2.1. Vertical cylinder

We first calculate the static equilibrium position of a small-scale cylindrical body floating with its axis perpendicular to the air–water interface. Static equilibrium refers to the state when there is no net force and torque acting on the body. In other words, the cylinder is statically floating at the air–liquid interface. Figure 2 shows the schematic of a small-scale cylinder floating in the vertical orientation at the air–liquid interface. Under the conditions of static equilibrium, the weight of the floating cylinder $F_w = {\rm \pi}R^2 l \rho _s g$, is balanced by the sum of the buoyancy force $F_b = ({\rm \pi} R^2 h \rho _l g + {\rm \pi}R^2 h_c \rho _l g)$ and the vertical component of the surface tension force $F_{st} = 2{\rm \pi} R \gamma \sin (\phi )$, which is mathematically expressed as

(2.1)\begin{equation} {\rm \pi}R^2 l \rho_s g = {\rm \pi}R^2 h \rho_l g + {\rm \pi}R^2 h_c \rho_l g + 2{\rm \pi} R \gamma \sin(\phi), \end{equation}

where $\phi$ denotes the angle of inclination of the surface tension force with respect to the horizontal at the TPCL, $h_c$ represents vertical deformation of the liquid–air interface from the undisturbed interface at the TPCL (see figure 2) and $h$ denotes the wetted length of the cylinder below the TPCL. The buoyancy force is equal to the weight of the liquid displaced by the floating body. The weight of the liquid displaced by a small-scale floating body is given by the generalised Archimedes principle (Mansfield et al. Reference Mansfield, Sepangi and Eastwood1997; Keller Reference Keller1998). As per the generalised Archimedes principle, the weight of the liquid displaced by the small-scale floating body is equal to the sum of the weight of the liquid displaced in the region ‘abcd’ and region ‘cdef’ shown in figures 2(a) and 2(c). The volume of liquid displaced in the region ‘abcd’ is equal to the volume of the cylinder, which would fill the volume bounded by the wetted surface of the body and circles ‘ab’ at the bottom and ‘cd’ at the top (cylindrical volume bounded by circles ‘ab’ at the bottom and ‘cd’ at the top). The volume of liquid displaced in the region ‘cdef’ is equal to the volume of the vertical cylinder passing through the TPCL ‘cd’ or the circumference of circle ‘cd’), and the original horizontal undisturbed free surface (surface bounded by circle ‘ef’). The region ‘gde’ shows the volume of liquid displaced by the curved air–liquid interface ‘gd’. The weight of the liquid displaced by the curved air–liquid interface ‘gd’ is equal to the vertical component of surface tension force acting along the TPCL. Rearranging (2.1) gives the wetted length $h$ of the cylinder as a function of $\rho _s$, $\rho _l$, $l$, $h_c$, $\gamma$ and $\phi$, which is expressed as

(2.2)\begin{equation} h = l\frac{\rho_s}{\rho_l} -h_c - \frac{2\gamma \sin(\phi)}{R \rho_l g}. \end{equation}

Figure 2.

Schematic showing the orientation of a cylinder with its longitudinal axis perpendicular to the liquid surface (a) under static equilibrium and (b) under small angular displacement $\beta$. (c) The three-dimensional projection of volume displaced by the small-scale floating cylinder as per the generalised Archimedes principle. (d) Comparison of TPCL before and after the perturbation of the small-scale floating cylinder. Here, $\gamma$ denotes the surface tension, $\phi$ represents the angle of inclination of the surface tension with respect to the horizontal and $\theta$ is the contact angle.

From figure 2, the inclination angle $\phi = (\theta - {\rm \pi}/2)$. The vertical deformation of the air–liquid interface $h_c$ around the cylinder is obtained from the solution of the Young–Laplace equation. The Young–Laplace equation for an axisymmetric air–liquid interface is given by (Fowkes & Hood Reference Fowkes and Hood1998; Norbury, Sander & Scott Reference Norbury, Sander and Scott2005; Anderson, Bassom & Fowkes Reference Anderson, Bassom and Fowkes2006; Finn Reference Finn2012)

