How loud is that whale, anyway?

When we're thinking about how loud a noise needs to be to damage a whale's hearing, it's important to consider how loud the sounds are that the whale makes itself. If a singing whale makes a sound itself that is at a certain sound level, it is pretty unlikely that it will be deafened by being exposed to a sound of the same level.  This is something like how you are unlikely to be deafened by the sound of your own voice. This isn't to say that continual exposure to sounds at or quieter than their own won't affect whales. It will - just like constant low-noise sounds such as traffic can damage human health.

Calibrating hydrophone sensitivity
curves is very pleasurable.

The first step in figuring out how loud your whale is to drop a hydrophone in the water. Hydrophones convert pressure into voltage (if you're interested in how, I explain here). Hydrophones also have a property called sensitivity, which is a measurement of how quiet of a sound that the hydrophone can hear. The difference in sound sensitivity between a very sensitive hydrophone and a very insensitive hydrophone is like the difference in touch sensitivity between your tounge and your heel. A light touch on the heel might not make any neurons fire, but...

As an aside, in the book "Last Chance to See," Douglas Adams puts a microphone in a condom and uses it to try and listen to Yangtzee river dolphins in China. (Insert nerdy bio-acoustition joke here).

Hydrophone sensitivity is usually listed as a single number, something you will notice if you are a buyer of hydrophones or microphones. This number is negative, and the less negative, the more sensitive the hydrophone. Listing a single number actually oversimplifies things. For example, let's look at the sensitivity of the Reson TC 4024 hydrophone at a range of frequencies between 0 and about 90 kHz. If you remember, frequency determines how shrill a sound is. Barry White, for example, has a low frequency voice, and the Chipmunks squeak their songs at high frequencies. OK, so the "typical receiving sensitivity of this hydrophone" is listed as -173 dB. But, as you can see, the sensitivity actually varies quite a lot with frequency:

Because of this, most people only use their hydrophones to record sounds in the "flat frequency response" region, which, for this hydrophone, is between about 4 Hz and 45 kHz. This flat frequency region is pretty much about equal to -173 dB (and if you want to be really specific about things you can always go back and correct for the variation later - as long as your hydrophone is calibrated!)

You'll probably also need to know your gain. Gain means amplification. Generally, the signals coming out of the hydrophone are still too quiet to use, so we amplify them again.

Finally, we can calculate how loud the whale is. So here we go:

Sound Pressure Level = Sensitivity - Gain + 20 * log (Voltage from the Hydrophone)

It's actually pretty simple once you know what all the numbers mean...

... except for the fact that, by convention, the Sound Pressure Level is measured at 1 m from the sound source. So, if we are working with wild animals, we will either have to get incredibly lucky, or we have to figure out a way to deal with attenuation of the sound in water. Attenuation is a fancy word that means "makes quieter." For example, sound-proof walls attenuate the sound of your neighbor's Eminem music.  I'll deal with that one a little bit later.

You almost never get this lucky. Photo by Flip Nicklin.

Tweet

We Should Have Studied Dolphins in High School Math

When I look back at my educational career, it's really striking how poor of a student I was.  That is, I was a pretty mediocre student in the "hard" sciences until I took Marine Biology class during my senior year of college.  Most of the time, I think I was just really bored.  It wasn't always because the material wasn't challenging or useful (although sometimes this was the case). It was often because I had no idea what the point of it was.  OK, so in physics we could calculate where a ball would land based on the angle and force.  I only played baseball when forced to in gym class, and when I did I had no control over the force and angle of my wild swings.  I didn't have a home catapult, and I just didn't get how this stuff could really matter to ME.

It wasn't until I got really interested in marine biology, and specifically whale and dolphin bioacoustics, that I started seeing how amazing and cool math & physics can be.  If you read my blog, you know that bioacoustics is the study of the sounds of living things.  Whales and dolphins use sound to communicate and find food underwater, so even if you can't see them, you can very often hear them.  Some whales, such as beaked whales, are actually easier to find by listening than by looking with your eyes.  In addition, sound travels further underwater than light, so people at the surface can even hear whales that are thousands of feet underwater, or even hundreds of miles away. We can use these sounds to track the location of the animal, and we can do it using high school math.  Super cool.

