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If I add a 3 ohm ballast in series, that should bring the resistance back up close to the original 4.5 of the stock one?

Am I missing a potential problem using this setup?

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If I add a 3 ohm ballast in series, that should bring the resistance back up close to the original 4.5 of the stock one?

Am I missing a potential problem using this setup?

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2,999 Posts

Resistor Pack for 3-ohm Coils, SOHC4shop.com

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Also, from the referenced site...."

This Resistor Pack reduces this current to a safe level (2.8 amps) for the bike's stock wiring, while still allowing for a nearly 3-fold spark voltage at the sparkplugs from using these high-output coils.".

Does this imply that a 1.6 ohm primary coil is necessarily "high-output"? The e-bay coil I have is certainly not a Dyna or Accel!

Does this imply that a 1.6 ohm primary coil is necessarily "high-output"? The e-bay coil I have is certainly not a Dyna or Accel!

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83 Posts

This starting bypass is built into the GL1000 wiring.

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What I am wondering is if the 1.6 ohm coil will have sufficient output for normal running with reduced voltage supplied to the primary due to the 3 ohm resistor?

Doing some math, using the 3 ohm resistor should leave only 4.16 volts (with 2.6 amps) at the primary; a 2 ohm would leave 5.33 volts (with 3.33 amps), and a 1.6 ohm would leave 6 volts (with 3.75 amps)…..assuming there are 12 volts available to start with...….also assuming the math is right! ( I=V/R for the total circuit, and V=IR for each component)

So the question seems to be,,,,does 2.6 amps to the primary produce the same result regardless of voltage used?

Stock 4.5 ohm coil at 12V produces 2.66 amps

Aftermarket 1.6 ohm (with 3 ohm resistor) coil at 4.16V also produces 2.6 amps?

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27,018 Posts

A stock 4.5 Ohm coil @ 12Volts draws 2.66Amps of current flow, expending/using/requiring 31.3 Watts of power......

A 1.6 Ohm coil @ 12V draws 7.5 Amps and uses up 90 Watts of power......

The addition of the 3 Ohm resistor is to reduce the amperage draw, and thus the power usage..........

The slightly diminished Voltage you read after a resistance is because unless the engine is running fast enough that the alternator can replenish the battery, its power is finite and being used-up......

Wattage is the factor showing power usage, Amps show the speed at which the power reserve (the battery) is being reduced......

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NO doubt with me it would be SOONER!

So maybe a momentary on bypass would be good....

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Not sure I follow you here....the 3 ohm ballast should "use up" 7.8 of the 12 volts available in the circuit containing a 1.6 amp coil.

A stock 4.5 Ohm coil @ 12Volts draws 2.66Amps of current flow, expending/using/requiring 31.3 Watts of power......

A 1.6 Ohm coil @ 12V draws 7.5 Amps and uses up 90 Watts of power......

The addition of the 3 Ohm resistor is to reduce the amperage draw, and thus the power usage..........

The slightly diminished Voltage you read after a resistance is because unless the engine is running fast enough that the alternator can replenish the battery, its power is finite and being used-up......

Wattage is the factor showing power usage, Amps show the speed at which the power reserve (the battery) is being reduced......

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Got it on the relay.....

What's your take on the 3 ohm ballast with a 1.6 ohm primary?

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OK if you say so.....but you had me going with the relay instruction paragraph then!

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I took a year of electronics school in the early '80s while working full time, and that's the last time I worked with the formulas that figure out the amperage draw and voltage drops, etc... but I understand the basic flow of current and did a decade of car accessories after working on bikes and cars for 20 years prior, so using relays and 12v specialty wiring is easy enough for me. I once had an '84 Monte Carlo SS wired with 2 keyless/alarm modules on a 4 button transmitter that controlled power locks and alarm, window roll-up/roll-down, remote trunk release, remote valet control for the auto-arming alarm and remote start - yes, on a carbureted V-8 - using a trunk actuator to pull the throttle to either set the choke when cold or clear the manifold when hot, holding the engine at 1500 rpm for 7 seconds on a timer... so I've wired up my share of Rube Goldberg stuffOK if you say so.....but you had me going with the relay instruction paragraph then!

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I can tell we are looking at this differently....In the interest of me learning something new, and to be on common ground, let's use the commonly portrayed electronic/hydraulic analogy........Not sure I follow you here....the 3 ohm ballast should "use up" 7.8 of the 12 volts available in the circuit containing a 1.6 amp coil.

Volts represents the water pressure available (potential) from a finite tank (the Battery) .....Amps represent the volume of water flowing through.....Ohms represent the pipe diameter resistance to flow rate....And Watts represents the amount of power (work) it takes to push a given volume through........

Are we agreed on the analogy so far?

IF so, lets add a valve at the far end of our pipe ...A Closed valve = open circuit as NO volume flows (NO Amps), pressure equalizes throughout the system (at the greatest Voltage potential the "tank"/Battery is currently capable of), NO resistance to flow as there is no flow (NO Ohms), and NO power being used (NO Watts).....

Still in agreement?

