SiO2 = 3(1.05 + 1) / 4
Or:
SiO2 = 6.15 / 4 = 1.5375
But as such a complete fraction is not necessary it may be stated as 1.54 equivalent. The formula would therefore be:
RO, Al2O3 .35, SiO2 1.54
The RO content should not be too fusible. Lead oxide is desirable up to about .5 equivalent and it is an advantage to use feldspar so that K2O may be introduced. Calcium oxide is also good but zinc oxide must be used sparingly as it is apt to suffer if overfired. The high content of alumina necessitates a good deal of clay and as this, if used raw, would make the glaze too plastic and cause it to crack, it is best to calcine a part of it, thus removing the combined water and changing the equivalent weight from 258 to 222. The calculation will then proceed as in the case of a bright glaze.
| PbO | .50 | } | ||||
| CaO | .35 | } | Al2O3 | .35 | SiO2 | 1.54 |
| K2O | .15 | } | ||||
| —— | } | |||||
| 1.0 | } |
| PbO | CaO | K2O | Al2O3 | SiO2 | |||
| .50 | .35 | .15 | .35 | 1.54 | |||
| Addition | .50 | White Lead | .50 × 258 = 129 | ||||
| Subtraction | .0 | .35 | .15 | .35 | 1.54 | ||
| Addition | .35 | Whiting | .35 × 100 = 35 | ||||
| Subtraction | .0 | .15 | .35 | 1.54 | |||
| Addition | .15 | .15 | .90 | Feldspar | .15 × 557 = 83 | ||
| Subtraction | .0 | .20 | .64 | ||||
| Addition | .15 | .30 | Calcined Kaolin | .15 × 222 = 33 | |||
| Subtraction | .05 | .34 | |||||
| Addition | .05 | .10 | Kaolin | .05 × 258 = 13 | |||
| Subtraction | .0 | .24 | |||||
| Addition | .24 | Flint | .24 × 60 = 14 | ||||
| Subtraction | .0 |
The mix, therefore, is: