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Effects of hysteresis of static contact angle (HSCA) and boundary slip on the hydrodynamics of water striders

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Abstract

It is known that contact lines keep relatively still on solids until static contact angles exceed an interval of hysteresis of static contact angle (HSCA), and contact angles keep changing as contact lines relatively slide on the solid. Here, the effects of HSCA and boundary slip were first distinguished on the micro-curvature force (MCF) on the seta. Hence, the total MCF is partitioned into static and dynamic MCFs correspondingly. The static MCF was found proportional to the HSCA and related with the asymmetry of the micro-meniscus near the seta. The dynamic MCF, exerting on the relatively sliding contact line, is aroused by the boundary slip. Based on the Blake–Haynes mechanism, the dynamic MCF was proved important for water walking insects with legs slower than the minimum wave speed \(23\,\hbox {cm}\cdot \hbox {s}^{-1}\). As insects brush the water by laterally swinging legs backwards, setae on the front side of the leg are pulled and the ones on the back side are pushed to cooperatively propel bodies forward. If they pierce the water surface by vertically swinging legs downwards, setae on the upside of the legs are pulled, and the ones on the downside are pushed to cooperatively obtain a jumping force. Based on the dependency between the slip length and shear rate, the dynamic MCF was found correlated with the leg speed U, as \(F\sim C_{1}U+C_{2} U^{2+\varepsilon }\), where \(C_{1}\) and \(C_{2}\) are determined by the dimple depth. Discrete points on this curve could give fitted relations as \(F\sim U^{b}\) (Suter et al., J. Exp. Biol. 200, 2523–2538, 1997). Finally, the axial torque on the inclined and partially submerged seta was found determined by the surface tension, contact angle, HSCA, seta width, and tilt angle. The torque direction coincides with the orientation of the spiral grooves of the seta, which encourages us to surmise it is a mechanical incentive for the formation of the spiral morphology of the setae of water striders.

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Acknowledgements

The project was supported by the National Natural Science Foundation of China (Grant 11502097), the Nature and Science Foundation of Jiangsu Province (Grant BK20130478), and the Foundation of Senior Talent of Jiangsu University (Grant 1281130025). We thank the profound questions of the reviewers of this paper. Speculations on their questions further enhanced the quality and completeness of this paper.

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Correspondence to J. Zheng.

Appendix

Appendix

In CSYS XsYsZs shown in Fig. 3b, the surface of the seta could be expressed using

$$\begin{aligned} x_s^2 =a^{2}-y_s^2 ,\hbox { }y_s =y_s ,\hbox { }z_s =z_s, \end{aligned}$$
(A1)

where a is the radius of the seta. Therefore, the tangent of the seta could be obtained

$$\begin{aligned} {{{\varvec{l}}}}_{b1} =\left( {\frac{\hbox {d}x_s }{\hbox {d}y_s },\frac{\hbox {d}y_s }{\hbox {d}y_s },\frac{\hbox {d}z_s }{\hbox {d}y_s }} \right) =\left( {-\frac{y_s }{x_s },1,0} \right) \end{aligned}$$

or

$$\begin{aligned} {{{\varvec{l}}}}_{b2} =(-y_s ,x_s ,0), \end{aligned}$$

then the unit vector of the tangent is

$$\begin{aligned} {{{\varvec{e}}}}_b ={{{{\varvec{l}}}}_{b2} }/{l_{b2} }={(-y_s ,x_s ,0)}/{\sqrt{x_s^2 +y_s^2 }}={(-y_s ,x_s ,0)}/a\nonumber \\ \end{aligned}$$
(A2a)

or

$$\begin{aligned} {{{\varvec{e}}}}_b ={(-y_s ,x_s ,0)}/a=(-\sin \varphi ,\cos \varphi ,0), \end{aligned}$$
(A2b)

where angle \(\varphi \) is shown in Fig. 3b. In CSYS XsYsZs, the contact line is the intersection line of WA interface and the seta. The contact line c is in a cutting plane which is approximately parallel to the WA interface, and the WA interface is approximately parallel to the tangent plane of the leg as Figs. 1b and 2a schematically show. Therefore, the cutting plane or WA interface is inclined with the axis of the seta as Fig. 3a shows. The equation of the cutting plane or WA interface could be determined by its normal and its center. The normal of the cutting plane is easier given

$$\begin{aligned} {{{\varvec{n}}}}_c= & {} (0,-\sin (90^{\circ }-\alpha ),\cos (90^{\circ }-\alpha ))\\= & {} (0,-\cos \alpha ,\sin \alpha ). \end{aligned}$$

