Let us play around a little with the reduced single pulse MPEr assuming it is the most
Dimensions and definitions:
Counting time for pulses:
for a point source
Expression for MPE of the standard:
Remember C5:=N^-0.25, when N = number of pulse during T2.
The duration function is given as
duration function with:
a see above
d = beam diameter
T = cycle time of the figure = 1/f
r = radius of the moving dot
in the spectator's area
s = 1 second
We take a safety factor because of the higher intensity close to the center of the beam.
From this you can calculate the Watts in the beam of diameter d in the spectator's plane.
To follow this simplification keep in mind the factor n*t with n= T2/T
and the speed of the dot, given by 2*p*r/T.
As you can see the formula is independant of the cycle time T (hence the
frequency) of the figure. The result will be Watts, the total power of the beam.
Now assume some data:
This value comes to the same order of magnitude as in the example you sent to here.
Now I try to come to 10 W???
Bad laser quality leads to C6 = 3 ... 5 possibly:
A very long path through the room (corresponding to seldom hits) leads to a big r
Finally we take a big beam, making long pulses but spreading the power over a
wide range (big dot of light):
or take a beam of diameter of 20 cm
If you can make sure the single hits are only relevant during 0.25 s in 100 s
T2 is limited to 0.25 s of course and you get:
fnmper(C62, d3, r2) = 8.9 W