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ContinuousWave: Small Boat Electrical
SONAR Depth Sounding and Boat Speed
|Author||Topic: SONAR Depth Sounding and Boat Speed|
posted 05-31-2009 11:18 AM ET (US)
When measuring water depth using echo sounding techniques, we often employ a moving boat. When the echo sounder emits a burst of ultrasonic energy or "ping" toward the sea bottom, the boat will move forward before the echo returns. I thought it interesting to investigate the distances and times involved in these measurements.
Speed of Sound In Water
Most influential in this problem will be the speed of sound in water. One source cites the speed of sound travel in water as
This is about 4,911.4 feet/second. Let us assume a boat is in water that is 100-feet deep. The distance the sound will travel is 200-feet. The time required is thus
200-feet x 1-second/4911.4-feet = 0.0407-seconds.
Speed of Boat
Assume the boat making the echo soundings is moving at 25-MPH, a reasonable cruising speed. This is rate of speed of
25-miles/1-hour x 1-hour/3600-seconds x 5,280-feet/1-mile = 36.66 feet/second
We can now deduce how far the boat moves in the time it takes our SONAR echo to travel to the bottom and return:
36.66 feet/second x 0.0407-seconds = 1.5-feet
In general the sensitivity of a SONAR transducer is rated in terms of a particular cone angle. We now determine the angle off-axis that the bottom echo returns. This is a problem of the familiar right-triangle, with sides of 100-feet and 1.5-feet. The angle THETA is thus
THETA = SIN-1(1.5/100)
Since most SONAR transducers are specified with a cone angle of at least 20-degrees, we see that for our case of a boat moving at 25-MPH in 100-feet of water the returning echo is very much within the transducer's main lobe.
posted 05-31-2009 11:43 AM ET (US)
[Corrected analysis follows--jimh]
THETA = SIN-1(2b/a)
If we examine this relationship we see that the off-axis angle is always going to be a function of the ratio of the boat speed to the sound speed, no matter what the water depth, thus
THETA = SIN-1(2 x boat speed/sound speed)
If we assume a symmetrical cone angle for the transducer of 20-degrees, we can calculate the boat speed needed to get the return echo off-axis by 10-degrees to see what sort of speed we need to reach before the boat begins to out run the bottom echoes:
SIN(10) = 0.1736
The boat speed maximum is this
boat speed = 0.1736 x sound speed x 0.5
Converting to miles-per-hour
426.5-feet/second x 1-mile/5280-feet x 3600-second/1-hour = 290-MPH
It does not appear we have anything to worry about with our boat speed becoming too great to out run the SONAR echoes.
posted 05-31-2009 12:00 PM ET (US)
How many "Cycles" are required before a sonar display will actually transmit a depth reading to the visual interface for the human driver to interpret?
I'm asking because, the deeper water I'm in, the slower I need to go for my sounder to register a depth on a reliable basis. For instance, in less than 400 feet of water, my sounder is very efficient in providing a reading at cruising speeds, but when the depth doubles to 800 feet or more, the soundings become erratic and sporadic - I "lose bottom" for several seconds, and then the display will show the depth before "loosing bottom" again.
Just curious. This has happened on every trip to Isle Royale, where the depth during the crossing is very deep.
Is it possible that the software requires a set number of solid returns before transmitting the information as actual depth? Might you "out-run" those when traveling at speed in very deep water?
posted 05-31-2009 12:18 PM ET (US)
I think Dave's problem may be more related to the signal level than to the boat speed.
As sound travels though water it is attenuated. The greater the distance traveled the more the attenuation. WIthout investigating this very scientifically, I suspect that the typical distances involved in echo sounding are not so great that the attenuation becomes a critical factor in determining the sensitivity of the echo sounder. Instead, I suspect that the most significant change in sensitivity occurs because of attenuation (or loss) in the coupling between the transducer and the water. Dave's case is one of a transom mounted transducer. As boat speed varies there may be more attenuation in the signal path due to effects related to the transducer-water coupling caused by flow of aerated water along the interface.
As boat speed increases there may be more aerated water created by the hull. This aerated water likely increases attenuation, resulting in less signal return from the bottom.
As for how a particular echo sounder processes return echoes before displaying a depth, I think this varies from unit to unit. Observing the rasterized display of the echoes on my LOWRANCE unit, I often see what appears to be only weak an intermittent bottom echo returns, yet the device continues to display a depth sounding reading in its digital indication. In that instance I am assuming the echo sounder is making some sort of integration of recent echo timings to get the displayed depth. The raster display shows only a faint trace of the bottom.
posted 05-31-2009 12:42 PM ET (US)
Dave, is your SONAR dual-frequency?
Mine is 50kHz/200kHz; 50kHz uses a wider beam and is better suited to deep water.
If you have a lower frequency option, try using it next time you experience the phenomenon you described.
posted 05-31-2009 12:58 PM ET (US)
I'm afraid mine is a much simpler unit than that David.
I was just curious, because I get "strong readings" all the way to about 400 feet deep even at speed, but noticed it was a significant problem out in Lake Superior where the depths really drop off and the prizes double.
A new sounder is on the list - I'd like to buy the new Garmin unit and attach it to my chart plotter - but little things like a new house have siphoned significant cash flow that used to be directed toward toys.
posted 05-31-2009 06:08 PM ET (US)
In Dave's situation the depth of the water may be causing too much attenuation of the echoes, so the bottom is lost from weak echo returns.
posted 05-31-2009 08:01 PM ET (US)
Interesting. You would think a modern SONAR would be good for more than 400', especially in fresh water.
posted 05-31-2009 11:50 PM ET (US)
When we were in the Pacific Ocean in 2003, in the Strait of Juan de Fuca, our LOWRANCE 200-kHz depth sounder was mainly of no use because it never could get an echo from the bottom which was over 350-feet below us. Once in a while we got into some shoal areas, that is, areas with depths less than 325-feet, and we got a faint bottom echo. Otherwise for about two weeks we just looked at the depth sounder and saw a flashing display (which meant no bottom echo).
posted 06-01-2009 12:23 AM ET (US)
From my SONAR user guide:
Sounds like you may need a dual frequency SONAR.
