Formula Needed

[crappie cowboy]

I'm trying to figure out how much lake bottom area a 20 deg. transducer will show say in 10 fow. Is it 2 foot? Don't sound right to me, but I only have a GED education.

crappie cowboy

[BuckCreekMike]

I may very well be wrong, but I came up with approx. 41s.f. I used the formula tan10=x/10' basically it is a circle about 7.25 feet across. Someone else let me know if I did this right.

[brookpe]

tan= opp/adj

10 degrees is half the cone

tan 10 degrees is .1763

then oppossite at 10' is

.1763*10= 1.763

then *2 = 3.53'

Brook

[BuckCreekMike]

I noticed that I wrote it wrong, but I actually used tan20=x/10.

[BuckCreekMike]

But what I did just post is wrong b/c it would not be a right triangle. What I wrote initially was right, but I figured it with tan20.

[brookpe]

You are right.

The cone angle is measured from the center out.

So you do use 20 degrees, not split in half

Brook

[Pat]

My old Garman says 20 degrees , coverage approx one third water depth........>>>
10 ft about 3. 3 ft

[crappieken]

If you don't want to do all the math work, the distance across the bottom with a 20 degree cone is approximately 1/3 of the depth. Thus, at a depth of 15 feet you are covering about 5 feet of bottom.

[crappieken]

Sorry to repeat what you said Pat, I just didn't scroll down far enough to read your post.

[Phantom309]

Yeah, 20 deg cone = divide depth by three .....