HELP starting problems

[mhb]

hello there
I have a strange intermittent starting problem, the engine does half a turn then nothing, it usually does this once then starts ok, but lately it has become a more consistent problem, it took about 4 attempts over about 30 mins this morning,

can this be battery problems, is it the immobiliser cutting in, or some other electronic fault.
I would welcolme advice or help on this frustrating problem.

regards
mike

 

[television]

Hello Mike, by the symptons ,Do you mean that the starter motor turns the engine ½ a turn then nothing while the key still turned, but if the key is released it does the same again.

This being the case it would be the battery low or needs replacing. If the battery is 4 years + then change it.

What car ?


malcolm

 

[mhb]

thanks malcolm that excatly what happens , it does look to me like a battery problem , the only strange thing is that the car has done since i bought it 5 years ago, not as bad as this though, the battery is the original one from 97, perhaps I am chancing things, its a 97 c240 estate

mike

 

[television]

Much against normal belief, a bad battery really hammers the starter motor, causing very high current to flow.

Yes you must change the battery, do get a good make,not a Halfords special.

If you have a Macro or the like card, the Exide and Bosch batteries are cheap.

malcolm

 

[mhb]

thanks malcolm
i'll get one to morrow
mike

 

[Hibbo]

Much against normal belief, a bad battery really hammers the starter motor, causing very high current to flow.

Please educate!....

 

[television]

Please educate!....

The higher the voltage, the less the current.

Malcolm

 

[Hibbo]

The higher the voltage, the less the current.

Malcolm

Surely not! Higher voltage = higher current (I=VxR)

and surely a knackered battery = lower voltage = lower current?

What am I missing here?

 

[Hibbo]

Are you maybe getting confused with 'a higher current is needed to produce the same amount of power at a lower voltage'? (ie 1kW @ 250v = 4A but 1kW @ 100v = 10A)

This doesn't apply in this case as the resistance of the starter does not change. (a starter designed to run at 6v would need twice the current of a 12v one for the same power)

 

[television]

Are you maybe getting confused with 'a higher current is needed to produce the same amount of power at a lower voltage'? (ie 1kW @ 250v = 4A but 1kW @ 100v = 10A)

This doesn't apply in this case as the resistance of the starter does not change. (a starter designed to run at 6v would need twice the current of a 12v one for the same power)

One thing,I am not confusued

The starter motor is shunt wound, therefor it tries to give full power as in your first equastion. The current rises rapidly as the voltage drops.

Ω law does not apply directly to this, owing to how the motor is wound.

Malcolm

 

[Hibbo]

One thing,I am not confusued

The starter motor is shunt wound, therefor it tries to give full power as in your first equastion. The current rises rapidly as the voltage drops.

Ω law does not apply directly to this, owing to how the motor is wound.

Malcolm

Easy tiger! Just trying to understand this my self!

In a shunt wound motor the current through the stator and field coils depends upon the voltage across them (which is the same). I don't see how current can possibly increase as voltage decreases.

Do you mean as the voltage drops due to the internal resistance of the battery or the effective voltage across the coil reduces due to back EMF as the motor spins up?

I thought Ohm's Law ALWAYS applies, but effective EMF must be used, not supply voltage.


PS. I assumed starter motors where series wound due to the high starting torque series wound motors produce.

 

[television]

I am not getting into this argument, can I re phrase this to excess current flows when the battery is low due to the long time that the starter is in mesh, and output is lost in terms of heat disapated within the field coils, the starter is designed for momentry operation. Left in mesh the current rises beyond its design limits, and it is this that I mean by low voltage will increase the CCA Cold cranking amps and I do not mean it with relation to Ω law.

Ω law does not apply to motors since the resistance stays the same (cant change) but as the motor is loaded the current will increase, with a starter motor power lost can be in terms of heat (wasted energy) but the energy (W) has to come from some where.

Motors is a world all on its own.

Malcolm

 

[BlackC55]

I think I know what the problem is.

The starter and ring gear needs lubricating. There is a small plate on the g/box which allows acess to the ring gear. Asmall amount of good quality aluminium grease applied here will sort this. I stress SMALL amount.

 

[stargazerplus]

I think I know what the problem is.

The starter and ring gear needs lubricating. There is a small plate on the g/box which allows acess to the ring gear. Asmall amount of good quality aluminium grease applied here will sort this. I stress SMALL amount.

This is a common problem on the V6 engines and relates to rust forming on the flywheel gear. The starter will turn but suddenly stop. WIS lists this under troubleshooting "Starter Does not Start" and recommends removing the starter motor and coating the ring gear with paste. I had this done for one hours labour.

The older the V6 gets the more frequently we will see this occuring. It has nothing to do with battery voltage. The symptoms are sometimes accompanied by the rev counter jumping up as the starter stops spinning.

 

[Hibbo]

can I re phrase this to excess current flows when the battery is low due to the long time that the starter is in mesh, and output is lost in terms of heat disapated within the field coils, the starter is designed for momentry operation. Left in mesh the current rises beyond its design limits, and it is this that I mean by low voltage will increase the CCA Cold cranking amps and I do not mean it with relation to Ω law..

Malcolm

I see.

Thank you Malcolm,

 

[Parrot of Doom]

I think what Malcolm is saying is that the starter motor spends more time in a position where it draws higher levels of current from the battery, because the potential difference across the motor isn't as much as it should be to push it further around - thus contributing to the poor charge level of the battery.

A bit like a weak hand not being strong enough to turn the tap off.

 

[stargazerplus]

Screen capture from WIS