Savage Marvel
@savagemarvel • Joined May 11, 2023
At a glance
Savage Marvel ranks #8,185,474 globally for subscribers (1).
Subscribers
1
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
Welcome to SavageMarvel, the channel where you'll find some of the most epic and savage replies from your favorite Marvel superheroes! From witty comebacks to snarky remarks, we've got it all. Join us as we explore the lighter side of the Marvel universe and showcase some of the most hilarious and savage moments from …
- Handle
- @savagemarvel
- Joined
- May 11, 2023
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.