The Better Me
@the_better_me-v8m • Joined April 26, 2016
At a glance
The Better Me ranks #2,652,652 globally for subscribers (1.59K), #4,521,122 globally for views (15.5K), 119 videos published from US.
Subscribers
1,595
Views
15,528
Videos
119
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
130
Total views ÷ total videos
Views per subscriber
10×
Total views ÷ subscribers
Subs per video
13
Subscribers ÷ total videos
Analytics tables
About
心に響く感動ストーリーや家族愛、夫婦愛、親子の絆、人生の奇跡を朗読形式でお届けするチャンネルです。涙なしでは見られない実話風ストーリーや、胸が温まる感動エピソードを毎日更新しています。寝る前やリラックスしたい時間に、ぜひごゆっくりお楽しみください。
- Handle
- @the_better_me-v8m
- Joined
- April 26, 2016
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.