Amelia
@ameliasmr • Joined January 7, 2015
At a glance
Amelia ranks #549,550 globally for subscribers (80.2K), #2,953,840 globally for views (169K), 33 videos published.
Subscribers
80,200
Views
169,196
Videos
33
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
5.12K
Total views ÷ total videos
Views per subscriber
2×
Total views ÷ subscribers
Subs per video
2.43K
Subscribers ÷ total videos
Analytics tables
About
𝕎𝕖𝕝𝕝𝕔𝕠𝕞𝕖 𝕥𝕠 𝕞𝕪 𝕔𝕙𝕒𝕟𝕖𝕝🦋✨ Setiap hari Open Request Makan sesuai emoji dengan syarat Subscribe,Like, dan Comment🌹🌹🌹 BANTU AKU YAAA SALAM SEHAT & HAPPY WATCHING☺️🎧
- Handle
- @ameliasmr
- Joined
- January 7, 2015
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.