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Transducer Beam Spread
Beam Characteristics | Olympus IMS
For half-angle equivalent to dB loss of 6 and 20 dB:
FAQ: What factors influence the beam spread of an ultrasonic probe in the far zone?
For absorption coefficient
https://upload.wikimedia.org/wikipedia/commons/8/89/Atmospheric_sound_absorption_coefficient_2.svg
For hearing ranges of animals
https://upload.wikimedia.org/wikipedia/commons/5/5d/Animal_hearing_frequency_range.svg
For damage guidelines
Guidelines for the Safe Use of Ultrasound: Part II - Industrial and Commercial Applications: Safety Code 24
## Damage Geometry ##
The absorption coefficient for a pure 100 kHz tone in standard atmosphere is between .5 and 4 dB/m depending on humidity. Use 3 dB/m to reduce spoon count.
The half-angle of the beam spread equivalent to a 6 dB loss is
half_theta_6dB = arcsin ( .56 * V / ( D* F) ) = 11 degrees
V = 343.2 m/s (speed of sound in air)
D = .01 m
F = 100e3 Hz
For 20 dB loss it is
half_theta_20dB = arcsin ( 1.08 * V / ( D* F) ) = 22 degrees
For the diffraction limit
half_theta_lim = arcsin( 1.22 * V / (D*F)) = 25 degrees (the maximum beam-width. beyond this cone, there is no substantive effect)
According to damage guidelines, eardrums likely burst and thin regions of skin may burn at ~160 dB. Death is likely at ~180 dB. Pain is somewhere around 130 dB. From these guidelines, and the previous loss mechanisms, determine the geometry of the death, consequence, and pain regions.
Along the axial direction of the cylinder
Death = 4 m (192 - 4*3 = 180 dB)
MC =10 m (192-10*3 = 162 dB)
Pain = 20 m (192-20*3 = 132 dB)
Along the 6dB half angle
Death = 2 m (192 - 2*3 - 6 = 180 dB). Radius = tan(theta)*2 = .4 m
MC = 8 m (192 - 8*3 - 6 = 162 dB). Radius = 1.6 m
Pain = 18 m (192 - 18*3-6 = 132 dB). Radius = 3.5 m
Along the 20dB half
Death - not possible
MC = 4 m (192 - 4*3 - 20 = 160 dB), 1.6 m
Pain = 14 m (192- 14*3 - 20). Radius = 5.6 m
## Heating Limit ##
The initial power flux coming out of the orifice is
Pref = 1e-12 W/m^2
P (192 dB) = 10^(19.2) * Pref = 1.58e7 W/m^2
With an orifice diameter of 1 cm, this corresponds to 1244.7 W emitted from the exit of the device
This should pose no problems to the atmosphere over the time scales this device would be activated. Note that the solar pressure at the surface of the Earth is approximately 1e3 W/m^2.
The device likely resembles a blow torch in the vicinity of the orifice.