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phenom880

side image question

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sidescan.JPG.7aa58acafa3ad36aab8b5ccbcd2ee485.JPG

so i understand that the tree in the center of picture is in about 20 fow and should be right under the boat since it spit the image..How ever if i was to move that curse, to the left to the base of the tree it would say 20 feet to the left of the boat...now if i was to move the cursor 5 feet further to the left it would say 25 feet to the left but wouldn't it really just be 5 feet from the tree that is right under my boat? This is something i have never understood with side image...someone help?

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I'm not a super expert, but I've always assumed that these types of things are a result of the sonar cones.  

coneangle.thumb.jpg.96bba7f486da4166f6657914b8a79e47.jpg

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Think of the image "folded.  Directly under the boat is that dark spot, or middle (/\) of what I typed below.  The sides are the flat.  You're in 20" water, so the tree occupies that 20' wide footprint in the (/\).  If *I* were to mark the point, I'd use the DI if I wanted the most accuracy.

 

SI is like this:

           /\

_____/  T\_____

 

The T is you basically your tree.  The owner's manual is probably a better bet for a perfect explanation.

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Believe it would depend on how they are calculating cursor position, and what true SI coverage is. I've seen suggestions that the split J Francho diagramed above doesn't really exist on some units. They might also be calculating distance from transducer (angular measure or additive measure) as opposed to horizontal distance from boat or centerline. A little  more time on the water and a few marker buoys would get you an answer next trip out 😎

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Based on what you are saying and what I'm seeing in the picture,  I think it has to be the distance from the transducer.  On my Humminbird,  that would be the case.  Team9nine brings up an interesting issue of the split or width of the V in J Francho's diagram. Is it wide like this _/\_ or narrow or non existent like this _|_ ?  On your picture,  it clearly exist because you have the herring school in the water column on the left only.  If it did not exist then the water column on both sides would be the same.  I hope your user's manual gives more information than mine on how wide this is.  It makes a difference in how you interpret what you see.

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I'm going by the documentation for my pre Mega Helix.

It's probably narrower than my diagram, somewhere between yours and mine?

Maybe this will help?  the bait is below and to both sides, all the way to the surface.

 

IMG_1316-X2.jpg&key=b8166deacbaf413c3903

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Yes, and in fact, here's some video as I approach:

 

 

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The discussion on the coverage split is interesting because nearly every manufacturer shows the coverage split as depicted by J Francho, but some curious anglers have produced enough anecdotal evidence to suggest otherwise. In the OP's screenshot above, if you only look super close at the tree and the fish marking in its branches, both sides (images) appear to be practically the same, right down to branches and individual fish location among them. If there is a true "black hole" beneath the boat as manufacturers often depict, how could this (the image above) be? My guess is that in ideal settings/conditions, there is no "black hole." Ideally, we'd need something smaller but easily identifiable (perhaps a cinder block) to see if you could get even small objects to split onscreen. Doing so would suggest coverage overlap.

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Very interesting discussion.  So let say you are in 20ft water and you pass is a large fish to the right side. The fish is  5ft deep and 18ft from the transducer.  Where would the fish appear on the screen?  On my Humminbird Solix I think it would appear in the water column on the right and would appear to be 18ft deep.

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jfrancho let me ask you this then....if i sank a 10 foot telephone pole vertical in lake bed and the actual location was 10  feet to your left in this picture how would it show up?jfranco.thumb.JPG.a76423d409f4dc59694db57d8526d52d.JPG

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8 minutes ago, phenom880 said:

jfrancho let me ask you this then....if i sank a 10 foot telephone pole vertical in lake bed and the actual location was 10  feet to your left in this picture how would it show up?

Really good question! Very close to this spot are a set of submerged dock pilings.  Some are straight, some are crooked.  The straight ones should work just like your telephone pole test case.  The water is really green right now, so this will be a good test of my ability to read the returns.  I'll try to get some shots next time I'm out on "The City Dump."

