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Denali mini sound bomb!

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So as above I have bought a pair of these things for the FJR, my thinking being if you want loud go large. BUT what I did not realise is that the poxy things need 5.2volts to the horn NOT the 3.2volts that the bike provides for the standard horns through the loom! So you have to buy another bespoke loom from Denali to make them work (bugger). But my main problem is how do I make the existing horn button work with the new horns if I have to by-pass the existing loom? I have 4 connectors that go to the original horns and the horn button which is effectively just a breaker in the line to the horns, is it possible to use the horn button to power the new horns and if so how? Cos what I don't want is to have 2 horn buttons of which only one works cos a sure as shit when the time comes to use it I'm going to forget which one works and which one is redundant! Any insight to help me figure this out will be great-fully received. At the moment I can only get them to work if I connect them directly to the battery so they work ok with 12volts DC but sound like an anaemic donkey with the lower voltage delivered from the existing loom. Thanks all.

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I guess you could run a new feed into the horn button, from an existing switched feed, then connect the original horn cabling to your new ones. 

I'd take a feed off the supply to the rear brake switch (assuming that's 12v) or the side lights 

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Use the horn wiring, just put a relay where the horn still is or where you can conveniently access the connections from the loom somewhere behind the fairing.

Fit the relay here, using said relay you can have either new or old depending on switching, a switch you can put wherever aesthetics or need suggest.  The default position (non energised) leaves you bike in stock config. That way if the new stuff fails you always have the horn, (yep deliberate).

From that point just wire up your super horns with whatever magic gets the extra voltage? 

Using the brake light circuit to feed high power horns is not a good idea, if your hitting the horn. You are most likely braking, that means full load on that circuit plus whatever the horns pull. You may lose the horn when you need it most...… (?) …. Er.

Fresh line from the battery, or pick up on a non essential. Heated grips? Phone charger?. Lose those in a critical moment it makes no odds.

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Thanks lads, that sounds like a plan, I have to confess it never occured to me that the new horns would need to be any different. Just goes to show your never to old to make a tit of yourself! Yeah a relay sounds like it might work a treat and I think I will use a non essential bit of wiring just in case it does go pop. I think the horn button is on the Neg side of the switch so I might actually run a single lead with 2 connectors direct from the battery (+ side) if all else fails and as you say I can then just re-instate the original horns if they fail. The tank is empty so no probs just whipping it off and running some new wiring across the top attached to the original loom to keep it neat.

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Cynics right about the relay - thought I'd mentioned putting one in - too many beers! 

Yeah horns usually have a permanent live feed so you're switching the neutral. Which is generally frowned upon electrically but it seems the norm on bikes!

Just remember to fit an inline fuse between the battery and your relay, stop you starting any fires if something shorts. 

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So does a 'how to' not come with the new loom you bought

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2 minutes ago, jimmy said:

So does a 'how to' not come with the new loom you bought

He's done the blokey thing and binned it already 😂

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Yeah it comes with a little leaflet thingy that tells you it is a straight swap for your old horn and the + and - polarity! But sod all else of any actual use plus of course no one even suggests that it might not work cos the voltage is wrong!!! Sorry Jimmy but Denali say that they sell a "loom" which is in fact just 2 wires and some connectors, I can do better I think, plus I can measure only what I need and not have rolls of extra wiring dangling off the bike. Should be fairly simple and the idea of a relay and in-line fuse are good. Using the existing horn button is going to be a bonus as well.

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20 hours ago, finnerz89 said:

He's done the blokey thing and binned it already 😂

And it came in very small print so he couldn't read it anyway 🤣🤣

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I'm confused. Isnt the soundbomb 12v? Isnt the feed from the loom also 12v? There is no difference in voltage.

Like Cynic, I would relay it however, as the current draw will be larger.

 

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I’ve got the mini sound bomb and it’s plug and play. It was only with my BMW that I had to buy an additional wire for it to work with their canbus.

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On 1/1/2020 at 2:04 PM, finnerz89 said:

 

Yeah horns usually have a permanent live feed so you're switching the neutral. Which is generally frowned upon electrically but it seems the norm on bikes!

Electrical principles mate, its bad on AC systems, well blody lethal really. On DC automotive systems it reduces the load on the switch and reduces voltage loss to the device your switching as well as reducing arcing within the switch.

Switch a 10A dc load (heated gips say) on the live side your switching a 10A resistive load. Switch it neutral side and maybe 8 A have been burned up doing the heating so you only need a switch capable of 2A load to switch 10A, its not that literal but that's the logic to it.

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6 minutes ago, Cynic said:

Electrical principles mate, its bad on AC systems, well blody lethal really. On DC automotive systems it reduces the load on the switch and reduces voltage loss to the device your switching as well as reducing arcing within the switch.

Switch a 10A dc load (heated gips say) on the live side your switching a 10A resistive load. Switch it neutral side and maybe 8 A have been burned up doing the heating so you only need a switch capable of 2A load to switch 10A, its not that literal but that's the logic to it.

