Topic: Mega SI transducer setup on flat bottom AL boat problems.

[Kenai]

Hi,
I recently switch to G4N Helix and I got new fantastic and also enormous transducer. My clearance between transom and engine is not enough now, and I have basically just one side of SI working. Also to make maters worse I need to be able to easily switch my engine from 15hp outboard to electric trolling motor for electric only lakes.
I considered jack plate. However, I already know I will run in a lot of new troubles. For example, my boat is only 12 ft, so shifting 35 kg outboard more to aft, definitely will not help with weight distribution of that 60 kg boat. I have some trim issues already and been forced to add ballast in front. For such a small boat, only a mechanic jack plate may be considered. And by looking of my setup right now, I will be forced to bolt mount engine to the jack plate, because there simply wouldn't be enough space for another solution. This is unacceptable for an electric boat setup because I need to be able to put that outboard easily and fast down.

So I need an extra 6" (16cm) of clearance and have no idea how to get it.

I also consider Y cable and two SI transducers.  This setup will certainly work, but there will be most likely blind spot between them. Am I right ?
I'm not happy about 600Eur expense to get a blind spot about a meter big just side on the side of the boat. 

Can anybody suggest better solution than this two bad I just mentioned? And please correct me if I'm wrong in something, that all just theory for me.

I attached an old picture of my setup, with 598 for reference. The only difference is the new monstrous XM 9 HW MSI T instead.

edit: Added visualization of dual transducer as I understand it.

[rnvinc]

That diagram is misleading  ... as it shows the SI “measured cone angle”’and not the SI “actual cone coverage”  ...

This diagram shows the SI “actual cone coverage” of one SI piezo  ...



The SI “actual cone coverage” of each SI beam extends way past vertical under the boat  ...

You will not have a gap in coverage with 2 SI xducers  ...

Rickie

[Kenai]

Thank you for good news, sir. I'm very happy to hear that. Hoped you will have some quality information, because HB support doesn't. They have in FAQ there is about 5 deg. Gap in SI even with only one transducer. https://humminbird-help.johnsonoutdoors.com/hc/en-us/articles/4415341433495-When-I-use-a-Side-Imaging-product-am-I-missing-critical-information-directly-under-my-boat
That measured diagram is from which transducer? Just from curiosity. That must have been a pretty expensive lab exercise to measure that.
I respectfully bow before such a quality of work. 

As I'm thinking of it now. What about detector side (microphones). They need to be much more directional, am I right. I'm getting curious, what else could I learn from you.
I graduated in electrical engineering, specializing in sound technology. So, I am quite familiar with it, even if the frequencies are here 1000 times higher and my main goal was mostly to get rid of echos. 

[rnvinc]

I must defer those kudos to our member and friend Rickard from Sweden. He is the person responsible for calculating beam angles for us if he has accurate measurements of the piezoes  ...

I dissected the XM 9 20 MSI T and Rickard calculated the “beam form” from my measurements  ...

http://forums.sideimagingsoft.com/index.php?topic=10052.msg59071#msg59071
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—-

The piezoes are dual function  ... they convert electrical signals (from the transmitters) into vibration to send the sound pulse thru the water  ... and they also convert the return back into an electrical signal which is then sent to the processor  ...

https://assets-7484b3de52.cdn.insitecloud.net/d98fc1bb5148190/Theory-of-Operations-2014.pdf

Myself, I have no formal training  ... I just enjoy tinkering with this technology  ... I’ve been here a while learning from these great minds  ...

