## Abstract

The general theory of optically compensated varifocal systems is applied to the case of a four-component system consisting of four alternate stationary and movable components. The second and fourth components, counting from the object side, are interconnected and displaced in unison to change the over-all focal length of the system. An iteration method for the solution of the varifocal equations is developed which enables the determination of the Gaussian parameters of the system for any predetermined focal range in a matter of minutes (without the use of computers). Using this method, explicit expressions are found for the approximate values of the parameters of the optimum four-component systems as functions of the focal range. A numerical example is given to illustrate the iteration method.

© 1962 Optical Society of America

### References

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1. L. Bergstein and L. Motz, J. Opt. Soc. Am. 52, 353 (1962).
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2. L. Bergstein and L. Motz, J. Opt. Soc. Am. 52, 363 (1962).
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3. L. Bergstein, J. Opt. Soc. Am. 48, 154 (1958).
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4. The “relative focal range” ℛ is defined as the ratio of maximum to minimum over-all focal lengths of the system, i.e.,R=FmaxFmin=fmaxfmin.For any required focal range ℛ two lens types are possible. Either a lens system can be chosen which will have the maximum focal length when the movable components are in the extreme front position or a system can be chosen which will have its maximum focal length when the movable components are in the rear position. We refer to the first system as the P system and the latter as the N system. We define the “focal ratio”r≜F(0)F(1)=f(0)f(1).For the P system r=ℛ, whereas for the N system r=1/ℛ. We also introduce the normalized ranger≜f(0)-f(1)f(0)+f(1)=r-1r+1=±R-1R+1,where the positive sign applies to the P system, the negative sign to the N system. We note that 1<ℛ<∞, 0<r<∞, and −1<τ<+1.
5. In reference 3 it is shown that under optimum conditions the maximum value of the image-plane deviation of a four-component system with full compensation at both ends of the operating range is approximately 43 times the maximum value of the image-plane deviation obtained when full compensation at both ends of the operating range is not required. The difference in the level of the two maximum values is thus small.
6. The choice between the two optimum systems is governed by a number of considerations [L. Bergstein, J. Opt. Soc. Am. 48, 154 (1958)]. (The PNP system is generally preferable.)
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#### Other (2)

The “relative focal range” ℛ is defined as the ratio of maximum to minimum over-all focal lengths of the system, i.e.,R=FmaxFmin=fmaxfmin.For any required focal range ℛ two lens types are possible. Either a lens system can be chosen which will have the maximum focal length when the movable components are in the extreme front position or a system can be chosen which will have its maximum focal length when the movable components are in the rear position. We refer to the first system as the P system and the latter as the N system. We define the “focal ratio”r≜F(0)F(1)=f(0)f(1).For the P system r=ℛ, whereas for the N system r=1/ℛ. We also introduce the normalized ranger≜f(0)-f(1)f(0)+f(1)=r-1r+1=±R-1R+1,where the positive sign applies to the P system, the negative sign to the N system. We note that 1<ℛ<∞, 0<r<∞, and −1<τ<+1.

In reference 3 it is shown that under optimum conditions the maximum value of the image-plane deviation of a four-component system with full compensation at both ends of the operating range is approximately 43 times the maximum value of the image-plane deviation obtained when full compensation at both ends of the operating range is not required. The difference in the level of the two maximum values is thus small.

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### Figures (10)

Fig. 1

The four-component optically compensated varifocal system.

Fig. 2

The four-component optically compensated varifocal system and its Gaussian parameters, its over-all focal length f(z) and image-plane deviation y(z) as functions of the normalized displacement.

Fig. 3

The optimum distribution of the points (z1,z2,z3,z4) of full compensation of a four-component varifocal system with full compensation at both ends of the operating range as a function of the focal range τ. z1=0, z2=0.293−, z3=0.707−, z4=1.0.

Fig. 4

The value of (β3)y4(max) as a function of the focal range τ; y4(max) is the maximum value of the normalized image-plane deviation of a four-component varifocal system with full compensation at both ends of the operating range and an optimized image deviation function, β3≡(r−1)b3=[2τ/(1−τ)]b3 and b3 is a Gaussian bracket of the system. The image-plane deviation y(z) as a function of the displacement z of the movable component is shown above the graph of β3y4(max).

