Lecture Notes of Dose distribution calculation inside phantom with ...

2.9 Step 9: Output of results ()write(1,280) sposi280 FORMAT(/' Absorbed energy inside phantom for 1.253MeV photon'/*' Source position ',F10.1,' cm from phantom surface'/*' Within 1cm x 1 cm area after 5 cm air')write(1,290) ncases, xhbeam, yhbeam290 FORMAT(1X,I8,' photons normally incident from front side'/*' Half width of beam is ',G15.5,'cm for X and ',G15.5,'cm for Y')! ---------------------------------------! Calculate average and its uncertainties! ---------------------------------------do k=1,kmaxdo j=1,jmaxdo i=1,imaxirl=1+i+(j-1)*imax+(k-1)*ijmaxamass=(xbound(i+1)-xbound(i))** (ybound(j+1)-ybound(j))** (zbound(k+1)-zbound(k))*rhor(irl)dose(i,j,k)=depeh(i,j,k)/ncasesdepeh2(i,j,k)=depeh2(i,j,k)/ncasesdoseun(i,j,k)=dsqrt((depeh2(i,j,k)-* dose(i,j,k)*dose(i,j,k))/ncases)dose(i,j,k)=dose(i,j,k)*1.602D-10/amassdoseun(i,j,k)=doseun(i,j,k)*1.602D-10/amassend doend doend do!----------------------------------------------! Print out the results of central phantom!----------------------------------------------i=imax/2+1j=jmax/2+1do kkk=2,kmax-1depths=zbound(kkk)depthl=zbound(kkk+1)irl=1+i+(j-1)*imax+(kkk-1)*ijmaxwrite(6,300) depths,depthl,(media(ii,med(irl)),ii=1,24),* rhor(irl),dose(i,j,kkk),doseun(i,j,kkk)300 FORMAT(' At ',F4.1,'--',F4.1,'cm (',24A1,',rho:',F8.4,')=',* G13.5,'+-',G13.5,'Gy/incident')end do!-------------------------------------------------! Calculate average exposure and its deviation!-------------------------------------------------area=(xbound(i+1)-xbound(i))*(ybound(j+1)-ybound(j))faexpa=faexps/ncasesfaexp2s=faexp2s/ncasesfaexrr=dsqrt((faexp2s-faexpa*faexpa)/ncases)faexpa=faexpa*1.6E-10/areafaexrr=faexrr*1.6E-10/areafexpsa=fexpss/ncasesfexps2s=fexps2s/ncasesfexerr=dsqrt((fexps2s-fexpsa*fexpsa)/ncases)fexpsa=fexpsa*1.6E-10/areafexerr=fexerr*1.6E-10/areaif (faexpa.gt.0.0) thenbsfa=fexpsa/faexpabsferr=bsfa*dsqrt((faexrr/faexpa)**2.+(fexerr/fexpsa)**2.)write(6,310) faexpa,faexrr,fexpsa,fexerr,bsfa,bsferr9