Laser against Gaza strip terrorism
Evaluation of laser as means against Gaza’s incendiary balloons
David Judbarovski, systems engineering, Israel,
Ideas to collapse of air balloons by a laser beam aren’t new. On first look
it seems very attractively in application against sadly famous Gaza’s
incendiary balloons (GIB) terrorizing Israeli people and lands near a boundary
with Gaza strip last months. It would be very amusing to crush inside the Gaza
territory their fire was sent to Israel.
But on other hand, it is looked very worrisome, because military lasers
after huge milliards dollars investment and some ten years development didn’t
give a birth nothing serious except of a weapon of some ten ton with a range
about 1 km or so.
Moreover, the said GIB-s are transparent and a lion share of the said
laser power would go through the balloon isn’t hurt it.
Just below I will show my evaluation of the GIB problem
For the happiness, the Gaza strip is 4-7 km width and 40 km length, and
our lasers range can be 1-4 km or much less and some minutes to hit and not
more for returning all the GIB back to burn the Gaza themselves.
Nevertheless, IR laser of 2,000 nm wavelength must be up to some kW even
if the GIB absorbing all 100% of laser beam energy, and it is a device of 1.0
ton or so, while about 50 lasers being very expensive each would be needed. But
if being 95% transparency of the GIB, the power, the investments and the weight
would be 20 times bigger.
It being more hopeful that the band of 200-300 nm is practically zero
transparent for PE, PVC, PS films, and for PMMA it is 200-270 nm. Moreover, standard greenhouse
films generally blocks the major of UV light transmission at wavelength below
360 nm.
Really, if laser beam
profile being approximated as Gaussian distribution with “A” Joule -amplitude and “-
a*x2” for exponent with a base E = 2.71 being a base of natural logarithm, the
total laser power is W = “A”/”a” joules. “A” and “a” – can be calculated, if supposing
4.0 kJ/cm3 needed to
hurt the said GIB’s wall of 100 micron thickness.
D – is the laser beam
width at (1-1/E^2) of the total laser power, allows to calculate
a = 8/D^2.
If “e” << D, e.g.
e/D = 0.09, here “e” being a hole in GIB, hurt by the laser, so power at the “e”
is We = 0.016 * A/a = 0.002 A * D^2, and the hole area Se = 0.006 D^2, so
We/Se =A/3 = 4000 joules/cm3, so A = 120.0 J/cm2 for 100.0 micron plastic film
So the total power W =
A/a = 120 * D^2/ 8 = 15 * D^2, while
D is optimized and
depended on “L km” and our laser’s wavelength “wl” = 200-270 nm, let us for 200
nm.
L is chosen as actual
for our case: 1 km, 4 km vs. Chinese laser rifle ZKZM-500, 500-800 m, 3 kg, 1000
hits per 2 sec each, so about 500 watt (was estimated quite speculative).
D = (2.44 * L * wl) ^ (1/2)
If L = 1 km, D = 4.4 cm
and W =290.0 Joules per 1 sec = 290 watt
For 4 km = D = 8.8, so W = 1.16 kW
A big loses of the UV lasers
energy in the air, especially if the being polluted naturally or intentionally,
I didn’t consider here.
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