This invention relates to a method for producing a progressive ophthalmic lens having at least one progressive surface and to a progressive lens having at least one progressive surface.
Progressive ophthalmic lenses, also known as no-line bifocal or multifocal lenses, are known in general. They have a far vision part for observing objects at great distances, a near vision part for observing objects at closer distances and a progression zone situated between the far vision part and the near vision part. In the progression zone, the effect of the lens increases continuously from the value of the far vision part to the value of the near vision part. This increasing effect of the lens is known as addition. A change in curvature of at least one surface of the lens, referred to as the progressive surface, is required to create this progressive effect of the lens. However, a surface astigmatism, which has a negative effect on the optical imaging properties, is unavoidably associated with the progressive effect; this astigmatism is typically equal to zero only along a planar or curved line (principal line) and increases laterally to this line with twice the value of the gradient of the surface refractive power according to the Minkwitz law. Consequently there is impaired vision in the form of distortion and blurring. In dynamic vision, “swimming” and “jumping” of images perceived on the retina are especially annoying and under some circumstances can have a strongly negative effect on the wear properties of the progressive lens.
DE 43 42 234 C2 describes a progressive lens having a progressive surface, where the maximum value of the gradient of the average surface refractive value of the progressive surface lies in a part of the principal line situated in the progression zone and the gradient of the cylinder and/or astigmatism of the surface has a value lower than the product obtained by multiplying the addition times a coefficient kc max of a constant value over the entire progressive surface of the lens.
An object of the present invention is to provide a progressive ophthalmic lens which has improved wearing properties. Another object is to provide a method for manufacturing an inventive progressive lens.
According to the present invention which achieves these objects, a method of manufacturing is provided for a progressive ophthalmic lens having at least one progressive surface where the lens comprises a far vision part for seeing at great distances and having a far reference point, a near vision part designed for seeing at short distances and having a near reference point, and a progression zone situated between the far vision part and the near vision part where the effect of the lens increases from the value at the far reference point to the value at the near reference point along a principal line, this increase being by a value known as addition. A calculation and optimization step of the progressive lens is performed so that the absolute value of the rotation |rot {right arrow over (A)}| and/or the divergence |div{right arrow over (A)}| of a vectorial astigmatism {right arrow over (A)} is a small as possible, where the absolute value |{right arrow over (A)}| of the vectorial astigmatism {right arrow over (A)} is proportional to the absolute value of an astigmatism in the use position of the progressive lens or a surface astigmatism of the at least one progressive surface of the progressive lens, and the direction of the vectorial astigmatism {right arrow over (A)} is proportional to the cylinder axis of an astigmatism in the use position of the progressive lens or a surface astigmatism of the at least one progressive surface of the progressive lens.
The far vision part is usually situated in the upper part of the lens in the use position and the near vision part is located in the lower part of the lens. The far vision part is usually designed for seeing at an infinite distance and the near vision part is designed for reading in particular. In the case of ophthalmic lenses for special applications, e.g., lenses for pilot's glasses or glasses for working at a computer monitor, the far vision part and the near vision part may be arranged differently and designed for different distances. Furthermore, there may also be multiple near vision parts and far vision parts as well as progression zones.
The progressive lens may have one progressive surface (front or back surface) or two progressive surfaces. The progressive surface may be the side of the lens facing the eyes, i.e., the back side, or the object side, i.e., the front side of the lens, but the opposing surface usually has a spherical or toric surface, which does not contribute to the surface variation of the astigmatism of the lens. In a lens having two progressive surfaces, the derivations of the vectorial astigmatism |rot {right arrow over (A)}| and |div {right arrow over (A)}|, which are to be kept small, are formed due to the combined effect of the front surface and the back surface (i.e., the surface facing the eyes and the object surface). The value of the addition is calculated as the difference in the use values of the refractive force at the far reference point and at the near reference point.
According to this invention, a progressive lens whose properties are especially advantageous, i.e., as small as possible with respect to the spatial variations in the astigmatism including its directional information, is achievable. According to this invention, such directional variations are recognized as the cause of the change in visual impression known as the swimming or jumping that occurs with movements of the head or line of sight.
In particular, it has been recognized that the change or variation in the cylinder axis of the astigmatism is not only a secondary parameter but, on the contrary, is an important parameter. In comparison with that, lenses corresponding to the prior art have so far never been calculated taking into account the spatial variation in their surface astigmatism including its direction information.
