Orellana, Enrique; Figueiredo, Afonso

doi:10.5902/1980509828670

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Artigo • Ciênc. Florest. 27 (3) • Jul-Sep 2017 • https://doi.org/10.5902/1980509828670 copy

THE USE OF THE PARAMETER PREDICTION METHOD TO PROJECT THE DIAMETER DISTRIBUTION IN NATURAL FOREST WITH THE WEIBULL FUNCTION

Authorship SCIMAGO INSTITUTIONS RANKINGS

ABSTRACT

The objective of this study was to use the 3P (Three Parameters) Weibull Parameter Prediction method to predict the diameter distribution in a Mixed Ombrophilous forest fragment, comparing the results with predictions generated by the Transition Matrix and Movement Ratio methods. The data come from 100 permanent plots of 2,500 m² (50 m x 50 m) established in the National Forest of Irati, in Parana state, measured in the years of 2002, 2005, 2008 and 2011. For the Parameter Prediction method it has been fitted (stepwise) models for predictions for 3, 6 and 9 years, using as dependent variables, the 3P Weibull parameters b and c (a parameter was set at 10.02 cm corresponding to the minimum value of diameter measured) and, as the independent variables were tested N.ha^-1, G.ha^-1, d_g, d_min, percentile diameters (d_P), d_max and their transformations (inverse, logarithmic and quadratic). A survival model was adjusted to predict the future number of trees. The predictions obtained by the Parameter Prediction method for 2008 and 2011 have been compared with the predictions generated by the Transition Matrix and Movement Ratio methods using the Kolmogorov-Smirnov (KS) and Reynolds index (RI). The models were evaluated by the Schlaegel index (I.A.) and relative standard error of estimate in percentage (S_YX%). The results indicated good fits for the 3P Weibull Parameter Prediction method. In comparison with the Transition Matrix and Movement Ratio approach, this method showed the best results according to the KS test, but according to IR, the best performance was found by the Transition Matrix method. All three methods evaluated had adherence between observed and theoretical distributions according to the KS test.

Keywords:
transition matrix; movement ratio; diametric structure prediction

Variáveis dependentes	Variáveis Independentes	Variáveis independentes transformadas
b e c	G(ha), N(ha), d_g(cm), d_min(cm), d_P0,05(cm),d_P0,1(cm), d_P0,2(cm),d_P0,3(cm), d_P0,4(cm),d_P0,5(cm), d_P0,6(cm), d_P0,07(cm), d_P0,8(cm), d_P0,9(cm), d_P0,95(cm), d_max (cm)	lnG, 1/G, G², lnN, 1/N, N², lnd_g, 1/d_g, d_g², lnd_min, 1/d_min,d_min², lnd_P0,05, 1/d_P0,05, d_P0,05²,lnd_P0,1, 1/d_P0,1, d_P0,1² , lnd_P0,2, 1/d_P0,2, d_P0,2², lnd_P0,3, 1/d_P0,3, d_P0,3², lnd_P0,4, 1/d_P0,4, d_P0,4², lnd_P0,5, 1/d_P0,5, d_P0,5², lnd_P0,6, 1/d_P0,6, d_P0,6², lnd_P0,7, 1/d_P0,7, d_P0,7², lnd_P0,8, 1/d_P0,8, d_P0,8², lnd_P0,9; 1/d_P0,9, d_P0,9², lnd_P0,95, 1/d_P0,95, d_P0,95², lnd_max, 1/d_max, d_max².

