| x1 | |||
| x2 | (9.52) | ||
| x3 |
etc. Then the sequence of equations is solved by using the values from all the preceding steps at each step in the solution, and the procedure is repeated iteratively.
An example where the method is effective is in the calculation of the lifted condensation level (LCL) of an unsaturated air parcel. This is needed, for example, in the calculation of equivalent potential temperature for an unsaturated air parcel. The LCL is the point at which the parcel, lifted adiabatically, reaches saturation with respect to water. During this ascent, potential temperature and mixing ratio remain constant, so finding the LCL is equivalent to finding the temperature and pressure for which a saturated air parcel would have the same potential temperature and mixing ratio as the observed parcel. If the observed temperature, pressure, and dewpoint are T, p, and Td, and the temperature and pressure at the LCL are TL and PL, these conditions are equivalent to
| (9.53) |
| (9.54) |
If the last equation is solved for TL (given pL),
for example by Lagrange interpolation, then this system of equations readily
converges to the conditions at the LCL. The sequence for T=20,
Td=10, and p=800 mb is:
| TL | pL | |
| 293.15 | 800.0 | |
| 283.26 | 709.51 | |
| 281.37 | 693.09 | |
| 281.03 | 690.13 | |
| 280.96 | 689.59 | |
| 280.95 | 689.49 | |
| " | 689.47 | |
| " | " |