Fig. 5

Optimum four-component varifocal systems. (Pa) and (Pb) show the two optimum PNP systems, and (N) shows the optimum NPN systems. Also shown is the focal range of these systems.

Fig. 6

The normalized final image distance (l)01≡[l′(0)]01 and the values of the parameter (β3)01≡[(r−1)b3]01 of the optimum four-component varifocal systems as functions of the focal range τ. (l)01 and (β3)01 are the values of ll1′(0) and β3≡(r−1)b3 for systems in which in the position z=0 the distances between the principal planes of the second and first, and the third and second components are s21=and s32=1.0, respectively.

Fig. 7

The normalized focal lengths (f1)01, (f2)01, (f3)01, and (f4)010 of the components and the normalized maximum over-all focal length (fmax)010 of the optimum four-component varifocal systems as functions of the focal range τ. The index 01 indicates that the particular parameter was found under the assumption that s21=0, and s31=1.0, the index 010, that the parameter in question was found under the assumption that s21=0, s32=1.0 and s43=0; s21, s32, and s43 are the distances between the principal planes of the components 2 and 1, 3 and 2, and 4 and 3, respectively, in the position z=0 of the movable components.

Fig. 8

The maximum value y4(max)01 of the normalized image-plane deviation of the optimum four-component varifocal systems with full compensation at both ends of the operating range (and an optimized image deviation function) as a function of the focal range τ. The distances between the principal planes of the components in the position z=0 are assumed to be s21=0 and s32=1.0, respectively.

Fig. 9

The maximum value YT(max)010 of the image-plane deviation of optimum four-component varifocal systems with full compensation at both ends of the operating range (and an optimized image deviation function) as a function of the focal range τ. The distances between the principal planes of the components in the position z=0 are assumed to be s21=0, s32=1.0, and s43=0, respectively. FT(max) is the maximum over-all focal length of the system and Zm is the maximum displacement of the movable components.

Fig. 10

The two optimum four-component systems of the example of Sec. III, their image-plane deviation and focal range.

