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ESSC

Portugal@fabiosmatosessc  •  Joined January 25, 2019

Live
Sub 5.38K  |  View 0  |  Video 0
#1,818,525
#3,344Portugal
#8,640,503
#10,954Portugal

At a glance

ESSC ranks #1,818,525 globally for subscribers (5.38K).

Subscribers

5,389

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

Views per subscriber

Total views ÷ subscribers

Revenue estimate

Details →

View data needed for estimate.

Based on estimated $0.7–$2 CPM.

Analytics tables

About

Liga ESSC, a maior liga da Europa de Free Fire, criada em 2021 e com Seasons Mensais e prémios de 500€ a 600€ + Troféus Inscrições & Regulamento em WWW.ESSC.PT

Joined
January 25, 2019

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.