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## glaze course lesson four part two july 24

### hal mc whinnie on mon 24 jul 00

ON LINE CERAMIC GLAZE COURSE
LESSON FOUR PART TWO
The seger method of glaze calculation
Sample
A seger formula for a cone 10 wheat blue glaze [blue-black]
Empirical formula
Na2o . 20 al2o3 . 30 sio2 3.00
Mgo . 20
Zno . 15
Cao . 15
Bao . 20
Cobalt .10 note1 note 2

Note1 subtract the value of the al2o3 from the silica required
Note2 subtract from the amount required of sio2 the silica content of the
spar

To convert a grams based glaze receipt you multiple the value for each glaze
element by the molecular equilvent of that element

Element mole in formula value of mole in material final amount
Na2o .20 x 106 = 21.2 soda spar
Mgo .20 x 84= 16.8 magnesium carb
Zno .15 x 81 = 12.1 zinc
Cao .15 x 100= 15 g of whiting
Bao .20 x 196 = 39.5 g barium
Al2o3 .30 x 258 = 77 g kaolin
Sio2 2.40 x 60= 144 g flint
Coo .10 x 58= 5.8 g cobalt

Final receipt for wheat blue
39 barium carb
144 flint
77 kaolin
16 magnesium carb
21 soda spat
15 whiting
12 zinc oxide
6 cobalt carb

Melter power in normal range
Melter amount low

Silica high
Alumna normal
Melters low

"If glazes in this category contain reasonable strong melters they will be
vitreous. Their high silica content encourages durability and transparency.
Their normal alumna content discourages glaze flow even if strong melters
are present. This is an excellent category for transparent glazes." From
computer analysis of Richard zakin

For review let us look at another glaze

Blue-green dry Matt cone 6 reduction

Mgo . 23 al2o3 .50 sio2 2.00
Fe2o3 . 01
Cuo . 01
Cao . 70

Mgo .23 x 84 =19g magnesium
Fe2o3 .01 x 159 = 15 g iron
Cuo .01 x 79 = 7 g copper
Cao .70 x 56 = 39 g whiting
Al2o3 .50 x 258 = 109 g kaolin
Sio2 2.00 x 60 = 120 g flint

Final formula for blue green Matt

120 flint
15 iron oxide
109 kaolin
19 magnesium carb
39 whiting

Melting power in normal range
Melter amount low
Sio2 high
Al2o3 normal
Melters low
"If glazes in this category contain reasonably strong melters they will be
vitreous. Their high silica content encourages durability and transparency.
Their normal alumna content discourages glaze flow even if strong melters
are present. This is an excellent category for transparent glazes." Zakin
analysis

Now for a glaze problem

Beckman red cone 8-10 reduction

Bao . 20 al2o3 .50 sio2 3.00
Cao . 40
Knao . 10
Zno . 30
Cuo 2%

Final formula consists of

Element total 100%total
Sio2 189.01 56.137
Al2o3 51.51 15.289
Cao 21.73 6.454
Tio2 2.58 0. 766
Bao 30.81 9.15
Li2o . 65 0. 193
Zno 24 7.128
P2o5 16.4 4.870

Sio2 56.14
Al2o3 15.3
Melter 28.56

Enough numbers for one lesson