Measuring capacitance by using two thresholds
First find at what time we reach 0.5 Vcc:

\$ 1 - e^{-t/R C} = 0.5 \$

That's for \$t\$ = 0.693147 \$R C\$. If we subtract 0.5 \$R C\$ from that we get

\$ V = 1 - e^{-(0.693147 - 0.5)/R C} = 0.175639 \text{  V}_{CC}\$ 

So the voltage varies from 0.1756 \$V_{CC}\$ to 0.5 \$V_{CC}\$ in 1/2 \$R C\$.

(I usually don't work with 6 significant digits, but here the discussion is between 0.17 and 0.18, and then it's relevant.)

edit
Note that you can use a current source instead of a voltage source + resistor to charge the capacitor.  



The blue line shows the charging over a resistor, the purple line at constant current. The more shallow the line, the larger the difference in time for the same voltage difference, and hence the larger the error. That's why you with the exponential curve you better don't measure above 80 % of \$V_{CC}\$. At constant current you have a constant precision over the full measurement range.