If we set the origin of the CSYS XsYsZs on the center of this cutting plan, then the vector in the cutting plane is \({{\varvec{r}}}=(x_{{s}} ,y_\mathrm{{s}} ,z_{{s}})\), so that the cutting plane is given by

$$\begin{aligned} {{{\varvec{r}}}}\cdot {{{\varvec{n}}}}_c =-y_s \cos \alpha +z_s \sin \alpha =0. \end{aligned}$$
(A3)

Then the contact line is determined by the surface of seta (Eq. (A1)) and cutting plane (Eq. (A3)):

$$\begin{aligned} x_s^2 =a^{2}-y_s^2 ,\hbox { }y_s =y_s ,\hbox { }z_s =y_s \cot \alpha . \end{aligned}$$
(A4)

Then we can obtain the tangent of the contact line c

$$\begin{aligned} {{{\varvec{l}}}}_{c1} =\left( {\frac{\hbox {d}x_s }{\hbox {d}y_s },\frac{\hbox {d}y_s }{\hbox {d}y_s },\frac{\hbox {d}z_s }{\hbox {d}y_s }} \right) =\left( {-\frac{y_s }{x_s },1,\cot \alpha } \right) \end{aligned}$$

or

$$\begin{aligned} {{{\varvec{l}}}}_{c2} =(-y_s ,x_s ,x_s \cot \alpha ), \end{aligned}$$

then the unit vector of the tangent is obtained

$$\begin{aligned} {{{\varvec{e}}}}_c= & {} \frac{1}{\sqrt{x_s^2 +y_s^2 +x_s^2 \cot ^{2}\alpha }}(-y_s ,x_s ,x_s \cot \alpha )\nonumber \\= & {} \frac{1}{\sqrt{a^{2}+x_s^2 \cot ^{2}\alpha }}(-y_s ,x_s ,x_s \cot \alpha ). \end{aligned}$$
(A5)

Then the angle \(\phi \) shown in Figs. 3b or 4a is determined by

$$\begin{aligned} \cos \phi= & {} {{{\varvec{e}}}}_c \cdot {{{\varvec{e}}}}_b\nonumber \\= & {} \frac{1}{\sqrt{a^{2}+x_s^2 \cot ^{2}\alpha }}(-y_s ,x_s ,x_s \cot \alpha )\cdot \frac{1}{a}(-y_s ,x_s ,0)\nonumber \\= & {} \frac{a}{\sqrt{a^{2}+x_s^2 \cot ^{2}\alpha }}>0. \end{aligned}$$
(A6)

Then

$$\begin{aligned} \sin \phi =\pm \sqrt{1-\cos \phi }=\pm \frac{|x_s |\cot \alpha }{\sqrt{a^{2}+x_s^2 \cot ^{2}\alpha }}. \end{aligned}$$

It could be easily observed in Fig. 3b that \(\phi >0\), where \(x_{{s}}>0\) and \(\phi <0\), where \(x_{{s}}<0\), therefore it is convenient to write

$$\begin{aligned} \sin \phi =\frac{x_s \cot \alpha }{\sqrt{a^{2}+x_s^2 \cot ^{2}\alpha }}. \end{aligned}$$
(A7)

The element of the contact line c could be given

$$\begin{aligned} \mathrm{d}l= & {} \sqrt{\hbox {d}x_s^2 +\hbox {d}y_s^2 +\hbox {d}z_s^2 }\nonumber \\= & {} |\hbox {d}y_s |\sqrt{\left( {\frac{\hbox {d}x_s }{\hbox {d}y_s }} \right) ^{2}+1+\left( {\frac{\hbox {d}z_s }{\hbox {d}y_s }} \right) ^{2}}, \end{aligned}$$
(A8)

using Eq. (A4), we have

$$\begin{aligned} \mathrm{d}l= & {} |\hbox {d}y_s |\sqrt{\left( {\frac{y_s }{x_s }} \right) ^{2}+1+\cot ^{2}\alpha }\nonumber \\= & {} \frac{|\hbox {d}y_s |}{|x_s |}\sqrt{y_s^2 +x_s^2 +x_s^2 \cot ^{2}\alpha }\nonumber \\= & {} \frac{|\hbox {d}y_s |}{|x_s |}\sqrt{a^{2}+x_s^2 \cot ^{2}\alpha }. \end{aligned}$$
(A9)