Rickie

[Kenai]

Thank you sir. That makes a lot of sense. Finally, many things clicked to place in my head. Thank you again.
The only thing bothers me now is why HB declare there is 5 degree gap. That is not something to fabric out as a sales argument. So I believe it is, or at lest, it was true at some point. Maybe it is an SW thing, because of the way data are processed or displayed.
Now I'm just thinking loud. If Y cable is used, head unit have no information about it. Data will be processed the same way as with one transducer. However, there will be a gap between left and right side piezos. SI most likely just use return time to determinate angle and therefor horizontal distance from boat. However, this can only work if the return came just from one side. 90 degree under boat can't work, there is too little delay. 85 is most likely the maximum what is possible to calculate as difference between depth and distance. I hope this is understandable. English is not my first language, and this is not easy to explain even if it was.  Basically, the delay between time of straight return from depth (90 degree) and the time of return from object throe SI piezo determinate angle to the side and subsequently position of object.
Has anybody tested this in real life ? Is there a gap or not? 5 deg to each side is a lot in deeper water. In 10m (32 ft) it will be 2m (6.5 ft) gap under the transducer. That will be even bigger if Y cable is used (sorting like 3m 10 ft). I notice just it looks like if I drive exactly over something I barely see it on IS if that is small enough object. However, the same object is perfectly visible, if I drove just a meter or two to the side.

[rnvinc]

We must remember that sound thru water does not “stop” at any line or degree of measurement  ... sound thru water continues until it either reflects off of an object or continuously dissipates until absolute silence  ...

Any brand’s angle “measurement” is simply a measurement in that sound pulse’s travel of how much the sound has dissipated to a certain sound intensity level.

Humminbird’s chosen angle “measurement” is -10db  ... which means “the angle measurement when the sound pulse has dissipated to 1/10 the intensity of the initial sound pulse intensity emitted from the piezo” ...

Targets outside this angle ‘measurement” do still reflect the sound pulse  ... and the piezoes do still receive echoes from those targets outside the angle “measurement”  ... it’s the processors job to discern if those reflected echoes “outside the angle measurement” are large enough (or reflective enough) to display those targets in the image  ...

Rickie

[rnvinc]

The “gap” indicated in Humminbird documentation (and in most people’s mind) comes from this Humminbird diagram :



The 86° angle is not where the sound pulse “stops”  ... 86° is where the sound pulse travel is ”measured” when the sound pulse intensity has dissipated to 1/10 of its original pulse strength when emitted from the piezo  ...

The sound pulse continues past this 86° (as indicated in the beam form diagram” in my first reply to this thread)  ...

Humminbird uses this diagram because the “measurement” is indeed factual  ... but documenting “sound pulse dissipates continuously to absolute silence” could not be drawn into a diagram that the general public could interpret  ...

Rickie

[rnvinc]

And BTW  ... your understanding of the AS SILRY Cable is correct  ...

The AS SILR Y cable is nothing more than wires = redirecting where signal paths go  ...

The head unit has no indication that anything has changed in the hardware  ... and transmits signals to piezoes thru wire channels  ... those wire channels (transducer cable and/or Y cables) physically determine where those transmitted signals end up  ...
——-
——-

Note : the older AS SIDBY Cable and the AS SILRY Cable are for the older core numbered series units and the HELIX G1 units   ...

Any HELIX newer requires the 9M SIDB Y cable or the 9M SILR Y cable because of the addition of a dedicated DI circuit to a dedicated DI piezo in any HELIX imaging unit G2N or newer  ...

ONIX/SOLIX/APEX are different animals and must be discussed in a different context  ...

Rickie

[Kenai]

Thanks for explaining. Maybe I'm somehow confused by microphones and reproductions. Possibly I'm trying to apply here something which not belong in sonar technology. All you described to me until now is also true in audio technic. With one difference. Microphone is one device and reproducer the other. If I understood you correctly, the transducer uses the same piezo for both functions, emitting and also receiving sound. This is basically true also for reproducer, reproducer small enough could be also used as microphone, really unsensitive one, but it will work.
However, the ting I do not understand is how is position of object (echo return) degerminated in left-right plane. If transducer emits ping than waits for reflection and this reflection is processed.
This reflection has I believe 3 measurable parameters. Amplitude, frequency drift, and time of return. So how could be Left right plane determined ?

