c$Author: fredfox $ c$Date: 2001-10-11 22:46:53 $ c$Revision: 1.1 $ c$Source: /weru/cvs/weps/weps.src/hydro/propsaxt.for,v $ subroutine propsaxt( sandf, clayf, sat, fc, pwp ) c Reference: K.E. Saxton et al., 1986, Estimating generalized soil-water c characteristics from texture. Soil Sci. Soc. Amer. J. 50(4):1031-1036 c + + + ARGUMENT DECLARATIONS + + + real sandf, clayf, sat, fc, pwp c + + + ARGUMENT DEFINITIONS + + + c sandf - fraction of soil mineral portion which is sand c clayf - fraction of soil mineral portion which is clay c sat - saturated volumetric water content c fc - 1/3 bar volumetric water content c pwp - 15 bar volumetric water content c + + + LOCAL VARIABLES + + + real sand_2, clay_2, percent_sand, percent_clay, acoef, bcoef c + + + LOCAL DEFINITION + + + c percent_sand - sand fraction expressed as percent c percent_clay - clay fraction expressed as percent c acoef - intermediate expression c bcoef - intermediate expression percent_sand = sandf * 100.0 percent_clay = clayf * 100.0 c equations are only valid in this range. This makes any outside c stay at the boundary value. percent_sand = max( min( percent_sand, 95.0), 5.0) percent_clay = max( min( percent_clay, 60.0), 5.0) sand_2 = percent_sand * percent_sand clay_2 = percent_clay * percent_clay acoef = exp(-4.396 - 0.0715 * percent_clay - & 4.88e-4 * sand_2 - 4.285e-5 * sand_2 * percent_clay) bcoef = - 3.140 - 0.00222 * clay_2 & - 3.484e-5 * sand_2 * percent_clay sat = 0.332 - 7.251e-4 * percent_sand & + 0.1276 * log10(percent_clay) if ((acoef .ne. 0.0) .and. (bcoef .ne. 0.0)) then fc = (0.3333/ acoef)**(1.0 / bcoef) pwp = (15.0 / acoef)**(1.0 / bcoef) end if c if (sat .ne. 0.0) then c ksat = exp((12.012 - 0.0755 * percent_sand) + c & (- 3.895 + 0.03671 * percent_sand c & - 0.1103 * percent_clay + 8.7546e-4 * clay_2) / sat) c end if return end