林业科学  2016, Vol. 52 Issue (11): 142-147 PDF
DOI: 10.11707/j.1001-7488.20161117
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#### 文章信息

Ma Yan, Xu Honggang, Yang Chunmei, Xu Shixiang

Volume Calculation Method and Benefit Analysis of Small-Diameter Wood Sided Sliced Veneer

Scientia Silvae Sinicae, 2016, 52(11): 142-147.
DOI: 10.11707/j.1001-7488.20161117

### 作者相关文章

Volume Calculation Method and Benefit Analysis of Small-Diameter Wood Sided Sliced Veneer
Ma Yan, Xu Honggang, Yang Chunmei , Xu Shixiang
Forestry and Woodworking Machinery Engineering Technology Center, Northeast Forestry University Harbin 150040
Abstract: 【Objective】 This paper proposed three measures: slicing double-sided, cutting feed optimization and reasonable selection, in order to solve the inefficiency, low-productivity, more deficiency and other problems of traditional methods for producing veneer. 【Method】 Firstly, this study proposed to improve volume ratio and production efficiency by the use of sided-slicing processing for small-diameter wood (sides-slicing refers to the continuous longitudinal chipping on both sides of small-diameter wood, it would not stop until 20 mm thickness was remained for a fixed core slice; then slicing the fixed core that remained after multiple processing). Secondly, we gave detailed formulas for both short and long trail direction of sliced veneer volume by conducting mathematical modeling and practical calculating on the ideal small-diameter wood, by which the theoretical basis for reasonable cutting feed was founded. Finally, this paper conducted mathematical modeling on the cross-section of small-diameter wood, and gave the corresponding volume ratio formulas, the actual parameters of the small-diameter wood were measured, and the volume ratio was calculated according to the formula. 【Result】 By the actual analysis of the small-diameter wood's volume ratio, the conclusion came that the increase of the size of small-diameter wood (The object is ideal small-diameter that has no knots, no radial bending), the sliced veneer volume was gradually increased, the larger size small-diameter wood should be chosen when slicing veneer, so that the volume ratio of small-diameter wood could be improved to a certain degree. 【Conclusion】 When the veneer is prepared, the sides slicing technology can not only save time, but also reduce manpower, as a result, greatly improve the processing efficiency and volume ratio. On the basis of mathematical modeling for small-diameter wood, sliced veneer volume formula in short and long trail direction and the volume ratio formula, the waste of timber resources could be decreased and the economic efficiency would be improved by reasonable selecting, cutting feed and slicing the wood.
Key words: sides slicing     veneer     volume     volume ratio     profit

1 双面刨切工艺

1.1 刨切下刀图

 图 1 双面刨切下刀 Fig.1 The sides slicing's cutting feed

1.2 双面纵向刨切的优点

1) 双面刨切可连续生产，同时对小径木相对的2个平面进行刨切，提高了机械化和自动化程度，小径木只需1次装夹即可完成整体的刨切，节省了反复装卡木材的时间，节省了人力物力，大大提高了生产效率(马岩，2006)。

2) 因为每次刨切结束后剩余的固定芯板需要进行再次刨切，每次完整加工剩余的废料仅为两侧板皮以及中间20 mm×20 mm的固定芯材，所以大大提高了小径木的出材率，从而提高了经济效益。

3) 因为每次刨切的方向都是顺着小径木的纤维方向进行的，相对于垂直木纤维方向的刨切，所需剪切应力小，而且不会造成横向刨切时产生的单板裂纹。双面纵向刨切时噪声小，工作平稳，安装调试容易。

1.3 中间固定芯板的处理工艺

 图 2 固定芯板双面刨切 Fig.2 The fixed cores' sides slicing 1.限位块Limit block；2.夹紧钉Clamping nail；3.刨刀Planer.

