The following test case calculates the inelastic and transfer cross sections for the reactions induced by F on Si at 60.0 MeV. A Triton-Oxygen cluster structure is assumed for F, with the potentials for the triton and the oxygen core being folded together to give a diagonal Watanabe potential for elastic F scattering as well as the couplings to the inelastic 1/2+ state. These elastic and inelastic channels are coupled together to all orders, then a finite-range transfer step is calculated for the O + P channel, thus performing a CCBA calculation.
Test run: CCBA calculation 28Si(19F,16O) 31P, cluster form factors. NAMELIST &FRESCO hcm=.10 rmatch=25 rintp=0.50 hnl=0.100 rnl=3. jtmin=0. jtmax=80 absend=.01 kqmax=1 thmin=0. thmax=60.0 thinc=2.5 iter=1 iblock=2 nnu=30 chans=1 listcc=2 smats=1 elab=60.0 / &PARTITION namep='19-F' massp=19. zp=9 nex=2 namet='28-SI' masst=28 zt=14 / &states Jp=0.5 Bandp=+1 Ep=0.0 cpot=1 Jt=0.0 bandt=+1 Et=0.0 / &states Jp=2.5 Bandp=+1 Ep=0.2 / &PARTITION namep='16-O' massp=16. zp=8 nex=1 namet='31-P' masst=31 zt=15 qval=6.199/ &states Jp=0.0 Bandp=+1 Ep=0.0 cpot=2 Jt=0.5 bandt=+1 Et=0.0 / &partition / &pot kp=1 At=28.0 Ap=19.0 Rc=1.2 / &pot kp=2 At=31.0 Ap=16.0 Rc=1.35 V=31.20 rv=1.45 av=0.47 W=15.10 rw=1.270 aw=0.310 Vso=0.75 rso=1.24 aso=0.37 / &pot kp=3 At=19.0 Rc=1.25 ac=0.65 V=115.00 rv=1.250 av=0.650 Vso=6.30 rso=1.25 aso=0.650 / &pot kp=4 At=31.0 Rc=1.25 ac=0.65 V=99.000 rv=1.250 av=0.650 Vso=6.30 rso=1.25 aso=0.650 / &pot kp=5 At=28.0 Rc=1.2 V=27.00 rv=2.42452 av=0.460 W=11.66 rw=2.14456 aw=0.238 / &pot kp=6 At=28.0 Rc=1.25 ac=0.65 V=160.00 rv=1.07 av=0.72 W=37.0 rw=1.35 aw=0.88 Vso=10.0 rso=0.85 aso=0.850 / &pot / &OVERLAP kn1=1 ic1=1 ic2=2 in=1 kind=0 nn=4 l=0 sn=.5 j=0.5 kbpot=3 be=11.73 isc=1 / &OVERLAP kn1=2 ic1=1 ic2=2 in=2 kind=0 nn=4 l=0 sn=.5 j=0.5 kbpot=4 be=17.899 isc=1 / &OVERLAP kn1=3 ic1=1 ic2=2 in=1 kind=0 nn=3 l=2 sn=.5 j=2.5 kbpot=3 be=11.53 isc=1 / &overlap / &COUPLING icto=2 icfrom=1 kind=7 ip1=0 ip2=0 ip3=0/ &cfp in=1 ib=1 ia=1 kn=1 a=1.00 / &cfp in=1 ib=2 ia=1 kn=3 a=1.00 / &cfp in=2 ib=1 ia=1 kn=2 a=1.00 / &cfp / &COUPLING icto=1 icfrom=2 kind=3 ip1=4 ip2=1 ip3=0 kfrag=6 kcore=5/