At a glance
RazeYT ranks #5,738,294 globally for subscribers (51), #7,666,132 globally for views (46), 2 videos published.
Subscribers
51
Views
46
Videos
2
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
23
Total views ÷ total videos
Views per subscriber
1×
Total views ÷ subscribers
Subs per video
26
Subscribers ÷ total videos
Analytics tables
About
𝘽𝙤𝙣𝙟𝙤𝙪𝙧 𝙡'𝙖𝙢𝙞 𝙟 𝙚𝙨𝙥𝙚𝙧𝙚 𝙦𝙪𝙚 𝙩𝙪 𝙫𝙖𝙨 𝙗𝙞𝙚𝙣 𝙗𝙞𝙚𝙣. 𝘽𝙞𝙚𝙣𝙫𝙚𝙣𝙪𝙚 𝙨𝙪𝙧 ma 𝙘𝙝𝙖𝙞𝙣𝙚 𝙮𝙤𝙪𝙩𝙪𝙗𝙚. nouveau sur ma chaine? : 𝗻𝗼𝘂𝗯𝗹𝗶𝗲 𝗽𝗮𝘀 𝗱𝗲 𝘁'𝗮𝗯𝗼𝗻𝗻𝗲𝗿. je joue a : _ 𝕭𝖗𝖆𝖜𝖑 𝕾𝖙𝖆𝖗𝖘 _ 𝓡𝓸𝓫𝓵𝓸𝔁 _ ąʍօղց մʂ (bientot!) _ET BIEN PLUS ENCORE. ---------------…
- Handle
- @raze_off
- Joined
- April 18, 2020
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.