# Multimeter Ohms exponential scale; formula?

1. Dec 12, 2011

### RoGuE_StreaK

1. The problem statement, all variables and given/known data
Hi, I'm asking the usual "what formula gives me these results" question, but with a twist; I'm attempting to make a "virtual" analogue multimeter that replicates a physical piece of equipment, for training purposes. The idea is that a random MegaOhm value will be generated, and displayed by the needle; the user must then determine whether it falls within the pass or fail mark for a piece of equipment being tested.

The problem is that the scale is non-linear. It's exponential of some sort. But after a day of intensive online searching, and resorting to throwing numbers at various equations, I can't come up with a formula that gives a reasonable approximation, or even anything remotely close!

NB: ANY formula that fits the points is fine by me; I've severely over-engineered this problem, but I'm determined to get it working!

2. Relevant equations
Essentially the multimeter needle range is 90degrees of motion, going from zero MOhms to Infinite MOhms in an exponential manner. The following are data points relating MOhms to degrees, very roughly measured from a photo of the multimeter:

value, angle (degrees)
0, 0
0.1, 5.5
0.2, 11.25
0.5, 28
1, 39.7
2, 46.5
5, 61.8
10, 68.65
20, 79.2
50, 86
100, 88.35
200, 89.4
infinity, 90

[EDIT] Here's an example multimeter to show the exponential scale (different from mine, but you get the idea)
http://www.opamp-electronics.com/tutorials/images/dc/50035.jpg

3. The attempt at a solution
I've tried plotting these in Excel and using a trendline, but nothing comes remotely close.
Tried creating dynamic formula that I could change various key values of and see the results for all input values, but could never get a curve that came close; always curved too sharply too late, and I just don't know what terminology I should be searching for to fix it!
I can SEE a definite trend in the plotted points, but Excel doesn't seem to see what I see.

This is intended just as a "cool" factor for some training material, and I've spent way too much time on it, so it's highly frustrating that something that should be so simple has had me (and several of my colleagues) climbing up the walls! Any help would be greatly appreciated!

Last edited: Dec 12, 2011
2. Dec 13, 2011

### fzero

Try

$R(x) = R_0 \tan\left(a \frac{\pi x}{180}\right)$

and determine the best-fit values of $R_0,a$.

3. Dec 13, 2011

### LCKurtz

Doesn't that have the wrong concavity?

4. Dec 13, 2011

### RoGuE_StreaK

Thanks guys, but it's been almost 20yrs since I've done stuff like this; can someone give a brief terminology of "R(x)","R0", and "a", as I don't know the terms to start googling to get a better understanding. Once I'm in the ballpark at I'm sure I should be able to stumble my way along to a solution, but I first have to FIND the ballpark! ; )
Is it something to do with radians rather than degrees?

In the meantime I think I'll just have to fudge a few random angles/MOhms to display, as really that is all that is required, but I would like to get it working as intended at some stage as I think it would be a cool/useful tool for some future projects.

5. Dec 13, 2011

### Staff: Mentor

Why don't you just set 1Meg as the highest value, then

log (10^6) = 6

Use 6 as your max deflection angle (90 degrees), and ratio other logarithmic values to that.

So the angle for 1k would be:

(90 degrees / 6) * log (10^3)

And so on...

6. Dec 13, 2011

### RoGuE_StreaK

hi Berkeman, the issue is that I'm trying to replicate a physical piece of equipment that is in use, so I want to replicate reasonably faithfully the dial used on it. Wouldn't be an issue if I was making one up from scratch.

7. Dec 13, 2011

### Staff: Mentor

Oh, I see now. The dial is not log scale after all. Yuck.

Maybe just do a look-up table with interpolation. The more points in your look-up table, the closer the accuracy... Might be the quickest way...

8. Dec 13, 2011

### fzero

I guess if you plot things the way he's listed them, yes. x is the angle, R(x) is the resistance.

In that formula:

$x$: angle in degrees.

$R(x)$: resistance in $\textrm{M}\Omega$ for the angle $x$.

$R_0$: scale factor in units of $\textrm{M}\Omega$. The value should turn out to be close to 1.9 $\textrm{M}\Omega$.

$a$: another scale factor introduced to improve the fit. This should turn out to be very close to 1.

The conversion between radians and degrees is taken care of by the factor of $\pi/180$ in the argument of tan.

9. Dec 13, 2011

### RoGuE_StreaK

Thanks for that, I'll get back onto it in a few days and see if I can tweak to get something about right, for the moment I'm just slapping it together in fudged format for a quick presentation tomorrow.