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<br />3 <br />free-pipe travel times as seen on the log, use them. When <br />interpreting the log, a decrease in travel time (faster <br />times) with simultaneous reduction of amplitude may show a <br />de-centered tool. A 4 to 5 micro-second (µs) decrease in <br />travel time corresponds to about a 35$ loss of amplitude. A <br />decrease in travel time more than 4 to 5 µs is unacceptable. <br />PART V - LOG INTERPRETATION <br />Do not rely on the service company charts for amplitudes <br />corresponding to a good bond. These amplitudes depend on <br />many factors: type of cement used, fluid in the hole, etc. <br />To estimate bond index, choose intervals on the log that <br />correspond to 0$ bond and S00$~bond. Read the amplitude <br />corresponding to 100$ bond from the best-bonded interval on <br />the log (NOTE: the accuracy of this amplitude reading is <br />very critical to the bond index calculations). Next, find . <br />the amplitude corresponding to 0•$ bond. Some bond logs may <br />not include a section with free pipe. In this instance, <br />choose the appropriate free-pipe travel time from the <br />service company charts for your specific tool, or from the <br />generalized chart.(TABLE 2) at the end of this guidance. To <br />calculate a bond index of 80$, use the following equation: <br />A =ZO~(0.2)log(Ao)-(0.8)log(Aloo)I <br />80 <br />where: <br />A80 =Amplitude at 80$ bond (mV) <br />AO = Amplitude at 0$ bond (mV) <br />Aloo = Amplitude at i00$ bond (mV) <br />EXAMPLE <br />As an example, consider a bond log showing the following <br />conditions: <br />- Free pipe (0$ bond) amplitude at 81 mV. <br />- 100 $ bond amplitude at 1 mV. <br />Substituting the above values into the equation results in: <br />A =10((0.2)logcal)-(o.8)log(1)] <br />80 <br />A80=2 .41mV <br />