Intervalo	Equações ajustadas	I.A.	S _YX%
3 anos	$\ln b = 6,9506 - 8,7158 ∙ (\frac{1}{{d p}_{0,7}}) - 22,0974 ∙ (\frac{1}{{d p}_{0,3}}) - 18,4190 ∙ (\frac{1}{d_{g}}) - 0,1681 ∙ \ln d_{m a x} - 0,2105 ∙ \ln {d p}_{0,95} - 0,0028 ∙ {d p}_{{0,3}^{2}} + 136,14 ∙ (\frac{1}{{N . h a}^{- 1}}) - 5,8566 ∙ (\frac{1}{{G . h a}^{- 1}}) + 0,0213 ∙ {d p}_{0,2}$	0,872	2,28
	$c = - 0,5684 + 0,1857 ∙ {d p}_{0,3} + 12,6799 ∙ (\frac{1}{{d p}_{0,95}}) + 13,9688 ∙ (\frac{1}{d_{m a x}}) - 3,8290 ∙ (\frac{1}{{d p}_{0,2}}) + 37,2261 ∙ (\frac{1}{{N . h a}^{- 1}}) + 6,2581 ∙ (\frac{1}{{d p}_{0,9}}) - 0,0050 ∙ {d p}_{{0,3}^{2}} - 4,6385 ∙ (\frac{1}{{d p}_{0,05}})$	0,769	6,02
	$N_{2} = 11,67047 + 0,96808 ∙ N_{1}$	0,971	3,66
6 anos	$\ln b = 12,5384 - 12,6299 ∙ (\frac{1}{{d p}_{0,7}}) - 60,9325 ∙ (\frac{1}{{d p}_{0,3}}) + 71,3996 ∙ (\frac{1}{{d p}_{0,5}}) + 44,7529 ∙ (\frac{1}{{N . h a}^{- 1}}) - 3,5269 ∙ \ln {d p}_{0,3} + 0,000001 ∙ {({N . h a}^{- 1})}^{2} - 0,7328 ∙ (\frac{1}{{d p}_{0,8}}) - 11,0032 ∙ (\frac{1}{{d p}_{0,6}}) - 0,0004 ∙ {d p}_{{0,6}^{2}} - 0,0009 ∙ {d p}_{{0,95}^{2}} + 0,1170 ∙ {d p}_{0,95} - 0,0008 ∙ {N . h a}^{- 1} - 3,5323 ∙ (\frac{1}{{d p}_{0,9}})$	0,832	2,51
	$\ln c = 2,4450 + 0,8378 ∙ \ln {d p}_{0,3} + 21,7439 ∙ (\frac{1}{d_{g}}) - 7,1688 ∙ (\frac{1}{{d p}_{0,7}}) + 2,6663 ∙ (\frac{1}{{G . h a}^{- 1}}) - 40,7430 ∙ (\frac{1}{{d p}_{0,05}}) - 0,0133 ∙ {d p}_{{0,05}^{2}}$	0,767	6,71
	$N_{2} = 15,07460 + 0,95256 ∙ N_{1}$	0,951	4,89
9 anos	$\ln b = 21,0212 - 3,8375 ∙ (\frac{1}{{d p}_{0,5}}) - 11,0769 ∙ (\frac{1}{{d p}_{0,7}}) - 6,2253 ∙ (\frac{1}{{d p}_{0,1}}) - 7,3508 ∙ (\frac{1}{{d p}_{0,9}}) - 71,0906 ∙ (\frac{1}{{d p}_{0,3}}) - 4,5623 ∙ \ln {d p}_{0,3} + 42,3126 ∙ (\frac{1}{{N . h a}^{- 1}})$	0,786	2,66
	$c = 0,0110 + 10,6306 ∙ (\frac{1}{d_{m a x}}) - 4,7234 ∙ (\frac{1}{{d p}_{0,3}}) + 30,2111 ∙ (\frac{1}{{d p}_{0,95}}) - 7,3469 ∙ (\frac{1}{{d p}_{0,05}}) - 0,0042 ∙ {d p}_{0,8} - 0,0001 ∙ {({G . h a}^{- 1})}^{2} - 5,80832 ∙ (\frac{1}{{d p}_{0,4}}) + 0,0002 ∙ {N . h a}^{- 1} + 0,4452 ∙ \ln {d p}_{0,95}$	0,748	5,54
	$N_{2} = 31,1493 + 0,9121 ∙ N_{1}$	0,945	5,19

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E-mail: cienciaflorestal@ufsm.br

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[TFN1] Em que: b e c = parâmetros da função Weibull; G = área basal; N = número de árvores; dg = diâmetro quadrático; dmin = diâmetro mínimo; dP = diâmetro percentil; dmax = diâmetro máximo; ln = logaritmo neperiano.