[TowerTalk] Fwd: Tower base/Ufer ground

Mon May 19 02:21:35 EDT 2014

On 5/18/2014 7:44 PM, Jim Lux wrote:
> On 5/18/14, 7:39 AM, Roger (K8RI) on TT wrote:
>>
>>  From the dictionary:  There is no need to go any farther. Forget your
>> calculations, this is the definition.
>>
>> Electrical resistivity quantifies how strongly a given material opposes
>> the flow of electric current. A low resistivity indicates a material
>> that readily allows the movement of electric charge. Resistivity is
>> commonly represented by the Greek letter ρ.
>
>
> The SI unit of electrical
>> resistivity is the ohm⋅metre although other units like ohm⋅centimetre
>> are also in use.
>
> Seems like what I was saying, no?

Well...not quite,  the S/I unit Om/meter is not the same as ohms per 
cubic meter./ One is squared and the other cubed.
It's an unfortunate choice of units.  They "can" be interchangeable "AT 
TIMES" and not at others.

Given a reistor which is for all intensive purposes, dimensionless. It 
can be a 1/2" long and a 1/4" in diameter, ot it can be a f18" long ans 
an inch in diameter.  It's value is ohms. You could add wattage, but 
that's not relevant in this discussion and The resistance and 
resistivity are the same.  It was the use of resistivity in the case 
that causes learning problems.

But instead of the resistor, let's look at the material (other than 
wire).  A carbon composition resistor depends not only the material, but 
the heat and pressure to get the desired "Resistivity" of the material.  
Another is concrete. In both cases we are using the resistivity of a 
unit volume.  We have added a dimension and the equations use for 
oms/meter no longer work because we have a third dimension.  You can not 
get from a linear resistance to a unit volume because we are short one 
dimension.  It's easy going the other way

>
>  *As an example, if a 1 m × 1 m × 1 m solid cube of
>> material has sheet contacts on two opposite faces, and the resistance
>> between these contacts is 1 Ω, then the resistivity of the material is 1
>> Ω⋅m.* (that is misleading as the formula actually  gives ohms per cubic
>> meter, not ohms per meter.
>
> Actually not.. if the resistance between the two faces is R ohms, then 
> the resistivity (rho) is rho=R * Area/length which has units of Ohms * 
> meters
>

If you need the resistivity of a material such as DI cooling water, the 
easiest way is to measure the resistance between two 1 cm square plates, 
with 1 cm spacing. This happens to be a spot where the two systems are 
the same and I could give a tech the probe and just take the meter 
reading directly in ohms per cubic cm. Just sticking two meter probes in 
will not directly give the resistivity needed to know if the water is 
still good although this is often referred to as resistivity.
>>
>> Please note Resistivity is in ohms per cubic Centimeter or ohms per
>> cubic meter not ohms per meter, or centimeter as is often stated which
>> is incorrect.
>
> Yes, I agree, there's no "ohms/meter"

Actually there is when purchasing wire. CAT5 and 6 have a resistance per 
unit length, but that's straying from the discussion.

>
> The resistance has to be proportional to length (distance between 
> electrodes), because if you double the length, the resistance doubles. 

I can get a wire wound twice as long, with the same wire and the same 
resistance. The resistor is longer, but the wire is the
But that is of concern in making the resistors, not using them, or 
dimensions are a second order of importance. We ned resistance, we need 
to dissipate power and we may be limited at to available weight and volume.
>
> Likewise, the resistance is inversely proportional to the cross 
> sectional area (put two identical bars in parallel, and the resistance 
> is half the resistance of a single bar).
>
> Therefore, so if you have a property called "resistivity" which turns 
> into "resistance" by factoring in the geometry, it has to be
>
> R = Length/Area * rho.
>
> Since Length/Area has units of 1/length (1/meters), and R is in ohms, 
> then rho has to be ohms * meters, so that when you do ohms * meters * 
> (1/meters) it comes out as ohms.

But formal resistivity is in ohms per cubic centimeter, or meter. If you 
sened a request to a lab for material resistivity it will come back, 
typically as ohms per cubic unit, not in ohms
>
>
> I note that a lot of software and tables mis-identify conductivity 
> (not conductance) and resistivity.  More than one table shows 
> resistivity as ohms/meter, which is, of course, in correct.  Ditto for 
> conductivty, you'll see Siemen-meter or mho-cm, when it should be 
> Siemen/meter or mho/cm
>

> And that doesn't even get into when you look at older texts before 
> units were systematized. Take a look at NBS pubs from the early 20th 
> century or handbooks of the era and you see a lot of weird units (for 
> instance, inductance specified in terms of length). For any consistent 
> system of units you get 3 that are nice easy numbers, and one oddball 
> one with weird constants (usually involving 4*pi or its inverse).
>
>> When calculating you must use ohms cc^3 or M^3 for the units to properly
>> cancel.
>>

I did at he beginning of the previous post,

"The SI unit of electrical resistivity is the ohm⋅metre although other 
units like ohm⋅centimetre are also in use. As an example, if a 1 m × 1 m 
× 1 m solid cube of material has sheet contacts on two opposite faces, 
and the resistance between these contacts is 1 Ω, then the resistivity 
of the material is 1 Ω⋅m."  They calculated a 3 dimensional case and 
presented the results in two.  They are adding the third dimension back 
by multiplying Ohms (two dimensionally derived)by m... It would be a 
whole lot simpler to just say M^3
> I don't think so.. can you show an example using cubic units?
>
> I can give one using ohm-meters as resistivity.
>
> Calculate the current through a concrete bar that is 10 ohm-meter 
> resistivity. The bar is 1 meter long and 10x10 cm in cross section. A 
> voltage of 1000 Volts is applied to the long axis of the bar.

If the bar dimensions are known, it can be treated as a simple resistor. 
Again, this is a case where the two systems coincide.

>
> R = 10 ohm-meter * 1 meter / (0.1 * 0.1 meter^2) = 10/0.01 = 1000 ohms.
> I = V/R, so current is 1 Amp.
> The current density is 100 Amps/square meter = 1 Amp/ 0.01 square meters
>
> If one makes the bar 20x20 cm, what is the current?

Are you defining a bar 20 X 20 cm sq?  4M^2?

> R = 10*1 /0.04 = 250 ohms
> The current density is 4 amps / 0.04 square meters or the same 100 
> Amps/square meter, which makes sense, because the voltage between the 
> ends is the same (1000V) and the current is flowing "along" the bar, 
> not crossways
>
>
> A more complex question.. Given that the soil has a conductivity 
> (sigma) of 5 mS/m, what is the current flow through a 2 meter trough 
> which is 30x30cm with 10,000 volts along the length it?
>
> The conductance G is sigma * Area/Length  = 0.005 * 0.09/2 Siemens = 
> 0.000225 Siemens (or mhos).
>
> I = G * V
>
> So, I = 2.25E-4 * 1E4 = 2.25 Amps
>
> so the current is 4 Amps
>
>
>
>> Too many texts use the simplified versions which are misleading.
>
>
>>
>> 73
>>
>> Roger (K8RI)
>>
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