Another step forwards in solving the problem...now, Lambda sensors info please?

Okay according to the calculation Power = Volts x Amps

Example you need 900 watts to turn the starter @13.5 volts That = 66 amps

NOW take it that the battery has dropped to 10.5 volts the amperage is now 85 amps (as it takes the same amount of power to turn the starter i.e. 900w)

A BIG difference for a drop of 3 volts? Turning it over with a reduced charge in the battery puts a helluva current surge thru the electrics!

Not sure about your calculations. V=IR, For a given current at a given voltage you need a specific resistance, ie the starter motor. If you drop the voltage, the resistance being fixed, means the current drops not increases, as you have dropped the voltage in your calculations, by about 1/3 then the current also drops by 1/3 ie down to around 44 amps, not up to 85 amps. (using your values, as I have not measured the resistance or taken into account the back emf etc.) That's why the starter would then turns over slowly.
 
I was using Power = volts x amps

I am not a mathematician of full blown electronics engineer I have a working knowledge of each

IF I understand If the starter requires 900watts to turn the engine over that will be 900 watts whether the voltage is 10 or 12 The only variable will be the current? N'est pas?


Not sure about your calculations. V=IR, For a given current at a given voltage you need a specific resistance, ie the starter motor. If you drop the voltage, the resistance being fixed, means the current drops not increases, as you have dropped the voltage in your calculations, by about 1/3 then the current also drops by 1/3 ie down to around 44 amps, not up to 85 amps. (using your values, as I have not measured the resistance or taken into account the back emf etc.) That's why the starter would then turns over slowly.
 
I've got a couple of old HESs here, and will be ordering some Rockwell sensors to make them up to 'as new'......that's going to take some time though....It'll be nice to have one around even if it isn't the cause of this saga :thumb

Off out to remove the right hand throttle body now, take it apart and see if there is anything I can clean.

I'm kind of hoping I find a dessicated mouse in there, but life's not that simple :beerjug:
 
Fanum pardon the remark BUT Stop Fannying around just go see Mr Steppers and get a known good HES off him and fit it

When he was saying about not being able to get a balance on throttles it clicked with me as I had similar issues it was like it was sparking thru 10 or 15 degrees of variance giving a lumpy idle that you couldn;t balance out

If you have good compression Fuel going in and a spark ?? the only two things that bollix up the equation is the ecu as it determines when the spark and the fuel squirt are and the HES as they tell the ECU when it should squirt and spark!

ECU's are not much more than a big calculator ....So .... Leaves ..

The end result
I've got a couple of old HESs here, and will be ordering some Rockwell sensors to make them up to 'as new'......that's going to take some time though....It'll be nice to have one around even if it isn't the cause of this saga :thumb

Off out to remove the right hand throttle body now, take it apart and see if there is anything I can clean.

I'm kind of hoping I find a dessicated mouse in there, but life's not that simple :beerjug:
 
I was using Power = volts x amps

I am not a mathematician of full blown electronics engineer I have a working knowledge of each

IF I understand If the starter requires 900watts to turn the engine over that will be 900 watts whether the voltage is 10 or 12 The only variable will be the current? N'est pas?

The current for a circuit is dependant on the voltage, this will then give a specific amount of current dependant on the resistance of the circuit.

If the starter requires 900watts to work then as its resistance remains the same it will need the correct voltage to achieve the right current to give this 900 watts Watts=Volts x Amps

Halve the voltage and as long as the circuit remains the same then the current also halves. Of course the power also halves therefore in your case the power would drop from 900watts to 450 watts.

If you double the voltage you would push double the current through the circuit. This would mean the circuit would consume 1800 watts, it would spin the motor over quickly but would overheat and not last long.
 
Took off the RHS throttle body today.

Cleaned it in petrol with a new paintbrush, paying particular attention to the cam channel that the cable sits in.

Butterfly looks good, no rattling on the shaft, spring action smooth and after a good clean, not the slightest bit gritty.

Put it all back together (Looks as bright and shiny as a bright shiny new one :D) and guesstimated the position on the cable for balance......fired it up and even without any more balancing (which will be completely out unless I was extremely lucky) it is running a bit better :thumb

After doing that, tracked down the noise on that side to a loose 'zorst manifold bolt that was finger tight :blast
(Don't think this is the underlying problem, but it can't have been helping matters!)

Tomorrow's job....build a quick 'n dirty manometer, clean LH throttle body in same way then check TPS and try to balance both sides.

I didn't get a chance to do Steptoe's test for a faulty HES today, but I did pay attention to tacho and it's fine, no jumpiness.
 
brand spankers !


That's good to know, thanks Chad :beerjug:

So this one has a MAXIMUM of 2 thousand miles on it :thumb2

(Although yes, if Jay's theory that they can be damaged by starting with a weak battery is correct, it has been through that situation a bit)
 
(Although yes, if Jay's theory that they can be damaged by starting with a weak battery is correct, it has been through that situation a bit)


lets hope not mate ,,, looking for fingers crossed smiley;)
 
10 pager yet...



Sorry, interesting thought process and fault finding, which I sincerely hope you get to the bottom of.