Take theoretic example to simplify things. We have something like this: Infinite depth, so no return except from an iron hollow ball floating in water. It is in spaces covered by left and also right beam. Therefore, return echo will be from both left and right piezo pings. Also, those echos returns will be received by both piezoes. I suspect left and right piezo is not pinging in the same phase. Maybe they use also slightly difference frequency. So unit can determinate which echo is from which piezo for sure. However, both piezoes will detect that ball because it is in area coverage by both piezoes. Return from left piezo will be most likely a little fainter because of energy drop shown by emitting diagram you posted. Anyway, that don't rule out two balls from different material exactly in same depth and exactly in the same distance from transducer in left right plane. Even worse, if there will be only one piezo I still can't see how it will be possible for the head unit to tell on which side of the boat the ball is.  I believe had unit can determinate only distance and amplitude of return, may be frequency drift. But not position. So if there are two possible solutions for the same distance, how can it tell which one is correct?

[rnvinc]

Your suspicion is correct in that each piezo is “pinged” in sequence  ... and each ping must wait for the return echo before the next “ping” (from a different piezo in the sequence) is sent  ...

The above fact, and the fact that the SI piezoes are physically “aimed” to the side (30° from horizontal) give the unit definitive discernment of whether a target is left or right  ...



Rickie

[Kenai]

Yes Sir, that is exactly what I was thinking. But if that's so, there simply can't be any overlapped area. Because there will be no way to tell if the object is right or left from piezo which covers space on both sides. That fact I was trying to show on my drawings. 
So If the beam diagram is correct, there must also be something else happening. Probably some filtering of used beam angle. That is most likely the reason, why SI don't cover space directly under the transducer and why HB said SI have 5 degree to each side of blind space under the boat.  It is also shown in SI animations, where is visible a small uncovered area under the boat.

[rnvinc]

This will be lengthy so bear with me to explain a “transmit cycle” to its basics  ...

Let’s back up and explain “ping sequence” in a little more step by step process  ... in the terms of what is a “transmit cycle” ...

(For explanation purposes we will say there are 4 piezoes in the “ping sequence” = 2d piezo, DI piezo, left SI piezo, right SI piezo)  ... (there are actually 7 piezoes in a MEGA transducer but only the 4 listed are used in a given “ping sequence”)  ...

Step 1. Processor tells 2d transmitter to fire  ...
Step 2. 2d transmitter sends electrical signal to 2d piezo  ...
Step 3. 2d piezo coverts electrical signal to mechanical signal (vibration)  ...
Step 4. Mechanical signal vibration emits from transducer into water (sound pulse)  ...
Step 5. 2d piezo listens for return echoes (no other piezoes are listening for return echoes) ...
Step 6. Return echoes return to 2d piezo  ...
Step 7. 2d piezo coverts return echo vibrations back into electrical signals  ...
Step 8. Electrical signals are sent to to processor  ...

Step 9. Processor tells DI transmitter to fire  ...
Step 10. DI transmitter sends electrical signal to DI piezo  ...
Step 11. DI piezo coverts electrical signal to mechanical signal (vibration)  ...
Step 12. Mechanical signal vibration emits from transducer into water (sound pulse)  ...
Step 13. DI piezo listens for return echoes (no other piezoes are listening for return echoes) ...
Step 14. Return echoes return to DI piezo  ...
Step 15. DI piezo coverts return echo vibrations back into electrical signals  ...
Step 16. Electrical signals are sent to to processor  ...

Step 17. Processor tells left SI transmitter to fire  ...
Step 18. Left SI transmitter sends electrical signal to left SI piezo  ...
Step 19. Left SI piezo coverts electrical signal to mechanical signal (vibration)  ...
Step 20. Mechanical signal vibration emits from transducer into water (sound pulse)  ...
Step 21. Left SI piezo listens for return echoes (no other piezoes are listening for return echoes) ...
Step 22. Return echoes return to left SI piezo  ...
Step 23. Left SI piezo coverts return echo vibrations back into electrical signals  ...
Step 24. Electrical signals are sent to to processor  ...

