1.16 求上升交点的简便方法（16K，15P）
求上升交点的简便方法：取奥古斯都纪元的整年数，乘以 19 1/3。每过一埃及月加 1° 35'，每日加 3'。以 360° 整圆除之。然后将 <除法的> 余数自 b 起沿周日运动的方向 <即自东向西> 数去，/29P/ 每一星座给 30°。计数终止之处即为上升交点。举例：哈德良 4 年，/30K/ Phamenoth 19。自奥古斯都以来的整年数为 148；此数乘以 19 1/3 得 2862。自 Thoth 至 Phamenoth 为 10°，合计 2872。我以 360° 除之，得 7，余 352。将此余数自 b 起沿周日运动方向数去，得 n 8°。所求的黄道点即在此处，降交点则在与之对冲之点。
论断要点
必须考察吉星是否与这些位置成相位，尤其是与上升交点成相位。若有，则本命昌盛而有效验。即便本命被断为平常或趋于减损，命主仍将上升而跻身高位。凶星则预示变故与控告。
由 <太阴历元与日行> 表，可如下求得上升交点及其纬度所属星座：举例来说，取前述本命，哈德良 4 年，Phamenoth 19。自历元至本命日期为 204。历元旁所载纬度为 12;18。204 旁所载纬度为 11;37。合计 23;55。以此乘以 15°，结果为 358° 45'。自 n 起沿自身运动的方向 <即自西向东> 数去，得 b 28° 45'。
17K.1
[Another more concise method: the 23;55 is counted
from n, 2 given to each sign. The count stops at v,
having allotted 22, with 1;55 remaining. This I multiply
by 15◦, and the result is 28◦ 45’ of b.)
Next in every case I take the degrees from c to the
previously determined degree; the distance is very nearly
89◦. I subtract this amount from the d’s degree-position
(which is C 7◦), and come to n 8◦, the ascending node.
It will be necessary to do the same calculation for the rest
of the nativities.
If I wish to know the sign of the latitude, I will calcu-
late as follows: the latitude entered next to the epoch is
12;18. I multiply only this by 15◦ and the result is 184◦
30’. I count this off from n and stop at N4◦ 30’. Next
the “degrees of latitude” entered next to 204 is 11;37. I
multiply this figure by 15◦, and the result is 174◦ 15’. I
add N 4◦ 30’ to this and count the sum off from the
same place. The result is b28◦ 45’. /30P/ By using this
method for the rest of the epochs we will find the sign of
the latitude.]/31K/
1这似乎是在 Kroll 本中一个未加标题的新章节的开端。