Why didn't we do this in my high school trigonometry class?

It's really pretty simple.  First, you need something in the water making a sound.  Let's say... a whale.

Then you'll need a boat and something to record the sound.  We're going to use a hydrophone array, which is made up of three underwater microphones (hydrophones) strung in a line, with 15 meters between each hydrophone.  

When the whale makes a sound, it travels in a straight line through the water* to each hydrophone, arriving at the closer hydrophones sooner and arriving at the further hydrophones last.  Let's say the sound arrives at our recordings at time 0 for hydrophone 1, 0.006 seconds for hydrophone 2, and 0.0007 seconds for hydrophone 3.

How on earth do we figure out the distance to a whale when all we know is the difference between when we heard it on hydrophones 1, 2, and 3?  Easy - there's a simple equation for that!

The sound speed in water is 1500 m/s (which is why I put the hydrophones 15 m apart - to make the math easier).

We've replaced some of the words with numbers, something I HATED in high school.  Oh wait, I still hate it, so I'll just explain what the equations mean: The time difference between hydrophone 1 and 2 (t21) is equal to the difference between the distance between the whale and hydrophone 1 and the distance between the whale and hydrophone 2 (d2-d1) divided by the speed of sound.  Holy mother of god, this is just geometry!  I could have been doing this when I was 15! Some of you over-achiever types probably took geometry in 8th grade, so you could have done this when you were 14. Not me!

With a little algebraic rearrangement (which I am not going to bore you with, since you can look it up here), we get:

**

Sx and Sy are the x and y locations of the whale. C the speed of sound in salt water, 1500 m/s.  dh is the distance between the hydrophones, which in this case is 15 m, and all the ts refer to the time difference between the hydrophones (the small numbers tell you which two hydrophones: t21 is the difference between the arrival at hydrophone 2 and 1, etc).  All you have to do now is plug in the numbers. Edited 5/12/2012: To make it even easier, I've provided a spreadsheet here.

**

Then you can graph the location of the whale (-16.7, ±241.7), along with the locations of your hydrophones at (0, 0), (0, 15), and (0, 30).  I've just made designated hydrophone 1 at (0, 0) to make things easy, but in real life you'd have a GPS on the boat and you'd have to add the length of the hydrophone cable to the location of the boat's gps.

You might notice something funny about this equation - it actually gives two solutions - one on either side of the line of hydrophones.  That's a big problem with using three (or four, five, etc.) hydrophones in a row - you're not really sure what side of you the whale's on.  The reason for this is that there are two possible solutions for Sy. However, you can figure out which side the whale is on by turning the boat.  The location that stays in relatively the same place regardless of where the hydrophones are is the correct one.  Other ways to fix the left-right ambiguity? Make your hydrophones into a 2D shape (hard to do when dragging them behind a boat because 2d shapes are not very streamlined)*** or use DIFAR.

I wish I had done this kind of math in high school - it is totally within the realm of a high school student!  I think I might have paid better attention to something cool like finding whales.  On the other hand, maybe not.  That boy across the room WAS pretty dreamy.  Not all kids will be interested enough in whales and dolphins to get interested in math and physics, but the cool thing about math and physics is that, regardless of what you're into, I bet it applies.  From video games to raising miniature ponies, there's math in there somewhere.  Sometimes it just takes the right spark to get people interested.

P.S. If I made any mistakes above, please call me out.  Like I said, I didn't do so well in math class.  Also, if you're interested in a copy of the xcel spreadsheet with the equations already in it, shoot me a message and I'll send you a copy.

* This isn't always the case - sometimes the sound bounces off the bottom or the surface to get to the hydrophone, but I'm ignoring those cases here for simplicity.

** Edit: May 9, 2012. I made a HUGE mistake the first time I posted this and completely ignored the equation for t1 (I thought t1 was a typo for t12).  This is now fixed.

*** Edit: May 8, 2012.  Changed "3D" to "2D" because of a comment from Sam Denes. I had thought a straight line was already 2D, but apparently a straight line is 1D and adding another point off the line makes something 2D.

Note: I got to talk to a bunch of high school math teachers about this math.  Recap here.

  Tweet