AKKA 악카
@akka0407 • Joined July 20, 2015
At a glance
AKKA 악카 ranks #684,295 globally for subscribers (54.7K), #114,186 globally for views (227M), 457 videos published.
Subscribers
54,773
Views
227,003,583
Videos
457
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
496K
Total views ÷ total videos
Views per subscriber
4,144×
Total views ÷ subscribers
Subs per video
120
Subscribers ÷ total videos
Analytics tables
About
난생 처음으로 덕질시작 그거슨 악뮤 편집에 대한 피드백? 얼마든지 받습니다 하지만 악뮤에 대한 피드백? 원치 않습니다 무조건적인 응원, 지지, 사랑, 박수갈채, 일방적이고 편향적인 칭찬만 부탁드립니다 현생보다 팬튜브 편집🐰💚 1위 건강 1.2위 팬튜브 (20241117 수현님 말씀)
- Handle
- @akka0407
- Joined
- July 20, 2015
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.