Question about 360 (and side scan) image interpretation

[doctorxring]

Assuming the boat is pinned with two power poles --


Would the spot marked "A" (or any point on this arc) be
right below the transducer ?

If I wanted to throw a marker to land right at the spot marked
"B", which is at the absolute edge of the image, would I throw
the maker in that direction 25.6 feet from the position of the transducer ?

thanks, Chris

[Wayne P.]

Number one thing about the 360, it is Side Imaging that rotates.

A=bottom below the transducer, same as Side Imaging since 360 is Side Imaging that rotates.

B= distance from the transducer. Slant range, not horizontal range, same as Side Imaging.

[Bob B]

At 4 feet, the horizontal range would only be slightly less than the slant range ...... but you need to do some Google searches about slant range to understand it. It is a very important concept in understanding and interpreting SI.

[doctorxring]

Thanks guys. So is this what is represented on this 360 scan, as per the original question ?













And plugging this data into a right triangle calculator will give me the distance for my marker toss,
which as Bob said would be close to the same. Close enough for sure, as there is no way I would
toss it perfect anyway. But when the water depth percentage of the screen is larger, then it
will become more exaggerated.

With the boat icon being placed where it is on 360 and side scan it's easy for people to loose track
of where these factors are placed.

[Wayne P.]

Yes, almost. The slant distance is 30' so the horizontal distance would be 25.6' at that depth. You have the number on the wrong line.

[doctorxring]

Yes, almost. The slant distance is 30' so the horizontal distance would be 25.6' at that depth. You have the number on the wrong line.

So the "slant" distance, or "range", is the sum of the bottom depth and the horizontal distance. Nothing to do with right triangle computations.

Then my original proposition is correct, is it not ?

That was my understanding of 1:53 on this video --







.

[Bob B]

It gets more complicated if the object is not on the bottom .....but I think you have the basic concept which is really what is important.

[Wayne P.]

So the "slant" distance, or "range", is the sum of the bottom depth and the horizontal distance. Nothing to do with right triangle computations.

Then my original proposition is correct, is it not ?


.

No that is not correct either. Your 25.6 is wrong. The only known distances is the hypotenuse (slant) is 30' and B is the depth 4.4'

The horizontal distance is 29.7'

[Wayne P.]

This calculator will also give you each angle.

http://www.cleavebooks.co.uk/scol/calrtri.htm

[doctorxring]

It gets more complicated if the object is not on the bottom .....but I think you have the basic concept which is really what is important.

I should have made that clear on my first post. I'm talking about a target that lies on the bottom at the very perimeter of the arc on the screen.

The main thing I need to know is if my "range" is set to 30 feet and the water depth is 4.4 feet, is the distance from a point on the bottom directly under the transducer to a point on the bottom at the edge of the scan 25.6 feet away.


.

[doctorxring]

I'm thinking the answer is not represented by a triangle, but an L shaped, two dimensional sum.

If there was zero depth then the "range" figure (in feet) would represent the distance from
the transducer to the edge of the scan. If there is depth, then the depth is
subtracted from the "range" setting if you want a figure that represents the distance
from a point on the bottom directly under the transducer out to the edge of the scan.

What is confusing for people, I believe, is to have a screen image that is 1/3 water column
and 2/3 distance over the bottom that has a boat icon in the middle and distance
range lines equally spread over the screen. If you have the range set at 100 feet but
the water depth is 90 feet you are going to have only 10 percent of your screen
showing the bottom. And you will have "range lines" equally spread out over the
screen. How many people realize that what is represented on the screen is only
a 10 foot wide bottom section beginning right underneath the boat and going out
only 10 feet to the right or left of the boat. The other 90 feet of the "range" is
directly underneath the boat, straight down. But the "range" is set at 100 feet
per side.

[Wayne P.]

It is not confusing at all if you realize 360 is Side Imaging. Range lines, water column, bottom display, and stuff above the bottom are the same. Just SI produces sharper images since everything is scanned at the same rate.
The 360 scans at different speeds (faster as the range increases).

[doctorxring]

Here is another example.

Question -- How far out is the bottom being scanned on each side from
the middle of the boat. The transducer is on the mid-line of the boat.
Take the latest data at the top of the image, which is a depth of 23.6 feet.

thanks, Chris

[Fishton]

Not sure if this will help ... https://www.facebook.com/13119687048...2850561495482/

[Wayne P.]

Here is another example.

Question -- How far out is the bottom being scanned on each side from
the middle of the boat. The transducer is on the mid-line of the boat.
Take the latest data at the top of the image, which is a depth of 23.6 feet.

thanks, Chris

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