Schmitt trigger low threshold calculation
Where does the number 11 come from?


Figure 1. The current path when the left transistor is on.
If we ignore the voltage drop across the left transistor when hard on then we have a potential divider given by the two resistors.
$$ V_{U2} = \frac {470}{4k7 + 470} {V_{CC} } = \frac 1 {11} V_{CC} $$

Why does one need to add the emitter voltage to the BE-voltage?

Since Q1's emitter is at 0.8 V we'll need to raise the base by 0.6 V to turn it on. Thus any voltage above 1.4 V will keep Q1 turned on and steal the bias for Q2. Only as you pull Q1's base below 1.4 V will it start to turn off and allow Q2 to turn on.