2 双面刨切单板的材积计算

 $F\left({x, y, z} \right) = \left\{ {\begin{array}{*{20}{c}} {{c_1}{x^2} + {c_2}{y^2} \le {c_4} - {z^{c3}}, }\\ {{c_5} \le z \le {c_6}, }\\ {{c_1}{c_2} \ge 0。} \end{array}} \right.$ (1)

 ${v_x} = 2\int_{{x_1} - s}^{{x_1}} {{\rm{d}}x} \int_0^L {{\rm{d}}z} \int_0^{\sqrt {\frac{1}{{{c_2}}}\left({{c_4} - z - {c_1}{x^2}} \right)} } {{\rm{d}}y} 。$ (2)
 图 3 短径方向单板数学模型 Fig.3 The mathematical model of veneer's short trail direction

 ${r_n} = \frac{1}{{\left({2n + 1} \right)!\left[ {2\left({n + 1} \right) + 1} \right]}} \approx 0.00119。$ (3)

 $\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {{v_{1x}} = \frac{{D_{{\rm{a2}}}^2D_{{\rm{b}}1}^3L}}{{8{D_{{\rm{a1}}}}\left({D_{{\rm{b}}1}^2 - D_{{\rm{a}}1}^2} \right)}}\left\{ {\frac{{2{x_1}}}{{3{D_{{\rm{b2}}}}}}\left({5 - \frac{{8x_1^2}}{{D_{{\rm{b2}}}^2}}} \right)} \right.}\\ {\sqrt {1 - \frac{{4x_1^2}}{{D_{{\rm{b2}}}^2}}} + \frac{{2{x_1}}}{{{D_{{\rm{b2}}}}}} + \frac{{4x_1^3}}{{3D_{{\rm{b2}}}^3}} + \frac{{4x_1^5}}{{15D_{{\rm{b2}}}^5}} - } \end{array}}\\ {\frac{{2\left({{x_1} - s} \right)}}{{3{D_{{\rm{b2}}}}}}\left({5 - \frac{{8{{\left({{x_1} - s} \right)}^2}}}{{D_{{\rm{b2}}}^2}}} \right)\sqrt {1 - \frac{{4{{\left({{x_1} - s} \right)}^2}}}{{D_{{\rm{b2}}}^2}}} }\\ {\begin{array}{*{20}{c}} { - \frac{{2\left({{x_1} - s} \right)}}{{{D_{{\rm{b2}}}}}} - \frac{{4{{\left({{x_1} - s} \right)}^3}}}{{3D_{{\rm{b2}}}^3}} - \frac{{4{{\left({{x_1} - s} \right)}^5}}}{{15D_{{\rm{b2}}}^5}} - }\\ {\left[ {\frac{{2{x_1}}}{{3{D_{{\rm{b2}}}}}}\left({\frac{{5D_{{\rm{a1}}}^2}}{{D_{{\rm{a2}}}^2}} - \frac{{8x_1^2}}{{D_{{\rm{b2}}}^2}}} \right) + \frac{{2D_{{\rm{a1}}}^3{x_1}}}{{D_{{\rm{a2}}}^3{D_{{\rm{b}}2}}}} + \frac{{4{D_{{\rm{a1}}}}x_1^3}}{{3{D_{{\rm{a}}2}}D_{{\rm{b}}2}^3}}} \right.}\\ { + \frac{{4{D_{{\rm{a2}}}}x_1^5}}{{15{D_{{\rm{a1}}}}D_{{\rm{b}}2}^5}} - \frac{{2\left({{x_1} - s} \right)}}{{3{D_{{\rm{b2}}}}}}\left({\frac{{5D_{{\rm{a1}}}^2}}{{D_{{\rm{a2}}}^2}} - \frac{{8{{\left({{x_1} - s} \right)}^2}}}{{D_{{\rm{b2}}}^2}}} \right)}\\ {\sqrt {\frac{{D_{{\rm{a1}}}^2}}{{D_{{\rm{a2}}}^2}} + \frac{{4{{\left({{x_1} - s} \right)}^2}}}{{D_{{\rm{b2}}}^2}}} - \frac{{2D_{{\rm{a1}}}^3\left({{x_1} - s} \right)}}{{D_{{\rm{a2}}}^3{D_{{\rm{b2}}}}}} - } \end{array}}\\ {\left. {\left. {\frac{{4{D_{{\rm{a1}}}}{{\left({{x_1} - s} \right)}^3}}}{{3{D_{{\rm{a2}}}}D_{{\rm{b}}2}^3}} - \frac{{4{D_{{\rm{a2}}}}{{\left({{x_1} - s} \right)}^5}}}{{15{D_{{\rm{a1}}}}D_{{\rm{b}}2}^5}}} \right]} \right\}。} \end{array}$ (4)

