As a supplement to the 5th teaching day, I sat down at home and reconstructed some of the format settings for the table using the punchtape.

Noted on the punchtape is “Franz Tabelle / 11¼ set Gill 16 Cicero”. This makes the format width 18-4 (18 quads and 4 units). We decide on the first column to be 10 Cicero (11-7) wide, the second 6 Cicero (6½-6). 11-7 + 6½-6 gives 18-4. The difference of the widths 11-7 – 6½-6 is 4½-1, this is needed when setting the second column. The format scale on the keyboard is now set to 11-7 and the first column is set.

I will now use the example of the line “Helium …………… 5.26” to describe the punchtape. At **A**, the line before is terminated, **B **and **C **belong to the first column. **D **is incorrectly set text that is terminated with the line killer. **E**, **F **and **G **belong to the second column and end the line.

Lets begin with the first column: It starts at **B **with “Helium”, you can easily recognise the first and the last letter by their high unit value (=large width). The column shall be ended with punctuation, and I have therefore used fixed width spaces. That makes 2x G2 (6-unit fixed space) and 15 half-quad dots. For fun, we can calculate the unit widths by looking in the “Unit Arrangements Of Monotype Composition Matrices”: H(14)+e(9)+l(5)+i(5)+u(10)+m(15)+2×6+15×9 = 205 = 11-7, which is exactly our column width. At **C **the column is terminated “top” then “bottom” (which, strictly speaking, is unnecessary, but resets the pointer at the format scale).

The second column is narrower than the first, and before it is set, the width at the pointer is at first reduced by the difference between the column widths, 4½-1. To do this (**E**), the upper justififcation is keyed four times with the 15th row and once with the 9th row. The Gill has an S5 set wedge (5-6-7-8-9-9-10-11-12-13-14-15-18), there the 15th row has a value of 18 units, and the 9th has 10 units. Doing so we have reduced the row width by exactly 4½-1. After that (**F**), come two 18 unit quads, two 9 half-quads, the “5”, the dot (5 units), then “2” and “6”. The digits all have 9 units. Then one quad, one half quad and two 5-unit spaces. So the total width is 2×18+2×9+9+1×5+2×9+18+9+2×5 = 123 = 6½-6, exactly the width of the column. At the end (**G**), first single “top”, then double “top/bottom” justification is punched, to conclude the line.