### Equations (116)

$f k ≜ F k Z m , d k , k - 1 ≜ D k , k - 1 Z m , s k , k - 1 ≜ S k , k - 1 Z m ,$
$z ≜ Z / Z m .$
$ψ 1 ( 4 , 1 ) ≡ b 1 = d 43 - d 32 + d 21 , ψ 2 ( 4 , 1 ) ≡ b 2 = - d 43 d 32 + d 43 d 21 - d 32 d 21 + f 3 2 + f 2 2 , ψ 3 ( 4 , 1 ) ≡ b 3 = - d 43 d 32 d 21 + d 43 f 2 2 + f 3 2 d 21 ; ψ 1 ( 4 , 2 ) ≡ a 1 = d 43 - d 32 , ψ 2 ( 4 , 2 ) ≡ a 2 = - d 43 d 32 + f 3 2 .$
$f ( z ) = f 1 f 2 f 3 f 4 z 3 + b 1 z 2 + b 2 z + b 3 .$
$x 1 ′ ( z ) = f 1 2 z 2 + a 1 z + a 2 z 3 + b 1 z 2 + b 2 z + b 3 ,$
$y ( z ) = z 4 + c 1 z 3 + c 2 z 2 + c 3 z + c 4 z 3 + b 1 z 2 + b 2 z + b 3 ,$
$c 1 = b 1 - ( x ′ - y 0 ) , c 2 = b 2 - ( x ′ - y 0 ) b 1 + f 1 2 , c 3 = b 3 - ( x ′ - y 0 ) b 2 + f 1 2 a 1 , c 4 = y 0 b 3 ;$
$x ′ ≡ x 1 ′ ( 0 ) = f 1 2 a 2 b 3 .$
$y ( z ) = ( z - z 1 ) ( z - z 2 ) ( z - z 3 ) ( z - z 4 ) z 3 + b 1 z 2 + b 2 z + b 3 = z 4 - γ 1 z 3 + γ 2 z 2 - γ 3 z + γ 4 z 3 + b 1 z 2 + b 2 z + b 3 ,$
$γ 1 = z 1 + z 2 + z 3 + z 4 , γ 2 = z 1 z 2 + z 1 z 3 + z 1 z 4 + z 2 z 3 + z 2 z 4 + z 3 z 4 , γ 3 = z 1 z 2 z 3 + z 1 z 2 z 4 + z 1 z 3 z 4 + z 2 z 3 z 4 , γ 4 = z 1 z 2 z 3 z 4 .$
$γ 1 = 1 + z 2 + z 3 , γ 2 = z 2 + z 3 + z 2 z 3 , γ 3 = z 2 z 3 , γ 4 = 0.$
$b 1 - ( x ′ - y 0 ) = - γ 1 , b 2 - ( x ′ - y 0 ) b 1 + f 1 2 = γ 2 , b 3 - ( x ′ - y 0 ) b 2 + f 1 2 a 1 = - γ 3 ,$
$y 0 = γ 4 / b 3 .$
$( r - 1 ) b 3 = 1 + b 1 + b 2 .$
$f 3 + d 32 + f 2 ≡ s 32 > s 32 ( min ) , f 2 + d 21 + f 1 ≡ s 21 > s 21 ( min ) ,$
$f 4 = s 43 - d 43 - f 3 ,$
$s 21 ( min ) ≈ 0 , s 32 ( min ) ≈ 1.0 , s 43 ( min ) ≈ 0.$
$∣ y ( 0 ) ∣ = ŷ 1 , 2 = ŷ 2 , 3 = ŷ 3 , 4 = ∣ y ( 1 ) ∣ .$
$[ ( r - 1 ) b 3 ] y ( z ) ≈ ( z - z 1 ) ( z - z 2 ) ( z - z 3 ) ( z - z 4 ) ( 1 - τ ) / 2 τ + z ,$
$ŷ 1 , 2 = ŷ 2 , 3 = ŷ 3 , 4 .$
$[ ( r - 1 ) b 3 ] y ( z ) ≈ z ( z - z 2 ) ( z - z 3 ) ( z - 1.0 ) ( 1 - τ ) / 2 τ + z .$
$( z i ) 0 = 1 2 { 1 - cos [ ( 2 i - 1 ) 22.5° ] cos 22.5° } ,$
$( z 2 ) 0 = 1 2 ( 1 - tan 22.5° ) = 1 - 2 2 = 0.293 ,$
$( z 3 ) 0 = 1 2 ( 1 + tan 22.5° ) = 2 2 = 0.707 ,$
${ 0 < ( z 2 ) P < 0.293 0 < ( z 3 ) P < 0.707 } and { 0.293 < ( z 2 ) N < 1.0 0.707 < ( z 3 ) N < 1.0 } .$
$( z 2 ) N = 1 - ( z 3 ) P , ( z 3 ) N = 1 - ( z 2 ) P .$
$z 2 = 1 - 2 2 - ∊ ′ = 0.293 - ∊ ′ , z 3 = 2 2 - ∊ ″ = 0.707 - ∊ ″ ,$
$z ˆ 1 , 2 ≈ 1 2 z 2 = 1 2 ( 1 - 1 2 2 - ∊ ′ ) , z ˆ 2 , 3 ≈ 1 2 ( z 2 + z 3 ) = 1 2 ( 1 - ∊ ′ - ∊ ″ ) , z ˆ 3 , 4 ≈ 1 2 ( z 2 + 1.0 ) = 1 2 ( 1 + 1 2 2 - ∊ ″ ) ,$
$[ ( r - 1 ) b 3 ] ŷ 1 , 2 ≈ ( ∣ τ ∣ 8 ) ( 0.293 - ∊ ′ ) 2 ( 1.121 + ∊ ′ - 2 ∊ ″ ) ( 1.707 + ∊ ′ ) 1 - 0.707 τ - τ ∊ ′ , [ ( r - 1 ) b 3 ] ŷ 2 , 3 ≈ ( ∣ τ ∣ 8 ) ( 1 - ∊ ′ - ∊ ″ ) ( 0.414 + ∊ ′ - ∊ ″ ) 2 ( 1 + ∊ ′ + ∊ ″ ) 1 - τ ( ∊ ′ + ∊ ″ ) , [ ( r - 1 ) b 3 ] ŷ 3 , 4 ≈ ( ∣ τ ∣ 8 ) ( 1.707 - ∊ ″ ) ( 1.121 + 2 ∊ ′ - ∊ ″ ) ( 0.293 + ∊ ″ ) 2 1 + 0.707 τ - τ ∊ ″ .$
$∊ ≡ 1 2 ( ∊ ′ + ∊ ″ ) ≈ ∊ ′ ≈ ∊ ″ ≈ 0.100 τ × ( 1 + 0.154 τ 2 + 0.274 τ 4 ) .$
$[ ( r - 1 ) b 3 ] y ( max ) ≈ [ ( r - 1 ) b 3 ] 1 3 ( ŷ 1 , 2 + ŷ 2 , 3 + ŷ 3 , 4 ) .$
$[ ( r - 1 ) b 3 ] y ( max ) ≈ 0.0209 ( ∣ τ ∣ ) ( 1 + 0.15 τ 2 + 0.25 τ 4 ) .$
$- ( d 43 - d 32 + d 21 ) + f 1 2 ( d 3 2 - d 43 d 32 ) - γ 4 d 43 f 2 2 + f 3 2 d 21 - d 43 d 32 d 21 = γ 1 , ( f 3 2 + f 2 2 - d 43 d 32 + d 43 d 21 - d 32 d 21 ) - ( d 43 - d 32 + d 21 ) · f 1 2 ( f 3 2 - d 43 d 32 ) - γ 4 d 43 f 2 2 + f 3 2 d 21 - d 43 d 32 d 21 + f 1 2 = γ 2 , - ( d 43 f 2 2 + f 3 2 d 21 - d 43 d 32 d 21 ) + ( f 3 2 + f 2 2 - d 43 d 32 + d 43 d 21 - d 32 d 21 ) · f 1 2 ( f 3 2 - d 43 d 32 ) - γ 4 d 43 f 2 2 + f 3 2 d 21 - d 43 d 32 d 21 - ( d 43 - d 32 ) f 1 2 = γ 3 , ( r - 1 ) ( d 43 f 2 2 + f 3 2 d 21 - d 43 d 32 d 21 ) - ( f 3 2 + f 2 2 - d 43 d 32 + d 43 d 21 - d 32 d 21 ) - ( d 43 - d 32 + d 21 ) = 1 , f 3 + d 32 + f 2 = s 32 , f 2 + d 21 + f 1 = s 21 ,$
$b 1 - x ′ = - γ 1 ,$
$b 2 - x ′ b 1 + f 1 2 = γ 2 ,$
$b 3 - x ′ b 2 + f 1 2 a 1 = - γ 3 ,$
$( r - 1 ) b 3 - b 2 - b 1 = 1.$
$x ′ b 3 = f 1 2 a 2 .$
$b 1 = a 1 + d 21 ,$
$b 2 = a 2 + d 21 a 1 + f 2 2 ,$
$b 3 = d 21 a 2 + f 2 2 d 43 .$
$f 2 + d 21 + f 1 = s 21 .$
$l ≡ l 1 ′ ( 0 ) ≡ l 1 ′ = x ′ + f 1 .$
