Steven 'Bo' Keeley - American adventurer. Article “Steven 'Bo' Keeley” in English Wikipedia has 59.4 points for quality (as of July 1, 2025). The article contains 65 references and 11 sections. The article also contains templates indicating quality issues, therefore its score was reduced by 3.12 points.
Since the creation of article “Steven 'Bo' Keeley”, its content was written by 53 registered users of English Wikipedia and edited by 53 registered Wikipedia users in all languages.
The article is cited 10 times in English Wikipedia and cited 10 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 | English: #654649 01.2015 |
Global: #1140927 01.2015 |
| Best Rank | English: #29820 07.2011 |
Global: #162523 07.2011 |
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 English Wikipedia the most popular articles on that day were: 98th Academy Awards, Michael B. Jordan, Paul Thomas Anderson, Jessie Buckley, Sinners (2025 film), Amy Madigan, Autumn Durald Arkapaw, Maya Rudolph, Sean Penn, Academy Awards.
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
$$Quality=1/c∑↙{i=1}↖c nm_i-RS=1/5(56.82+52.85+28.57+78.57+95.6)-3.12$$
where:
The results of the calculations may be slightly different because of rounded values of normalized metrics in formulas