Now we can calculate the integration

$$\begin{aligned}&\int _{c_a } {\sin \phi {{{\varvec{e}}}}_b \hbox {d}l}\nonumber \\&\quad =\int _{c_a } {\left( {\frac{x_s \cot \alpha }{\sqrt{a^{2}+x_s^2 \cot ^{2}\alpha }}{{{\varvec{e}}}}_b \frac{|\mathrm{d}y_s |}{|x_s |}\sqrt{a^{2}+x_s^2 \cot ^{2}\alpha }} \right) }\nonumber \\&\quad =\cot \alpha \int _{c_a } {\left( {\frac{x_s }{|x_s |}{{{\varvec{e}}}}_b |\hbox {d}y_s |} \right) } . \end{aligned}$$
(A10)

On the advancing side, in Fig. 3b, we can see \(x_{{s}}>0\) and \(y_{{s}}\) grows in the anticlockwise direction of the contact line c, therefore

$$\begin{aligned} \int _{c_a } {\sin \phi {{{\varvec{e}}}}_b \mathrm{d}l}= & {} \cot \alpha \int _{c_a } {{{{\varvec{e}}}}_b \hbox {d}y_s } =\cot \alpha \int _{-\frac{\uppi }{2}}^{\frac{\pi }{2}} {{{{\varvec{e}}}}_b \hbox {d}a\sin \varphi } \\= & {} a\cot \alpha \int _{-\frac{\uppi }{2}}^{\frac{\uppi }{2}} {{{{\varvec{e}}}}_b \cos \varphi \hbox {d}\varphi } . \end{aligned}$$

Substituting Eq. (A2b) into the above equation, we have

$$\begin{aligned} \int _{c_a } {\sin \phi {{{\varvec{e}}}}_b \mathrm{d}l}= & {} a\cot \alpha \int _{-\frac{\uppi }{2}}^{\frac{\uppi }{2}} {{{{\varvec{e}}}}_b \cos \varphi \hbox {d}\varphi }\\= & {} a\cot \alpha \int _{-\frac{\uppi }{2}}^{\frac{\uppi }{2}} {(-{{{\varvec{i}}}}_s \sin \varphi +{{{\varvec{j}}}}_s \cos \varphi )\cos \varphi \hbox {d}\varphi } , \end{aligned}$$

which gives

$$\begin{aligned} \int _{c_a } {\sin \phi {{{\varvec{e}}}}_b \mathrm{d}l}= & {} -{{{\varvec{i}}}}_s a\cot \alpha \int _{-\frac{\uppi }{2}}^{\frac{\uppi }{2}} {\sin \varphi \cos \varphi \hbox {d}\varphi }\\&+\,a{{{\varvec{j}}}}_s \cot \alpha \int _{-\frac{\uppi }{2}}^{\frac{\uppi }{2}} {\cos \varphi \cos \varphi \hbox {d}\varphi } \\= & {} a{{{\varvec{j}}}}_s \cot \alpha \int _{-\frac{\uppi }{2}}^{\frac{\uppi }{2}} {\frac{\cos 2\varphi +1}{2}\hbox {d}\varphi } =\frac{\uppi }{2}a\cot \alpha {{{\varvec{j}}}}_s, \end{aligned}$$

i.e.,

$$\begin{aligned} \int _{c_a } {\sin \phi {{{\varvec{e}}}}_b \mathrm{d}l} =\frac{\uppi }{2}a\cot \alpha {{{\varvec{j}}}}_s =\frac{\uppi }{4}d\cot \alpha {{{\varvec{j}}}}_s . \end{aligned}$$
(A11)

