Arashi two
@arashitwo6334 • Joined July 21, 2015
At a glance
Arashi two ranks #5,393,603 globally for subscribers (45).
Subscribers
45
Views
0
Videos
0
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Views per subscriber
0×
Total views ÷ subscribers
Analytics tables
About
My kachburator! :3 ►PC Spect -CPU: Intel Core i3 6100 3.7 GHz Motherboard: MSI 970 GAMING Memory: Kingston HyperX Predator 2x 4GB Graphics card: NVIDIA GeForce GTX 960 2GB Case: EuroCase MC X202 ►Myš : Razer Deathadder Chroma / A4tech Bloody V3 ►Klávesnice : Razer Blackwidow Chroma X Tournament Edition ►Monitor : 24" …
- Handle
- @arashitwo6334
- Joined
- July 21, 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.