Table Size  
Figure 2

Figure 2To enlarge Figure 2 shows a ray reflecting off one side of the pavilion and directly back to the table; it would be reflected back into the stone and then out at the same angle it entered. To prevent this, the table size should be limited; such rays will then be refracted 3 through the bezel (dotted line) 6. The table size which corresponds to the extreme ray shown hitting the table at symmetrical positions, is ... (T/W*)max = tanC*tanP.

For diamond, using the pavilion angle P=40.75o... (T/W*)max=0.392.

Note that this is not the final table/girdle ratio (T/W); the girdle width shown will be reduced by the bezel). 7

A more valid reason for limiting the table size 8 is also shown in Figure 2, where the innermost ray at the critical angle is shown, drawn from the culet. Angles greater than this are possible (qmax=33.32o) and would be reflected back from the table - to leak out through the pavilion. Such rays hit beyond the point shown, so the addition of the bezel from this point can prevent this problem. This point on the table is exactly the same as developed above for Tolkowsky's reason.

Tolkowsky mentioned this problem in his discussion of pavilion angles, but never showed or developed a formula for either criterion; he probably did not know that the simple relationship solved both problems.

  Bezel Angle  

(see footnote 7)

The bezel angle is determined by analyzing the light which enters one side of the table and exits through the opposite bezel; that which enters the bezel and leaves via the table is simply light following the same path in the reverse direction. There is also some light which enters one bezel and leaves via the other; this is small by comparison and is probably negligible 9 (Tolkowsky never mentions it).

Several criteria could be used to define the 'best' bezel angle; Tolkowsky chooses that angle which passes the most light from the table through the bezel (and thus vice- versa). 10

  General Method Of Determining Bezel Angle  

To reduce the complexity of the problem, Tolkowsky divides the light which enters the table into three groups of equal intensity (I) 3. For each group he determines the mean direction of entry (q) and the corresponding direction (B) at which it approaches the bezel. He determines the amount of light (L) in each group as the product of its intensity normal to the table and the cross-sectional area of the beam (A), which should be measured parallel to the table (entering beams). Finally he calculates the mean direction of all this light as it approaches the bezel and calls this the bezel angle (B) so that the bezel will be perpendicular to the mean and thus pass the most light (no consideration given to dispersion here10). The formula for this angle is therefore ...

B = (sum of light:angle products) / (sum of light)
= (S(BiLi) / S(L) ) = S(BiIiAi) / S (IiAi)

Editors note 1

    ... where "S" means "sum of terms" to mathematicians, and the "i"s indicate subscripts denoting the various light groups (C, L, R).

  Light Entering The Table  
Click for enlarge
Figure 3

Light enters the table at external angles from f = -90o to +90o, and its intensity normal to the table varies according to the cosine of this angle 3. Figure 3To enlarge Figure 3 shows the distribution of intensity for these rays at any point on the table; the area under this curve is the net intensity at any point

I = sin (+90o) - sin(-90o) = 1 - (-1) = 2

 The angle dividing this into three equal areas is ...

I/3 = 2/3 = sin(+90o) - sinf = 1 - sinf ...for which... f = 19.47o(±)
The corresponding internal angle is ............... q = 7.93o(±)

  Central Group Of Light Rays  
Click for enlarge
Figure 4a

This group consists of internal rays from q = -7.93o to +7.93o. None of these are lost through the pavilion, for which the limits on internal rays are:

qmin= -16.32o, qmax= +33.32o;
thus this group has maximum intensity IC = 2/3 =0.667
The mean internal angle of this group is qC = 0o

All of this group reflects off both sides of the pavilion and approaches the bezel. According to Eq.(1c) ...

BC = q3 = 4(45o-P) -q1 = 4(45o-40.75o) -0o = 17o ..... (4b)

The cross-sectional area of this beam, measured parallel to the table is the projection of a pavilion facet from the culet to a distance X = T/2 from the axis, as shown in Figure 4aTo enlarge Figure 4a. The beam therefore has a triangular section whose area is ...
(T/2)2 / 2 = T2/8 = 0.125T2 ..... (4c)

Tolkowsky does not mention that some of this beam hits the table (10.8%, which should be subtracted from the total light).

Editors note 2

  Group Oblique From Left  
Click for enlarge
Figure 4b
(shows mean beam only)

 This group consists of internal rays from q = +7.93o to +24.43o, except that these angles must be treated as negative after the rays cross the axis of the gem.

