Just noticed there are changes, among many others, to the definition of arc flash boundary in the upcoming NFPA 70E Standard for Electrical Safety in the Workplace (see http://www.ecmag.com/section/safety/whats-new-whats-changed-2018-nfpa-70e-update for the comprehensive list of changes)...
Can anybody explain why 1.2 cal/cm^2 incident energy was selected in solving the equation for the arc flash boundary in IEEE P1584 and NFPA 70E? Does it imply that an exposure to less than the 1.2 cal/cm^2 is not enough to cause a second degree burn?