Of the properties of the surfaces of progressive lenses, so far in the literature at most the second derivatives of the depths of camber have usually been discussed. In some cases, such as in DE 4 342 234 C2, gradients of the surfaces refractive value ∇D(x, y) or the surface astigmatism ∇A(x, y), i.e., third derivations of the depths of camber of the surface have been discussed. The aforementioned patent refers in several places to the relevance of these gradients for dynamic vision and thus for the site conditions associated with movement of the head or line of site or combinations thereof.
However, astigmatism is always interpreted here only as a two-dimensional scalar function which is usually defined as being proportional to the difference |1/r1−1/r2| of the main radii of curvature. The conventional approach disregards the cylinder axis of the astigmatism and concentrates only on its absolute value or as in DE 4 342 234 C2, concentrates on the gradient of the absolute value of the surface astigmatism.
The present invention is based on the discovery that with a spatial change in a quantity which also has a cylinder axis as is the case with astigmatism, it is not only the change in the absolute amount that is of interest but also the change in its directional information. However, any information regarding the change in cylinder axis of the astigmatism which is critical for the optical correction is lost when the vector field of the astigmatism is reduced to its absolute value.
The role played by the cylinder axis of astigmatism can be illustrated on the basis of a determination of the refraction, for example. The determination of the cylinder axis of the astigmatism is performed here in sequence before the determination of the absolute value of the astigmatism. The reason for this is that in a determination of the cylinder axis, e.g., by the cross-cylinder method, a cylinder whose absolute value is estimated is modified with regard to its cylinder axis with the help of the cross-cylinder until it has been defined optimally. This is followed by a final determination of the absolute value of the cylinder. An attempt to determine the absolute value of the cylinder even before defining the axis would result in skewed cylinders in the course of the determination of absolute value. This means that a resulting cylinder whose absolute value depends on the size of the correction cylinder and the size of the axial error would be obtained. One would thus have two constantly changing parameters which would have to be repeatedly corrected again in alternation. However, if the cylinder axis of the astigmatism is already known, then the correction cylinder is always changed by the full amount, keeping the cross cylinder in reserve. Finally, if the axis of the correction cylinder has been determined correctly, it coincides with that of the refraction deficit and the resulting cylindrical effect will be zero, as described, for example, in H. Diepes, Refraktionsbestimmung [Determining Refraction], Verlag Heinz Postenrieder, Pforzheim (1972), page 322.
In addition to having a negative effect on vision, any astigmatism also results in distortion, as already indicated above. The subjective perception of the resulting distortion of the retinal images also depends on the direction of the cylinder axis as described, for example, by J. Reiner in Auge und Brille [Eye and Lens], 4th edition, Ferdinand Enke Verlag Stuttgart (1987), page 27. In the case of an astigmatism according to the rule or against the rule, the result is distortion, which does not usually cause much interference for the wearer of the lens, because the distortion is only in the vertical or horizontal direction. Distortion of this type is often caused by viewing from a lateral position so it is not unknown. Problems occur for the users of ophthalmic lenses mainly in astigmatism with an oblique cylinder axis because the image of vertical and horizontal object elements on the retina is skewed. Distortion of this type is much less tolerable because vertical and horizontal lines no longer appear at right angles in space but instead form angles deviating from 90° (greater than or smaller than 90°). This results in fusion problems, which make it difficult to become adjusted to lenses. According to this invention, it has been found that such stresses are to be expected for sensory fusion when the cylinder axis of the astigmatism changes in the course of lateral eye movements. If astigmatic effects occur, which remain the same in terms of absolute value but now have an oblique cylinder axis which also changes depending on the point of viewing through the lens, this can result in intolerance of the lens in the peripheral area.
To be able to be describe spatial variations in astigmatism, first a suitable vectorial astigmatism is defined. The spatial distribution of this vectorial astigmatism defines a vector field accordingly. When the change in absolute value and direction (corresponding to the absolute value and the cylinder axis of the astigmatism) of this vectorial astigmatism is integrated into the calculation, i.e., optimization of a lens and in particular a progressive lens, new aspects which lead to improved imaging properties and thus improved wearing properties are yielded, as will be explained below. It is especially expedient for describing the derivations of the vectorial astigmatism {right arrow over (A)} to consider its rotation and divergence, i.e., the vector operators rot {right arrow over (A)}={right arrow over (∇)}×{right arrow over (A)} and div {right arrow over (A)}={right arrow over (∇)}·{right arrow over (A)}.
Although the present invention refers to both the consideration of the surface astigmatism of a progressive lens as well as astigmatism in the use position, surface astigmatism of the progressive surface is considered below as an example. Therefore, a detailed definition of a vectorial surface astigmatism is given below, and then a brief similar definition of a vectorial astigmatism in the use position is also given.
