I'd start with the engine earth/s. I use the plural because although there's usually one primary earth connection to the block (commonly a braided tinned copper strap) there are usually several other extraneous return paths ('earths') via the wiring to other electrical engine accessories or structural connections back to the bodywork.
OK now consider that the cranking current for a petrol engine can easily exceed 200 Amps whilst the key is engaged and the starter is churning and that when it first starts to turn against the stationary load of the engine compression and friction the initial inrush current can reach 500Amps:!: That is one hell of a current to draw through wiring from a battery. I won't get too far into volts ohms and amps here (if I can help it) but to cut a long story short; at these current levels
even very slight increases in the resistance of the electrical path or the components within it can have serious repercussions on the actual power (the ability to do work) that is received at the business end, i.e. the starter motor's windings.
Right here we go; volts, ohms, amps and electrical 'POWER'
the basics... where the electrical 'power rating' of an electrical device comes from.
POWER (in Watts) = VOLTS (unit = volt) x CURRENT (unit = Amp)
So if I have a heater blower motor I can clip a meter round it's live wire and measure the current flowing down it into the motor. This would probably be around, oh I dunno, let's say 2 Amps. We know the volts, (12v obviously) so we can work out the electrical power of the blower motor as power = volts x amps. Put those numbers into the P=VxI equation...
12 volts x 2 amps = 24 Watts ...sounds about right...
Similarly you can switch that little formula around so that if you know the power of a device you can work out the current to expect in its wiring when switched on and this is where it gets useful in fault finding. the voltage is always a nominal (12v) for most purposes. You just
DIVIDE THE POWER BY VOLTS
here I'm going to use leccy terminology; current is written as the letter 'I'
so we know that P = V x I power(P) = volts(V) x current(I)
so transposing it to solve for the current you divide P by V and the formula can be re-arranged as
I (the current we want to find) = P/V
Your headlamp bulb says 50W on the box, how much current will a pair of them draw from your battery or charging system?
...just divide that power figure by your voltage (12) so
50W divided by 12 = 4.2A
There's two of them so they'll draw 8.4A from the battery. If you had an 84 Amp/hour battery and you left them on it would (theoretically) flatten it in about 10hrs! (10 x 8.4 = 84A/hrs)
RESISTANCE (measured in OHMS)
When a voltage is applied to a circuit the resulting current that flows depends on the resistance of that circuit. For 'ontheball247's starter motor prob there will be several elements that make up the whole circuit and each one will add in its own bit of electrical resistance including even the crimp lugs! When the key is turned to start, a relay pulls in the hefty starter solenoid (an even bigger relay), this closes its equally hefty pair of contacts which apply 12v straight from the battery's positive (+ve) terminal onto the terminal of the starter motor, through the carbon brushes, through the windings and the current then returns to the battery negative terminal via the frame of the starter motor to which the other ends of the windings are 'commoned', via the conductive metal mass of the engine block, through the earth strap, into the bodywork to the point where the negative cable is bolted, it then travels up that cable back into the battery's negative (-ve) terminal, through THE INTERNAL RESISTANCE of the battery matrix where the chemical energy of the battery drives it round again in a constant loop. Hence the term circuit!
Resistance is an opposition to the flow of electrical current, a bit like standing on a hose pipe. It's measured in Ohms and abbreviated to the Greek letter omega but I can't find it in the font set so I'll have to write the unit as 'r'.

The relationship with voltage and current is such that when a voltage is applied to any circuit the magnitude of the current that is allowed to flow through that cct due to the 'pressure' of the voltage is
inversely related to the resistance of the circuit. The
more resistance there is the
less current can flow, that hosepipe analogy again. Let's put that relationship into real numbers.
OHMS LAW
That inverse relationship between current (I) and resistance (r) is described by the formula...
I = V/R Current = Volts divided by the Resistance
If I apply 12v to a circuit with a resitance of say 2r (ohms) I would expect a current of about 6A to flow I = 12v / 2r = 6A
If the cct had a resistance of 400r then the current would be in the region of...
I = 12 / 400 =0.03 amps or alternatively 30milliamps [you get the idea]
As with the power formula before you can transpose Ohms Law in several ways; if you know two of the three quantities you can re-arrange the formula to find the third figure you want.
V = I x R (if you have the current and the resistance figures by measuring them for instance and want to work out a voltage) more applicable to electronics really!
or...
R = V / I (if you know the voltage [12v] and have a current reading you can work out the resistance of a solenoid coil or whatever and for instance compare it to a value in a shop manual.)
OK; BACK TO THE STARTER MOTOR PROBLEM
The starter motor needs 200A to do its job. Automotive starting batteries are designed to deliver that with no major trauma although it is quite a feat of chemical engineering when you think about it, to supply that much sheer cranking power from a 10" squared box!
Sticking some numbers into that Ohms Law formula; to get 200A from a 12 volt battery
the resistance of the WHOLE circuit must be no greater than...
R = V / I
R = 12 div. by 200
R =
0.06 ohms
That is 60 milliohms and that is not very much at all, pretty much what in simple terms could be termed a 'short circuit'. A smear of insulating engine oil on the main earth clamp could easily add a few ohms at least to the value of the overall cct resistance. Let's say for instance another half an Ohm or 500 milliohms (but it could very easily be more!) Stick that new resistance number into the Ohms law formula for finding current...
I = V / R
I = 12v div. by 0.56R (560 m/Ohms)
I = 21.4 Amps!!!!!!!
So just half an Ohm of extra electrical resistance has cropped the power (the ability to do work)
by a factor of TEN!
A layer of carbonized cr@p on the solenoid's internal contacts could easily account for that, a loose engine earth strap, that crappy, oil encrusted, wrench damaged crimp lug that's hanging on by only three strands of conductor....
If you've got two or more of the above in series then they all add series resistance and all turn cranking power into heat not tourque.
Hope it helps in understanding why the integrity of the engine earth is so important.
Also if that strap is open circuited then that cranking current will try to find alternative routes back to the battery through wiring which simply will not cope and very quickly overheat = ner ner, ner ner...