Acoustic Absorption Coefficient Calculator
Find how many decibels per kilometer seawater absorbs from a sound at your frequency, and how much intensity survives over your range.
🔊 What is the Acoustic Absorption Coefficient?
The acoustic absorption coefficient of seawater is the rate at which sound energy is converted to heat as a sound wave travels through the water. It is normally quoted in decibels per kilometer (dB/km), so a coefficient of 2 dB/km means that after one kilometer the sound intensity has fallen by 2 dB, and after ten kilometers by 20 dB. Because the loss is a constant number of decibels per unit distance, the intensity decays exponentially with range, and the pressure amplitude decays at the equivalent rate in nepers per meter.
Seawater absorbs sound far more strongly than fresh water, and the effect depends heavily on frequency. Viscous friction contributes at all frequencies and grows as the square of frequency. On top of that, dissolved salts take part in pressure-driven chemical relaxation: boric acid with a relaxation frequency near 1 kHz, and magnesium sulfate with one near 60 kHz. The result is that a 100 Hz signal can cross an ocean basin with a loss of a few decibels, while a 200 kHz fish finder pulse is nearly gone after a few hundred meters. This single fact drives the choice of frequency in sonar design, echo sounders, acoustic modems and ocean tomography.
This calculator uses Thorp's empirical formula, a simple four term fit of absorption against frequency that is widely used in sonar equations and textbooks. It is a deep-water generic fit: it contains no temperature, salinity, depth or pH terms, and it was originally fitted at roughly 0.1 to 3 kHz for deep ocean water and is commonly extended to about 50 kHz. Temperature, salinity and pH dependence is not modelled here. Full models such as Francois and Garrison (1982) or Ainslie and McColm (1998) add those variables when you need survey-grade numbers, but Thorp is accurate enough for planning, teaching and first estimates.
It is important not to confuse absorption with the other losses a sound pulse suffers. Spherical spreading loss, 20 log10 of the range in meters, comes from the energy spreading over a growing wavefront and does not depend on frequency. Absorption is additional, linear in distance, and strongly frequency dependent. Use this tool to see at what range absorption alone, ignoring spreading, scattering and boundary losses, takes 3 dB, 20 dB or 60 dB from your signal, and to see how the percentage of intensity left shrinks as the range or frequency increases.
📐 Formula
The boric acid term dominates below about 10 kHz, magnesium sulfate between roughly 10 and 100 kHz, and viscosity above that. Spreading loss, which this calculator does not add, is a separate 20 log10 r term.
📖 How to Use This Calculator
Steps
💡 Example Calculations
Example 1 — 1 kHz Sonar over 10 km
f = 1 kHz, R = 10 km
Example 2 — 12 kHz Echo Sounder over 5 km
f = 12 kHz, R = 5 km
Example 3 — 200 kHz Fish Finder over 0.5 km
f = 200 kHz, R = 0.5 km
Example 4 — 100 Hz Whale-Band Signal over 1000 km
f = 0.1 kHz, R = 1000 km
❓ Frequently Asked Questions
🔗 Related Calculators
What is the sound absorption coefficient of seawater?
It is the loss of acoustic intensity per unit distance caused by the water itself, usually in dB/km. At 1 kHz it is about 0.07 dB/km, at 12 kHz about 1.6 dB/km and at 200 kHz about 51 dB/km according to Thorp's empirical formula.
What is Thorp's formula for absorption in seawater?
With f in kHz, alpha in dB/km = 0.11 f^2/(1+f^2) + 44 f^2/(4100+f^2) + 2.75e-4 f^2 + 0.003. The first term is boric acid relaxation, the second magnesium sulfate relaxation, the third viscosity of water and the last a small constant.
Why does sound absorption increase with frequency?
Absorption comes from viscous friction and from chemical relaxation of dissolved salts, mainly boric acid and magnesium sulfate. Each process removes more energy per cycle at higher frequency, so loss per kilometer climbs from almost nothing at 100 Hz to tens of dB/km at 100 kHz.
How do I convert dB/km to nepers per meter?
Divide by 8.686 to convert dB to nepers (1 Np = 20 log10(e) dB), then divide by 1000 to go from per kilometer to per meter. So 1.645 dB/km equals 1.894e-4 Np/m. Nepers refer to amplitude, which is what appears in the wave equation.
How far does a 12 kHz echo sounder signal travel before losing 20 dB to absorption?
At 12 kHz Thorp's formula gives 1.645 dB/km, so absorption alone reaches 20 dB at about 12.2 km of one-way path. For an echo that goes down and back, the same 20 dB is reached at about 6.1 km depth, before counting spreading loss.
What does a 3 dB loss mean in terms of intensity?
A 3 dB drop corresponds to a factor of 10^(-0.3), about 0.501, so almost exactly half of the intensity is left. The calculator shows the distance at which absorption alone has halved the intensity at your frequency.
Is absorption the same as transmission loss?
No. Transmission loss in the sea is spreading loss plus absorption (plus scattering, boundary losses and ducting effects). Spherical spreading follows 20 log10 of range and is independent of frequency, while absorption is linear in range and strongly frequency dependent.
Does temperature or salinity change sound absorption?
Yes, but this calculator does not model it. Thorp's fit has no temperature, salinity, depth or pH terms. Full models such as Francois-Garrison and Ainslie-McColm include them, and pH in particular changes the low-frequency boric acid term.
What frequency range is Thorp's formula valid for?
It was originally fitted to low-frequency data (roughly 0.1 to 3 kHz) in deep ocean water and is commonly extended to about 50 kHz with its two small extra terms. This calculator accepts 0.1 to 500 kHz so you can read the trend, but values above 50 kHz are extrapolations and should be treated as estimates.
How much does absorption reduce a fish finder signal at 200 kHz?
At 200 kHz absorption is about 51 dB/km, or 0.051 dB per meter. A 500 m path costs about 25.5 dB, leaving only 0.28 percent of the intensity, which is why high frequency sounders are limited to a few hundred meters.
Why does low frequency sound travel so far in the ocean?
At 100 Hz absorption is only about 0.004 dB/km, so a signal can cross many hundreds of kilometers before absorption removes a few dB. Whale calls and some ocean acoustic tomography sources exploit this.