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tcclab

银虫 (正式写手)

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注册: 2008-02-22
专业: 理论和计算化学

[交流] 已知三点坐标求夹角已有2人参与

最近需要处理大量数据，需要对化学键键角批量输出。
我已经把原子坐标以xyz的形式给出。
现在搞不定如何把夹角以degree（度数）的方式给求出来。
哪位知道怎么弄？
本人很菜，别笑话。
#It is 4 Bond Angles
for XYZ in `ls *.xyz`
do
#read in the atom Nr.
AN1=`echo $[$1+2]`
AN2=`echo $[$2+2]`
AN3=`echo $[$3+2]`
echo $AN1 $AN2 $AN3
#The first atom
A11=`awk "NR==$AN1" $XYZ |awk '{print $1}'`
A1x=`awk "NR==$AN1" $XYZ |awk '{print $2}'`
A1y=`awk "NR==$AN1" $XYZ |awk '{print $3}'`
A1z=`awk "NR==$AN1" $XYZ |awk '{print $4}'`
#The second atom
A21=`awk "NR==$AN2" $XYZ |awk '{print $1}'`
A2x=`awk "NR==$AN2" $XYZ |awk '{print $2}'`
A2y=`awk "NR==$AN2" $XYZ |awk '{print $3}'`
A2z=`awk "NR==$AN2" $XYZ |awk '{print $4}'`
#
A31=`awk "NR==$AN3" $XYZ |awk '{print $1}'`
A3x=`awk "NR==$AN3" $XYZ |awk '{print $2}'`
A3y=`awk "NR==$AN3" $XYZ |awk '{print $3}'`
A3z=`awk "NR==$AN3" $XYZ |awk '{print $4}'`
#
echo -e "$A11\t$A1x\t$A1y\t$A1z\t"
echo -e "$A21\t$A2x\t$A2y\t$A2z\t"
echo -e "$A31\t$A3x\t$A3y\t$A3z\t"
#
TT=`echo -e "$A11\t$A1x\t$A1y\t$A1z\t$A21\t$A2x\t$A2y\t$A2z\t$A31\t$A3x\t$A3y\t$A3z\t" `
echo $TT
#
A1A2=`echo $TT | awk '{print $6-$2,$7-$3,$8-$4}' `
echo A1A2 $A1A2
A1A2X=`echo $TT | awk '{print $6-$2}' `
A1A2Y=`echo $TT | awk '{print $7-$3}' `
A1A2Z=`echo $TT | awk '{print $8-$4}' `
A2A3=`echo $TT | awk '{print $10-$6,$11-$7,$12-$8}' `
echo A2A3 $A2A3
A2A3X=`echo $TT | awk '{print $10-$6}' `
A2A3Y=`echo $TT | awk '{print $11-$7}' `
A2A3Z=`echo $TT | awk '{print $12-$8}' `
A1A2A2A3=`echo $A1A2  $A2A3 `
echo A1A2A2A3 $A1A2A2A3
#乘积A1A2*A2A3=(x2-x1)*(x3-x2)+(y2-y1)*(y3-y2)+(z2-z1)*(z3-z2)
TA1A2A2A3=`echo $A1A2A2A3 | awk '{print $1*$4+$2*$5+$3*$6}'`
echo TA1A2A2A3 $TA1A2A2A3
#
A1A2A1A2=`echo $A1A2 | awk '{print $1^2+$2^2+$3^2}'`
A2A3A2A3=`echo $A2A3 | awk '{print $1^2+$2^2+$3^2}'`
echo A1A2A1A2 $A1A2A1A2
echo A2A3A2A3 $A2A3A2A3
#|P1P2|=根号[(x2-x1)2+(y2-y1)2+(z2-z1)2] |P2P3|=根号[(x3-x2)2+(y3-y2)2+(z3-z2)2]
#var absA1A2A2A3=A1A2*A2A3
absA1A2=`echo $A1A2A1A2 | awk '{print sqrt($1)}'`
echo absA1A2 $absA1A2
absA2A3=`echo $A2A3A2A3 | awk '{print sqrt($1)}'`
echo absA2A3 $absA2A3
#
#cos(A1A2,A2A3)=A1A2*A2A3/(|A1A2|*|A2A3|)
A1A2A3=`echo $TA1A2A2A3 $absA1A2 $absA2A3 `
echo A1A2A3 $A1A2A3
# 前面检查，读入和输出，应该是正确的，但下面这部分搞不定了
cosA1A2A3=`echo $A1A2A3 | awk '{print $1/($2*$3)}'`
#弧度=角度乘以π后再除以180 角度=弧度除以π再乘以180
#pi=3.1415926535898
cosA1A2A3=`echo $cosA1A2A3 | awk '{print cos($1)}'`
Angle=`echo $acosA1A2A3 | awk '{print $1*180/3.1415926535898}' `
echo $A1A2 $A1A2XX $A1A2YY $A1A2ZZ $A1A2A1A2 $cosA1A2A3
echo $cosA1A2A3
echo $Angle
done

xyz 文件如下：
36

  Fe  4.84655858507584    0.56633277215833    0.34035878855785
  Al  4.79235130609276    2.90413572930704    0.18293815072370
  H 4.28535603237677    3.97006317825069    1.30657195333100
  O 1.96706095735722    1.03530980080275    0.77397530855226
  O 4.93691132336707    -2.35361098202682    0.67632768640727
  O 6.77764534437593    1.25909183704602    2.45505687074638
  O 5.76193037993391    0.63511115108140    -2.45907090865639
  N 2.60533354124266    4.09822722851467    -1.75829932419780
....

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