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5genTexan

Depth Finder

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5genTexan

We were talking the other day...is the depth measured from the bottom of the boat, or is there some sort of figuring done, to actually give you the lake level. For instance, if it read 3.7 feet, is that from the bottom of the boat or from the top of the water? This could be good information based on the shallow places we ride.

MERRY CHRISTMAS!!!

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onamission

From where ever the transducer in mounted, usually the bottom of the boat. Just 1 foot for the depth.

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Pistol Pete

FWIW.

The depth guage will not work on dry land. It will give an errounous reading beacuse the sonar is not travelling through water. Just so you know.

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SacRiverRat

Thanks Pete - I'll keep my boat on the wet land, where it seems to work better Yes.gif

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sheeprides

Ouch - tough crowd.

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Pistol Pete

No problem, Troy.

Just saving you that embarrassing phone call to Lowrance like I made a couple of years ago. Wait till I tell you how I fell off the porch yesterday.

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Boomer

The aftermarket depth-finder I had in my last boat (Humminbird) had an offset setting where you could enter how far under the water line the transducer was, so it would give real depth readings.

Unless you have that, it will read how far the lake bottom is down from the bottom of the boat. I don't believe the Malibu depth finder offers an offset.

I liked having the offset, but in reality what really matters is how far the boat bottom is from the lake bottom. :)

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VinRLX
FWIW.The depth guage will not work on dry land. It will give an errounous reading beacuse the sonar is not travelling through water. Just so you know.

Thanks Pete - I'll keep my boat on the wet land, where it seems to work better Yes.gif

What happens if the land was once wet, but now frozen? This is a major issue for me, due to location. I definitely don't want any errounous readings--that's gotta hurt, whatever it means. Heck, I don't even want to deal with erroneous readings--I detest inaccuracy. :)

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VinRLX
I liked having the offset, but in reality what really matters is how far the boat bottom is from the lake bottom. :)

Actually, shooting from the bottom, with no offset, is like a built-in safety margin.

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aneal000

Another thing to keep in mind is the transducer is often mounted on an angled part of the boat. Rarely is it pointing directly downward or even when underway it's actually pointing forward due to the running angle of the boat at speed. At 8 mph it might actually be something like this / - ok, maybe not that much, but either angle can cause the accuracy of the depth gauge to be off. Anyone good with trigonometry? Pointing the sensor toward the bottom of the lake at an angle (not straight down) creats a longer path or what could be know as the hypotenuse of a right triangle. Obviouly the difference in actual depth and the reading shown on your depth gauge will be relative to the angle and will have more error in deeper water as well.

Just wanted you guys to have all the facts. Biggrin.gif

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Malibudude
Thanks Pete - I'll keep my boat on the wet land, where it seems to work better Yes.gif

Well the nice posts didn't last. :lol: There is the sarcasm we know... Biggrin.gif

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JohnDoe

The pythagorean theorem:

(a*2) + (b*2) = c*2

*= square c=hypotenuse

Once you have figures that work for the formula, then you have to compute the angle of the hull. Once you know the relationship of angle, depth, and perceived depth, then you could determine actual depth as long as you knew the perceived depth and "relative" angle of the boat.

I could just do an example, but I am lazy.

Remember that all interior angles of a triangle = 180.

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chathamsolutions
...

I could just do an example, but I am lazy.

...

Obviously.

;)

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John I.

Gee, lots of interesting information! And possibly useful too, if you boat somewhere with a gradual, gentle sloping lake bottom. Unfortunately, several years ago I was done in by an extremely large, submerged rock at the very moment I was checking my depth gauge. It read 13.6 feet! Shocking.gif

By the time the gauge registered the depth change, the boat had ground to a halt. Cry.gif

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mrothwell

Even tho the transducer isn't pointing straight down, it is still fairly accurate. The transducer doesn't send a "beam" straight down, but rather more of a radio wave that encompasses a larger angle. The first response back to the transducer is what it will read as the depth. Of course, you still need to add that 1 foot, or whatever distance below water level the transducer is. At speed, the transducer is at water level. At rest or idling, the transducer is several inches below water line.

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aneal000

Trig 101

Say your depth gauge reads 50' and we have determined that the transducer is mounted on the hull at a 5 deg angle to port. Since we now know two variables we can caculate the rest.

