アオト
@aoto • Joined December 20, 2014
At a glance
アオト ranks #291,809 globally for subscribers (204K), #154,975 globally for views (156M), 1.03K videos published from JP.
Subscribers
204,312
Views
156,453,097
Videos
1,032
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
151K
Total views ÷ total videos
Views per subscriber
766×
Total views ÷ subscribers
Subs per video
198
Subscribers ÷ total videos
Analytics tables
About
やっほーみんな！アオトだぞ！ 持ち前の元気さで、見てくれた人みんなを笑顔にしたい！そんな想いで活動してます！ 🎮活動内容 ・腐男子全開の動画投稿 ・BL漫画あるある系 ・YouTubeライブでセ●ハラ 📍活動場所 YouTube ツイキャス：企画系だったりメンシプだったり TikTok：動画投稿＆配信で活動中！
- Handle
- @aoto
- Joined
- December 20, 2014
More features coming
We're working on new analytics and tools. Stay tuned.
More features coming
We're working on new analytics and tools. Stay tuned.
Frequently asked questions
How accurate are these YouTube channel analytics?
Subscriber, view, and video counts come from YouTube's public API and refresh regularly. Growth and revenue figures are estimates built from tracked historical data, so they can differ slightly from what a channel owner sees privately in YouTube Studio.
How is a channel's global and country rank calculated?
Rank compares this channel's subscriber count and view count against every other channel Socialcounts.org tracks, both worldwide and within its home country.
How is the revenue estimate calculated?
It's a rough public estimate based on a $0.7–$2.0 CPM (revenue per 1,000 views) applied to view counts — not actual creator earnings, which depend on ad rates, audience geography, and monetization settings YouTube doesn't expose publicly.
Can I check analytics for any YouTube channel, not just my own?
Yes — search any public YouTube channel by name, handle, or channel URL. No login is required to see the basics.
Is this YouTube channel analytics checker free?
Yes. You get 10 free channel and video analytics lookups per rolling hour without an account — logging in, which is also free, removes that limit entirely.