CHAPTER XX.
USEFUL FACTS AND FORMULAS.
One turn of the crank runs off 1 foot of film.
One foot of film contains 16 pictures.
The audience sees about 960 different pictures in every minute that a film is being run.
The relative number of revolutions made by two pulleys connected together by belting or friction is in proportion to their diameters.
The relative number of revolutions made by two gears connected together is in proportion to their number of teeth.
A theater seat occupies from 4 to 5 square feet.
One 25-watt tungsten lamp or its equivalent provided for every 20 seats will give fairly good illumination in a small theater.
3 to 5 watts per cubic foot will be required to heat small spaces by electricity.
_Ohm’s Law._
Direct current equals e.m.f. divided by resistance.
Alternating current equals e.m.f. divided by impedance.
Electromotive force, d.c., equals current time resistance.
Electromotive force, a.c., equals current time impedance.
Resistance equals e.m.f. divided by current.
Impedance equals e.m.f. divided by current.
The joint resistance of two conductors connected in parallel is equal to the product of their resistances divided by their sum:
_r_ = (_r_₁ × _r_₂) ÷ (_r_₁ + _r_₂})
The joint resistance of any number of resistances connected in parallel is the reciprocal of the sum of the reciprocals. The reciprocal of a number is 1 divided by that number:
1 _r_ = -------------------------- 1 1 1 ---- + ---- + ---- + ..... _r_₁ _r_₂ _r_₃
The total resistance of a number of resistances in series is equal to the sum of all of them:
_r_ = _r_₁ + _r_₂ + _r_₃ + .....
The heating of a rheostat is proportional to the square of the current it carries.
Drop in voltage is proportional to the product of the current and resistance for a direct-current circuit, and the product of current and impedance for an alternating-current circuit.
If a reactance is used in place of a rheostat in an alternating-current circuit, the loss of energy is greatly reduced.
If the same intensity of illumination is to be provided for pictures of different sizes, the amperage must vary about as:
(_W_ × 1.2)²
where _W_ is the width of picture on screen.
For a given power the current in conductors is:
_W_ for direct current _I_ = --- _E_
_W_ for single-phase a.c. _I_ = ------------- _E_ × _p. f._
_W_ for two-phase a.c. _I_ = .5 × ------------- _E_ × _p. f._
_W_ for three-phase a.c. _I_ = .58 × ------------- _E_ × _p. f._
in which _I_ is the current in amperes, W is the watts, E is the voltage, and p. f. is the power factor.
_Lens Formulas._
To find the size of a picture obtainable under given conditions and lens: _Multiply distance from center of lens to screen by one dimension of slide or film and divide by e.f. focal length of lens, taking all measurements in inches._
Example: Slide 2-3/4 inches. Length of throw 360 inches. e. f. 10 inches
2-3/4 × 360 ÷ 10 = 99 inches.
To find focal length needed for a given slide or film to produce a given size of picture: _Multiply slide or film dimension by length of throw and divide by dimension of picture, taking all measurements in inches._
Example: Same dimensions as above:
2-3/4 × 360 ÷ 99 = 10 inches.
To find length of throw needed to obtain a certain size of picture: _Multiply required picture dimension by focal length of lens and divide by slide or film dimension._
Example: Same dimensions as above:
99 × 10 ÷ 2-3/4 = 360 inches.
To find slide size necessary to produce a certain size of picture under fixed conditions: _Multiply focal length of lens by size of picture and divide by length of throw._
Example: Same dimensions as above:
10 × 99 ÷ 360 = 2-3/4 inches.
_Measurement of Surfaces._
To find the area of a parallelogram: _Multiply the base by the altitude._
To find the area of a triangle when base and altitude are given: _Multiply the base by the altitude and take half the product._
To find the area of any angular surface: _Divide it into triangles and find the area of the different triangles and add them together._
To find the circumference of a circle: _Multiply the diameter by π; or 3.1416._
To find the diameter of a circle when the circumference is given: _Divide the circumference by π, or 3.1416._
To find the area of a circle when the radius is given: _Multiply the square of the radius by 3.1416._ When diameter is given: _Multiply the square of the diameter by .7854._
To find the radius of a circle when the area is given: _Divide the area by 3.1416 and extract the square root of the quotient._
_Measurement of Solids._
To find the lateral area of a right prism: _Multiply the perimeter of the base by the altitude._
To find the lateral area of a right cylinder: _Multiply the circumference of the base by the altitude._
To find the volume of a cylinder or prism: _Multiply the area of the base by the altitude._
To find the lateral area of a right pyramid: _Multiply the perimeter of the base by the slant height and take half the product._
To find the lateral area of a cone: _Multiply the circumference of the base by the slant height and take half the product._
To find the volume of a pyramid or cone: _Multiply the area of the base by the altitude and take one-third of the product._
To find the surface of a sphere: _Multiply the square of the diameter by 3.1416._
To find the volume of a sphere: _Multiply the cube of the diameter by one-sixth of 3.1416 or .5236._
In a right-angled triangle, the sum of the squares of the base and the perpendicular equals the square of the hypotenuse. To find the length of the hypotenuse: _Extract the square root of the sum of the squares of the base and the perpendicular._
To find the base or the perpendicular: _From the square of the hypotenuse subtract the square of the other given side and extract the square root of the remainder._