Failing that, hope you have a substitute for Morocco. :thumb2
 
Doesn't seem five minutes since Bill turned up at mine with this bike when it was nearly new. We promptly broke it and dumped the entire 32 litres of fuel across my garage floor. :D
 
Doesn't seem five minutes since Bill turned up at mine with this bike when it was nearly new. We promptly broke it and dumped the entire 32 litres of fuel across my garage floor. :D


The garage floor wasn't what caused the panic though.....it was the hissing and spitting as it all flowed over the hot engine and 'zorst headers :eek:

:D
 
The current for a circuit is dependant on the voltage, this will then give a specific amount of current dependant on the resistance of the circuit.

If the starter requires 900watts to work then as its resistance remains the same it will need the correct voltage to achieve the right current to give this 900 watts Watts=Volts x Amps

Halve the voltage and as long as the circuit remains the same then the current also halves. Of course the power also halves therefore in your case the power would drop from 900watts to 450 watts.

If you double the voltage you would push double the current through the circuit. This would mean the circuit would consume 1800 watts, it would spin the motor over quickly but would overheat and not last long.

A good explanation of Ohms Law but you've forgotten the back EMF produced by the motor. The current draw will increase rapidly as the motor slows (perhaps because of low voltage) whilst the load remains the same. That's why a starter motor tested under no load conditions on the bench will only draw perhaps 30 or 40 Amps but bolted to an engine might take 3 or 4 hundred Amps. The 'constant resistance' theory just don't work:rob
 
Halve the voltage and as long as the circuit remains the same then the current also halves. Of course the power also halves therefore in your case the power would drop from 900watts to 450 watts.

The work done by the motor will be proportional to the power required. If the volts drop, the motor will spin slower (if brushed), the load (i.e. the engine will stay the same so the current will rise to compensate (assuming there is still enough volts to spin the engine fast enough to make compression)

To me, these problems sound like they are electrical in nature, certainly now especially if one cylinder head has been worked on.

  • Is the fault temperature related?
  • Is the fault vibration related?
  • Check there are no pinched wires in the loom, possibly try flexing the loom with the engine running.
  • Check the ground points (braided strap) to the battery terminal is good and not corroded, check the metal work is bright where the ground points meet.
  • Check battery voltage whilst you are testing, especially when the engine is running
  • Have you got access to an oscilloscope? Do you or Steppers know how to use one? Have a look at the injector pulse (immediately prior to connecting to the injector) to see if there is any noise when you have the system stuttering? See if it swings from a constant pulse width to a random collection of pulses when it stutters
  • Watch the Hall signal pulse to see if they are being received from the sensor? Is the Gap between the sensor set right? - Does the stuttering effect happen at a certain rev range? (i.e. if the engine is spinning above a certain rate are the sensors missing the pickup? )
  • Check the coils are firing correctly and not breaking down, if a twin spark can you run solely on the secondary plugs and check the motoronic is sending the fire pulses to the coils? (would be hard to easily measure on the stick coils but should fire at the same time as the secondaries?)
  • Check the fuel pump is maintaining pressure and the filter isnt blocked, is the fuel return clear? - Are there air leaks getting in to any of the fuel lines?

I would be amazed if it is not electrical and neil hasn't picked it up already, the tough faults with any piece of electromechanical kit normally fails on the electrics first (normally sensors and connectors). Check them with a meter if you can or sling a 'scope between the terminals to look at the pulses. You need something to spot the changes with a quick time base, most meters are too slow.


my 2p - Good luck.
 
A good explanation of Ohms Law but you've forgotten the back EMF produced by the motor. The current draw will increase rapidly as the motor slows (perhaps because of low voltage) whilst the load remains the same. That's why a starter motor tested under no load conditions on the bench will only draw perhaps 30 or 40 Amps but bolted to an engine might take 3 or 4 hundred Amps. The 'constant resistance' theory just don't work:rob

If look at my first post, because the complexity, I specifically said that I was "(using your values, as I have not measured the resistance or taken into account the back emf etc.)"
It was clear he was having issues with W=VA. So it seemed unfair to bring in another complicated factor for him.

I agree with you summation that the current would increase dramatically and indeed its why the contacts of the solenoid are often welded together with low battery voltage. Its also why I am a fan of AGM batteries as with low internal resistance it maintains a higher terminal voltage than Gel batteries in adverse conditions.

The constant resistance theory still works and limits the maximum current draw even at stall position though obviously much greater current than you would wish. (The BMW motor is specified as 1.2KW ) The maximum current is limited by the resistance of the field and one winding of the armature as there will be nil back EMF on stall. This is assuming the starter is a series motor which I believe it should be. This resistance would be sufficient to limit the current to high 2 figures or low 3 figures at the most.
As to draw 300 amps this resistance would need to be around 0.043 ohms which I believe is much lower than you will get on the starter.
Regards , Dave.
 
I agree with you summation that the current would increase dramatically and indeed its why the contacts of the solenoid are often welded together with low battery voltage. .

Whilst the rest of the post is "IF" "buts" and "maybes" and unsubstantiated

The crux of the matter as Grissom would say rests in Physical Evidence !!!
 
Whilst the rest of the post is "IF" "buts" and "maybes" and unsubstantiated

The crux of the matter as Grissom would say rests in Physical Evidence !!!

No "buts" and only one "if" and that was to say "If you look at my first post!" Unsubstantiated? thats strange from someone that does not know the relationship between watts amps and Volts, maybe its because you cant understand it??
 
Fcuk this I'm buying an air head, all this talk about stuff which seriously confuses me hopefully keeps Fanum on the straight and narrow....




BTW fanum, if you get in a total pickle on this, I'll consider lending my 1150 for your trip...no really I would. It needs an adventure :thumb2
 


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