Similarly we can calculate the integration on the retreating side of the contact line c

$$\begin{aligned} \int _{c_r } {\sin \phi {{{\varvec{e}}}}_b \mathrm{d}l}= & {} \cot \alpha \int _{c_r } {\frac{x_s }{|x_s |}{{{\varvec{e}}}}_b |\hbox {d}y_s |} \nonumber \\= & {} -\cot \alpha \int _{\frac{\uppi }{2}}^{\frac{3\uppi }{2}} {{{{\varvec{e}}}}_b |a\cos \varphi \hbox {d}\varphi |}\nonumber \\= & {} a\cot \alpha \int _{\frac{\uppi }{2}}^{\frac{3\uppi }{2}} {{{{\varvec{e}}}}_b \cos \varphi \hbox {d}\varphi } . \end{aligned}$$

Obviously, the integration also gives

$$\begin{aligned} \int _{c_r } {\sin \phi {{{\varvec{e}}}}_b \mathrm{d}l}= & {} a\cot \alpha \int _{\frac{\uppi }{2}}^{\frac{3\uppi }{2}} {{{{\varvec{e}}}}_b \cos \varphi \hbox {d}\varphi } =\frac{\uppi }{2}a\cot \alpha {{{\varvec{j}}}}_s\nonumber \\= & {} \frac{\uppi }{4}d\cot \alpha {{{\varvec{j}}}}_s . \end{aligned}$$
(A12)

So that

$$\begin{aligned} \oint _c {\sin \phi {{{\varvec{e}}}}_b \mathrm{d}l}= & {} \int _{c_a } {\sin \phi {{{\varvec{e}}}}_b \mathrm{d}l} +\int _{c_r } {\sin \phi {{{\varvec{e}}}}_b \mathrm{d}l} =\uppi a\cot \alpha {{{\varvec{j}}}}_s\nonumber \\= & {} \frac{\uppi d}{2}\cot \alpha {{{\varvec{j}}}}_s . \end{aligned}$$
(A13)

The integration in Eq. (37) is simpler for calculation. Obviously,

$$\begin{aligned} {{{\varvec{n}}}}_b =\cos \varphi {{{\varvec{i}}}}_s +\sin \varphi {{{\varvec{j}}}}_s . \end{aligned}$$
(A14)

Hence, the integration

$$\begin{aligned} \int _{c_a } {{{{\varvec{n}}}}_b \mathrm{d}l}= & {} \int _{c_a } {\left( {{{{\varvec{n}}}}_b \frac{|\hbox {d}y_s |}{|x_s |}\sqrt{a^{2}+x_s^2 \cot ^{2}\alpha }} \right) }\\= & {} \int _{c_a } {\left( {{{{\varvec{n}}}}_b \frac{|a\cos \varphi \hbox {d}\varphi |}{|a\cos \varphi |}\sqrt{a^{2}+x_s^2 \cot ^{2}\alpha }} \right) } \\= & {} \int _{c_a } {{{\varvec{n}}}}_b \sqrt{a^{2}+x_s^2 \cot ^{2}\alpha }\hbox {d}\varphi \\= & {} \int _{-\frac{\uppi }{2}}^{\frac{\uppi }{2}} {{{{\varvec{n}}}}_b \sqrt{a^{2}+a^{2}\cos ^{2}\varphi \cot ^{2}\alpha }\hbox {d}\varphi } . \end{aligned}$$

Substituting Eq. (A14) into the above equation, we have

$$\begin{aligned} \int _{c_a} {{{{\varvec{n}}}}_b \mathrm{d}l}&={{{\varvec{i}}}}_s \int _{-\frac{\uppi }{2}}^{\frac{\uppi }{2}} {\cos \varphi \sqrt{a^{2}+a^{2}\cos ^{2}\varphi \cot ^{2}\alpha }\hbox {d}\varphi }\\&\quad +\,{{{\varvec{j}}}}_s \int _{-\frac{\uppi }{2}}^{\frac{\uppi }{2}} {\sin \varphi \sqrt{a^{2}+a^{2}\cos ^{2}\varphi \cot ^{2}\alpha }\hbox {d}\varphi }. \end{aligned}$$

