23-07-2014, 10:23 PM
If that chart in the linked PDF can be accurately applied to the lamp in question - or if we are allowed to produce our own graph before we attempt to answer the question raised, then yes, it's relatively easy. The curve appears to be a straight line around the area of interest, so a y=ax+b equation can be derived, and then perhaps a simultaneous equation could be arrived at. Or would a bit of calculus be a better bet? Too hard in this heat? Just bung it into Spice, with the lamp simulated by a voltage-dependant resistance...
However, it's all pointless - just an academic exercise. The effect of 0.054 Ohms of cable is utterly insignificant in the real world - it will drop less than a quarter of a volt (~2%), and what change in brightness will that have (especially remembering that vision is quite logarithmic, like hearing). Working things out with any precision is folly when you consider the typical tolerance of filament windings and the effect that will have on the calculations. Not to mention the resistance of the connectors and switches.
But, in very rough terms, we see from the graph that a rise in voltage of 14% (12.6 to 14.4 volts) results in a current rise of 8%. Ignoring losses, if the lamp was taking 4.365 amps at 12.6V, then it would take 4.714 amps at 14.4 volts. OK, so what about the drop across the wire? First, let's forget about the potential division effect (because that's the bit that needs the clever maths), let's just calculate the drop for both of those currents: it's 0.236 volts and 0.255 volts respectively. The difference in voltage drop for the two currents is just 19mV. OK, this isn't the full solution, because there is a slightly lower voltage across the lamp thanks to these voltage drops, but that will tend to reduce the 19mV differential. At this point, I really don't think there is any need to take it any further.
Some 20 years ago I measured the voltage drop of my old Mk2 Golf, and was amazed to find that about 6 volts was being lost in the wiring, switches, connectors, etc. Yet the lights were perfectly bright enough to drive with. Hence my assertion that in the real world, this really is not a problem that is worth solving to 5 significant decimal points. I will say, however, that my DIY high-power loom (using a box full of relays from a Rover Metro IIRC) did make the lights a lot better. Well actually, it made them much whiter, but I don't think I was able to see that much better in practice. My eyes were younger then, and my night vision has always been quite good. Plus, all the other headlights out there were dimmer than today, and people seemed to understand that you should slow down when visibility got bad. But that's another thread entirely
I'm wary about saying any more as we don't know anything about the root of the question. This might be someone's homework - it's the sort of pointless question academics might set - in which case let's be upfront about it. If this is a real-world question - perhaps from someone trying to establish the "optimum" size cable to use - then you don't need to get clever, you simply consider the worst-case scenario (which would be to establish the highest current), and round up to the next available CSA.
However, it's all pointless - just an academic exercise. The effect of 0.054 Ohms of cable is utterly insignificant in the real world - it will drop less than a quarter of a volt (~2%), and what change in brightness will that have (especially remembering that vision is quite logarithmic, like hearing). Working things out with any precision is folly when you consider the typical tolerance of filament windings and the effect that will have on the calculations. Not to mention the resistance of the connectors and switches.
But, in very rough terms, we see from the graph that a rise in voltage of 14% (12.6 to 14.4 volts) results in a current rise of 8%. Ignoring losses, if the lamp was taking 4.365 amps at 12.6V, then it would take 4.714 amps at 14.4 volts. OK, so what about the drop across the wire? First, let's forget about the potential division effect (because that's the bit that needs the clever maths), let's just calculate the drop for both of those currents: it's 0.236 volts and 0.255 volts respectively. The difference in voltage drop for the two currents is just 19mV. OK, this isn't the full solution, because there is a slightly lower voltage across the lamp thanks to these voltage drops, but that will tend to reduce the 19mV differential. At this point, I really don't think there is any need to take it any further.
Some 20 years ago I measured the voltage drop of my old Mk2 Golf, and was amazed to find that about 6 volts was being lost in the wiring, switches, connectors, etc. Yet the lights were perfectly bright enough to drive with. Hence my assertion that in the real world, this really is not a problem that is worth solving to 5 significant decimal points. I will say, however, that my DIY high-power loom (using a box full of relays from a Rover Metro IIRC) did make the lights a lot better. Well actually, it made them much whiter, but I don't think I was able to see that much better in practice. My eyes were younger then, and my night vision has always been quite good. Plus, all the other headlights out there were dimmer than today, and people seemed to understand that you should slow down when visibility got bad. But that's another thread entirely
I'm wary about saying any more as we don't know anything about the root of the question. This might be someone's homework - it's the sort of pointless question academics might set - in which case let's be upfront about it. If this is a real-world question - perhaps from someone trying to establish the "optimum" size cable to use - then you don't need to get clever, you simply consider the worst-case scenario (which would be to establish the highest current), and round up to the next available CSA.
