 5.1, 5.2, and 5.4. The theoretical design can now be calculated from the solution of two simultaneous equations. The following example was taken from Ref. 6. Suppose we wished to design a 32-kft loop with a maximum loop resistance of 1300 S. If we were to use 24-gauge copper pair, Table 5.1 shows that we exceed the 1300-S limit; if we use 22 gauge, we are under the limit by some amount. Therefore, what combination of the two gauges in series would just give us 1300 S2? The loop requires five H66 load coils, each of which has a 9-S2 resistance. It should be noted that the 1300-S2 limit value does not include the resistance of the telephone subset. Let X be the kilofeet value of the length of the 24-gauge pair and let Y be the kilofeet value of the length of the 22-gauge pair. Now we can write the first equation: = 32 kft. Table 5.1 shows the resistance of a 24-gauge wire pair and of a 22-gauge wire pair as 51.9 S2/kft, and 32.4 S2/kft, respectively. We can now write a second simultaneous equation: 51.9X +32.4Y+5(9) =1300 S X = 11.2 kft of 24-gauge cable - X = 20.8 kft of 22-gauge cable. (See Appendix B for the solution of simultaneous equations.) We stated earlier that if the resistance design rules are followed, the North American 8-dB objective loss requirement will be met for all loops. However, to ensure that this is the case, these additional rules should be followed: Inductive load all loops over 18 kft long. Limit the cumulative length of all bridged taps on nonloaded loops to 6 kft or less. Reference 6 recommends H88 loading where we know the spacing between load coils is 6000 ft with a spacing tolerance of 120 ft. Wherever possible, it is desirable to take deviations greater than 120 ft on the short side SO that correction may later be applied by normal build-out procedures. The first load section out from the serving switch is 3000 ft for H66/H88 loading. In the measurement of this length, due consideration should be given to switch wiring SO that the combination is equivalent to 3000 ft. It should be r