Every earlier control gate lesson in this book (chapters one, four, and six) fixed the target at ten and only ever scanned the controls or the tolerance. This scan holds the same three controls (9, 10, 11) and tolerance (1) fixed and moves the target instead: target 9 excludes the highest control, target 10 admits all three controls exactly matching chapter six's own checked example, and target 11 excludes the lowest control. The target is not a fixed reference point; it is one more value the gate has to be checked against.
Every earlier control gate lesson fixed the target at ten and scanned the controls or the tolerance instead; this scan holds the controls and tolerance fixed and moves the target. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.
highlighted = computed this step
Every earlier control gate held target fixed at ten
Chapters one, four, and six all kept target 10 fixed and only ever changed the controls or the tolerance. Here the same three controls 9, 10, and 11 and tolerance 1 stay fixed, but the target itself shifts to 9. The highest control now falls outside, so the row pass bit is 0.
∣11−9∣=2>1
The target itself decides which controls fit the gate
The controls and tolerance never change across these rows. Moving the target alone flips which controls sit inside the inclusive window, not just how many.
target91011pass bits(1,1,0)(1,1,1)(0,1,1)A010
A target shifted the other way excludes the low control instead
At target 11, control 9 is now the one outside tolerance 1, while the other two controls still pass. The row pass bit is 0.