Article “Corps Franconia Jena” in German Wikipedia has 43.3 points for quality (as of July 1, 2025). The article contains 9 references and 14 sections.
In June 2025 the article “Corps Franconia Jena” was edited by 1 authors in German Wikipedia and written by 1 authors in all languages.
Since the creation of article “Corps Franconia Jena”, its content was written by 61 registered users of German Wikipedia and edited by 61 registered Wikipedia users in all languages.
The article is cited 200 times in German Wikipedia and cited 200 times in all languages.
The highest Authors Interest rank from 2001:
The highest popularity rank from 2008:
There is 1 language version for this article in the WikiRank database (of the considered 55 Wikipedia language editions).
The quality and popularity assessment was based on Wikipédia dumps from July 1, 2025 (including revision history and pageviews for previous years).
| Best Rank | German: #172677 11.2011 |
Global: #1108042 12.2008 |
| Best Rank | German: #11683 03.2020 |
Global: #187480 11.2006 |
On 16 March 2026 in multilingual Wikipedia, Internet users most often read articles on the following topics: 98th Academy Awards, Michael B. Jordan, Sinners, Jessie Buckley, Paul Thomas Anderson, Sean Penn, One Battle After Another, Amy Madigan, Hamnet, Timothée Chalamet.
In German Wikipedia the most popular articles on that day were: Oscarverleihung 2026, One Battle After Another, Blood & Sinners, Michael B. Jordan, Jessie Buckley, Jürgen Habermas, Sean Penn, Banksy, Liste der größten Auslegerbrücken, Amy Madigan.
The WikiRank project is intended for automatic relative evaluation of the articles in the various language versions of Wikipedia. At the moment the service allows to compare over 44 million Wikipedia articles in 55 languages. Quality scores of articles are based on Wikipedia dumps from July, 2025. When calculating current popularity and AI of articles data from June 2025 was taken into account. For historical values of popularity and AI WikiRank used data from 2001 to 2025... More information
$$Quality=1/c∑↙{i=1}↖c nm_i-RS=1/5(28.71+12.86+64.29+66.67+44.21)$$
where:
The results of the calculations may be slightly different because of rounded values of normalized metrics in formulas
