在网上看了一些queryNorm的计算公式，但是怎么都算不对，请问哪位能够具体说说是由哪些值得到的？
<str name="defType">edismax</str>
<str name="pf">
scale^50 contractname
</str>
<str name="qf">
scale contractname
</str>
"\n0.024871612 = (MATCH) sum of:\n
0.01114073 = (MATCH) sum of:\n
0.009402936 = (MATCH) max of:\n
0.0056990404 = (MATCH) weight(contractname:合同 in 8) [DefaultSimilarity], result of:\n
0.0056990404 = score(doc=8,freq=1.0 = termFreq=1.0\n), product of:\n
0.008392063 = queryWeight, product of:\n
1.8109303 = idf(docFreq=3, maxDocs=9)\n
0.0046341172 = queryNorm\n
0.67909884 = fieldWeight in 8, product of:\n
1.0 = tf(freq=1.0), with freq of:\n
1.0 = termFreq=1.0\n
1.8109303 = idf(docFreq=3, maxDocs=9)\n
0.375 = fieldNorm(doc=8)\n
0.009402936 = (MATCH) weight(scale:合同 in 8) [DefaultSimilarity], result of:\n
0.009402936 = score(doc=8,freq=2.0 = termFreq=2.0\n), product of:\n
0.008392063 = queryWeight, product of:\n
1.8109303 = idf(docFreq=3, maxDocs=9)\n
0.0046341172 = queryNorm\n
1.1204559 = fieldWeight in 8, product of:\n
1.4142135 = tf(freq=2.0), with freq of:\n
2.0 = termFreq=2.0\n
1.8109303 = idf(docFreq=3, maxDocs=9)\n
0.4375 = fieldNorm(doc=8)\n
0.0017377939 = (MATCH) max of:\n
0.0017377939 = (MATCH) weight(contractname:规范 in 8) [DefaultSimilarity], result of:\n
0.0017377939 = score(doc=8,freq=1.0 = termFreq=1.0\n), product of:\n
0.0046341172 = queryWeight, product of:\n
1.0 = idf(docFreq=8, maxDocs=9)\n
0.0046341172 = queryNorm\n
0.375 = fieldWeight in 8, product of:\n
1.0 = tf(freq=1.0), with freq of:\n
1.0 = termFreq=1.0\n
1.0 = idf(docFreq=8, maxDocs=9)\n
0.375 = fieldNorm(doc=8)\n
0.013730882 = (MATCH) weight(contractname:\"合同 规范\" in 8) [DefaultSimilarity], result of:\n
0.013730882 = score(doc=8,freq=1.0 = phraseFreq=1.0\n), product of:\n
0.013026181 = queryWeight, product of:\n
2.8109303 = idf(), sum of:\n
1.8109303 = idf(docFreq=3, maxDocs=9)\n
1.0 = idf(docFreq=8, maxDocs=9)\n
0.0046341172 = queryNorm\n
1.0540988 = fieldWeight in 8, product of:\n
1.0 = tf(freq=1.0), with freq of:\n
1.0 = phraseFreq=1.0\n
2.8109303 = idf(), sum of:\n
1.8109303 = idf(docFreq=3, maxDocs=9)\n
1.0 = idf(docFreq=8, maxDocs=9)\n
0.375 = fieldNorm(doc=8)\n"
