At a glance
명란젓 ranks #1,120,037 globally for subscribers (20.5K), #177,375 globally for views (129M), 214 videos published from KR.
Subscribers
20,573
Views
129,841,194
Videos
214
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
606K
Total views ÷ total videos
Views per subscriber
6,311×
Total views ÷ subscribers
Subs per video
96
Subscribers ÷ total videos
Analytics tables
About
일상 속 발생한 사건 사고 및 억울한 사연 웃기고, 놀라고, 때론 소름돋는 이야기까지 오늘도 당신의 지루한 하루에 한 스푼 '명란젓' 같은 자극을 드릴게요 재밌는 사연, 사건 사고등 일상생활에서 겪었던 이야기를 제보해 주시면 정성스레 영상 만들어 올려볼게요 사연 제보: [email protected] 구독과 좋아요는 사랑입니다🧡
- Handle
- @myeongranjeot
- Joined
- September 17, 2024
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.