ラクシャーナ
@rc_ラクシャーナ • Joined April 8, 2026
At a glance
ラクシャーナ ranks #578,936 globally for subscribers (77.3K), #1,282,248 globally for views (3.62M), 142 videos published.
Subscribers
77,399
Views
3,629,193
Videos
142
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
25.5K
Total views ÷ total videos
Views per subscriber
47×
Total views ÷ subscribers
Subs per video
545
Subscribers ÷ total videos
Analytics tables
About
地獄の使者を自称しているただのオタクでゲーマー 学生時代に隠していたオタク心が花開き、今は大好きなドールの姿で配信中！ とある目的の為REJECT掌握を狙っている。 本名は楽原 さな子（らくはら さなこ） REJECT Live2d 母上 しゅびぃ様 @shu__bii 父上 星宮みづき様 @Hoshi7Mizuki
- Handle
- @rc_ラクシャーナ
- Joined
- April 8, 2026
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.