$f 1 2 - 2 e 10 - e 11 l + e 12 l 2 - l 3 ( l + s 21 ) 2 f 1 - e 00 - e 01 l + e 02 l 2 - e 03 l 3 + l 4 ( l + s 21 ) 2 = 0 ,$
$e 10 = 1 2 [ ( γ 1 + 2 s 21 ) ( γ 2 + γ 1 - 1 r - 1 ) - γ 3 ] , e 11 = [ ( γ 1 + 2 s 21 ) ( γ 1 + 1 r - 1 ) + γ 1 - 1 r - 1 ] , e 12 = [ 2 γ 1 + 3 s 21 + 2 r - 1 ] , e 00 = [ r r - 1 ( γ 1 + 2 s 21 ) γ 3 ] , e 01 = [ ( γ 1 + 2 s 21 ) ( γ 2 + γ 1 - 1 r - 1 ) + r + 1 r - 1 γ 3 ] , e 02 = [ ( γ 1 + 2 s 21 ) ( γ 1 + 1 r - 1 ) + γ 2 + 2 r - 1 ( γ 1 - 1 ) ] ,$
$e 03 = [ 2 γ 1 + 2 s 21 + 2 / ( r - 1 ) ]$
$x ′ = l - f 1 , b 1 = x ′ - γ 1 , b 2 = x ′ b 1 - f 1 2 + γ 2 , b 3 = [ 1 / ( r - 1 ) ] ( 1 + b 1 + b 2 ) , a 1 = ( 1 / f 1 2 ) ( x ′ b 2 - b 3 - γ 3 ) , a 2 = ( 1 / f 1 2 ) ( x ′ b 3 ) .$
$d 21 = b 1 - a 1 , f 2 = s 21 - d 21 - f 1 , d 43 = ( 1 / f 2 2 ) ( b 3 - a 2 d 21 ) , d 32 = d 43 - a 1 , f 3 2 = a 2 + d 43 d 32 .$
$s 32 = f 3 + d 32 + f 2 ,$
$f max = ( 1 - τ 1 - ∣ τ ∣ ) ( f 1 f 2 f 3 f 4 b 3 ) = ( 2 τ 1 - ∣ τ ∣ ) ( f 1 f 2 f 3 f 4 ( r - 1 ) b 3 ) .$
$s 32 = s 32 ( l , τ ) = 1.0 ,$
$( β 3 ) 01 ≡ ( r - 1 ) ( b 3 ) 01 = 1.0 + ( b 1 ) 01 + ( b 2 ) 01$
$( l ) 01 ≈ 2.608 τ [ ( 1 - 0.290 τ 2 - 0.233 τ 4 ) + 0.277 τ ] ,$
$( β 3 ) 01 ≈ 1.614 τ 2 [ ( 1 - 0.435 τ 2 - 0.166 τ 4 ) + 0.063 τ ] .$
$( f 4 ) 010 ≡ ( f 4 ) 01 - s 43 = - ( d 43 ) 01 - ( f 3 ) 01$
$( f 1 ) 01 ≈ + 0.996 τ ( 1 - 0.295 τ 2 - 0.268 τ 4 ) + 0.118 ,$
$( f 2 ) 01 ≈ - 0.805 τ ( 1 - 0.287 τ 2 - 0.268 τ 4 ) + 0.138 ,$
$( f 3 ) 01 ≈ + 0.996 τ ( 1 - 0.256 τ 2 - 0.229 τ 4 ) + 0.119 ,$
$( f 4 ) 010 ≈ - 2.609 τ ( 1 - 0.220 τ 2 - 0.198 τ 4 ) + 0.276.$
$( f max ) 010 = 2 τ 1 - ∣ τ ∣ ( ( f 1 ) 01 ( f 2 ) 01 ( f 3 ) 01 ( f 4 ) 010 ( β 3 ) 01 ) .$
$( f max ) 010 ≈ 3.167 τ [ ( 1 + 0.951 τ 2 + 0.639 τ 4 ) - 0.109 τ ( 1 + 0.72 τ 2 + 2.30 τ 4 ) ] .$
$y ( max ) 01 = [ β 3 y ( max ) ] ( β 3 ) 01 .$
$Y T ( max ) 010 F T 2 ( max ) / Z m ≡ Θ ( max ) 010 = y ( max ) 01 ( f max ) 010 2 ,$
$y ( max ) 01 ~ ( 0.021 / 1.614 ) ∣ τ 3 ∣ = 0.013 ∣ τ 3 ∣ ,$
$Θ ( max ) 010 ~ ( 0.013 ) / ( 3.167 ) 2 ∣ τ 5 ∣ = 0.0013 ∣ τ 5 ∣ .$
