At a glance
바랄라 페어리즈 ranks #978,606 globally for subscribers (29.4K), #783,448 globally for views (11.4M), 243 videos published from KR.
Subscribers
29,467
Views
11,414,593
Videos
243
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
46.9K
Total views ÷ total videos
Views per subscriber
387×
Total views ÷ subscribers
Subs per video
121
Subscribers ÷ total videos
Analytics tables
About
바랄라 별빛섬에는 세개의 스타스톤으로 둘러싸인 마법의 성이 있어! 그런데 못된 요정이 스타스톤이 지닌 강력한 마법의 힘을 탐내 훔치려고 하다가 스타스톤이 떨어져 흩어지고 말았어! 스타스톤을 지키는 세 요정은 인간 세상에 내려와 마법에 걸린 세 소녀를 만나게 되는데... 과연 요정들과 어린 소녀들은 스타스톤을 되찾을 수 있을까?
- Handle
- @balalafairies_kr
- Joined
- July 21, 2023
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.