PKC21=PT(JT,10)
PKC22=PT(JT,11)
ENDIF
+C
+C Quenching requested ?
+C
IF(IHPR2(4).GT.0.AND.IHNT2(1).GT.1.AND.IHNT2(3).GT.1) THEN
IF(NFP(JP,10).EQ.0) THEN
PHI=-PHI2
PT(JT,4)=PT(JT,4)-DPE2
ENDIF
ENDIF
- 16 DPKC11=-(PP(JP,10)-PKC11)/2.0
- DPKC12=-(PP(JP,11)-PKC12)/2.0
- DPKC21=-(PT(JT,10)-PKC21)/2.0
- DPKC22=-(PT(JT,11)-PKC22)/2.0
+ 16 CONTINUE
+ PKC11=0.
+ PKC12=0.
+ PKC21=0.
+ PKC22=0.
+ DPKC11=0.
+ DPKC12=0.
+ DPKC21=0.
+ DPKC22=0.
+
WP=EPP+ETP
WM=EPM+ETM
SW=WP*WM
ENDIF
ENDIF
- MISS=0
- PKC1=0.0
- PKC2=0.0
- PKC11=0.0
- PKC12=0.0
- PKC21=0.0
- PKC22=0.0
- DPKC11=0.0
- DPKC12=0.0
- DPKC21=0.0
- DPKC22=0.0
+ IF(NFP(JP,10).EQ.1) THEN
+ PPJET=SQRT(PP(JP,10)**2+PP(JP,11)**2)
+ PKC1=PPJET
+ ENDIF
+ IF(NFT(JT,10).EQ.1) THEN
+ PTJET=SQRT(PT(JT,10)**2+PT(JT,11)**2)
+ PKC2=PTJET
+ ENDIF
C ********If jet is quenched the pt from valence quark
C hard scattering has to reduced by d*kapa
C
ENDIF
C ********maximun PT kick
C*********************************************************
-C
IF(NFP(JP,10).EQ.1.OR.NFT(JT,10).EQ.1) THEN
IF(PKC1.GT.PKCMX) THEN
PKC1=PKCMX