 $\begin{array}{*{20}{c}} {{v_x} = \frac{1}{4}{D_{{\rm{a1}}}}{D_{{\rm{b1}}}}L\left({\frac{{D_{{\rm{a1}}}^2 + tD_{{\rm{a2}}}^2}}{{D_{{\rm{a1}}}^2\left({1 + t} \right)}}} \right)\left[ {\frac{{2{x_1}}}{{{D_{{\rm{b2}}}}}} + \frac{{4x_1^3}}{{3D_{{\rm{b}}2}^3}}} \right. + }\\ {\frac{{4x_1^5}}{{15D_{{\rm{b}}2}^5}} - \frac{{2\left({{x_1} - s} \right)}}{{{D_{{\rm{b2}}}}}} - \frac{{4{{\left({{x_1} - s} \right)}^3}}}{{3D_{{\rm{b}}2}^3}} - }\\ {\frac{{4{{\left({{x_1} - s} \right)}^5}}}{{15D_{{\rm{b}}2}^5}} + \frac{{D_{{\rm{a2}}}^2\left({1 + t} \right)}}{{tD_{{\rm{a2}}}^2 + D_{{\rm{a1}}}^2}}\left({\frac{{2{x_1}}}{{{D_{{\rm{b2}}}}}}\sqrt {1 - \frac{{4x_1^2}}{{D_{{\rm{b}}2}^2}}} } \right.}\\ {\left. {\left. { - \frac{{2\left({{x_1} - s} \right)}}{{{D_{{\rm{b2}}}}}} \times \sqrt {1 - \frac{{4{{\left({{x_1} - s} \right)}^2}}}{{D_{{\rm{b}}2}^2}}} } \right)} \right]。} \end{array}$ (5)

 $\begin{array}{*{20}{c}} {{v_y} = \frac{1}{4}{D_{{\rm{a1}}}}{D_{{\rm{b1}}}}L\left({\frac{{D_{{\rm{a1}}}^2 + tD_{{\rm{a2}}}^2}}{{D_{{\rm{a1}}}^2\left({1 + t} \right)}}} \right)\left[ {\frac{{2{y_1}}}{{{D_{{\rm{a2}}}}}} + \frac{{4y_1^3}}{{3D_{{\rm{a}}2}^3}}} \right. + }\\ {\frac{{4y_1^5}}{{15D_{{\rm{a}}2}^5}} - \frac{{2\left({{y_1} - s} \right)}}{{{D_{{\rm{a2}}}}}} - \frac{{4{{\left({{y_1} - s} \right)}^3}}}{{3D_{{\rm{a}}2}^3}} - }\\ {\frac{{4{{\left({{y_1} - s} \right)}^5}}}{{15D_{{\rm{a}}2}^5}} + \frac{{D_{{\rm{a2}}}^2\left({1 + t} \right)}}{{tD_{{\rm{a2}}}^2 + D_{{\rm{a1}}}^2}}\left({\frac{{2{y_1}}}{{{D_{{\rm{a2}}}}}}\sqrt {1 - \frac{{4y_1^2}}{{D_{{\rm{a}}2}^2}}} } \right.}\\ {\left. {\left. { - \frac{{2\left({{y_1} - s} \right)}}{{{D_{{\rm{a2}}}}}} \times \sqrt {1 - \frac{{4{{\left({{y_1} - s} \right)}^2}}}{{D_{{\rm{a}}2}^2}}} } \right)} \right]。} \end{array}$ (6)
3 双面刨切单板的出材率

