1_FPl=(2*r_e)/((6*f_e)/π)^(1/3)

July 15, 2011 at 9:15 pm | Posted in Questions, Quotabull, Uncategorized | Leave a comment

1_FPl=(2*r_e)/((6*f_e)/π)^(1/3)  

This is the simplified version of the equations at https://votedavedowling.wordpress.com/2011/07/14/defined-by-the-electron/

1_FPl(One very small length)=(2*the Electron radius)/((6*frequency of the electron)/π)^(1/3) 

copyright 2011 Dave Dowling

Advertisements

Leave a Comment »

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at WordPress.com.
Entries and comments feeds.

%d bloggers like this: