firewood profit

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Will someone share with me an easy way to determine the number of ricks in a log laying on the ground. I am trying to find a way so I do not have to stack to determine how many ricks I have available. Right or Wrong I consider a rick to be 16" X 4' X 8' or 1/3 of a cord. I have a large order pending and it sure would be nice not to have to stack it. Any help you veterans can give me would be greatly appreciated.
Thanks
 
Measure the length and small end of the log. Then using a log scale or a volume calculator tool, figure the board feet international scale. Approximately 500 board feet to a full cord. At least that's how I would do it.
 
Is there any firewood processers in the US. Here there is nothing else
Here the wood byers have mostly 3m length, up to 40cm diameter.
One truckload is about 70kubic meters/35ton wood, that is prcessed on a day.
My idea was to sell the 3m wood to coustemer, and if they want it cut and split, I have people for that too. So far this year I have sold 250kubik meters.
At almost no profit since I bought the timber expencive.
Next year will be different.
 
If you figure volume,while in the log,remember a cord of wood,is 4' by 4' by 8 'or 128 cu ft,wood,and an amount of air space.When I sold the stuff,it was by the stacked pick up load,whice is about 1/2 cord.A rick,is usualy taken,as one cut[16 "] 4 by 8 ,but I.ve seen a lot of variations.Many states have passed laws that says all fire wood sales will be based on cord measurements or fractions there of.It also has to do with the preperation of the wood,a well split stack would contain more wood then a stack of 18" rounds etc.
 
When I sold firewood cut and split I sold /ton.
If the wood was hard to estimate, or if it was a coustemers recuest I also sold /ton. Even a traind eye can have problems at times, so instead of guessing, I sold /ton. This way everyone knew exactly how much there was and no problems.
 
Mange said:
When I sold firewood cut and split I sold /ton.
If the wood was hard to estimate, or if it was a coustemers recuest I also sold /ton.
What do you mean? You sold it by the ton or ? Just not sure what you mean...
 
Mange said:
Yes.

Since I can not just enter yes, I ramble a little
10 character rule?

I thought you said that you sell it using two methods. Is that right or do you only sell it by the ton?
 
I used to by wood growing, fall then cut and split, now I by 3m logs and sell almost everything without laying a hand on it. Just redirect.
There is a lot of people here that wants to cut and split their own wood.
 
I cannot get the math to work if a cord of wood is about 500 board feet. A board foot is 1" X 1' X 1'. When I multiply the length X height X width or 8' X 4' X 48" I get 1536 board feet in one cord of wood. What am I doing wrong?
 
I think,what he was refering to,is a log that would scale 500 bdf,would contain 1 cord of wood,by volume.Remember,a saw log,has a certain deduction for kerf,non usable etc.A cord,as has been pointed out,is 92 cu ft of wood,and 36 cu ft of air.I had never heard of the 500 bdf thing,but it makes sense.
 
What are you doing wrong?&nbsp; Nothing, except using inconsistent units.&nbsp; When you convert the volume "one cord" to "one board-foot", the ratio is 1536:1.&nbsp; Trouble is, a cord is not 128 ft&sup3; <i>displacement</i> and a board-foot is not 12 in&sup3; <i>displacement</i>.&nbsp; Both measurements take "waste" into account, but do so differently, so they're really not relative units.

Erik gave a figure above of 92 ft&sup3; total displacement in a cord.&nbsp; That figure is one of several available.&nbsp; Figuring the amount of cordwood in a log might go something like this:

&nbsp; &nbsp; ((average_diameter [feet] &divide; 2)&sup2; &times; pi &times; length [feet]) &divide; 92

or make it a little easier:

&nbsp; &nbsp; ((average_diameter [feet] &divide; 2)&sup2; &times; length [feet]) &divide; 30

16" log, 10' long:

&nbsp; &nbsp; ((1.33 &divide; 2)&sup2; &times; 10) &divide; 30

&nbsp; &nbsp; (0.44 &times; 10) &divide; 30

&nbsp; &nbsp; 0.44 &divide; 3

&nbsp; &nbsp; 0.15 cord

Glen
 
Thanks for the clarification Al. That's exactly what I meant. When I said volume calculator I was referring to a tool that scales the board feet in the tree, a different kind of volume when compared to how much water is in this bucket sort of thing. :D

Of course that convoluted explanation I just offered probably didn't clarify anything.
 
Sometimes I think that I am denser than Hedge, but then that is another story. Here is my problem and I hope someone can put me on the right path.
According to Doyles Rule a log that measures 28" X 14' contains 504 board feet, and this is allowing waste for the taper and the bark. I have read that a cord of firewood contains 500 board feet, so in theory this log should have at least one cord of firewood.

Now, using the formula "((average_diameter [feet] ÷ 2)² × pi × length [feet]) ÷ 92"

((28"÷12)÷2)² x 3.14 x 14÷ 92

1.3689 x 3.14 x14 ÷ 92 = 0.654 cord

I would think regardless of which approach you take, the results should indicate the same amount of cordwood in a log, but these two approaches do not, what am I doing wrong.
 
I've got a Hewlett Packard graphing calculator which works with units of measurement.&nbsp; Using it, I convert from built-in units of ft&sup3; to board-feet (1 board foot = .002359737216 m&sup3;<tt></tt>).&nbsp; In terms of pure displacement, 92 ft&sup3; = 1104 board feet.

I'd say what you're doing wrong is going by the rule that a log which will produce 500 board feet will produce one cord.&nbsp; Unless you're taking into consideration that every bit of the material goes to making cordwood while you're throwing away a fair percentage (half, by volume?) making the boards.

What's your dilemma?&nbsp; Are you trying to decide quickly in the field which way a log will be more valuable to you?

Glen
 
I have a customer for firewood. He would like the firewood to be in logs. This is my dilemma. How do I know when I have cut enough logs to make seven cords. As you know each log is different and I am trying find a formula I can use in the field to determine how many cords are in a log, and the easiest measurements are average diameter and length.

I have heard repeatedly that there are 500 board feet in a cord of wood, so according to Doyles Rule you should yield a cord of wood from a log that is 28" in diameter and 14' long. It is my understanding the Doyles Rule takes into consideration, the taper of the log, the bark on the log and allows 1/4" kerf for each pass for the sawblade when cutting lumber. Now with all this waste using Doyles Rule, one would think that if you are using the entire log you should yield more cords of firewood than you would yield cords of lumber, but when I calculate using the formula "Radius Squared * Pi * Length / 92 I get less yield in cords of firewood per log which to me is not logical.

Bottom line is I do not want to short my customer, but at the same time I do not want to give lumber away and short myself.

A log 28" x 14' yields 504 board feet (More than a cord) according to Doyles Rule with a lot of waste for the kerf, bark and taper.

A log 28" x 14' yields only 0.65 of a cord using the formula "((average_diameter [feet] ÷ 2)² × pi × length [feet]) ÷ 92" and there is no waste for kerf, bark, and taper. To me if you do not waste anything you should yield more cords of firewood then you would lumber out of the same log.

I sure wish there was a smiley face of one scratching ones head.
 
28" on the small end? When determining BF in a log the measurement inside the bark at the small tip is the "diameter".
 

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