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2.4 Battery upgrade

Keep going guys - I love a good argument, even if it has strayed a long way from the OP's problem!
 
Jon_G said:
Well yes, but right up until my final technical post I merely thought I was being helpful, not arguing...it came as something of a surprise!
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I guess we've all been on forums where arguments can get very nasty :( :angry: :lol:

But this forum is nice because people don't argue, I reckon that this is the first "argument" on this forum :lol:

I think that Jon is right, not on the grounds of impedance, but on back emf

i.e. I said that if the CCA of the battery is 330 Amps, then on a cold winter's day I ***ume that the starter will draw about 300 Amps initially, and ***uming initial voltage across the starter terminals is 12V, then using Ohms Law the resistance is 0.04 Ohms. I don't know if this is true, if it is, then it will be for the first few milliseconds I would think, after which the current would reduce due to back emf. Also, an unloaded starter motor (out of car) will not draw much current, so again Jon is onto something. However, the battery's rated CCA of 330 Amps does imply a very large initial current draw.

It would be worth measuring the current over the first few seconds, but to do this one would need to remove the cable and measure its resistance very very very accurately, then put it back and measure the voltage across the cable with an oscilloscope. Even that would be difficult, because the leads may pick up induced emf "noise" and give false readings. :eek:

I'd love to do it though .... just need to find somewhere that can give an accurate reading for the cable's resistance :rolleyes:

But reagrding the original post, it is kind of relevant, because it begs the question, how much resistance can you lose on the clamp before it starts to cause problems. If CCA of 330 Amps is needed, then you cannot lose much resistance on loose terminals.
 
OK, it seems that I can't stop myself replying...

freddofrog said:
i.e. I said that if the CCA of the battery is 330 Amps, then on a cold winter's day I ***ume that the starter will draw about 300 Amps initially, and ***uming initial voltage across the starter terminals is 12V, then using Ohms Law the resistance is 0.04 Ohms. I don't know if this is true, if it is, then it will be for the first few milliseconds I would think, after which the current would reduce due to back emf. Also, an unloaded starter motor (out of car) will not draw much current, so again Jon is onto something. However, the battery's rated CCA of 330 Amps does imply a very large initial current draw.
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Impedance and resistance are both measured in ohms, therefore I'm happy with your (very rough) estimate of what you are calling resistance (which is actually a combination of inductance and resistance). But please bear in mind that the starter cannot only be resistive, if it were then the only result of passing current through it would be a heating effect. A magnetic field is also set up in the windings which interacts in such a way as to cause rotation... this effort does not come for free, the inrush current is mostly due to initiating this magnetic field (which is why the cabling has to be so heavy duty), this inrush current is instantaneous and will only last a few milliseconds at most (then falls as the magnetic field is established and 'back EMF' kicks in). It will be higher with a stationary armature, but each time the spinning brush/commutator components re-make the circuit, then there will be a smaller instantaneous current (but smaller than when the armature was initially stationary). So yes, there is a high initial current drawn from the battery!

Please note that I am not disagreeing with you, merely expanding upon your thoughts.

freddofrog said:
But reagrding the original post, it is kind of relevant, because it begs the question, how much resistance can you lose on the clamp before it starts to cause problems. If CCA of 330 Amps is needed, then you cannot lose much resistance on loose terminals.
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Not much (fractions of ohms)

I apologise in advance if my explanations aren't as detailed as you'd like (with links to "prove" my points). While I am an experienced electronics engineer, I am neither an auto-electrician or a lecturer!
 
No need to apologise on either side I hope. I too am an electronics engineer, but I may be older than you (so rusty), and, I switched into digital electronics in the 80's.

A nice explanation, kind of what I was thinking (though I wouldn't be able to write it that clearly), and I didn't quite get your previous stuff (hence I went off about reactive currents).
 
Glad that's sorted out then. I just had the feeling it might develop into a willy-waving contest.

I'm not going to argue over who is right with a couple of electronics engineers. Now, if we get into chemical engineering that's a different matter..........
 
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