    Those rays that hit the near side of the pavilion are within the limits of qmin= -16.32o and qmax= +33.32o, thus none are lost. Most of these are reflected back to the table; only those from q = +7.93o to 8.5o (q2 = 90o) can hit the far bezel, and then only a few of those which are near the edge of the table.

    The rays that hit the far side of the pavilion directly change sign, and those from -qmin= +16.32o to +24.43o are lost through the pavilion. The remainder, from q = +7.93o to -qmin= +16.32o, are reflected back to the near-side pavilion and then again to the near-side bezel at angles which are (q1+q3) = 17o greater, thus from +24.93o to +33.32o.

    Tolkowsky discarded this entire group, saying only that they never hit the near side of the pavilion (and thus could not be reflected to the far bezel). This is incorrect, and those rays reflected back to the near-side bezel should properly be included (because of symmetry). Fortunately their angles are not far from the final mean of 35o, so that their neglect does not significantly affect the final answer.

  Group Oblique From Right  
Click for enlarge
Figure 4c

This group consists of internal rays from q = -7.93o to -24.43o. Rays from qmin= -16.32o to q = -24.43o are lost through the pavilion. Those which remain have external angles from f = -19.47o to fmin= -42.8o; their net intensity is therefore ...
IR = sin(-19.47o) - sin(-42.8o) = 0.347 = 0.521 IC (5a)

Tolkowsky got IR = 0.493 IC, mistakenly integrating the intensity lost 11.

The mean angle of this light is that which halves its area in Figure 3To enlarge Figure 3 ...
IR/2 = 0.1735 = sin(-19.47o) - sinf = -0.3333 - sinf
therefore ............... sinf = 0.5068 ...thus..... f = -30.5o
The corresponding internal angle is ............ q = -12.1o

This beam reflects off the near side of the pavilion and then most of it goes directly to the bezel. According to Eq.(1a) ...
BR = q2 = 2P+q1 = 2(40.75o)+(-12.1o) ............... BR = 69.4o       (5b)

    The cross-sectional area of this beam, measured parallel to the table, is the projection of a pavilion facet from Y to Z in Figure 4cTo enlarge Figure 4c - considering only that which hits the bezel directly; the remainder of this beam is used later to design the girdle facets because it hits in the corresponding area. The area of the beam which hits the bezel directly is ...
AR Y2/2 - Z2/2 ............... (5c)
or ............... AR/T2 (Y/T)2/2 - (Z/T)2/2

    Distance Y is defined by the ray q1 = -12.1o from the near edge of the table ...

thus ............... (Y/T)2/2 = (0.621)2/2 = 0.193

    Distance Z is defined by the ray q2 = +69.4o from the far edge of the girdle ...

... etc., until ...
Z / T = 0.196 + (0.263 / 0.861 + tanB)

(simplest form, least prone to error)

    This cannot be solved until bezel angle B is known, but is required to solve for B. It is not possible to rearrange the equations to solve for B explicitly because one term is just the angle B, while the other is tanB. Trial-and-error is the best approach in such cases. Taking values of B in the region where the answer is known to exist ...
q tanB Z/T (Z/T)2/2 AR/T2
35o 0.700 0.365 0.067 0.126
34o 0.674 0.368 0.068 0.125
33o 0.649 0.371 0.069 0.124

Obviously AR / T2 doesn't change significantly and can be assumed to be constant at 0.125 to simplify solution.......

AR/T2 0.125 ............... (5c)


    * 6 In Tolkowsky's final selection of bezel angle (34.5o), this ray is not refracted out of the gem; it hits the bezel at 26.68o incidence and is therefore reflected back into the gem! He apparently never checked-back to confirm his original premise.

    * 7 Had Tolkowsky used his original logic (mean internal ray approaching facet at 17o off-normal), he would have made the bezel angle 34o and been done except to figure the adjusted girdle width W and related stone dimensions (see next section "Final Angles and Proportion`s"). The rest of this section can be ignored.

    * 8 This alternative reason was seen and offered by this author, B.L.Harding.

    * 9 It is shown, in "Faceting Limits", that there are no bezel-to-bezel rays with the bezel angle B=34.5o that Tolkowsky finally chooses.

    * 10 This is the breakdown in Tolkowsky's logic. Earlier he chose the pavilion angle P=40.75o to produce 45o refraction of table-to-table rays, to get a compromise between brilliance and dispersion; now he is trying to put the bezel perpendicular to the internally-reflected light rays for maximum light return !

    * 11 This was his only mathematical error, which is only about 5% less than correct value.

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