Svg_destroyer bannerSvg_destroyer banner
Svg_destroyerSvg_destroyer

Svg_destroyer

Unknown@svg_destroyer6395  •  Joined June 1, 2022

Live
Sub 0  |  View 0  |  Video 0
#9,629,461
#5,181,287Unknown country
#8,554,266
#6,070,672Unknown country

At a glance

Subscribers

0

Views

0

Videos

0

Charts

Subscribers, views & videos over time

Hover or tap the chart to see values

Subscribers gained (weekly) · 6 mo

Views gained (weekly) · 6 mo

Engagement

Not enough data yet.

Revenue estimate

Details →

View data needed for estimate.

Based on estimated $0.7–$2 CPM.

Analytics tables

About

So the reason I wanted to start this channel is because I got wrongfully Perma banned so I made this account to try and get riots attention on this new account

Joined
June 1, 2022

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.