Definition of the vectorial surface astigmatism {right arrow over (A)}
The depth of camber of a progressive surface is given by the functionz=f(x, y)where the coordinates (x, y) are in the plane of the projection. The depth of camber is understood to refer to the distance of a point with coordinates (x, y) from the tangential plane of the surface vertex.
The first basic form
      g    ⁡          (              x        ,        y            )        =            (                                    E                                F                                                F                                G                              )        =          (                                                  1              +                              f                x                2                                                                                        f                x                            ⁢                              f                y                                                                                                        f                x                            ⁢                              f                y                                                                        1              +                              f                y                2                                                        )      and the second basic form
      1    ⁢          (              x        ,        y            )        =            (                                    K                                L                                                L                                M                              )        =                            -          1                                      1            +                          f              x              2                        +                          f              y              2                                          ⁢              (                                                            f                xx                                                                    f                xy                                                                                        f                xy                                                                    f                yy                                                    )            are needed to calculate the main curvatures, where the derivations are abbreviated by
            f      x        =                  ∂        x            ⁢              f        ⁡                  (                      x            ,            y                    )                      ,          ⁢            f      y        =                  ∂        y            ⁢              f        ⁡                  (                      x            ,            y                    )                      ,          ⁢            f      xx        =                  ∂        x        2            ⁢              f        ⁡                  (                      x            ,            y                    )                      ,          ⁢            f      yy        =                  ∂        y        2            ⁢              f        ⁡                  (                      x            ,            y                    )                      ,          ⁢            f      xy        =                            ∂          x                ⁢                              ∂            y                    ⁢                      f            ⁡                          (                              x                ,                y                            )                                          .      The main curvatures k1 and k2 and the main curvature directions are solutions to the general eigenvalue problem
                                                        (                                                                    K                                                        L                                                                                        L                                                        M                                                              )                        ⁢                          (                                                                                          x                      1                                                                                                  x                      2                                                                                                                                  y                      1                                                                                                  y                      2                                                                                  )                                -                                    (                                                                    E                                                        F                                                                                        F                                                        G                                                              )                        ⁢                          (                                                                                          x                      1                                                                                                  x                      2                                                                                                                                  y                      1                                                                                                  y                      2                                                                                  )                        ⁢                          (                                                                                          k                      1                                                                            0                                                                                        0                                                                              k                      2                                                                                  )                                      =        0.                            (        1        )            
The eigenvectors (x1, y1) and (X2, y2) calculated in this way are the projections of the vectors in the tangential plane of the lens surface and indicate the main directions of curvature there.
To calculate the main direction of curvature, the eigenvectors (x1, y1) and (X2, y2) must be transformed to the system of the tangential plane after their calculation. This transformation can best be determined by formulating the entire eigenvalue problem itself in the tangential plane. This is done by bringing it to the form
                                                                        (                                                                            K                                                              L                                                                                                  L                                                              M                                                                      )                            eff                        ⁢                          (                                                                                          u                      1                                                                                                  u                      2                                                                                                                                  v                      1                                                                                                  v                      2                                                                                  )                                -                                    (                                                                                          u                      1                                                                                                  u                      2                                                                                                                                  v                      1                                                                                                  v                      2                                                                                  )                        ⁢                          (                                                                                          k                      1                                                                            0                                                                                        0                                                                              k                      2                                                                                  )                                      =        0.                            (        2        )            because the eigenvectors (u1, v1) and (U2, V2) must lie in the tangential plane since the first basic form in this presentation is the unit matrix.
It seems simplest to multiply the original problem (1) simply from the left times
            (                                    E                                F                                                F                                G                              )              -      1        .However, this would lead to an asymmetrical matrix
      (                            K                          L                                      L                          M                      )    effwhose eigenvectors would no longer be orthogonal because the (u, v) coordinate system would then be skewed. To retain symmetry, the first basic form g must first be transformed to a diagonal form:WTgW=gd⇄g=WgdWT=(Wgd1/2)(Wgd1/2)T where gd is a diagonal matrix with the eigenvalues of g and W is a matrix whose columns have the eigenvectors of g,. The transformation (Wgd1/2)T transforms a vector (x, y) to the tangential plane by way of
      (                            u                                      v                      )    =                    (                  Wg          d                      1            /            2                          )            T        ⁢          (                                    x                                                y                              )      because the length of a vector in the (u, v) plane is given by
                                          (                                                            u                                                  v                                                      )                    ⁢                      (                                                            u                                                                              v                                                      )                          =                              (                                                            x                                                  y                                                      )                    ⁢                      (                          Wg              d                              1                /                2                                      )                    ⁢                                    (                              Wg                d                                  1                  /                  2                                            )                        T                    ⁢                      (                                                            x                                                                              y                                                      )                                                  =                              (                                                            x                                                  y                                                      )                    ⁢                                    g              ⁡                              (                                                                            x                                                                                                  y                                                                      )                                      .                              
