舊頻道-【公認】rpr中文翻譯頻道
@rprcntl • Joined August 28, 2021
At a glance
舊頻道-【公認】rpr中文翻譯頻道 ranks #1,196,844 globally for subscribers (19.3K), #972,891 globally for views (7.13M), 237 videos published from TW.
Subscribers
19,325
Views
7,135,460
Videos
237
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
30.1K
Total views ÷ total videos
Views per subscriber
369×
Total views ÷ subscribers
Subs per video
82
Subscribers ÷ total videos
Analytics tables
About
本頻道已遷移至新頻道【公認】rpr中文翻譯精華 網址為https://www.youtube.com/channel/UC0L38unB_r-IhFza0VoUhog 請大家多多支持謝謝 只在YT更新 禁止搬運 喜歡的話歡迎訂閱和按讚 這將會是我們很大的動力
- Handle
- @rprcntl
- Joined
- August 28, 2021
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.