of increase in magnification that the lens leads to if it is of medium focal length.

Examples of how suprafocal distance (e) is calculated:

Example 1: A telephoto lens with a focal length of 4 inches on a 35 millimeter camera, what is the superfocal distance if the focal number = 2 f?

the answer :

The suprafocal distance can be determined by any of the following three methods:
(a) Calculating the suprafocal distance by means of Equation No. 3 (p. 199).

... The average focal length of a 35 mm camera lens = 2 inches

... the magnification power of a lens whose focal length is 4 inches = 2/4 = 2 times

... e = 1000 x 4/f x v . . . (From Equation No. 3).

...e = 1000 x 2/4 x 2 = 4000 in

(b) Solve the previous example using Equation No. 7, p. 200

Mixing circle diameter = focal length averaged over 1000 = 2/1000 = 1/500 of an inch.

... e = h2 over f x f = 4 x 4 over 2 x 1/500 = 4 x 4 x 500 over 2 = 4000 inches.


(c) Solve the previous example by suprafocal distance table p. 201:

With reference to table p.

However, since a lens with a focal length of 4 inches is considered a long focal length for a camera whose size is 35 millimeters, so the above focal distance must be multiplied by the lens magnification power. The result will be as follows:

Magnification power = 4/2 = 2 times.

... supfocal distance = 2000 x 2 = 4000 inches.

Example 2:
Assuming that the size of the camera was not 35 mm as in the previous example, but was only 16, and with the focal length of the lens remaining constant (i.e. 4 inches) and with its focal number also constant (i.e. 2 f), what is the superfocal distance?

the answer :

The focal distance can be known in any of three immediate ways:

(A) Calculation of the suprafocal distance by Equation No. 3
(p. 199):

... The average focal length of a 16 mm camera = 1 inch.

... the magnification power of a lens whose focal length is 4 inches = 1/4 = 4 times.

... e = 1000 x p/v x t . . . (Equation No. 3).

...E = 1000 x 2/4 x 4 = 8,000 in.


(b) Calculation of the suprafocal distance by Equation No. 7 (p. 200)

... x (i.e. the diameter of the mixing circle) = the focal length averaged over 1000

... x = 1 1000th of an inch.


... e = h2 over f x f = 4 x 4 over 2 x 1/ 1000 = 16 x 1000 over 2 = 8000 inches.

(C) Calculation of the suprafocal distance by means of the special table (pg. 201)

Referring to table p. 201, we find that if the lens (medium focal length) has a focal length of 4 inches, and its focal number is 2 f, then its focal distance is 50 meters = 166 3/2 feet (i.e. 2000 inches).

However, since a lens with a focal length of 4 inches is considered a long focal length for a camera whose size is 35 millimeters, so the above focal distance must be multiplied by the lens magnification power. The result will be as follows:

Magnification power = 4/2 = 2 times.

... supfocal distance = 2000 x 2 = 4000 inches.

Example 2:
Assuming that the size of the camera was not 35 mm as in the previous example, but was only 16, and with the focal length of the lens remaining constant (i.e. 4 inches) and with its focal number also constant (i.e. 2 f), what is the superfocal distance?

the answer :

The focal distance can be known in any of three immediate ways:

(A) Calculation of the suprafocal distance by Equation No. 3
(p. 199):

... The average focal length of a 16 mm camera = 1 inch.

... the magnification power of a lens whose focal length is 4 inches = 1/4 = 4 times.

... e = 1000 x p/v x t . . . (Equation No. 3).

...E = 1000 x 2/4 x 4 = 8,000 in.


(b) Calculation of the suprafocal distance by Equation No. 7 (p. 200)

... x (i.e. the diameter of the mixing circle) = the focal length averaged over 1000

... x = 1 1000th of an inch.


... e = h2 over f x f = 4 x 4 over 2 x 1/ 1000 = 16 x 1000 over 2 = 8000 inches.

(C) Calculation of the suprafocal distance by means of the special table (pg. 201)

Referring to table p. 201, we find that if the lens (medium focal length) has a focal length of 4 inches, and its focal number is 2 f, then its focal distance is 50 meters = 166 3/2 feet (i.e. 2000 inches).

However, since a lens whose focal length is 4 inches is considered a telephoto lens for a 16 mm camera, then the previous result must be multiplied by the magnification power.

... the magnification power = 1/4 = 4 times.

... supfocal distance = 2000 x 4 = 8000 inches.