With each transformation (Wgd1/2)T to the (u, v) plane, (Wgd1/2R)T is also such a transformation when
  R  =      (                                        cos            ⁢                                                  ⁢            φ                                                sin            ⁢                                                  ⁢            φ                                                                          -              sin                        ⁢                                                  ⁢            φ                                                cos            ⁢                                                  ⁢            φ                                )  is a rotation within the (u, v) plane. It is practical now to setup the rotation R in such a way that the u axis coincides with the intersection formed by the tangential plane with the horizontal plane. This can be achieved by an angle φ for which the lower left matrix element of (Wgd1/2R)T disappears. As can be confirmed by calculation, this is the case for
      φ    0    =      arctan    ⁢                  ⁢                  f        y                    f        x              ⁢                  1        +                  f          x          2                +                  f          y          2                    and the resulting transformation is then given by
                    T        =                              (                                          Wg                d                                  1                  /                  2                                            ⁢                              R                ⁡                                  (                                      φ                    0                                    )                                                      )                    T                                        =                                            Sign              ⁢                                                          ⁢                              (                                  f                  x                                )                                                                                      (                                      1                    +                                          f                      y                      2                                                        )                                ⁢                                  (                                      1                    +                                          f                      x                      2                                        +                                          f                      y                      2                                                        )                                                              ⁢                      (                                                                                (                                          1                      +                                              f                        y                        2                                                              )                                                                    0                                                                                                                        -                                              f                        x                                                              ⁢                                          f                      y                                                                                                                                  1                      +                                              f                        x                        2                                            +                                              f                        y                        2                                                                                                                  )                              and
      T          -      1        =                    Sign        ⁡                  (                      f            x                    )                                      1          +                      f            y            2                                ⁢          (                                                                  1                +                                  f                  x                  2                                +                                  f                  y                  2                                                                          0                                                                              f                x                            ⁢                              f                y                                                                        1              +                              f                y                2                                                        )      
The eigenvalue problem can now be written by inserting g=TTT in the form:
                    (                                            K                                      L                                                          L                                      M                                      )            ⁢              (                                                            x                1                                                                    x                2                                                                                        y                1                                                                    y                2                                                    )              -                  T        T            ⁢              T        ⁡                  (                                                                      x                  1                                                                              x                  2                                                                                                      y                  1                                                                              y                  2                                                              )                    ⁢              (                                                            k                1                                                    0                                                          0                                                      k                2                                                    )              =            0      ↔                          ⁢                                                  (                              T                                  -                  1                                            )                        T                    ⁢                      (                                                            K                                                  L                                                                              L                                                  M                                                      )                    ⁢                      T                          -              1                                ⁢                      T            ⁡                          (                                                                                          x                      1                                                                                                  x                      2                                                                                                                                  y                      1                                                                                                  y                      2                                                                                  )                                      -                              T            ⁡                          (                                                                                          x                      1                                                                                                  x                      2                                                                                                                                  y                      1                                                                                                  y                      2                                                                                  )                                ⁢                      (                                                                                k                    1                                                                    0                                                                              0                                                                      k                    2                                                                        )                                =                  0        ↔                                  ⁢                                                            (                                                                            K                                                              L                                                                                                  L                                                              M                                                                      )                            eff                        ⁢                          (                                                                                          u                      1                                                                                                  u                      2                                                                                                                                  v                      1                                                                                                  v                      2                                                                                  )                                -                                    (                                                                                          u                      1                                                                                                  u                      2                                                                                                                                  v                      1                                                                                                  v                      2                                                                                  )                        ⁢                          (                                                                                          k                      1                                                                            0                                                                                        0                                                                              k                      2                                                                                  )                                          =      0.      
