The procedure for using the tables is as follows:

- Determine the fractional Julian day number, , corresponding to the date and time at which the ecliptic longitude is to be calculated with the aid of Tables 27-29. Form , where is the epoch.
- Calculate the ecliptic longitude, , and radial anomaly, , of the sun using the procedure set out in Sect. 5.1.
- Enter Table 58 with the digit for each power of 10 in and take out the corresponding values of and . If is negative then the corresponding values are also negative. The value of the mean longitude, , is the sum of all the values plus value of at the epoch. Likewise, the value of the mean anomaly, , is the sum of all the values plus the value of at the epoch. Add as many multiples of to and as is required to make them both fall in the range to . Round to the nearest degree.
- Enter Table 59 with the value of and take out the corresponding value of the equation of center, , and the radial anomaly, . It is necessary to interpolate if is odd.
- Form the epicyclic anomaly, . Add as many multiples of to as is required to make it fall in the range to . Round to the nearest degree.
- Enter Table 60 with the value of and take out the corresponding values of , , and . If then it is necessary to make use of the identities and .
- Form .
- Obtain the values of and from Table 5. Form .
- Enter Table 17 with the value of and take out the corresponding values of and . If then it is necessary to use the identities and .
- Form the equation of the epicycle, .
- The ecliptic longitude, , is the sum of the ecliptic longitude of the sun, , and the equation of the epicycle, . If necessary convert into an angle in the range to . The decimal fraction can be converted into arc minutes using Table 31. Round to the nearest arc minute. The final result can be written in terms of the signs of the zodiac using the table in Sect. 2.6.

*Example 1*: May 5, 2005 CE, 00:00 UT:

From Cha. 8,
JD,
, and
. Making use of
Table 58, we find:

(JD) | ||

+1000 | ||

+900 | ||

+50 | ||

+.5 | ||

Epoch | ||

Modulus | ||

Given that , Table 59 yields

so

It follows from Table 60 that

Now,

However, from Table 5, and , so

According to Table 17,

so

Finally,

Thus, the ecliptic longitude of Venus at 00:00 UT on May 5, 2005 CE was 23TA45.

*Example 2*: December 25, 1800 CE, 00:00 UT:

From Cha. 8,
JD,
, and
. Making use of
Table 58, we find:

(JD) | ||

-70,000 | ||

-2,000 | ||

-600 | ||

-90 | ||

-.5 | ||

Epoch | ||

Modulus | ||

Given that , Table 59 yields

so

It follows from Table 60 that

Now,

so

According to Table 17,

so

Finally,

Thus, the ecliptic longitude of Venus at 00:00 UT on December 25, 1800 CE was 8AQ5.