How is this calculated?
The math continuously tracks how strongly this post is connected to the rest of the network. Every tag forms a 2-way link. The base stats determine personal node strength, and the pie charts below show this node's share against its direct neighbours.
// 1. Base variables (floored at 1 to prevent zero-multiplication math errors)
$inbound = max(1, 2) = 2
$outbound = max(1, 2) = 2
// 2. Node Base Values (Local connection strength)
Base_Strength (PV) = $inbound * $outbound = 2 * 2 = 4
Base_Influence (IV) = $inbound / $outbound = 2 / 2 = 1
// 3. Exponential Network Values (accumulating 20 direct neighbours)
Network_Strength (CV) = Base_PV * (Neighbour_1_PV * Neighbour_2_PV * ...)
= 4 *
( 1 [Chris August] *
64 [MercyMe] *
1 [Bart Millard] *
1 [Kendall Combes] *
1 [Mike Scheuchzer] *
1 [Nathan Cochran] *
1 [Robby Shaffer] *
1 [Trent Austin] *
144 [Whitecross] *
1 [Quinton Gibson] *
9 [Tracy Ferrie] *
1 [Benny Ramos] *
1 [Michael Feighan] *
1 [Mike Elliot (2)] *
1 [Rex Carroll] *
1 [Rick Armstrong (2)] *
1 [Scott Harper (4)] *
1 [Scott Wenzel (2)] *
1 [Whitecross - Elements of Rock 2013] *
1 [Whitecross - Elements of Rock 2017]
)
= 331.8K
Network_Influence (TV) = Base_IV * (Neighbour_1_IV * Neighbour_2_IV * ...)
= 1 *
( 1 [Chris August] *
1 [MercyMe] *
1 [Bart Millard] *
1 [Kendall Combes] *
1 [Mike Scheuchzer] *
1 [Nathan Cochran] *
1 [Robby Shaffer] *
1 [Trent Austin] *
1 [Whitecross] *
1 [Quinton Gibson] *
1 [Tracy Ferrie] *
1 [Benny Ramos] *
1 [Michael Feighan] *
1 [Mike Elliot (2)] *
1 [Rex Carroll] *
1 [Rick Armstrong (2)] *
1 [Scott Harper (4)] *
1 [Scott Wenzel (2)] *
1 [Whitecross - Elements of Rock 2013] *
1 [Whitecross - Elements of Rock 2017]
)
= 1
Outbound
2
Tags on post
Inbound
2
Posts tagging this
Connections
20
Total nodes
Base Node Strength
4
Base Node Influence
1
Strength Share (vs Direct Neighbours)
Dominant nodes (excluded from chart)Whitecross 60.50%MercyMe 26.89%
Influence Share (vs Direct Neighbours)
Connected Network Hierarchy
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Connection Health Audit (Red = broken 1-way link)