Answer
Using Equation 17.23, the parameter K must convert the units of W, ρ, A, and t, into the unit scheme for the CPR.
CPR = $\frac{KW}{pAt}$
Use unit conversions to show that K = 534 mpy and 87.6 mm/yr.
For CPR in mpy (mil/yr):
K= $\frac{W (mg) (1 g/1000mg)}{p(\frac{g}{cm^{3}}) (\frac{2.54 cm}{1 in.})^{3} [A (in^{2})] (\frac{1 in}{1000 mil}) [t (h)] (\frac{1 day}{24 h}) (\frac{1 yr}{365 days}) }$
K= 534.6
For CPR in mm/yr:
K= $\frac{W (mg) (1 g/1000mg)}{p(\frac{g}{cm^{3}}) (\frac{1 cm}{10 mm.})^{3} [A (cm^{2})] (\frac{10 mm}{cm})^2 [t (h)] (\frac{1 day}{24 h}) (\frac{1 yr}{365 days}) }$
K= 87.6
Work Step by Step
Using Equation 17.23, the parameter K must convert the units of W, ρ, A, and t, into the unit scheme for the CPR.
CPR = $\frac{KW}{pAt}$
Use unit conversions to show that K = 534 mpy and 87.6 mm/yr.
For CPR in mpy (mil/yr):
K= $\frac{W (mg) (1 g/1000mg)}{p(\frac{g}{cm^{3}}) (\frac{2.54 cm}{1 in.})^{3} [A (in^{2})] (\frac{1 in}{1000 mil}) [t (h)] (\frac{1 day}{24 h}) (\frac{1 yr}{365 days}) }$
K= 534.6
For CPR in mm/yr:
K= $\frac{W (mg) (1 g/1000mg)}{p(\frac{g}{cm^{3}}) (\frac{1 cm}{10 mm.})^{3} [A (cm^{2})] (\frac{10 mm}{cm})^2 [t (h)] (\frac{1 day}{24 h}) (\frac{1 yr}{365 days}) }$
K= 87.6