Now, let's open the valve (close the circuit).......Unless the water in the tank is being replaced as quickly as we are draining it out of the valve, the pressure WILL slowly drop as the water level decreases in gravity fall height, (likewise our voltage reading will reduce slightly).....Since the tank represents the battery, (the power used to pump the water, and the lesser volume of water pumped back to the tank are represented as P/I), and unless otherwise replaced, the remaining energy (stored as Voltage potential) will diminish ....We will still have whatever the remaining tank pressure entering the system is, as our new maximum pressure (Voltage potential)......

Still good?......

Here's where our points of thought differ.....You posit that upon entering/passing through the 3 Ohm resistor the VOLTAGE is reduced by 7.8 Volts.....This would leave only 4.3 Volts to enter the coil, (nothing in between to increase the voltage) and IF it drew @ 2.625 Amps, there would be NO remaining power (P=V x A and you've reduced one of those to zero) to carry through the last leg back to the battery......This is easily disproved as a simple voltage and/or amperage reading between coil output (ground side) and B- will show......

I posit that there is (nominally) 12 Volts everywhere in the circuit and the resistances change the current flow speed (amperage) and thus the Wattage used......Employing the wattage used and amperage of the components, you can calculate the net voltage potential loss at the Battery, which while small is cumulative and is why a "total-loss" system usually lasts less than an hour.........(remember, Amps are actually Ampere Hours)

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You are still ahead of me then...….I had 9 months of electronics in the service but that was in the late 60's and most of that is gone now....I remember the basic theories, but have to look up the details to get refreshed.I took a year of electronics school in the early '80s while working full time, and that's the last time I worked with the formulas that figure out the amperage draw and voltage drops, etc... but I understand the basic flow of current and did a decade of car accessories after working on bikes and cars for 20 years prior, so using relays and 12v specialty wiring is easy enough for me. I once had an '84 Monte Carlo SS wired with 2 keyless/alarm modules on a 4 button transmitter that controlled power locks and alarm, window roll-up/roll-down, remote trunk release, remote valet control for the auto-arming alarm and remote start - yes, on a carbureted V-8 - using a trunk actuator to pull the throttle to either set the choke when cold or clear the manifold when hot, holding the engine at 1500 rpm for 7 seconds on a timer... so I've wired up my share of Rube Goldberg stuff

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Exactly my point earlier... you gotta exercise it or it goes away...but that was in the late 60's and most of that is gone now...

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Ok, I'm with you mostly on the analogy, but I think we are looking at 2 different visions.I can tell we are looking at this differently....In the interest of me learning something new, and to be on common ground, let's use the commonly portrayed electronic/hydraulic analogy........

Volts represents the water pressure available (potential) from a finite tank (the Battery) .....Amps represent the volume of water flowing through.....Ohms represent the pipe diameter resistance to flow rate....And Watts represents the amount of power (work) it takes to push a given volume through........

Are we agreed on the analogy so far?

IF so, lets add a valve at the far end of our pipe ...A Closed valve = open circuit as NO volume flows (NO Amps), pressure equalizes throughout the system (at the greatest Voltage potential the "tank"/Battery is currently capable of), NO resistance to flow as there is no flow (NO Ohms), and NO power being used (NO Watts).....

Still in agreement?

Now, let's open the valve (close the circuit).......Unless the water in the tank is being replaced as quickly as we are draining it out of the valve, the pressure WILL slowly drop as the water level decreases in gravity fall height, (likewise our voltage reading will reduce slightly).....Since the tank represents the battery, (the power used to pump the water, and the lesser volume of water pumped back to the tank are represented as P/I), and unless otherwise replaced, the remaining energy (stored as Voltage potential) will diminish ....We will still have whatever the remaining tank pressure entering the system is, as our new maximum pressure (Voltage potential)......

Still good?......

Here's where our points of thought differ.....You posit that upon entering/passing through the 3 Ohm resistor the VOLTAGE is reduced by 7.8 Volts.....This would leave only 4.3 Volts to enter the coil, (nothing in between to increase the voltage) and IF it drew @ 2.625 Amps, there would be NO remaining power (P=V x A and you've reduced one of those to zero) to carry through the last leg back to the battery......This is easily disproved as a simple voltage and/or amperage reading between coil output (ground side) and B- will show......

I posit that there is (nominally) 12 Volts everywhere in the circuit and the resistances change the current flow speed (amperage) and thus the Wattage used......Employing the wattage used and amperage of the components, you can calculate the net voltage potential loss at the Battery, which while small is cumulative and is why a "total-loss" system usually lasts less than an hour.........(remember, Amps are actually Ampere Hours)

I don't think I'm wrong, but IF I am, somebody please correct me.......and explain why......

You seem to be looking at the system as it is moving, i.e. the battery charge is changing, and I am looking at a point in time, i.e. the instant points close and the current begins to flow.

Theoretically, as your example, checking voltage between the minus side of the coil and B- should be "0" (assuming there is nothing in between).

And if you check amperage anywhere in the loop, it will be the same (2.6 amps)

Both of the above checks apply to the points being closed.

If the points are open, then you would have 12 volts to ground anywhere in the circuit, and no amps anywhere......Right?

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