Step 25. Processor tells right SI transmitter to fire  ...
Step 26. Right SI transmitter sends electrical signal to left SI piezo  ...
Step 27. Right SI piezo coverts electrical signal to mechanical signal (vibration)  ...
Step 28. Mechanical signal vibration emits from transducer into water (sound pulse)  ...
Step 29. Right SI piezo listens for return echoes (no other piezoes are listening for return echoes) ...
Step 30. Return echoes return to right SI piezo  ...
Step 31. Right SI piezo coverts return echo vibrations back into electrical signals  ...
Step 32. Electrical signals are sent to to processor  ...

What we learn from this explanation of piezo “transmit cycles” and how each fits into the “ping sequence” is  ...:

A. There is only 1 piezo operating (in its own “transmit cycle”) at a time in the “ping sequence”  ...

B. The 8 steps in a certain piezo’s “transmit cycle” must complete before the processor will tell the next transmitter to initiate its “transmit cycle”  ...

C. Therefore the unit “knows” exactly which piezo received a certain return echo  ...

Rickie

[rnvinc]

The “overlap” under the boat of the left SI beam coverage and the right SI beam coverage are factual  ... sort of  ...

If … the left SI piezo “transmitted” at the same time as the right SI piezo “transmitted” = both would indeed overlap at the same time   ... and the unit could not discern where a certain target was under the boat (as you supposed)  ...

But because each piezo is operating indendently and only in its section (“transmit cycle”) of the “ping sequence” = the beam form diagram is correct  ... SI beams do extend past vertical under the boat  ...

But also because, the left SI “transmit cycle” and right SI “transmit cycle” are not operating at the same time = the unit is not confused on which piezo received which echo return  ...

Rickie

[rnvinc]

And just an added small note  ...

The sequence of steps I listed are not exactly in the sequence of 2d piezo, DI piezo, left SI piezo, right SI piezo  ...

I listed the piezos in that sequence for ease for understanding “ping sequence”  ...

But the fact remains that only one piezo is operating at a time in its queue of the “ping sequence”  ...

Rickie

[Kenai]

Thank you sir, so far it is clear. However, I'm still confused about that over 90 degree coverage of SI piezo.  I will write it as simply as I can. English is not my first neither second language and technical terms are some problem for me. So bear with me.
I will simplify this example to just one piezo.
1. Let's say it is DI like piezo firing directly down with cone-shaped coverage. There is no way to tell on which side of that cone is fish (or some other return)  present. We can just tell, if it is in the center or more to the side of the cone, by the strength of that return. We can't tell to which side. But we know exactly the depth of that fish in water column (distance of return) The same is true for 2D. Correct? 
2. If that cone is oval-shaped as SI and still firing directly down (hypothetically) the same is true. We can't tell anything about fish/return position, but we narrowed it in one dimension (front aft) And we know exactly the depth of that fish in water column (distance of return) Correct ?
3. If we tilted that SI oval beam to the side, in the way, it covers the angle from vertical to horizontal on the right side (as it should in reality) our depth will change to distance from the boat. (in reality it is not horizontal distance, but this is good enough for our example)  We still don't have any new information about return's position.  The only way how we can tell it is on the right is by knowing that the beam covers only the right side. Time elapsed from ping to return is the function of distance.  Shortest possible distance (and time) = directly under boat. Longest =  parallel with water surface. Correct ?
That is the way I understand how SI works. Basically by beam shape and orientation. And the head unit interprets time of return as distance to side, instead of depth.

4. Now, how is possible to tell where the position of return is, if the cone is oriented like my pictures shows?   From 60 to horizontal.  (past 90 degree) So it covers more than right side only.

I believe if the SI coverage will be bigger than 90 degrees (from vertical to horizontal) then will be all echos with the same return time displayed on screen on top of each other blended together.
Correct ?

[rnvinc]

You are correct on all points  ...

The target cannot be discerned as to where it is in the thin slice beam of SI  ... only distance to the target and strength of the return echo  ...

Here is a diagram that represents what you mention as targets the same distance from the transducer being “stacked together in the SI image  ...



We see in the upper section of the diagram that fish 1 and fish 2 are the same distance from the transducer  ... and we see in the lower section of the diagram ( SI image depiction) that fish 1 and fish 2 are shown stacked in the water column  ...