$y ( max ) 01 = 0.01337 ( ∣ τ 3 ∣ ) [ ( 1 + 0.090 τ 2 + 1.912 τ 4 ) - 0.071 τ ( 1 - 0.66 τ 2 + 4.82 τ 4 ) ] ,$
$Θ ( max ) 010 = 0.001284 ( ∣ τ 5 ∣ ) [ ( 1 - 1.224 τ 2 + 0.778 τ 4 ) + 0.147 τ ( 1 - 1.10 τ 2 + 0.88 τ 4 ) ] .$
$Y T ( max ) 010 = 0.000072 F T 2 ( max ) / Z m ,$
$Y T ( max ) 010 = 0.000060 F T 2 ( max ) / Z m .$
$[ Y T ( max ) 01 ] P = 0.018 mm ,$
$[ Y T ( max ) 01 ] N = 0.015 mm .$
$( F 1 ) 01 = + ( 1.544 ) ( 40.00 ) mm = + 61.76 mm , ( F 2 ) 01 = - ( 1.019 ) ( 40.00 ) mm = - 40.76 mm , ( F 3 ) 01 = + ( 1.577 ) ( 40.00 ) mm = + 63.08 mm ,$
$( F 4 ) 010 = - ( 3.618 ) ( 40.00 ) mm = - 144.72 mm .$
$( F max ) 010 = + ( 6.988 ) ( 40.00 ) mm = + 279.52 mm ,$
$[ L 1 ′ ( 0 ) ] 01 = + ( 4.485 ) ( 40.00 ) mm = + 179.40 mm .$
$( F 1 ) 01 = - ( 1.307 ) ( 40.00 ) mm = - 52.28 mm , ( F 2 ) 01 = + ( 1.295 ) ( 40.00 ) mm = + 51.80 mm , ( F 3 ) 01 = - ( 1.339 ) ( 40.00 ) mm = - 53.56 mm , ( F 4 ) 010 = + ( 4.167 ) ( 40.00 ) mm = + 1.66.68 mm ; ( F max ) 010 = - ( 8.057 ) ( 40.00 ) mm = - 322.28 mm ,$
$[ L 1 ′ ( 0 ) ] 01 = - ( 3.036 ) ( 40.00 ) mm = - 121.44 mm .$
$( r ) P = ℛ = 4.0 and τ = + 0.6.$
$∊ ′ = ∊ ″ = 0.065 ,$
$z 1 = 0 , z 2 = 0.228 , z 3 = 0.641 , z 4 = 1.000.$
$γ 1 = 1.870 , γ 2 = 1.015148 , γ 3 = 0.146148 , γ 4 = 0 ,$
$e 10 = 1.146275 + 1.304815 s 21 , e 11 = 4.405827 + 4.404667 s 21 , e 12 = 4.404667 + 3.0 s 21 ; e 00 = 0.364201 + 0.398728 s 21 , e 01 = 2.682277 + 2.609630 s 21 , e 02 = 5.710641 + 4.404667 s 21 , e 03 = 4.404667 + 2.0 s 21 .$
$f 1 2 + 2 K 1 f 1 - K 0 = 0 ,$
$K 1 ≡ 1 ( l + 0.15 ) 2 ( - 1.341998 + 5.066527 l - 4.854667 l 2 + l 3 ) ,$
$K 0 ≡ 1 ( l + 0.15 ) 2 ( 0.422660 - 3.073724 l + 6.371342 l 2 - 4.704667 l 3 + l 4 ) .$
$f 4 = - 3.5467 , s 43 = 0.1500 , f 3 = + 1.6187 , s 32 = 1.1537 , f 2 = - 0.9820 , s 21 = 0.1500 , f 1 = + 1.5847 , } or , with Z m = 40.0 mm , F 4 = - 141.87 mm , S 43 = 6.00 mm , F 3 = + 64.75 mm , S 32 = 46.15 mm , F 2 = - 39.28 mm , S 21 = 6.00 mm , F 1 = + 63.39 mm .$
$L 1 ′ ( z ) ≈ ( 4.562 - z ) ( 40.00 ) mm = ( 182.48 - 40.0 z ) mm ,$
