关于GSM44018里starting time的判断，不明白是什么意思

tsdododo

44018里关于starting time的定义如下

The Starting Time IE can encode only an interval of time of 42 432 frames, that is to say around 195.8 seconds. To remove any ambiguity, the specification for a reception at time T is that the encoded interval is (T-10808, T+31623). In rigorous terms, if we note ST the starting time:
        if 0 <= (ST-T) mod 42432 <= 31623, the indicated time is the next time when FN mod 42432 is equal to ST;
        if 32024 <= (ST-T) mod 42432 <= 42431, the indicated time has already elapsed.
The reception time T is not specified here precisely. To allow room for various MS implementations, the limit between the two behaviours above may be anywhere within the interval defined by:
        31624 <= (ST-T) mod 42432 <= 32023.

问题1. 当前的接收时间T 为什么是在(T-10808, T+31623)里面，10808和31623是怎么来的？协议里没有这方面的介绍。

问题2. 判断starting time到没到，根据 (ST-T) mod 42432 的值来判断，spec给了三个区间，这三个区间是怎么定义的？看了半天没明白啊？

求指点。谢谢啊