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Table IV
Review of the literature about the relationships between wood density and Hounsfield numbers obtained with medical CT scanners.
Wood density | |||||||||||
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Reference | Species | Moisture content | (kg m−3) | Slope | Intercept | n | R 2 | Scanner | kVp | mA | |
Radiata pine, Douglas fir, | |||||||||||
Benson-Cooper | Eucalyptus delegatensis | Technicare | 120 | 40–200 | |||||||
et al. (1982) | Tasmanian blackwood, Red beech | Green wood | ~500–1150 | 0.910 | 1002 | 40 | 0.92 | Delta2020 | (during 2 s) | ||
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Pine, Mahogany, Poplar, | Ohio-Nuclear | 50 | |||||||||
Mull (1984) | Maple, Ash, Teak | 367–798 | 7 | 2010 CT | 120 | ||||||
(+ graphite) | scanner | (during 4 s) | |||||||||
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2–96% | ~600–1000 | (a) | 0.876 | 968 | 35 | 0.98 | |||||
Hattori and | Red meranti | (b) | 1.002 | 1018 | 35 | 0.98 | |||||
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Kanagawa | Oven-dry wood | ~550–620 | (b) | 0.973 | 1040 | 35 | 0.74 | Toshiba | 120 | 230 | |
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(1985)1 | 2–27% | ~430–500 | (b) | 0.894 | 966 | 35 | 0.94 | TCT-20A | |||
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Agathis | (a) | 0.855 | 948 | 35 | |||||||
Oven-dry wood | ~425–450 | (b) | 0.894 | 966 | 35 | ||||||
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Lindgren | 6–117% | ~350–1000 | 0.993 | 1015 | 50 | GE 9800 | |||||
(1991b) | Oven-dry wood | 352–619 | 1.052 | 1053 | 11 | Quick | |||||
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Balsa, Treated pine, Radiata | |||||||||||
pine, Oregon, Meranti, | |||||||||||
Davis and | Cypress pine, Merbau, | ||||||||||
Wells (1992) | NZ pencil pine, Red gum | ~100–1100 | 1.006 | 1035 | 13 | EMI CT1010 | 120 | 20–30 | |||
Jarrah, Gray box, Red box, | |||||||||||
Red iron bark | |||||||||||
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Taylor | Air-dried wood | Medical CT | |||||||||
(2006) | West African hardwood | (~12%) | ~350–1100 | 1.044 | 1044 | 25 | 0.986 | scanner |
(a) Without granulated sugar surrounding the wood samples. (b) With granulated sugar surrounding the wood samples.
1
At the opposite of other authors, Hattori and Kanagawa modelled the Hounsfield numbers as a function of density, i.e. H = a′ ρ + b′ instead of ρ = aH + b. We computed the parameters a and b, presented in Table , from their reported values of a′ and b′ as: a = 1 / a′ and b = − b′ / a′. It is slightly different from what would have been obtained by estimating directly a and b parameters by regressions performed on the raw data. In this case, the slope parameter would have been equal to R2 × 1 / a′.