Mine is actually quite old (~7 years) and it's maximum depth is 800 meters.
posted 06-01-2009 01:23 PM ET (US)
At 50 KHz the absorption coefficient is 8 db/kiloyard, at 200 KHz it is ~53 db/kiloyard. Numbers are for seawater at 20 deg C.
Urick, p 109.
posted 06-01-2009 03:21 PM ET (US)
The problems related to depth are caused by a lack of signal strength. The problems related to speed are caused by aeration of the tranducer, not by the the transducer outrunning the return signal.
posted 06-01-2009 03:56 PM ET (US)
My sounder is 200-kHz only. I believe that likely explains the challenges I have in finding bottom in very deep water when running on plane.
posted 06-01-2009 06:05 PM ET (US)
While I have no technical comments to add, I was in my boat this morning driving to work, and my deepest, accurate measured depth was 912-feet while 24-knots speed. (I watched the number move up and down a bit, but it was close to the charts a little deeper than expected.)
I thought to myself: "wow, I must have really good laminar flow over the transducer, because that is deep!" I have never had readings that good while on plane.
I was using an older Garmin 238 with 50/200 kHz transducer. (My other transducer went dead last weekend.) And it is saltwater ~50 F.
posted 06-01-2009 11:24 PM ET (US)
I am not familiar with the unit "kiloyard." I assume:
1-kiloyard = 1,000-yards
In other words, a kiloyard is 3,000-feet. Is that correct?
Also, could we have a bit of the background on why such an odd unit of measurement is used in these acoustic attenuation factors.
posted 06-02-2009 07:19 AM ET (US)
Jim, you are correct, a kiloyard is 1 thousand yards. It comes from antisubmarine warfare where almost all underwater acoustics research has been done. It turns out that 1 KYD is almost exactly 1/2 a nautical mile, a convenient unit of measure at sea. That's also why most texts on sound propagation only deal with salt water, not many U-boats or Russian submarines made it into fresh water. The absorption coefficient in fresh water is much lower due to the absence of dissolved MgSO4.
posted 06-02-2009 11:27 PM ET (US)
David--Thanks for the information on the sound attenuation in seawater. With a SONAR frequency of 200-kHz, the attenuation is
-53 dB/kiloYard = -53 dB/3,000-feet
If a boat were using a SONAR 200-kHz echo sounder in water that was 150-feet deep, a ping from the boat to the bottom and back would travel through 300-feet of water. I then would assume the attenuation due to water would be
-53 dB/3000-feet x 300-feet = -5.3 dB
Is that correct?
posted 06-03-2009 12:01 PM ET (US)
Yes, for sea water.
posted 06-03-2009 05:53 PM ET (US)
Preview of Fisheries Acoustics
Sorry for the long url but this is a book preview on Google Books, "Fisheries Acoustics," and it is much more germane to the kind of questions we deal with, like where are the fish and what kind are they? Most of the SONARs we use were designed to be fishfinders, not fathometers.
posted 06-03-2009 09:20 PM ET (US)
A transmission loss of only -5.3-dB (for 300-feet of seawater) is not very much attenuation. Most of the signal loss must be from the poor reflection of the sound energy off the targets.
posted 06-03-2009 10:19 PM ET (US)
Right, but Buckda was talking about ~900 feet of water (300 yds), so the calculations become -53X600/1000=31.8 db, that is not insignificant. But he was in fresh water and the absorption loss will be less. Plus absorption is not the only transmission loss factor in the SONAR Equation, there is spreading, the loss as the sound wave expands out from the source and again from the target on its return path. The design of the transducer introduces the additional factor of Directivity Index, a measure of beam efficiency, that can greatly improve the signal strength of the transmitted signal and the antenna gain for the received echo.
posted 06-04-2009 11:16 PM ET (US)
Dave--Hold on a minute.
Bucka talked about 400-foot depths. That would be 800-feet of water path for the sound. The attentuation at 200-kHz is
so the attenuation for 800-feet would be
-53 dB/3,000-feet x 800-feet = -14 dB
That is much less than your calculation of -31.8 dB.
It is surprising to me how much less sound absorption takes place at the lower frequencies. This helps me understand how a 50-kHz transducer can work so much better in deeper water.
Also, thanks for the reference to the book Fisheries Acoustics. At $200 a copy, I probably will not be ordering one. The preview did contain some interesting information. There is much about hydroacoustics that is similar to radio propagation. I guess they're both waves.
Resolution capability is tied to wavelength. Let's look at the popular 200-kHz and 50-kHz frequencies for fish finders. The book says the wavelength is equal to
λ = c/f where c is the propagation velocity and f the frequency
If c = 1500-meters/sec and f = 200-kHz, then wavelength would be
λ = 1500/200,000 = 0.0075-meter or 7.5-millimeter
For f = 50-kHz, the wavlength would be
λ = 1500/50,000 = 0.03-meter or 30-millimeter
posted 06-05-2009 09:04 AM ET (US)
In my first advanced course on SONAR we used a RADAR text book. Way back we had a discussion about Lowrance's new HDS line of fishfinders. We were speculating that they were using FM signal processing to get high target resolution, good noise rejection, and low peak power. All techniques first used in RADAR. Coincidentally, Lowrance is now marketing HDS RADAR. I wonder how many of the digital components and software routines are common to both.
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