At a glance
iemon ranks #424,401 globally for subscribers (121K), #283,523 globally for views (65.5M), 481 videos published from JP.
Subscribers
121,963
Views
65,561,136
Videos
481
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
136K
Total views ÷ total videos
Views per subscriber
538×
Total views ÷ subscribers
Subs per video
254
Subscribers ÷ total videos
Analytics tables
About
昔は週1投稿してました。 今は見る影もありません。 ※当チャンネルのゆっくりを用いた動画は、【東方プロジェクト】を基にした二次創作です※ ●ファンレターやプレゼントはこちらまで● 〒 104-0032 東京都中央区八丁堀二丁目25番10号 三信八丁堀ビル5F 株式会社ブルーオーシャン iemon宛 所属事務所:Buber 【お問い合わせ&仕事依頼】 [email protected]
- Handle
- @iemon2353
- Joined
- March 27, 2013
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.