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DIY Moisture Meter

Started by lowpolyjoe, June 07, 2013, 12:28:44 PM

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lowpolyjoe

Hey Guys,

I while ago I read about using a multimeter as a moisture meter.   Generally, a multimeter cannot be used as-is because the resistance of the wood is high relative to the internal resistance of the meter itself.    This website article i found explains an approach using a 9Volt battery and some math to measure resistance and then some tables to convert to moisture content.  I'm a fan of saving money whenever i can, so i was eager to investigate this approach and not spend a few hundred dollars on a decent meter.

http://woodgears.ca/lumber/moisture_meter.html

Unfortunately this approach requires measuring the internal resistance of the meter you will be using.  This means you will need a second meter (or you could use a known resistor and a battery i suppose).  I happen to have 2 meters, so i measured it directly.  The meter you will use to measure the wood should be set to DC Voltage mode and the other set to Resistance mode (obviously :) ).  My meter's internal resistance was about 1MOhm (compared to the 10MOhm in the article) as shown on the left meter.



Testing my recently planed Basswood board near the edge gave the following reading with the battery setup.  The type of nails used (diameter), the distance driven into the wood and spacing between them all impact the measurement.  I used some finishing nails i had laying around.  They were spaced 1 1/4" and driven to a depth of 5/16".



The equation for calculating the resistance of the wood is:

Rwood= Rmeter * ( (Vbattery / Vmeter) - 1)

In this case i had

Rwood= 1,000,000 * ( (9 / 0.11) -1  ) ~= 80,800,000 Ohms ~= 81MegaOhms
Using the table in the linked article for Basswood, that maps to near 11% Moisture


Some oak i sawed



Rwood= 1,000,000 * ( (9 / 0.17) -1  ) ~= 51,940,000 Ohms ~= 52 MegaOhms
Using the table in the linked article for Oak (red?), that maps to near 13.5% Moisture

Another oak measurement near the center of the board instead of the edge




Rwood= 1,000,000 * ( (9 / 0.27) -1  ) ~= 32,330,000 Ohms ~= 32 MegaOhms
Again using the table in the linked article for Oak (red?), that maps to 14% Moisture

This was a consistency check... i expected a little more moisture at the center, and that's what i got (correct me if i'm wrong).  And i expected more voltage measured to result in lower resistance calculated and higher moisture content, which is what i got. edit :  i think What i meant to say was that more voltage across the meter's internal resistance means less voltage across the nails, meaning  lower resistance in the wood and higher moisture content... I think that's right anyway ???  :)

Finally, some sort of conifer.  I think i asked for an ID on the tree a while ago but don't remember what was said?  I'll have to dig it up or ask again if i can find pics of the tree before it fell.  Need a type so i can lookup on the moisture table.







kelLOGg

I don't see anything amiss with your approach. You might use hardened nails mounted in a suitable "electrode holder" to make insertion and removal easier and consistent.
Bob
Cook's MP-32, 20HP, 20' (modified w/ power feed, up/down, loader/turner)
DH kiln, CatClaw setter and sharpener, tandem trailer, log arch, tractor, thumb tacks

lowpolyjoe

Thanks Bob - yeah i'm thinking about a little holder for the nails.  Maybe to hold the battery too.  The article i linked to originally used a simple block of wood with holes drilled through it that allowed you to hammer the nails into the wood the correct amount each time. 

GeneWengert-WoodDoc

I notice that you used 9 volts, but most new batteries of the type you used are actually 9.48 v.  Lithium would be even more.  So, this might help you refine you calculation.

We have used nails with commercial moisture meters for years when we needed the MC of large pieces of wood...the standard needle is usually 2" in length.  We measure the MC of bridge timbers with nails back in the late 60s.  We found that spacing was not an issue and diameter was also not important...the reason is that the resistance to MC relation ship is logarithmic, so things that make a factor of 10 difference are important.  One company (maybe more) also sold nails to use during kiln drying.

A word of caution...the current will change with time, so only use a momentary application of voltage rather than continuous.

Most multimeters have an ampere measurement ability.  Why can't you use that scale so you do not need to know the meter resistance or voltage drop?

If we have a piece of oak at 9% MC or drier, I do believe that the technique you referenced will not work too well, as there is not enough resolution in the voltage measurement...the current is really small.  Is this correct?

I recall a moisture meter from long time ago (circa 1950s) that actually had you adjust the resistance of one leg of the bridge circuit until there was essentially no current and this adjustable resistance was calibrate on a dial for moisture content.  Another one used a flasher and when the bridge was not balanced, the light flashed, so you adjusted the variable resistor until the light shut off, as I recall.
Gene - Author of articles in Sawmill & Woodlot and books: Drying Hardwood Lumber; VA Tech Solar Kiln; Sawing Edging & Trimming Hardwood Lumber. And more

lowpolyjoe

Good points Gene.