Well I have to disagree, but current is the same throughout the circuit (Kirchoff's Law). It's a legacy of automotive electrics to switch to earth - I think it was because having both sides of the switch at a potential of zero when it the contacts are closed causes fewer malfunctions.

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Yesssss forum discussion. Not been one of these for ages.

The way i see it current is the measure of energy, the energy after the above example has to be less due to the energy expended as heat.

One side of the heated grips will have less energy than the other,  incoming side from the action of heating needing energy. My theoretical grips have 96watts, that's work done, power, current drawn and radiated as heat. If the output from the grips was the same as the incoming  that's grips heated without using any energy.

How does volt drop figure due to load and resistance. If you powered a light bulb through 10 miles of lighting flex there would not be enough energy left to light the bulb. 

Newton versus Kirchhoff. Kirchhoff is talking about a very specific point in a circuit. Not the whole circuit including load and losses.

Purely on experience switching post load needs less capacity and results in smaller switches being needed.

 

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Current is the same anywhere in the circuit. But the voltage will be lower in the situation you're describing.

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So why have a fuse block with 5a 10 a and 20 a fuses, by that line with 30a of load why do the fuses not pop. 

The current in the wiring of my house should be the same as the secondary of the step down transformer that feeds my estate then?

 

I know im 20 years, er more like 30 out of date on the math, I just cant see it.

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53 minutes ago, Cynic said:

Yesssss forum discussion. Not been one of these for ages.

The way i see it current is the measure of energy, the energy after the above example has to be less due to the energy expended as heat.

One side of the heated grips will have less energy than the other,  incoming side from the action of heating needing energy. My theoretical grips have 96watts, that's work done, power, current drawn and radiated as heat. If the output from the grips was the same as the incoming  that's grips heated without using any energy.

How does volt drop figure due to load and resistance. If you powered a light bulb through 10 miles of lighting flex there would not be enough energy left to light the bulb. 

Newton versus Kirchhoff. Kirchhoff is talking about a very specific point in a circuit. Not the whole circuit including load and losses.

Purely on experience switching post load needs less capacity and results in smaller switches being needed.

 

you cant treat whole parts of a circuit as a node using Kirchoff's. Quite normal to do so, as Finnerz says current flow is conserved and measurable at any point in a path.

If you think about the long piece of flex, the potential difference at that distance would not be able to drive a current flow (the resistivity would be too high) - you might get microamps of flow [DC has other effects that occur over long distance that arent always desirable). It wouldn't be that there would be energy to light the bulb, there wouldn't be a measurable and usable flow to allow any conversion of energy.

What you have to remember is horns, heated grips, bulbs are all electrical devices which are specifically designed to convert electrical energy into other forms (sound, heat and light).

You can switch a DC circuit anywhere - general convention is that you do it at the most positive part of the circuit, automotive circuits buck this by switching close to earth. As you stated mains AC is different mainly because the Lives is alternating between a positive and negative potential and there is a risk that any thing which is connected could create a dangerous potential difference, hence equipotential bonding.

2 minutes ago, Cynic said:

So why have a fuse block with 5a 10 a and 20 a fuses, by that line with 30a of load why do the fuses not pop. 

The current in the wiring of my house should be the same as the secondary of the step down transformer that feeds my estate then?

 

I know im 20 years, er more like 30 out of date on the math, I just cant see it.

Not sure i get the first question.

The second though is like this: the substation transformer that feeds you estate (and probably others) has 3 live lines out each out of phase by 120 degrees. A group of houses will get a live line each so the transformer, so the windings in the transistor will drive the current for all those houses in that group. Typical house in very typical conditions drawing several amps (between 2-20A unless cooker or electric shower on), so it's a summation (Kirchhoffs) of the current draws of all the houses. 

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3 hours ago, Cynic said:

So why have a fuse block with 5a 10 a and 20 a fuses, by that line with 30a of load why do the fuses not pop. 

The current in the wiring of my house should be the same as the secondary of the step down transformer that feeds my estate then?

 

I know im 20 years, er more like 30 out of date on the math, I just cant see it.

If you imagine a simple circuit with a battery, switch, and a light bulb. With the circuit made an ammeter will record the same current pull, before the switch, before the bulb and after. 

Your example is parallel circuits so the pull on one circuit won't affect the other. 

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Ok thanks folks I never thought that something so simple could get so complicated. I have a mate who is an auto electrical engineer so I think I might just hand the job off to him to come up with a workable scheme. Then I will post what he's done on here so we can at least have a reference for when it happens to others. I did enjoy the posts tho, not often now that we get to have a chat about bikes and the problems we come up with and the solutions that we all have that differ from each other, a fused line and a relay being just 2 examples that I had completely forgotten to even think of! 🤦‍♂️

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It still doesent compute in my mind slice but the last time I saw the inside of a class room as a student Maggie had the big chair at no10.

Paul and Merv  have far more current (geddit) technical knowledge and I know when to shut up.

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