Mathematical models to estimate leaf area of melon plant cultivar Hale’s Best Jumbo at different phenological stages using linear measurements

• Main Authors: R. A. F. Machado, M. R. Zanuzo, C. V. Vieira, R. S. Boeno
• Format: Article
• Language: English
• Published: Universidade Federal de Rondonópolis 2014-07-01
• Series: Scientific Electronic Archives
• Subjects: cucumis melo, growth analysis, leaf area integrator, nondestructive method, linear regression
• Online Access: https://sea.ufr.edu.br/index.php?journal=SEA&page=article&op=view&path=120
• ISSN: 2316-9281; 2316-9281

Melon plants cv. Hale's Best Jumbo Cantaloupe were cultivated under field conditions in Sinop - MT in order to develop mathematical models based on linear measurements which allow the estimation of leaves area. Were collected four times leaves, first collection at 20 days after sowing (DAS) with 45 leaves, second collection at 30 DAS with 40 leaves, third collection was performed 40 DAS with 40 leaves was performed 50 days with 33 sheets. 45 leaves in the first test at 30 days after sowing (DAS), 40 sheets in the 2nd collection at 40 DAS, 40 leaves on the 3rd collection at 50 DAS and 33 leaves on the 4th collection at 60 DAS. Linear measurements were taken along the length of leaf central vein (C) and the largest width (W) perpendicular to the central vein. At each time was determined leaf area (LA) using a LI-COR LI-3100 area integrator. The data length and width and and product of C by L were considered the independent variable (x) and real leaf area (AF) as dependent variable (y). Were performed linear regression analyzes for each of the seasons and the length and width of the sheet and the product CL. From the results it is concluded that: a) the length of the central rib does not allow to accurately estimate the leaf area of melon, b) the greatest width of the sheet provides a better estimative of leaf area, c) the phenological stage 3 provides the best estimate of the foliar area and d) the model that best estimates leaf area of melon is Y = 10.284 x - 16.404 value of r = 0.8627.