I didn't think to check the voltage of the battery itself but i will next time.

It is certainly lucky for us that the resistance is logarithmic so small variations in measurement aren't a big deal.   Nail size probably won't matter too much unless you're comparing a brad to a  giant framing nail or something.  Maybe i'll try that next time and see what the difference is.

You made me stop and think about the current measurement approach.  If i'm looking at resistances in the wood between 10M and 100M, a 9V battery ought to get me something like:

From
I = V/R = 9/ 10,000,000  = 0.9mA  => about 1mA or 1,000uA
To
I = V/R = 9/ 100,000,000= 0.09mA => about 1/10th mA or 100uA

I think those are measureable with my meter, and get me to about 12% or 13% MC on many species.  Lower MC give resistances of 1,000M to 10,000M or more

From
I = V/R = 9/ 1,000,000,000  = 0.009mA  => about 1/100th mA or 10uA
To
I = V/R = 9/ 10,000,000,000= 0.0009mA => about 1/1000th mA or 1uA

Those i don't think my meter can resolve since it's lowest DCA setting is "200ua".  Better meters might be OK.   

I'm neglecting internal meter resistance in current-measurement mode.  I have no idea what it is, but i'm guessing it's negligable in these cases.

Your 9% oak case at about 1,000M would fall in the 10uA case above for current measurement.  If i juggle the voltage measurement equation around i get:

Vmeter = (Vbattery)/( 1 + (Rwood/Rmeter ) = 9 / (1 + (1,000,000,000 / 1,000,000)) = 0.00899v ~= 0.009V, or on the order of 10mV

Looking at my meter in the pic, it has 200mV as its lowest voltage  reading setting, so 10mV would be a stretch similar to the current measuring technique.

I've been taking measurements with the meter on the right in the side-by-side pics... it's a harbor freight special :).     The meter on the left is from Radio Shack and is probably a *little* nicer but the leads are broken.  It's auto-range-sensing so i'm not sure what the lowest resolution settings are.

Seems like i'll suffer resolution losses at lower MC's with either voltage or current technique, but at least i have a ballpark idea of where i'm at.  I'll only be building shelves or small tables with this lumber but i certainly don't want to use something that's still up in the teens as far as MC goes.    I will target under 8-10%.  Hopefully i'll have the patience to wait that long  :D


Thanks
Joe

KnotBB

To check the voltage of the battery it needs to be under load.   Free standing voltage is not what will be applied in a working circuit.  High resistance volt meters do not load the batter enough to give an accurate reading.  And the "older" the battery is the more the voltage drop will be. 

A "dead" car battery may read 12 volts but it won't turn the engine over because of the voltage drop under load.  That's why they use a load tester.
To forget one's purpose is the commonest form of stupidity.

GeneWengert-WoodDoc

What is stated about measuring a battery voltage under load is indeed correct.  However, with a moisture meter circuit, as illustrated, the load of wood resistance of many megohms is almost zero (current is microamps), so a high resistance volt meter would be ok...that is the way I see it, but maybe I missed something.

I just tried to measure the voltage of a 9 volt battery that was brand new and had a load (not real big load) and got 9.4 volts, as close as I can read it, but my meter is not perfect.
Gene - Author of articles in Sawmill & Woodlot and books: Drying Hardwood Lumber; VA Tech Solar Kiln; Sawing Edging & Trimming Hardwood Lumber. And more

lowpolyjoe

Yeah that's an important point about battery load when testing.  I read up on it once and i think it said something about current draw causing a voltage drop across an internal resistance in the battery that increases as it ages and drains ???  If there is no load to draw current, that internal resistance is neglected.  Something like that   :)

In any case, i think the draw on the tester battery i'm using is pretty low since the resistances in the circuit are so high and it's not connected for too long.  I didn't have any time this weekend to do any more testing.

If my Basswood is really around 11% MC as my measurements suggest, can i bring it into my garage for further drying?  It's currently outside under an overhang of the house but somewhat exposed to the elements.   I want it dry as soon as possible  :)  and  I'm trying to figure if the wind and open air outside will dry it faster or is the sheltered garage a better bet?  Garage is not that well insulated and not heated/cooled.

I've already got a bunch of bad splits, so i'm thinking it may already be drying faster than is good for it anyway sitting outside.   I was ignorant and didn't seal the ends :(

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