The eigenvalue problem transformed in this way to the tangential plane contains a symmetrical representation of the second basic form:
                                          (                                                            K                                                  L                                                                              L                                                  M                                                      )                    eff                =                ⁢                                            (                                                          ⁢                              T                                  -                  1                                            )                        T                    ⁢                      (                                                            K                                                  L                                                                              L                                                  M                                                      )                    ⁢                      T                          -              1                                                              =                ⁢                  1                                    (                              1                +                                  f                  y                  2                                            )                        ⁢                                          1                +                                  f                  x                  2                                +                                  f                  y                  2                                                                                                    ⁢                  (                                                                                                                                                                                                                            -                                                              f                                x                                                                                      ⁢                                                          (                                                              1                                +                                                                  f                                  y                                  2                                                                                            )                                                                                +                                                                                    f                              x                                                        ⁢                                                                                          f                                y                                                            (                                                                                                                                    -                                                                          f                                      x                                                                                                        ⁢                                                                      f                                    y                                    2                                                                                                  +                                                                                                                                                                                                                                                                              2                            ⁢                                                                                          f                                xy                                                            ⁡                                                              (                                                                  1                                  +                                                                      f                                    y                                    2                                                                                                  )                                                                                                              )                                                                                                                          1                    +                                          f                      x                      2                                        +                                          f                      y                      2                                                                                                                                                                                      f                        x                                            ⁢                                              f                        y                        2                                                              -                                                                  f                        xy                                            ⁡                                              (                                                  1                          +                                                      f                            y                            2                                                                          )                                                                                                                        1                      +                                              f                        x                        2                                            +                                              f                        y                        2                                                                                                                                                                                                                                    f                        x                                            ⁢                                              f                        y                        2                                                              -                                                                  f                        xy                                            ⁡                                              (                                                  1                          +                                                      f                            y                            2                                                                          )                                                                                                                        1                      +                                              f                        x                        2                                            +                                              f                        y                        2                                                                                                                                          -                                      f                    y                                                                                )                    and thus yields the main curvatures k1 and k2 and the respective main directions of curvature (u1, v1) and (U2, v2).
For definition of a vectorial astigmatism, i.e., a vector field for the astigmatism, the absolute value of which is given by the quantity |k1−k2|, it is advantageous to double the angle; this is usually also done to check the cylinder axis of the TABO scheme in polar coordinates as proposed, for example, in WO 01/81979. If the direction of the first cylinder axis is given approximately by the angle ψ=arctan v1/u1, then a vectorial astigmatism can be defined as follows:
                                                                        A                →                            =                            ⁢                                                (                                      n                    -                    1                                    )                                ⁢                                                                                              k                      2                                        ·                                          k                      1                                                                                        ⁢                                  (                                                                                    cos                                                                                              2                          ⁢                          ψ                                                                                                                                    sin                                                                                              2                          ⁢                          ψ                                                                                                      )                                                                                                        =                            ⁢                                                n                  -                  1                                                                      (                                          1                      +                                              f                        y                        2                                                              )                                    ⁢                                                            (                                              1                        +                                                  f                          x                          2                                                +                                                  f                          y                          2                                                                    )                                                              3                      /                      2                                                                                                                                                            ⁢                                                (                                                                                                                                                                        -                              2                                                        ⁢                                                          f                              x                                                        ⁢                                                          f                                                              xy                                ⁢                                                                                                                                                                                        ⁢                                                          f                              y                                                        ⁢                                                          (                                                              1                                +                                                                  f                                  y                                  2                                                                                            )                                                                                +                                                                                                                    f                                xx                                                            ⁡                                                              (                                                                  1                                  +                                                                      f                                    y                                    2                                                                                                  )                                                                                      2                                                    +                                                                                                                                                                                                                                                        f                                x                                2                                                            ⁡                                                              (                                                                                                      -                                    1                                                                    +                                                                      f                                    y                                    2                                                                                                  )                                                                                      ⁢                                                          f                              yy                                                                                -                                                                                    (                                                              1                                +                                                                  f                                  y                                  2                                                                                            )                                                        ⁢                                                          f                              yy                                                                                                                                                                                                                    2                          ⁢                                                                                    1                              +                                                              f                                x                                2                                                            +                                                              f                                y                                2                                                                                                              ⁢                                                      (                                                                                          f                                xy                                                            +                                                                                                f                                  xy                                                                ⁢                                                                  f                                  y                                  2                                                                                            -                                                                                                f                                  x                                                                ⁢                                                                  f                                  y                                                                ⁢                                                                  f                                  yy                                                                                                                      )                                                                                                                                )                                ,                                                                        (        3        )            where n is the refractive index.
It should be pointed out here that the second cylinder axis does not lead to a new astigmatism but instead causes only a global plus or minus sign: the second cylinder axis is perpendicular to the first and the right angle is expanded to 180° by doubling the angle.
