Reciprocal, not logarithmic

Thread Starter

rspuzio

Joined Jan 19, 2009
77
In the chapter on Ohmmeters (8.6), it is repeatedly said that the
scale on an ohmmeter is logarithmic and, in particular, there is the
statement

"With a logarithmic scale, the amount of resistance spanned for
any given distance on the scale increases as the scale progresses
toward infinity, making infinity an attainable goal."

This is not correct --- per Ohm's law, the scale on an ohmmeter
varies as the reciprocal of the current, not as its logarithm. Whilst
the reciprocal of zero is infinity, the logarithm of no finite number
is infinity so infinity is not attainable with a logarithmic scale
although it is attainable with a reciprocal scale.

Call me a mathematical stickler, but this sloppiness sticks out at
me like a sore thumb when I read this otherwise excellent
exposition --- one does not make an ohmmeter by pasting a slide
ruler on an ammeter movement :) Please consider changing the
terminology so as to fix this inaccuracy.
 

Thread Starter

rspuzio

Joined Jan 19, 2009
77
> Interesting reasoning, but incorrect.

Could you kindly please point out what you see as
incorrect in my post?

> If you don't like our Ebook,

Liking or not liking the book is not the issue.
As I understand it, this forum is the
place to post corrections and suggestions for
improving the book "All About Circuits".
 

steveb

Joined Jul 3, 2008
2,436
>
Could you kindly please point out what you see as
incorrect in my post?
I have to say I agree with rspuzio. The inverse function is not the same a logarithm function. The two funcitons are similar, and related, but not the same. The derivative of a logarithm is the inverse function, or you can say that the logarithm function is the integral of the inverse function.

It is clear in the example given that resistance in inversely proportional to current. Further, the D'Arsonval meter movement has a needle angular displacement that is proportional to current. Hence, the OP is correct.
 

Ratch

Joined Mar 20, 2007
1,070
To the Ineffable All,

This is the equation for the example given below, resistance = 9V/(k*0.001A)-9000 ohms where k is the rotation of the meter needle. k = 0 is the leftmost position and 1.0 is the rightmost position. http://www.allaboutcircuits.com/vol_1/chpt_8/6.html .

Looking at the plot in the attachment we see that it is a displaced hyperbola which is a form of 1/x or inverse function. Also when you look at the ohmmeter scale, you can tell it is not logarithmic, because log scales don't have a zero, and the displacement from 1 to 2 is not the same as the displacement from 2 to 4 like a log scale would be.

Ratch
 

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someonesdad

Joined Jul 7, 2009
1,583
The function is θ = f(R) where R is resistance and θ is angular deflection. The function should either be written down for the example given or just call it "nonlinear". End of discussion. :)
 

Ratch

Joined Mar 20, 2007
1,070
someonesdad,

The function is θ = f(R) where R is resistance and θ is angular deflection. The function should either be written down for the example given or just call it "nonlinear".
Well, you identified the parameter and the result. What is the range of the result θ? 0° to what? What is the definition of the function "f"?

...or just call it "nonlinear".
Would that help someone who is manufacturing a meter scale? Or give you a good grade on a test?

Ratch
 

Thread Starter

rspuzio

Joined Jan 19, 2009
77
> The function should either be written down for the example given
> or just call it "nonlinear"

Agreed. This is the correction I was asking for. Changing the term
"logarithmic" in the figure caption, the text, and the review point
to "non-linear" would fix the problem.

While the function isn't written explicitly in the book, it's value is
computed for several values of θ, which serves the same purpose.
 
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