Gauley_vlog49
@gouley-vlog49 • Joined January 14, 2023
At a glance
Gauley_vlog49 ranks #3,402,680 globally for subscribers (630), #2,654,348 globally for views (291K), 143 videos published.
Subscribers
630
Views
291,488
Videos
143
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
2.03K
Total views ÷ total videos
Views per subscriber
463×
Total views ÷ subscribers
Subs per video
4
Subscribers ÷ total videos
Analytics tables
About
𝐻𝑒𝑙𝑙𝑜 𝑔𝑢𝑦𝑠🖐 𝑖𝑡'𝑠 𝑚𝑒 𝐺auley_𝑣𝑙𝑜𝑔𝟒𝟗 𝑎𝑛𝑑 𝑖 𝑢𝑠𝑒𝑑 𝑡𝑜 𝑢𝑝𝑙𝑜𝑎𝑑 𝑠ℎ𝑜𝑟𝑡 𝑣𝑖𝑑𝑒𝑜𝑠 𝑎𝑛𝑑 𝑚𝑜𝑟𝑒 𝑣𝑖𝑑𝑒𝑜 𝑖𝑛 𝑡ℎ𝑖𝑠 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 𝑠𝑜 𝑚𝑎𝑘𝑒 𝑠𝑢𝑟𝑒 𝑡𝑜 𝑠𝑢𝑏𝑠𝑐𝑟𝑖𝑏𝑒 𝑡ℎ𝑒 𝑐ℎ𝑎𝑛𝑛𝑒𝑙.😊
- Handle
- @gouley-vlog49
- Joined
- January 14, 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.