By analogy with vectorial surface astigmatism, an astigmatism in use position is defined vectorial as |k2−k1|(cos 2ψ, sin 2ψ) where k1,2 are not the eigenvalues of the second basic form of the refractive surface, however, but instead the emergent wave front. The quantity (n−1) which depends on the refractive index is contained in the eigenvalues k1,2—in contrast with the situation with the surface astigmatism in equation (3)—and therefore need not be multiplied as an additional factor. The angle ψ is intended in the tangential plane at the wave front and has as the reference direction the line of intersection of this plane with the horizontal plane. For perpendicular incidence of a wave front from the infinite, the factor given in equation (3) is obtained for the vectorial astigmatism in the use position; otherwise deviating values are obtained.
In the following discussion, the astigmatism {right arrow over (A)} is understood to be the vectorial surface astigmatism defined according to equation (3).
A first approximation of the vectorial astigmatism is obtained by assuming that the first derivatives fx, fy of the depth of camber are small in comparison with 1, i.e.,
            1      +              f        x        2              ≈    1    ,          ⁢            1      +              f        y        2              ≈    1    ,etc. For the vectorial astigmatism it then holds that:
                              A          →                =                              (                          n              -              1                        )                    ⁢                      (                                                                                                      f                      xx                                        -                                          f                      yy                                                                                                                                        2                    ⁢                                          f                      xy                                                                                            )                                              (        4        )            
It follows from equation (4) that the vectorial astigmatism is an expression of the second derivatives of the depth of camber.
On the whole, this yields four independent derivatives of the vector field {right arrow over (A)}, namely
            ∂              A        x                    ∂      x        ,            ∂              A        x                    ∂      y        ,                    ∂                  A          y                            ∂        x              ⁢                  ⁢    and    ⁢                  ⁢                            ∂                      A            y                                    ∂          y                    .      They are all of the same order of magnitude.
Although they are all of interest, the following discussion is limited to two especially customary combinations of these four derivatives of the vectorial astigmatism {right arrow over (A)}, namely its divergence
                              div          ⁢                                          ⁢                      A            →                          =                                            ∇              →                        ⁢                          ·                              A                →                                              =                                                    ∂                                  A                  x                                                            ∂                x                                      +                                          ∂                                  A                  y                                                            ∂                y                                                                        (        5        )            and the z component of its rotation
                                          (                          rot              ⁢                                                          ⁢                              A                →                                      )                    z                =                                            (                                                ∇                  →                                ⁢                                  ×                                      A                    →                                                              )                        z                    =                                                    ∂                                  A                  y                                                            ∂                x                                      +                                                            ∂                                      A                    x                                                                    ∂                  y                                            .                                                          (        6        )            
For reasons of simplicity, rot {right arrow over (A)} is written below instead of (rot {right arrow over (A)})z because rot {right arrow over (A)}=(0, 0, (rot {right arrow over (A)})z).
The exact formulas for rar {right arrow over (A)} and div {right arrow over (A)} derived from equation (3) for the vectorial astigmatism {right arrow over (A)} are not given here for reasons of space but instead the exact results obtained from them are plotted graphically below. When the approximation obtained according to equation (3) for the vectorial astigmatism is inserted in equations (5) and (6), this yields the following approximations for the divergence and rotation of the vectorial astigmatism:div {right arrow over (A)}∝fxxx+fxyy and (rot {right arrow over (A)})z∝fyyy+fxxy   (7)
It can be seen from equation (7) that the rotation and divergence of the vectorial astigmatism {right arrow over (A)} characterize the derivations of the depth of camber. There are a total of four independent third derivatives of the depth of camber, namely fxxx, fxxy, fxyy, fyyy.
Ophthalmic lenses manufactured by the inventive method have definitely improved wear properties in comparison with ophthalmic lenses produced by conventional methods without optimization with regard to the absolute value as well as the direction of astigmatism of the ophthalmic lens. It should be pointed out that the present application refers to the astigmatism inherent in a progressive lens which does not serve to correct the natural astigmatism of the eye (the latter is constant over the entire lens anyway and therefore has negligible derivations).
The calculation and optimization step is preferably performed in such a way that a global maximum of the absolute value |div {right arrow over (A)}| of the divergence of the vectorial astigmatism {right arrow over (A)} is outside of the zone of good vision of the lens, in which the absolute value of the vectorial astigmatism |{right arrow over (A)}| is less than 0.6 dpt, preferably less than or equal to 0.5 dpt, and is preferably located in the peripheral area of the lens.
As a result, the interfering maximums and minimums of div {right arrow over (A)} are shifted into a range which is not often used for vision purposes and therefore the wearing properties of the lens are improved. An astigmatism with a value of more than 0.5 dpt leads to an image that is perceived as blurred on the retina.
