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Author Topic: Hemlock weight after AD  (Read 3507 times)

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Offline GeneWengert-WoodDoc

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Re: Hemlock weight after AD
« Reply #20 on: May 24, 2013, 06:41:28 PM »
Flat sawn 1.250" x 8.00" Eastern hemlock green will be approximately  1.24" x 7.81" at 20% MC.
Gene - Author of articles in Sawmill & Woodlot and books: Drying Hardwood Lumber; VA Tech Solar Kiln; Sawing Edging & Trimming Hardwood Lumber. And more

Offline Jay C. White Cloud

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Re: Hemlock weight after AD
« Reply #21 on: May 24, 2013, 07:28:18 PM »
You,er the man Gene...

I don't know if you have time, but I was thinking it would be great, (especially for me ???) if I could see the math steps you took to arrive at that conclusion.  I am always second guessing myself, and would love to see a pro plug the numbers into the formula.  I kinda understand them, but alway think I am getting out of sequence; just not sure of myself.  Using Hillybillyhogs's question as the model?

Regards,

jay
"To posses an open mind, is to hold a key to many doors, and the ability to created doors where there were none before."

"When it is all said and done, they will have said they did it themselves."-teams response under a good leader.

Offline GeneWengert-WoodDoc

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Re: Hemlock weight after AD
« Reply #22 on: May 24, 2013, 10:19:39 PM »
To get the number, we have to make some assumptions.  First, at 20% MC, the entire piece is 20% MC...not the core at 30% and the shell at 10% MC, or similar.  Second, we assume that shrinkage satrts are 30% MC, whcih is true for a wood cell, but not for a piece of lumber.  Third, we assume that the large piece can actually shrink as much as it is suppose to and that there is no tension set.

So, now we use the shrinkage values from a book such as THE WOOD HANDBOOK.  In the 2010 edition, these values appear on page 13-16 and 13-17.   Note that it says these are for 6% to 14% MC, but if we use the total shrinkage values (p. 4-6). we will get the same answers.  For eastern hemlock, the values given are 0.00102 (radial shrinkage for each 1% MC change) and 0.00237 (tangential).

So, at 20% MC, we have the wood cells only shrinking from 30% MC to 20% MC, or 10%.  So, multiply the two numbers by 10.  The radial number is the thickness change for the flat sawn lumber.  So, 1.25 x .0102 = 0.01275".  So, the thickness shrinkage is 0.01".  For the 8" width, 8 x .00237 = 0.1896 = 0.19" shrinkage in width.  So, at 20% MC, the new size is [1.25 - 0.01] x [8.00 - 0.19] or 1.24" x 7.81".

OK?
Gene - Author of articles in Sawmill & Woodlot and books: Drying Hardwood Lumber; VA Tech Solar Kiln; Sawing Edging & Trimming Hardwood Lumber. And more

Offline Jay C. White Cloud

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Re: Hemlock weight after AD
« Reply #23 on: May 25, 2013, 12:05:55 AM »
Thanks Gene, you are great!!
"To posses an open mind, is to hold a key to many doors, and the ability to created doors where there were none before."

"When it is all said and done, they will have said they did it themselves."-teams response under a good leader.


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