$F ( z ) = [ 6.8515 1 + 1.3832 z + 0.8499 z 2 + 0.7669 z 3 ] 40.00 mm .$
$W = ( 6.8515 ) ( 40.0 ) / 100.00 = 2.7406.$
$S 10 = ( 1.30 ) ( 40.0 ) mm = 52.00 mm .$
$L 0 = 130.48 mm .$
$F 0 = + ( 130.48 ) / 1.7406 mm = 74.96 mm ,$
$L 0 ′ = ( 130.48 ) / 2.7406 = 47.61 mm .$
$S = ( 2.7537 ) ( 40.0 ) mm = 110.15 mm ,$
$L = S + L 0 ′ = 157.76 mm = 1.5776 F T ( max ) .$
$Y T ( z ) = ( 250.00 ) × [ 0.0163 z ( z - 0.228 ) ( z - 0.641 ) ( z - 1.000 ) 1 + 1.3832 z + 0.8499 z 2 + 0.7669 z 3 ] mm .$
$( r ) N = 1 / ℛ = 1 / 4 and τ = - 0.6.$
$∊ ′ = ∊ ″ = - 0.065 ,$
$z 1 = 0 , z 2 = 0.358 , z 3 = 0.773 , z 4 = 1.000.$
$γ 1 = 2.13100 , γ 2 = 1.408148 , γ 3 = 0.277148 , γ 4 = 0 ,$
$e 10 = - 0.244966 - 0.099852 s 21 , e 11 = + 0.191828 + 1.595333 s 21 , e 12 = + 1.595333 + 3.0 s 21 ; e 00 = - 0.196867 - 0.184765 s 21 , e 01 = - 0.674698 - 0.199704 s 21 , e 02 = + 0.091976 + 1.595333 s 21 , e 03 = + 1.595333 + 2.0 s 21 .$
$f 1 2 + 2 K 1 f 1 - K 0 = 0 ,$
$K 1 ≡ - [ 1 / ( l + 0.15 ) 2 ] ( - 0.259944 - 0.431128 l + 2.045333 l 2 - l 3 ) ,$
$K 0 ≡ 1 ( l + 0.15 ) 2 ( - 0.224582 + 0.704654 l + 0.331276 l 2 - 1.895333 l 3 + l 4 ) .$
$f 4 = + 4.2594 , s 43 = 0.1500 , f 3 = - 1.3079 , s 32 = 1.1496 , f 2 = + 1.3411 , s 21 = 0.1500 , f 1 = - 1.2760 , } or , with Z m = 40.0 mm , F 4 = + 170.38 mm , S 43 = 6.00 mm , F 3 = - 52.32 mm , S 32 = 45.98 mm , F 2 = + 53.64 mm , S 21 = 6.00 mm , F 1 = - 51.04 mm .$
$L 1 ′ ( z ) ≈ - [ 3.978 - ( 1.0 - z ) ] 40.0 mm = - [ 159.12 - 40.0 ( 1.0 - z ) ] mm ,$
$F ( z ) = - ( 8.2392 4.0 - 5.4486 z + 3.3130 z 2 - 0.8643 z 3 ) 40.00 mm .$
$W = - ( 8.2392 ) ( 40.00 ) / ( 100.00 ) = - 3.2957.$
$F 0 = ( 171.129 ) / ( 4.2957 ) mm = 39.84 mm ,$
$L 0 ′ = ( 171.12 / 3.2957 ) mm = 51.92 mm .$
$S = ( 2.7496 ) 40.0 mm = 110.0 mm ,$
$L = 161.90 mm = 1.619 F T ( max ) .$
$Y T ( z ) = ( 250.0 ) × [ 0.01273 z ( z - 0.358 ) ( z - 0.773 ) ( z - 1.000 ) 4.0 - 5.4486 z + 3.3130 z 2 - 0.8643 z 3 ] .$
$R=FmaxFmin=fmaxfmin.$
$r≜F(0)F(1)=f(0)f(1).$
$r≜f(0)-f(1)f(0)+f(1)=r-1r+1=±R-1R+1,$