How Much the Resistance it is Between the Two Points

The short answer is "no".

The resistance for soil will not be in ohms/m, the untis should be ohms/m^3 or something like that. (Where resistance is measured from two parallel faces across the unit volume.)

go on.

why?

For calculation convinient, we can accept different form of density,

for example we can use volume, area and linear density. we can say current density J has unit of A/M3, volume current density, or A/m2 area current density, and A/M, linear current density. as well as charge, weight etc.

I think this will be useful for your later works.

Only 5ohm.

haha,

you are only guess?

I vote for you, but Im still not very comrehensive.

I'm not specialized in the field (I'm just a SW guy) but I have a strong feeling that the question is just whether the 5km distance is enough to be taken as infinity. As an engineer, I wouldn't mind some lost accuracy. As a scientist though, I would say that the pattern of the current flow is not exactly spherical or cylindrical (depends on shape of rod), in any case, it's not symmetrical, therefore more current is forced to pass through a smaller cross section of the separating soil, thus making point-to-point resistance appear bigger than the expected 5Ω.

Anyway, I wouldn't bother for another reason: The grounding resistance has in practice been measured by probably applying a grounding triangle or using that 61% rule (if I remember well) or using these fancy gadgets that are around for some years. In any case, the separation of the measuring rods must have been much smaller than 5km, so the 1Ω and 4Ω you mention have already incorporated an error caused by the asymmetrical current flux. if it is so, then the resistance measured between the two distant point may be smaller than 5Ω!

I've been quite intuitive, no time to do maths or brush up my undergraduate knowledge, sorry! I wish somebody more experienced in the field gives us some insight.

I post the problem at other sites as well. but hant got a satisfacotry answer so far.

I wish experience specialist and somebody could contribute their skill.

btw, what is SW you said in your thread?

btw, what is SW you said in your thread?

I'ts a TLA (two letter acronym) for Software!

haha, it was this.

I hve three answers,

one,

greater than 5ohms

two,

equal to 5ohm,

three,

there are two solutions(solution is not unique).

1, from view of user, he has 4ohms between the two piints.

2, from the generator, only 1ohm betweent the two points.

Too many visitors visit, but only a few guys will to discuss and give thier viwepoint.

It seems I cannt get more help here.

Let me calculate myself and tommorow I shall post the claculation formula for futher discussion.

ah, hand script can show you the way of calculation. From this pic, you will know from A(generator terminal) to B (user) it will be of 1ohm and otherwise, will be 4ohms.

if we take account of total loop, adding two touch resistance will result in total of 5ohms+wire resistance.

Try KVL and KCL. They work for me.

5Ω is the answer indeed, given that 5km can be considered almost ∞. As a said previously I think it is to any engineering standards accuracy, nevertheless, to be mathematical accurate, the total resistance will be something slightly less than 1+4=5Ω.

Let me clarify here a previous mail of mine: I said that for distances less than infinity, the resistance will be higher than expected, due to the asymmetrical current flux. To be more precise, what I mean with 'expected' here is the value obtained from integrating - like what you do in your hand notes - up to 5km instead of ∞. In short, the measured resistance will be something slightly less than 5Ω and more than the integration of the resistances from the edge of the row up to 5km.

But the question I need to get answered is how do you know that the ground resistances are indeed 1Ω and 4Ω? Is it the result of an analytic calculation, or were they measured in field? (I fail to understand whether this is a thought experiment or a problem of real life). In the latter case, the measuring rods would have been much less than 5km apart (let alone ∞) therefore error induced in the measurement (dependent on method) will most probably be larger than the error caused by taking 5km ≈ ∞. So my advice is: Don't bother; take 5Ω and you will be fine.

The values I give out is measure value in practice. the method of measure is giving in my front thread( last year some time I forget). I know how to measure with a resistant meter.

I just hope to make clear from theory. the measure distance is 40M.

the value is not constant, vary a little with condition of soil and wet degree.

the value is not constant, vary a little with condition of soil and wet degree.

Of course soil permeability varies with humidity, but then the values 1Ω and 4Ω will also vary! Anyway, if we assume that the variation of soil permeability changes (after a rainfall for example) uniformly in all broader area then again the total resistance will be the sum of the grounding resistances (with new values this time) at the two ends.

BTW, in an idealistic world, where the two ends were infinite distance apart and the calculation of the two grounding resistances has been done analytically (no measuring rods used), then we wouldn't mind about permeability being uniform: the sum would hold in any case.