In a right triangle we can use the SIN function (SIN = OPP/HYP) so the HYP = 50'.

One way to do this is to first calculate the the OPP side, or what we would call the bottom of the lake distance between where the transducer actually measured and where we wish it measured.

In our right trangle we will call the angles A, B, and C. A = is at the bottom of the boat, B is directly below the boat (90 deg) and C is at the bottom of the lake where the transducer is measuring.

The sides of our triangle are a, b, and c. a being opposite angle A (bottom distance or OPP), b being opposite angle B (or the real depth), and c being opposite angle C or the measured depth (also know as the HYP).

Trust me, just draw it and it will make sense.

Here is what's known:

c = 50'

A = 5 deg

Since sin A = a/c, therefore a = c sin A. That gives you a. Next use the Pythagorean theorem to find b knowing a and c.

Lets run the numbers:

a = c sin A = 50 sin 5 = 4.358' (bottom distance)

Using Pythagorean theorem from the post above we get (4.358*2) + (x*2) = 50*2

or 18.99 + (x*2) = 2500

(x*2) = 2481

Square Root of 2481 = 49.81'

Therefore the actual water depth is 49.81'

See, math is useful Tongue.gif

Edited by aneal000

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gooddog
Trig 101

Say your depth gauge reads 50' and we have determined that the transducer is mounted on the hull at a 5 deg angle to port.  Since we now know two variables we can caculate the rest. 

In a right triangle we can use the SIN function (SIN = OPP/HYP) so the HYP = 50'. 

One way to do this is to first calculate the the OPP side, or what we would call the bottom of the lake distance between where the transducer actually measured and where we wish it measured.

In our right trangle we will call the angles A, B, and C.  A = is at the bottom of the boat, B is directly below the boat (90 deg) and C is at the bottom of the lake where the transducer is measuring. 

The sides of our triangle are a, b, and c. a being opposite angle A (bottom distance or OPP), b being opposite angle B (or the real depth), and c being opposite angle C or the measured depth (also know as the HYP).

Trust me, just draw it and it will make sense.

Here is what's known:

c = 50'

A = 5 deg

Since sin A = a/c, therefore a = c sin A. That gives you a. Next use the Pythagorean theorem to find b knowing a and c.

Lets run the numbers:

a = c sin A = 50 sin 5 = 4.358' (bottom distance)

Using Pythagorean theorem from the post above we get (4.358*2) + (x*2) = 50*2

or 18.99 + (x*2) = 2500

(x*2) = 2481

Square Root of 2481 = 49.81'

Therefore the actual water depth is 49.81'

See, math is useful  Tongue.gif

What about the submerged rock? :unsure:

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aneal000
What about the submerged rock? :unsure:

1) sonar

2) you hit it - been there done that

3) this formula would work for the submerged rock as well, only difference is you would want to use approriate numbers, for example Water depth as shown on the gauge would now be 0 instead of 50. The angle would still be 5 deg. See if you can calculate the actual depth...

:lol:

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gooddog

There is no entry for boat speed in the formula though... :(

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NorCaliBu
There is no entry for boat speed in the formula though... :(

The boat speed only determines how fast you need to do the other math. :lol:

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jetskipro550
Gee, lots of interesting information! And possibly useful too, if you boat somewhere with a gradual, gentle sloping lake bottom. Unfortunately, several years ago I was done in by an extremely  large, submerged rock at the very moment I was checking my depth gauge. It read 13.6 feet!  Shocking.gif

By the time the gauge registered the depth change, the boat had ground to a halt.  Cry.gif

a little off topic, but seeing how this thread has turned into a math lesson I don't think it matter much Biggrin.gif A few years ago I was at Shasta and my friend and I were out on our jet skis and there was a nautique full of girls driving around. The water was down 90ft and our of no where their boat hit a nice size rock just under the water about 1-2 feet it was about 10x20 feet and flat across the top. One second they were cruising along and the next I look over and they are sitting on top of this rock. It sucks cause it was probably 50+ feet about all around that rock. My friend had to go find the father of the girl who was driving and tell him:LOL:

Edited by jetskipro550

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Pistol Pete

Keeping with the off topic.

I now know that it's 4 feet from the top of my porch to my asphalt driveway. I did all the trig. that Tony gave us and the answer was 4 stitches to the back of the head.

Edited by Pistol Pete

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