 $双面刨切出材率 = \frac{合格单板材积}{小径木整体材积}。$ (7)

 ${v_{\rm{T}}} = 4\int_0^{{D_{{\rm{b1}}}}} {{\rm{d}}x} \int_0^L {{\rm{d}}z} \int_0^{\sqrt {\frac{1}{{{c_2}}}\left({{c_4} - z - {c_1}{x^2}} \right)} } {{\rm{d}}y} 。$ (8)

 $双面刨切单板出材率 = \frac{合格单板底面积}{小径木底面积}。$ (9)

 图 4 小径木横截面 Fig.4 The cross section of small-diameter wood

 $\;\frac{{{y^2}}}{{{{\left({{D_{{\rm{a1}}}}/2} \right)}^2}}} + \frac{{{x^2}}}{{{{\left({{D_{{\rm{b1}}}}/2} \right)}^2}}}1。$ (10)

 ${S_1} = 2\int_{\frac{{{D_{{\rm{b1}}}}}}{2} - 2}^{\frac{{{D_{{\rm{b1}}}}}}{2}} {} \sqrt {\frac{{D_{{\rm{a1}}}^2}}{4}\left({1 - \frac{{4{x^2}}}{{D_{{\rm{b1}}}^2}}} \right)} {\rm{d}}x。$ (11)

 $\begin{array}{*{20}{c}} {{S_1} = \frac{{{D_{{\rm{a1}}}}{D_{{\rm{b1}}}}}}{2}\left({\frac{1}{2}{\theta _x}\left| {\begin{array}{*{20}{c}} {\frac{{\rm{\pi }}}{2}}\\ {\arcsin \left({1 - \frac{4}{{{D_{{\rm{b1}}}}}}} \right)} \end{array} + } \right.} \right.}\\ {\left. {\left. {\left({\frac{1}{4}\sin 2{\theta _x} + {c_0}} \right)} \right|\begin{array}{*{20}{c}} {\frac{{\rm{\pi }}}{2}}\\ {\arcsin \left({1 - \frac{4}{{{D_{{\rm{b1}}}}}}} \right)} \end{array}} \right)。} \end{array}$ (12)

 $\sin {\theta _x} = \frac{{2x}}{{{D_{{\rm{b1}}}}}}。$ (13)

 ${S_2} = 20 \times 20 = 400\left({{\rm{m}}{{\rm{m}}^2}} \right)。$ (14)

 $S = {\rm{\pi }}{D_{{\rm{a1}}}}{D_{{\rm{b1}}}}/4。$ (15)

 ${S_3} = S - 2{S_1} - {S_2}。$ (16)

 $\begin{array}{*{20}{c}} {\eta = \frac{{S - 2{S_1} - {S_2}}}{S} \times 100\% = }\\ \begin{array}{l} \frac{{{\rm{\pi }}{D_{{\rm{a1}}}}{D_{{\rm{b1}}}}/4 - 400 - {D_{{\rm{a1}}}}{D_{{\rm{b1}}}}\left[ {\left({\frac{1}{2}{\theta _x} + \frac{1}{4}\sin 2{\theta _x} + {c_0}} \right)\left| {\begin{array}{*{20}{c}} {\frac{{\rm{\pi }}}{2}}\\ {\arcsin \left({1 - \frac{4}{{{D_{{\rm{b1}}}}}}} \right)} \end{array}} \right.} \right]}}{{{\rm{\pi }}{D_{{\rm{a1}}}}{D_{{\rm{b1}}}}/4}}\\ \times 100\% 。 \end{array} \end{array}$ (17)

4 结论

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