It’s the additional information (in the SI image) of the fish shadows cast against the bottom detail that discern where that fish is in the water column  ...

Rickie

[Kenai]

Thanks for the reassurance.
So now just to figure out how that over 90 degree of coverage of SI piezo could possibly work. I have no idea. Does anybody ever really test if that coverage is more than 90 degrees (120 in fact from that piezo diagram)? It must be somehow trimmed out to that 86 degrees or so.

For fun, this is what HB support know about that:
"I do apologize but we are not able to give examples of this. We do no have exact angles or any information on how big the blind spot will be."
 
If we can assist you further on this subject or any other Humminbird related questions please feel free to email us again.
 
Good Boating and Fishing!
 
Tiffanie
Technical Support Representative

[rnvinc]

Here is a SI image example that clearly depicts the fact that the SI beam extends past vertical  ...



Now here is a diagram of why that happens  ...



It is clear that the only explanation of the bait ball showing in the right SI and not casting a shadow on the right SI bottom detail is simply because there is only one baitball and that baitball is actually slightly left of vertical and being picked up by the right SI beam extending past vertical  ...

I assure you there is no gap or blind spot between the left and right SI beams  ...

Humminbird CS agents answer your question vaguely because they don’t have in front of them adequate info (on their canned answer sheet) of how to explain SI beam forms and how SI beams are measured vs how the SI beam actually travels thru the water  ...

Rickie

[Kenai]

Thank you sir. I really can't find any other explanation for this. I will think about it and I will try to come out with a real world experiment to confirm this. Right now, I think about what this means for the picture of bottom. It also has to be doublet. That must be visible on the SI screen unless it is cut out somehow.

[Kenai]

Here is a SI image example that clearly depicts the fact that the SI beam extends past vertical  ...
...
Rickie

So I made some testing, and it came out with even more questions than answers.
Pictures are quite self-explanatory. I idled repetitively along vertical pipe in various distances and directions. I got inconsistent and strange outcome. Sometimes the pipe can be seen on both sides (confirming SI overlap hypothesis) but sometimes it is not.
Also, the closer I went to the pipe, it shows less in the second SI beam. This should be the other way around. Could you please somehow bring some light to this?

I also made other experiment. Also, inconclusive. I run recording and on a stationary boat tied to a dock I tried lover underwater camera along transducer and I try to see if it will be visible on both sides. It wasn't. It wasn't visible at all until I move it quite far from the transducer to the side. I can see it perfectly on DI and 2D, but not on SI. Si start to pick it up when it was at lest 30 cm to the side from transducer, regardless of depth. It could be caused by interference in close range from transducer, or I was unable to place it on correct place in front aft plane. So I need to come with something else. I will most likely place some hollow sphere in water some 3 ft off bottom and I will try the same thing as I did with that pipe. I will do it in deep water >30ft to get decent SI overlap (if it exists) end I will idle around in various distances. This will hopefully solve this for good.

edit: one more thing. I have some shaded area in the left beam because of my trolling motor on the left side. So it could be, it is not past 90 beam cover (overlap). It could be simply reflection from motor casing. It is consistent with showing itself only when I pass it on starboard side.

[rnvinc]

I agree some of your images clearly indicate the off side beam picking up the pole and some images do not  ...

I have found vertical poles to be a challenge for SI imaging for a few reasons  ...

*Poles are usually not very wide to create a good echo target  ...
*Poles are usually round - which deflect more echos than reflect echos  ...

If your boat has a trolling motor - I would suggest getting the boat speed down to 1mph or less and also lower the SI Chart speed to 1  ...

This will allow more “successive” pings to hit a known target  ...

Rickie

[Kenai]

I will do. Just need to find adequate weather. For 1mph it needs to be zero wind otherwise I will drift all over the place.
I didn't make the underwater ball experiment, yet. I struggle to figure about how to make it retractable, not dangerous to the screws and not overengineered at the same time.

[Kenai]

Well, finally I tested it again with motor on jack plate. And as i suspected, there was no image of the pole on the opposite side. So as long I can tell, no overlap.