Density method for moisture (just for fun)

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Danno77

Minister of Fire
Hearth Supporter
Oct 27, 2008
5,008
Hamilton, IL
I have some black walnut. It's in about a 20" diameter x 15" high round right now. it weighs 82 lbs. I also have a round of oak(?) that is 17" diameter and 15" tall that weighs 37lbs. With those numbers (all guesses for now, except the weights, yep pulled out the bathroom scale for them) shouldn't I be able to find average density for this type of wood at X% of moisture and then figure out what my average moisture level is for this round? How handy would that be to be out in the woods and have just bucked a big tree and to be able to get a good idea of which tree you should split first for stacking in the place you hit last when pulling out firewood.

does that make sense? Like, if I burn my stack top to bottom over the winter the bottom has longer to dry, so it would be better if it was the wetter of my wood. And since I make multiple trips to the timber I'd bring back that really wet tree first and leave the dryer one for the next trip.

I know, I know. Bring my splitter with me and do a reading in the field. Blah. This way sounds more scientificified, well, at least more nerdy. Anyway, how cool would I be if I could hoist a piece of oak onto the trailer and say "gee this feels like about 33%." I'm sure you'd get good at it after a while.
 
Density measurement is a common method for wood testing. Normally a test piece is cut square for accurate measurement, and it's weighed. The piece is then oven dried and re-weighed. The difference is the amount of water.

A discrepancy occurs in measuring moisture content, and many lumber people will divide the weight of the water by the dry weight of the wood (0% moisture), so that moisture content of over 100% is not uncommon in green wood. Wood burners take the weight of the water and divide by the weight of the wet wood, so that 100% is not possible.

For your purposes, if you know the density of a given species at a given moisture content, then you should be able to tell if it's higher or lower (or make a chart of values) based on density, not accounting for variations in the same species (knots, base wood, rot, tree-to-tree variation).
 
ok, so let me work this out here. Black walnut looks to be 40lbs per cuft dry (should be 40/1728 to determine lbs/cu.in, right?). using basic volume of a cylinder my piece of wood comes to 4713cu.in. So dry weight would be 4713x.0231=109lbs (ok, i'm doing the math right but it's not working. I guess I need to figure exact measurements on my wood *insert teenage boy snicker*)

OK, Wife & BIL now find me to be nuts. Just called home and had her brother (visiting for holidays) measure the wood.

Walnut seems to be average of 14.5" diameter and 16.5" tall. volume for that is 2724.6508cu.in. dry weight for that volume should be 62.9394lbs. ok, now we are getting somewhere. so what's the next step? the difference is about 20lbs of water I imagine. do I just divide to get the percentage? like 19.0606/82 = 23% moisture (on average, of course: middle=wetter, outer=dryer)? Is that how I do it? that seems low, doesn't it? even for a tree that's been standing dead (bark on) for a few years.
 
similarly the volume of my oak(?) piece is 1978.9871cubic inches. 47 is dry weight per lb of oak this should weigh53.8lbs dry. I think I must be wrong on wood type.

This stuff I swear isn't pine. There isn't any pine around. I dunno, maybe it is. I mean I feel like it's got to be pine, but the wyes in it and branch sizes threw my. i'll cut it and post it I guess. Since when can't i tell the difference between pine and oak. go figure.

What else could be this light in the midwest? cedar?
 
(weight of wet wood that you have)-(weight of wood if oven dried) = weight of water

(Weight of water/ weight of wet wood)*100=moisture content.

So yah, 19# of water/ 82# log = 23% mc.
 
ok, ok, that stuff just has to be pine. I didn't want to believe it, but it is. Calculations of dry pine is 35.5lbs for the size of the log I have. It weighed 37, that gives it an average moisture content of about 4%. Again, this seems low, but I will venture a guess to say that the stuff IS seasoned to the point I can burn it. I threw a couple of big logs on last night and they caught quick and burned well. It's barkless and was standing dead for years (5+ at least)
 
Danno77 said:
Walnut seems to be average of 14.5" diameter and 16.5" tall. volume for that is 2724.6508cu.in. dry weight for that volume should be 62.9394lbs. ok, now we are getting somewhere. so what's the next step? the difference is about 20lbs of water I imagine. do I just divide to get the percentage? like 19.0606/82 = 23% moisture (on average, of course: middle=wetter, outer=dryer)? Is that how I do it? that seems low, doesn't it? even for a tree that's been standing dead (bark on) for a few years.
So, I checked this same round today. It still has it's bark on it and it weighs in at 61lbs. I split it up and checked in lots of places. at about 3 inches in from one end it was 19%, at about 3" from the other end it was 26%, and in the dead middle it was 30%. If I determined back then that the dry weight would be 62lbs, then how on earth does it weigh in at less than that and now have an average of mid 20's for MC?

interesting data, just thought I'd post it. i'll come back to this at some other point when my brain is working right (still sick!!!)
 
Density is mass (weight) divided by volume. The forumula for the volume of a cylinder is pi x radius squared x height. The formula for the volume of a triangular prism (roughly the shape of a split log) is .5 x lengh x width x height. So, you can figure out the density of the wood before drying and test the moisture level, and then compare that to a dried piece and see how much volume was lost.

I homeschool my kids, so I'm always trying to think of how to turn real life things into lessons. That would be fun!

~Rose
 
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