Garmin Striker 5dv

[sand850]

Just purchased a 5dv. Am trying to figure out how large an area on the bottom I'd be looking at relative to the water depth. Found a blog that says with a traditional 20 degree transducer you would be looking at approx 1/3 of the depth. So at a depth of 10 ft you'd be seeing about 3.5 ft of bottom.

My question is with DownV?: 260/455/800 kHz and a 20 degree transducer, how large an area of the bottom would I see ?

 

[Bondo]

Ayuh,.... Welcome Aboard,... It's a simple enough geometry question,....

20? by the depth of the water,....

The area will change with the water depth,....

 

[bruceb58]

Found a blog that says with a traditional 20 degree transducer you would be looking at approx 1/3 of the depth. So at a depth of 10 ft you'd be seeing about 3.5 ft of bottom.

That is exactly correct. 0.35 is the factor. Tangent of 10 degrees times 2.

 

[sam am I]

how large an area of the bottom would I see ?

The total "area" on the bottom as the trans (standard types) forms a cone shape radiating towards the bottom is projecting circle on the bottom surface (not perfect but a circle non the less).

So the "area" (you asked for area not the diameter at the bottom, right?) of a 20 degree cone at 10' is the diameter (10*.35) = d and d^2*pi*1/4 = 9.62 ft^2 = A (area)