I was watching a u tube video and the guy on it said that if you are in 20 feet of water, the area that you see on your display will be representing around 6 feet of bottom area. My question is this, If you are in 10 feet of water, would the bottom area that the display is showing be then representing around 3 feet?

What is Ifishfinder?

Could it be an eyeball attached to a long stick.

just fishfinder sorry,, im using an Eagle fishmark 480.

And no im NOT dunk, I rarely even Dink. !!!!

Depends on the transducer cone angle, but for a 20° cone angle at 20 ft. the coverage area would be 7.06 ft at the bottom. In simpler terms, the width of the coverage is approximately 1/3 the depth with a 20° cone transducer.

Gman

Thanks Gamblinman, your reply led me to some more useful information, According to eagle electronics, if the sensitivity is turned up on my fishmark 480, it increases the cone angle,, up to a 60 degree cone angle. Now things are getting interesting!!!

Soo If Im in 20 feet of water, and have my unit sensitivity set at max, and my cone angle is now around 60 degrees, the area im viewing on my LCD would be ????

In theory, increasing the sensitivity would widen the cone angle, but it also increases false returns to the point of not being able to accurately determine what you're seeing. There's a fine line between having the appropriate sensitivity, and having too much, and it has to be adjusted accoring to the depth you are fishing. That's where turning off the auto snsitivity and manually adjusting really pays off.

Gman

Guys, it's pretty simple geometry actually. Only problem is it's been a long time since I got out of school and therefore not real sure I remember the equations correctly.
The transducer straight down to the bottom of the lake is side 'a' of a right triangle. From the bottom of side 'a' going straight along the bottom of the water is side 'b', the side we want to find as the radius of the cone, and the hypotanouse of the triangle (going back up to the transducer) is side 'c'. The angle the two sides 'a' and 'c' make at the transducer is the 'cone angle' Now, here is where I'm having memory problems...I believe the equation is b=a/cos 20 if we have a 20 degree cone angle but, I could be wrong...someone can correct me here. Then double b to find the diameter of the cone on the bottom of the lake. Easy huh? Make sense?
Okie Bob

In laymans terms......

A transducer's cone angle is not a concrete number. The loss coeficients used to calculate the cone angle can vary across manufacturers and there are higher and lower quality transducers.

The cone angle of a transducer is measured by a certain loss in decibels. For example, if the transducer is putting out X then some manufacturers may say that the point that the power is 50% of X would be the effective angle of the cone. However, the sound waves continue to radiate beyond that value...all the way to 0% which could be a multiple of 2,3, or more of the stated cone angle depending on the quality of the transducer.

While the above manufacturer used a loss coeficient of 50% another manufacturer may use only 20%...for purposes of this example. In this case their "stated" cone angle would be much narrower but the effective cone angle could very well be the same.

This is where the sensitivity of the receiver(display) comes into play at reading those returns that are from the lower power edges of the useable cone.

So, to answer the original post...there is not hard fast measurement that can be used because it changes for every combination of transducer and display based on conditions and settings.

However, a general rule for a "normal" 20 degree ducer would be 1/3 the depth....21ft of water you would be looking at a 7ft diameter circle on the bottom....to say it a different way that would be 3.5ft in front of your ducer, 3.5ft behind, and 3.5ft to each side.

The highest quality transducers typically "waste" less power on the outer edges of the cone...rather, they "focus" all of the power...think of a spot light versus a flood ligth.

I went out on the web and found a calculator if anyone is interested. http://www.csgnetwork.com/righttricalc.html
From this, you need to know the cone angle (angle b) and the depth from the transducer to the bottom of the lake 'a'. Plug them in and hit 'calculate'. It works out that with a cone angle of 20 degrees and a depth of 10' gives a distance of 24.47' but, this is only half the distance of the cone on the bottom of the lake so double it and you see that at 10' deep, the diameter of the cone is almost 55' across. At 20' the diameter is 110' across. Doesn't seem correct but, the numbers don't lie.
Okie Bob

Baitfish, you are probably correct, I was only doing the trig and not factoring in power which is going to be logrythemic.
U da man.

So in ten feet of water what im seeing is only a 3.3.999999999999999999999999999999999999999999 ft cone?

As I stated above there isn't a concrete way to answer your question because every setup is different.

Based on the manufacturers definition of cone angle at 20 degrees you would be seeing about a 3-4ft diameter circle at the bottom. In reality though with a newer unit with higher sensitivity settings etc that number could easily be 3 times that as far as your "useable" cone goes.

To throw a monkey in the wrench...

The total area we are viewing (roughly):

Area = Pi x (radius)squared

Area = 3.14 X 12.25'
Area = 38.5'

So with a 20 degree transducer in 20' of water, we are scanning approximately 38.5'.

thanks for all of the replies!! I will from now on assume that when im looking at my Lcd, the bottom area Im seeing is roughly representing at least 1/3 of what the depth is. Maybe more.