The calculation and optimization step is also preferably performed in such a way that the x coordinate of the position of the global maximum of the absolute value |div {right arrow over (A)}| of the divergence of the vectorial astigmatism {right arrow over (A)} is greater than 6.0 mm and the y coordinate is less than −8.5 mm, where x is the horizontal axis and y is the vertical axis in the use position, and the zero point x=0, y=0 is 4 mm below the centering point of the lens.
The horizontal axis x is therefore parallel to the direction which is defined by two permanent marks on the rimless lens which are a distance of 17 mm laterally from the principal line. The vertical axis y is perpendicular to the horizontal direction. The zero point x=0, y=0 is 4 mm below the centering point of the lens and corresponds to the center of the lens in the case of non-predecentered lenses. For most lenses this is also the prism reference point, i.e., the point at which the prismatic effect of the lens is to be determined. The centering point is 4 mm above the midpoint of the lens and is prestamped on the lens in the form of a cross. In observation of the surface astigmatism of the progressive surface, x and y are in a tangential plane passing through the vertex of the progressive surface. Within the range y≧−8 mm, there are no mentionable extremes at all of the divergence |div {right arrow over (A)}| of the vectorial astigmatism {right arrow over (A)}. This feature is satisfied for all additions.
An extreme of div {right arrow over (A)} is considered “mentionable” if its absolute value is more than (0.1/mm) times the addition.
Furthermore, the calculation and optimization step is preferably performed so that all extremes of the amount |div {right arrow over (A)}| of the divergence of the vectorial astigmatism {right arrow over (A)} exceeding the value of (0.1/mm) times the addition are outside the range y≧−9 mm of the lens for all progressive surfaces with addition ≧2.0 dpt. This feature is satisfied for all basic curves.
In addition, it is also preferable if the calculation and optimization step is performed so that the absolute value |rot {right arrow over (A)}| of the rotation of the vectorial astigmatism {right arrow over (A)} in the near vision part and/or in the far part does not increase beyond a maximum value of |rot {right arrow over (A)}|max≈0.25 addition/dpt·dpt/mm.
The calculation and optimization step is especially preferably performed so that the absolute value |rot {right arrow over (A)}| of the rotation of the vectorial astigmatism {right arrow over (A)} in the horizontal section at y=−14 mm does not exceed a maximum value of |rot {right arrow over (A)}|max≈0.115 addition/dpt*dpt/mm, preferably |rot {right arrow over (A)}|max≈0.08 addition/dpt*dpt/mm.
Furthermore, the calculation and optimization step is preferably performed so that the absolute value |rot {right arrow over (A)}| of the rotation of the vectorial astigmatism {right arrow over (A)} in the horizontal section at y=+6 mm does not exceed a maximum value of |rot {right arrow over (A)}|max≈0.115 addition/dpt*dpt/mm, preferably |rot {right arrow over (A)}|max≈0.06 addition/dpt*dpt/mm.
The calculation and optimization step is preferably performed so that in the far vision part between y=3 mm and y=5 mm there is at least one horizontal section y=const along which the value |rot {right arrow over (A)}| of the rotation of the vectorial astigmatism {right arrow over (A)} increases monotonically from the principal line outward to a coordinate of |x|=16 mm.
Furthermore, it is preferable for the calculation and optimization step to be performed so that the divergence div {right arrow over (A)} of the vectorial astigmatism {right arrow over (A)} in the horizontal section at y=0 mm does not increase beyond a maximum value of (div {right arrow over (A)})max≈(0.11 addition/dpt+0.03) dpt/mm, preferably (div {right arrow over (A)})max≈(0.08 addition/dpt+0.03) dpt/mm.
The calculation and optimization step is preferably performed so that the divergence div {right arrow over (A)} of the vectorial astigmatism {right arrow over (A)} in the horizontal section at y=0 mm does not fall below a minimum value of (div {right arrow over (A)})min≈(−0.07 addition/dpt−0.11) dpt/mm, preferably (div {right arrow over (A)})min≈(−0.05 addition/dpt−0.08) dpt/mm.
A method for manufacturing a progressive lens that is currently especially preferred is one in which the calculation and optimization step is performed so that the divergence div {right arrow over (A)} of the vectorial astigmatism {right arrow over (A)} in the horizontal section at y=−14 mm does not increase beyond a maximum value of (div {right arrow over (A)})max≈(0.12 addition/dpt+0.06) dpt/mm.
