<p>I don't think swisseph has that capability, but skyfield.searchlib.find_discrete() is made for this sort of thing.
You would need to pass it a function which returns an integer indicating whether the node longitude at a given time is east or west of the value you're searching for.
<a href="https://rhodesmill.org/skyfield/searches.html" rel="nofollow noreferrer">The documentation</a> and an <a href="/a/54642">answer here</a> have relevant examples.</p>

CC BY-SA 4.0
2023-10-03T19:50:11.870Z
thank you Mike, i really appreciate it <3
dimitri33
CC BY-SA 4.0
2023-10-05T22:38:36.127Z
hey Mike G, can you give me another hint with this please ... i can't even sleep as i keep going over this thing :(  https://pastebin.com/NBUMFrtH
dimitri33
CC BY-SA 4.0
2023-10-06T04:00:05.713Z
1. observe from Earth instead of the Sun;
2. let function return ((observed_longitude - longitude) // 180).astype(int) % 2;
3. step_days = ~1/4 of the expected period;
4. filter longitude_times by in_state == 0
Mike G
CC BY-SA 4.0
2023-10-06T04:24:42.377Z
https://pastebin.com/dcUDALrB
dimitri33
CC BY-SA 4.0
2023-10-06T04:35:04.780Z
That code looks at the longitude of the Moon itself and should find 12 or 13 crossings per year. Now try looking at its [orbital elements](https://rhodesmill.org/skyfield/elements.html).
Mike G
CC BY-SA 4.0
2023-10-06T18:11:01.113Z
https://pastebin.com/z5S4HjCQ .... Not getting it, however I did played with the documentation that you posted first. I like that <3 however been a noob can be frustrating as hell  :)
dimitri33
CC BY-SA 4.0
2023-10-03T17:16:56.850Z
2023-10-03T17:16:56.850Z
Mike G
<p>It appears that that swetest can tell the exact longitude as long as you give it the date and time</p>
<pre><code>./swetest -pt -b17.11.2023 -ut22:43
date (dmy) 17.11.2023 greg.   22:43:00 UT        version 2.10.03
UT:  2460266.446527778     delta t: 69.109781 sec
TT:  2460266.447327659
Epsilon (t/m)     23°26'18.3245   23°26'10.2223
Nutation          -0° 0' 7.8330    0° 0' 8.1021
true Node         24°29'59.7465    0° 0' 0.0000    0.002502442   -0° 3'35.2933
</code></pre>
<p>Now my question is can one get the time and date for the nodes if it gives the longitude  lets say <code>24.5</code> same as the one in the swetest example .. can we calculate using swisseph, kind a like how solcross does  or mooncross ?  I only need a little hint on how to do it .. thank u &lt;3</p>

CC BY-SA 4.0
2023-10-02T02:17:34.230Z
The ``` has to go on a line by itself.
Mike G
<the-moon>
North Node cross longitude?
dimitri33
# North Node cross longitude?

(Asked by: username_0 on 2023-10-01)

It appears that that swetest can tell the exact longitude as long as you give it the date and time

```
./swetest -pt -b17.11.2023 -ut22:43
date (dmy) 17.11.2023 greg.   22:43:00 UT        version 2.10.03
UT:  2460266.446527778     delta t: 69.109781 sec
TT:  2460266.447327659
Epsilon (t/m)     23°26'18.3245   23°26'10.2223
Nutation          -0° 0' 7.8330    0° 0' 8.1021
true Node         24°29'59.7465    0° 0' 0.0000    0.002502442   -0° 3'35.2933

```

Now my question is can one get the time and date for the nodes if it gives the longitude lets say `24.5` same as the one in the swetest example .. can we calculate using swisseph, kind a like how solcross does or mooncross ? I only need a little hint on how to do it .. thank u <3

**Tags:** the-moon

> **Comments:**
> **username_1**: The ``` has to go on a line by itself.  

---

## 1 answers

### Answer by username_1 on 2023-10-03. Score: 1 | Accepted answer

I don't think swisseph has that capability, but skyfield.searchlib.find\_discrete() is made for this sort of thing.
You would need to pass it a function which returns an integer indicating whether the node longitude at a given time is east or west of the value you're searching for.
[The documentation](https://rhodesmill.org/skyfield/searches.html) and an [answer here](/a/54642) have relevant examples.

> **Comments:**
> **username_0**: thank you username_1, i really appreciate it <3  
> **username_0**: hey username_1 G, can you give me another hint with this please ... i can't even sleep as i keep going over this thing :(  https://pastebin.com/NBUMFrtH  
> **username_1**: 1. observe from Earth instead of the Sun; 2. let function return ((observed_longitude - longitude) // 180).astype(int) % 2; 3. step_days = ~1/4 of the expected period; 4. filter longitude_times by in_state == 0  
> **username_0**: https://pastebin.com/dcUDALrB  
> **username_1**: That code looks at the longitude of the Moon itself and should find 12 or 13 crossings per year. Now try looking at its [orbital elements](https://rhodesmill.org/skyfield/elements.html).  
> **username_0**: https://pastebin.com/z5S4HjCQ .... Not getting it, however I did played with the documentation that you posted first. I like that <3 however been a noob can be frustrating as hell  :)  

---

