No, this hasn’t been published yet. There is a little bit about this in Schütz and Pommerening (2013), see my reference list on pommerening.org.

]]>Thank you for your words of appreciation. Yes, it is correct, individuals of different species exert different competition signals but also handle the same signals but from different species in different ways. There are a number of approaches to this and would take too long to go into detail here. In ecosystems with many species it may be useful to go by trait groups.

]]>Any growth function can be used to model the growth processes in your forest.

]]>For MAI you just need to fit the growth function and divide any of the values of the growth function by t.

]]>The third parameter p is often not easy to estimate and different starting values lead to different end results. Therefore some researchers have set p to 2 or 3 based on von Bertalanffy’s theory.

]]>dweibull3 <- function(x, gamma, beta, alpha) {

(gamma/beta) * ((x -alpha)/beta)^(gamma – 1) * (exp(-((x – alpha)/beta)^gamma))

}

loss.w3 <- function(dia, p3)

sum(log(dweibull3(data, p[1], p[2], p[3])))

mle.w3.nm <- optim(c(gamma = 1, beta = 5, alpha = 10),

loss.w3, data = p4$dia, hessian = TRUE,

control = list(fnscale = -1))

> dweibull3

> loss.w3

> mle.w3.nm <- optim(c(gamma = 1, beta = 5, alpha = 10),

+ loss.w3, data = p4$dia, hessian = TRUE,

+ control = list(fnscale = -1))

Error in fn(par, …) : unused argument (data = p4$dia)

You mentioned that the empirical parameter p it is often restricted to a value of three for theoretical, biological reasons. I couldn’t find any information related. Could you tell me where what are the reasons or where I can find information. ]]>

I trust this email finds you well. I am Jerry a master student with Kyushu University, Japan. I am trying to use the Gompertz function to estimate CAI and MAI with my data. I have been able to estimate the CAI but challenged with the MAI.

Principle: I used the first derivative to estimate the CAI. With the MAI, I divide the function by t, equating it to the first derivative (CAI), {implying tbexp(a-bt)=1}

Gompertz Function: y=Aexp(-exp(a-bt))

From the first derivative, CAI (t1) =a/b

I get MAI value less than t1 always.

Am I doing something wrong?