## Abstract

In order to use high-speed calculating machines in ray tracing, it is advisable to replace trigonometrical formulas by formulas involving the solution of quadratic equations. It is shown that the tracing of skew as well as of meridian rays is accomplished by solving two quadratic equations and an iterative process (based on Newton’s method) is suggested for solving the equations speedily by machine calculation. A numerical example is given.

© 1951 Optical Society of America

### References

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1. T. Smith, Proc. Phys. Soc. (London) 27, 502 (1915); Proc. Phys. Soc. (London) 33, 174 (1921).(Note the last sentence.)
[CrossRef]

#### 1915 (1)

T. Smith, Proc. Phys. Soc. (London) 27, 502 (1915); Proc. Phys. Soc. (London) 33, 174 (1921).(Note the last sentence.)
[CrossRef]

#### Smith, T.

T. Smith, Proc. Phys. Soc. (London) 27, 502 (1915); Proc. Phys. Soc. (London) 33, 174 (1921).(Note the last sentence.)
[CrossRef]

#### Proc. Phys. Soc. (London) (1)

T. Smith, Proc. Phys. Soc. (London) 27, 502 (1915); Proc. Phys. Soc. (London) 33, 174 (1921).(Note the last sentence.)
[CrossRef]

### Cited By

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### Figures (1)

Fig. 1

Path of a ray traced through a system consisting of a sphere and a plane.

### Equations (24)

$x 2 = a x + b ,$
$x = a + b / x ,$
$x ′ = a + b / ( a + b a + ) ⋯ ,$
$- x ″ = b / [ a + ( b / a + b a + ⋯ ) ]$
$( x 1 + Δ ) 2 = a ( x 1 + Δ ) + b .$
$Δ = ( x 1 2 - a x 1 - b ) / ( a - 2 x 1 ) .$
$x 2 = x 1 + Δ = ( x 1 2 + b ) / ( 2 x 1 - a ) .$
$x 1 = - b / a .$
$∣ 4 b / a 2 ∣ < 1 ,$
$- x 1 = a b / ( a 2 + b ) .$
$ξ 2 + η 2 + ζ 2 = μ 2 , ξ ′ 2 + η ′ 2 + ζ ′ 2 = μ ′ 2$
$b = a + λ s ,$
$λ 2 = 2 ( r ζ - x ξ - y η ) μ 2 λ - x 2 + y 2 μ 2 .$
$X = x + λ ξ , Y = y + λ η , Z = z + λ ζ .$
$s ′ - s = γ [ k - ( b / r ) ] ,$
$γ 2 = 2 γ [ 1 / r ( ξ X + η Y + ζ Z ) - ζ ] + μ ′ 2 - μ 2 ,$
$ξ ′ = ξ - γ ( X / r ) , η ′ = η - γ ( Y / r ) , ζ ′ = ζ + γ [ 1 - ( Z / r ) ] .$
$λ 2 = 2 ( r ζ + x ξ ¯ + y η ¯ ) μ 2 ( λ ) - x 2 + y 2 μ 2 , X = x - λ ξ ¯ , Y = y - λ η ¯ , Z = λ ζ .$
$γ 2 = 2 γ [ 1 / r ( ζ Z - ξ ¯ X - η ¯ Y ) - ζ ] + μ ′ 2 - μ 2 ,$
$ξ ¯ ′ = ξ ¯ + γ ( X / r ) , η ¯ ′ = η ¯ + γ ( Y / r ) , ζ ′ = ζ + γ [ 1 - ( Z / r ) ] .$
$x ′ = X - ( d - z ) ξ ¯ ′ / ζ ′ , y ′ = Y - ( d - z ) η ¯ ′ / ζ ′ .$
$X = x γ 2 = - 2 γ ζ + μ ′ 2 - μ 2 Y = y ξ ′ = ξ Z = z η ′ = η ζ ′ = ζ + γ .$
$Given r = 14.30 μ = 1.0 μ ′ = 1.73400 y = 15.3239 - η = 0.608761 + x = 2.8912 + ξ = 0.0 ζ = 0.793353 + a = 0.413468 + b = 0.024318 - λ 0 = - b / a = 0.058815 + Y = 0.110009 - λ 1 = 0.070508 + X = 0.028912 + λ 2 = 0.071010 + = λ 3 Z = 0.056336 + a 2 = 1.898256 - b 2 = 2.006756 + γ 0 = - b 2 / a 2 = 1.057157 + η ′ = 1.190378 - γ 1 = 0.778637 + ξ ′ = 0.152858 + γ 2 = 0.756188 + ζ ′ = 1.251545 + γ 3 = 0.756040 +$
$d - z = 0.0475 - 0.056336 = 0.008836 - y 2 = 0.118413 - x 2 = 0.029991 +$