Obviously, because the integrand of the second term in the RHS is an odd function, therefore, only the first term is kept, then

$$\begin{aligned} \int _{c_a } {{{{\varvec{n}}}}_b \mathrm{d}l} ={{{\varvec{i}}}}_s \int _{-\frac{\uppi }{2}}^{\frac{\uppi }{2}} {\sqrt{a^{2}+a^{2}(1-\sin ^{2}\varphi )\cot ^{2}\alpha }\hbox {d}\sin \varphi } . \end{aligned}$$

Using variable replacement

$$\begin{aligned} q=\sin \varphi , \end{aligned}$$

we have

$$\begin{aligned} \int _{c_a } {{{{\varvec{n}}}}_b \mathrm{d}l} ={{{\varvec{i}}}}_s \int _{-1}^1 {\sqrt{a^{2}+a^{2}\cot ^{2}\alpha -a^{2}\cot ^{2}\alpha q^{2}}\hbox {d}q} , \end{aligned}$$
(A15)

which could be transformed to

$$\begin{aligned}&\int _{c_a } {{{{\varvec{n}}}}_b \mathrm{d}l} ={{{\varvec{i}}}}_s \sqrt{a^{2}+a^{2}\cot ^{2}\alpha }\\&\quad \times \, \sqrt{\frac{a^{2}+a^{2}\cot ^{2}\alpha }{a^{2}\cot ^{2}\alpha }} \\&\quad \times \int _{-1}^1{\sqrt{1-\left( {\sqrt{\frac{a^{2}\cot ^{2}\alpha }{a^{2}+a^{2}\cot ^{2}\alpha }}q} \right) ^{2}}\hbox {d}\left( {\sqrt{\frac{a^{2}\cot ^{2}\alpha }{a^{2}+a^{2}\cot ^{2}\alpha }}q} \right) } . \end{aligned}$$

Then we have

$$\begin{aligned}&\int _{c_a } {{{{\varvec{n}}}}_b \mathrm{d}l} ={{{\varvec{i}}}}_s a\frac{1+\cot ^{2}\alpha }{\cot \alpha }\int _{-1}^1 \sqrt{1-\left( {\sqrt{\frac{a^{2}\cot ^{2}\alpha }{a^{2}+a^{2}\cot ^{2}\alpha }}q} \right) ^{2}}\\&\quad \times \,\hbox {d}\,\sqrt{\frac{a^{2}\cot ^{2}\alpha }{a^{2}+a^{2}\cot ^{2}\alpha }}q . \end{aligned}$$

Using another variable replacement

$$\begin{aligned} \tau =\sqrt{\frac{a^{2}\cot ^{2}\alpha }{a^{2}+a^{2}\cot ^{2}\alpha }}q, \end{aligned}$$

we have

$$\begin{aligned} \int _{c_a } {{{{\varvec{n}}}}_b \mathrm{d}l} ={{{\varvec{i}}}}_s a\frac{1+\cot ^{2}\alpha }{\cot \alpha }\int _{\tau _1 }^{\tau _2 } {\sqrt{1-\tau ^{2}}\hbox {d}\tau }, \end{aligned}$$
(A16)

where

$$\begin{aligned} \tau _1= & {} -\sqrt{\frac{a^{2}\cot ^{2}\alpha }{a^{2}+a^{2}\cot ^{2}\alpha }}, \end{aligned}$$
(A17a)
$$\begin{aligned} \tau _2= & {} \sqrt{\frac{a^{2}\cot ^{2}\alpha }{a^{2}+a^{2}\cot ^{2}\alpha }}. \end{aligned}$$
(A17b)

Using the variable replacement

$$\begin{aligned} \tau =\sin \delta , \end{aligned}$$

we have the result of the integral in Eq. (A16)

$$\begin{aligned} \int _{c_a } {{{{\varvec{n}}}}_b \mathrm{d}l}= & {} {{{\varvec{i}}}}_s a\frac{1+\cot ^{2}\alpha }{\cot \alpha }\int _{-\varDelta }^{+\varDelta } {\cos \delta \cos \delta \hbox {d}\delta } \nonumber \\= & {} {{{\varvec{i}}}}_s \frac{d}{2}\frac{1+\cot ^{2}\alpha }{\cot \alpha }(\varDelta +\sin 2\varDelta ), \end{aligned}$$
(A18)

