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.
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
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.
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
I noticed that I wrote it wrong, but I actually used tan20=x/10.
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.
You are right.
The cone angle is measured from the center out.
So you do use 20 degrees, not split in half
Brook
My old Garman says 20 degrees , coverage approx one third water depth........>>>
10 ft about 3. 3 ft
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.
Ken
Sorry to repeat what you said Pat, I just didn't scroll down far enough to read your post.
Ken
Yeah, 20 deg cone = divide depth by three .....
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