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Capture1234.JPG.adf6a755e2461a98fa396ca8fbcfd4fc.JPG

So in red i circled what looks to be smaller rocks the boat went right over because they show up split on both sides......the blue marker is supposed to be a 10 foot telephone pole ...you can tell depth because its roughly half way up the water column....how far away from the boat to the left is it? If it is 22 feet away like the cursor would tell you then where is all the information from 22 ft to right under the boat where the rocks are? 

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52 minutes ago, Tennessee Boy said:

Very interesting discussion.  So let say you are in 20ft water and you pass is a large fish to the right side. The fish is  5ft deep and 18ft from the transducer.  Where would the fish appear on the screen?  On my Humminbird Solix I think it would appear in the water column on the right and would appear to be 18ft deep.

what if said fish was 18 feet deep and 5 foot from transducer...i think it would show in the same spot

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3 minutes ago, phenom880 said:

what if said fish was 18 feet deep and 5 foot from transducer...i think it would show in the same spot

The fish would appear in the same spot as long as it was on the right side of the boat and 18ft from the transducer as if you measured this distance with a tape measure from the fish to the transducer.  So it could be 18 ft deep and almost directly below the boat, it could be 18ft out to the side of the boat and almost at the surface of the water, or somewhere in between.  

My unit allows you to remove the water column and this is suppose to give you the horizontal distance to an object on the screen and not the angular distance (direct distance from the transducer).  I use this setting most of the time.

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1 hour ago, Tennessee Boy said:

Very interesting discussion.  So let say you are in 20ft water and you pass is a large fish to the right side. The fish is  5ft deep and 18ft from the transducer.  Where would the fish appear on the screen?  On my Humminbird Solix I think it would appear in the water column on the right and would appear to be 18ft deep.

I would agree...depth shown is distance from transducer, which is why you can mark fish that appear to be deeper than the water depth you're in.

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I am getting a headache but this topic is extremely interesting.

Thanks for sharing.

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58 minutes ago, phenom880 said:

Capture1234.JPG.adf6a755e2461a98fa396ca8fbcfd4fc.JPG

So in red i circled what looks to be smaller rocks the boat went right over because they show up split on both sides......the blue marker is supposed to be a 10 foot telephone pole ...you can tell depth because its roughly half way up the water column....how far away from the boat to the left is it? If it is 22 feet away like the cursor would tell you then where is all the information from 22 ft to right under the boat where the rocks are? 

Can't say for certain since you added it to the picture but it doesn't really exist. If it was in the actual screenshot, you would use its shadow to help tell height and location, as well as its base as a reference point along with which side of screen it showed up on.

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It's no wonder I don't catch fish, I need to bring along an abacus to figure them out.😁

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k...here is a post from another site on the same question i asked..

 

 

 

Correct ... Targets directly under the boat will show in both left SI and right SI ... 

Distance to targets in SI (out to the side) is calculated as "slant range" to targets ... (that geometry Pythagorean thing) ... 

 

slant.JPG

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It was just a matter of time before Pythagoras got involved.

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Just now, mattkenzer said:

It was just a matter of time before Pythagoras got involved.

ahahaha so true.....But it seems distance to side is not measured horizontally like i thought...so if i move my cursor

over to a brush pile and it says 34 feet its not 34 feet across the top of the water when im sitting in my boat. if i was in 22.1 FOW it would be 25.8 feet horizontally .....i think ahahahahaha

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9 minutes ago, phenom880 said:

k...here is a post from another site on the same question i asked..

 

 

 

Correct ... Targets directly under the boat will show in both left SI and right SI ... 

Distance to targets in SI (out to the side) is calculated as "slant range" to targets ... (that geometry Pythagorean thing) ... 

 

slant.JPG

Sounds like it confirms the "no dead zone" theory and we can ignore the split cone diagrams they all like to show 👍

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Just drop a way point on it and turn around.

:smiley:

A-Jay

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