where

$$\begin{aligned} \varDelta= & {} \hbox {asin} \sqrt{\frac{a^{2}\cot ^{2}\alpha }{a^{2}+a^{2}\cot ^{2}\alpha }}\nonumber \\= & {} \hbox {asin} \sqrt{\frac{\cot ^{2}\alpha }{1+\cot ^{2}\alpha }}=\hbox {a sin} (\cos \alpha )=\frac{\uppi }{2}-\alpha , \end{aligned}$$
(A19)

so that

$$\begin{aligned} \int _{c_a } {{{{\varvec{n}}}}_b \mathrm{d}l} ={{{\varvec{i}}}}_s \frac{d}{2}\frac{1+\cot ^{2}\alpha }{\cot \alpha }\left( \frac{\uppi }{2}-\alpha +\sin 2\alpha \right) . \end{aligned}$$
(A20)

Similarly, we can have

$$\begin{aligned} \int _{c_r } {{{{\varvec{n}}}}_b \mathrm{d}l} =-{{{\varvec{i}}}}_s \frac{d}{2}\frac{1+\cot ^{2}\alpha }{\cot \alpha }\left( \frac{\uppi }{2}-\alpha +\sin 2\alpha \right) , \end{aligned}$$
(A21)

so that

$$\begin{aligned} \oint _c {{{{\varvec{n}}}}_b \mathrm{d}l} =\int _{c_r } {{{{\varvec{n}}}}_b \mathrm{d}l} +\int _{c_a } {{{{\varvec{n}}}}_b \mathrm{d}l} =0. \end{aligned}$$
(A22)

When \(\alpha \) approaches to \(90^{\circ }\), we have

$$\begin{aligned}&\mathop {\lim }\limits _{\alpha \rightarrow \frac{\uppi }{2}} \int _{c_a} {{{{\varvec{n}}}}_b \mathrm{d}l} \nonumber \\&\quad ={{{\varvec{i}}}}_s \mathop {\lim }\limits _{\alpha \rightarrow \frac{\uppi }{2}} \frac{d}{2}\frac{1+\cot ^{2}\alpha }{\cot \alpha }\left( \frac{\uppi }{2}-\alpha +\sin 2\alpha \right) \nonumber \\&\quad ={{{\varvec{i}}}}_s \frac{d}{2}\mathop {\lim }\limits _{\alpha \rightarrow \frac{\uppi }{2}} \frac{1}{\sin \alpha \cos \alpha }\left( \frac{\uppi }{2}-\alpha +\sin 2\alpha \right) \nonumber \\&\quad ={{{\varvec{i}}}}_s \frac{d}{2}\mathop {\lim }\limits _{\alpha \rightarrow \frac{\uppi }{2}} \frac{2}{\sin 2\alpha }\left( \frac{\uppi }{2}-\alpha +\sin 2\alpha \right) \nonumber \\&\quad ={{{\varvec{i}}}}_s \frac{d}{2}\mathop {\lim }\limits _{\alpha \rightarrow \frac{\uppi }{2}} \frac{1}{\sin 2\alpha }(\uppi -2\alpha +2\sin 2\alpha ) \nonumber \\&\quad ={{{\varvec{i}}}}_s \frac{d}{2}\mathop {\lim }\limits _{\alpha \rightarrow \frac{\uppi }{2}} \left( \frac{\uppi -2\alpha }{\sin 2\alpha }+2\right) \nonumber \\&\quad ={{{\varvec{i}}}}_s \frac{d}{2}\left( {\mathop {\lim }\limits _{\alpha \rightarrow \frac{\uppi }{2}} \frac{\uppi -2\alpha }{\sin 2\alpha }+2} \right) \nonumber \\&\quad ={{{\varvec{i}}}}_s \frac{d}{2}\left( {\mathop {\lim }\limits _{\alpha \rightarrow \frac{\uppi }{2}} \frac{-2}{2\cos 2\alpha }+2} \right) ={{{\varvec{i}}}}_s \frac{3d}{2}. \end{aligned}$$
(A23)

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Zheng, J., Wang, B.S., Chen, W.Q. et al. Effects of hysteresis of static contact angle (HSCA) and boundary slip on the hydrodynamics of water striders. Acta Mech. Sin. 33, 40–61 (2017). https://doi.org/10.1007/s10409-016-0620-0

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  • DOI: https://doi.org/10.1007/s10409-016-0620-0

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