A morphological and geometric method for estimating the selectivity of gill nets

Abstract

We propose a new method for estimating gill net selectivity which estimates the probabilities leading to retention by analyzing both the fish morphology and the mesh geometry. This method estimates the number of fish approaching and contacting gill nets of different mesh sizes as an intermediate step towards computing the selectivity. Instead of assuming an underlying probability distribution as in indirect methods, we split the entire interaction between a fish and the gill net into several stages, each with its own probability. All the necessary parameters to compute these probabilities can be obtained from measurements of the fish, knowledge of the mesh geometry, and catch data from different mesh sizes. The framework offers three pathways for computing the total number of fish contacting the gill nets and has the capability to use both wedged and entangled fish in the analysis. As a proof of concept, the method is applied to catch data for cod (G. morhua) and Dolly Varden (S. malma) to estimate the number of fish contacting the gill nets in both cases. By estimating the number of fish contacting the gill net in addition to the selectivity, this method provides an important step towards deriving estimates of fish density in a particular fishery from gill net measurement.

Appendices

Appendix 1

If the saturation curve is described by function (17), the reduction of vacant net area is described by the following function:

$$f(t)^{*} = \exp \left( { - \frac{t}{{\tilde{\tau }}}} \right).$$

Let the number of fish caught by the net be proportional to the net area where fish are retained, and let $t = \tilde{\tau }$. Then, $\exp \left( { - \frac{{t = \tilde{\tau }}}{{\tilde{\tau }}}} \right) = \exp \left( { - 1} \right) \approx 0.368$. Next, we compare the vacant net area at time $\tilde{\tau }$ (its share is 0.368 from the total space) with the net area at the beginning of fishing where there are no fish in the net (share is 1). Then, 1/0.368 ≈ 2.717 ≈ e.

Appendix 2

According to “Appendix 1”, if $t = \tilde{\tau }$, then $N_{{AP,\tilde{\tau }}} = Q_{{\tilde{\tau }}} \cdot e + SL_{{\tilde{\tau }}}$, where $Q_{{\tilde{\tau }}} \cdot e = Q_{{\tilde{\tau }}} + B_{{\tilde{\tau }}}$. Then, for 1 h of fishing $N_{AP,1} = \frac{{Q_{{\tilde{\tau }}} \cdot e + SL_{{\tilde{\tau }}} }}{{\tilde{\tau }}}$, where $Q_{{\tilde{\tau }}} = N_{\lim } \left( {1 - \exp \left( { - \frac{{\tilde{\tau }}}{{\tilde{\tau }}}} \right)} \right) \approx N_{\lim } \cdot 0.63$. Hence, $\frac{{N_{\lim } \cdot 0.63 \cdot e + SL_{{\tilde{\tau }}} }}{{\tilde{\tau }}} \approx \frac{{N_{\lim } \cdot 0.63 \cdot 2.72 + SL_{{\tilde{\tau }}} }}{{\tilde{\tau }}} \approx \frac{{N_{\lim } \cdot 1.71 + SL_{{\tilde{\tau }}} }}{{\tilde{\tau }}}$—that is the angular coefficient k of $f(t) = kt$ (Eq. 18).