In addition it is preferable if the calculation and optimization step is performed so that the divergence div {right arrow over (A)} of the vectorial astigmatism {right arrow over (A)} in the horizontal section at y=−14 mm does not drop below a minimum value of (div {right arrow over (A)})min≈(−0.13 addition/dpt−0.05) dpt/mm. Furthermore, according to this invention, a progressive lens having at least one progressive surface is provided, whereby the lens has at least:                one far vision part designed for seeing at great distances and having a far reference point;        one near vision part designed for seeing at short distances and having a near reference point;        progression zone situated between the far and near vision parts where the effect of the lens increases along a principal line by a value (referred to as “addition” or “add power”) from the value at the far reference point to the value at the near reference point the global maximum of the absolute value |div {right arrow over (A)}| of the divergence of a vectorial astigmatism {right arrow over (A)} is situated outside of the zone of good vision of the lens in which the absolute value of vectorial astigmatism |{right arrow over (A)}| is less than 0.6 dpt, preferably less than or equal to 0.5 dpt and is preferably situated in the peripheral area of the lens; and/or        the absolute value |rot {right arrow over (A)}| of the rotation of the vectorial astigmatism {right arrow over (A)} in the near vision part and/or in the far vision part does not exceed a maximum value of |rot {right arrow over (A)}|max≈0.25 addition/dpt·dpt/mm, and further the absolute value |{right arrow over (A)}| of the vectorial astigmatism {right arrow over (A)} is proportional to the absolute value of an astigmatism in the use position of the progressive lens or a surface astigmatism of the at least one progressive surface of the progressive lens, and the direction of the vectorial astigmatism {right arrow over (A)} is proportional to the cylinder axis of an astigmatism in the use position of the progressive lens or a surface astigmatism of the at least one progressive surface of the progressive lens.        
The x coordinate of the position of the global maximum of the absolute value |div {right arrow over (A)}| of the divergence of the vectorial astigmatism {right arrow over (A)} is preferably greater than 6.0 mm and the y coordinate is preferably less than −8.5 mm, where x denotes the horizontal axis and y denotes the vertical axis in the use position, and the zero point x=0, y=0 is situated 4 mm beneath the centering point of the lens.
For all progressive surfaces with addition ≧2.0 dpt, it is also preferable for all extremes of the absolute value |div {right arrow over (A)}| of the divergence of the vectorial astigmatism {right arrow over (A)} exceeding the value (0.1/mm) times the addition to be outside of the range y≧−9 mm of the lens.
The absolute value |rot {right arrow over (A)}| of the vectorial astigmatism {right arrow over (A)} in the horizontal section at y=−14 mm especially preferably does not increase beyond a maximum value of |rot {right arrow over (A)}|max≈0.115 addition/dtp·dpt/mm, preferably |rot {right arrow over (A)}|max≈0.08 addition/dpt·dpt/mm.
The absolute value |rot {right arrow over (A)}| of the rotation of the vectorial astigmatism {right arrow over (A)} in the horizontal section at y=+6 mm preferably does not increase beyond a maximum value of |rot {right arrow over (A)}|max≈0.115 addition/dpt·dpt/mm, preferably |rot {right arrow over (A)}max≈0.06 addition/dpt·dpt/mm.
It is also preferable if, in the far vision part between y=3 mm and y=5 mm, there is at least one horizontal section y=const along which the absolute value |rot {right arrow over (A)}| of the rotation of the vectorial astigmatism {right arrow over (A)} increases monotonically from the principal line outward to a coordinate of |x|=16 mm.
It is also preferable if the divergence div {right arrow over (A)} of the vectorial astigmatism {right arrow over (A)} in the horizontal section at y=0 mm does not exceed a maximum value of (div {right arrow over (A)})max≈(0.11 addition/dpt+0.03) dpt/mm, preferably (div {right arrow over (A)})max≈(0.08 addition/dpt+0.03) dpt/mm.
The divergence div {right arrow over (A)} of the vectorial astigmatism {right arrow over (A)} in the horizontal section at y=0 mm preferably does not drop below a minimum value of (div {right arrow over (A)})min≈(−0.07 addition/dpt−0.11) dpt/mm, preferably (div {right arrow over (A)})min≈(−0.05 addition/dpt−0.08) dpt/mm.
It is also preferable if the divergence div {right arrow over (A)} of the vectorial astigmatism {right arrow over (A)} in the horizontal section at y=−14 mm does not exceed a maximum value of (div {right arrow over (A)})max≈(0.12 addition/dpt +0.06) dpt/mm.
It is especially preferable if the divergence div {right arrow over (A)} of the vectorial astigmatism {right arrow over (A)} in the horizontal section at y=−14 mm does not drop below a minimum value of (div {right arrow over (A)})min≈(−0.13 addition/dpt−0.05) dpt/mm.
USPTO Backgrounds
USPTO Backgrounds