(2.3)\begin{equation} \gamma \left[\frac{{\rm d}^2 z/ {\rm d}r^2}{[1+ ({\rm d}z/ {\rm d}r)^2]^{3/2}} + \frac{{\rm d}z/ {\rm d}r}{r[1+ ({\rm d}z/ {\rm d}r)^2]^{3/2}} \right] = g(\rho_l - \rho_a)z , \end{equation}

where $z$ and $r$ denote axial and radial coordinates, respectively, with the origin located at the point of intersection of the longitudinal axis ($z$) with the plane of the undisturbed interface, as shown in figure 2. We solve (2.3) numerically with boundary conditions

(2.4)\begin{equation} \frac{{\rm d}z}{{\rm d}r} = \tan \phi\ \text{at} \ r = R \quad \text{and}\quad \frac{{\rm d}z}{{\rm d}r} \rightarrow 0 \ \text{at} \ r \rightarrow \infty . \end{equation}

The solution of (2.3) gives the vertical position of the air–liquid interface $z$ as a function of $r$, from which we obtain $h_c$ by substituting $r = R$. Using $h_c$ in (2.2) gives the wetted length $h$ of the cylinder. From $h$ and $h_c$ we obtain the position of the centre of buoyancy $d_{cob} = (h_c - h)/2$ with respect to the TPCL. In the case of small-scale floating cylinders, the centre of buoyancy is defined as the centre of mass of the fluid displaced by the floating cylinder. The volume of liquid displaced by the cylinder is equal to the volume of the cylinder in the region $abfe$, as shown in figure 2.

After obtaining the equilibrium position, we perturb the body by giving a small angular displacement $\beta$. After that, we calculate the resulting net moment acting on the body due to the surface tension, buoyant and gravitational forces. In our approach, we calculate the restoring torque by taking moments of the forces about the centre of the TPCL. As we consider the cylinder surface to be perfectly smooth, the TPCL slides on the curved surface of the cylinder without pinning when we perturb the vertical cylinder with infinitesimal angular displacement $\beta$ (Singh & Joseph Reference Singh and Joseph2005; Liu et al. Reference Liu, Feng and Wang2007; Gao & Feng Reference Gao and Feng2011). Now, when the cylinder is tilted infinitesimally in the clockwise direction, the TPCL moves axially upwards in the right half and downwards in the left half relative to the cylinder surface, as shown in figure 2(b). The sliding of the TPCL in opposite directions relative to the lateral surface of the cylinder in the right and left halves of the cylinder changes the shape of the TPCL from circular (represented by length $dc$ in figure 2d) to slightly elliptical (represented by length $uv$ in figure 2d). Moreover, the slide of the TPCL results in an equal depth of the TPCL $h_c$ around the cylinder, keeping the TPCL planar. As shown in figures 2(b) and 2(d), the TPCL before perturbation ($dc$) is equal to the projection of the TPCL after perturbation ($uv$) on the plane inclined at an angle $\beta$, which gives $dc = uv \cos (\beta )$. For small $\beta$, $uv \simeq st$, which means that the shape and length can be assumed to remain constant after the vertical cylinder is perturbed by infinitesimally small angular deflection. The infinitesimally small angular displacement and the assumption of no contact line pinning means that the TPCL remains horizontal with the depth of the TPCL ($h_c$) remaining fixed. Therefore, we assume that the vertical force balance remains unaffected when the cylinder is given a small angular displacement.

As the contact angle is fixed, the inclination angle of the surface tension force with respect to horizontal $\phi$ decreases in the right half of the cylinder and increases in the left half of the floating cylinder. The change in inclination angle varies from $\pm \beta$ at points on the contact line that are normal to the axis of angular displacement to zero at points parallel to the axis of angular disturbance. Figure 3 illustrates the variation of the inclination angle along the TPCL when the floating cylinder is perturbed by smaller angular displacement $\beta$. The inclination angle varies from $\phi -\beta$ at point $A$ in figure 3 to $\phi$ at point B along the TPCL in the anticlockwise direction. In the left half, the inclination angle varies from $\phi +\beta$ at point $C$ to $\phi$ at $B$ along TPCL in the clockwise direction. This asymmetry in the magnitude of vertical component of the surface tension force about the centre of the TPCL gives rise to an unbalanced moment (see Appendix) given by

(2.5)\begin{equation} m_s = 4{\int_0}^{{\rm \pi}/2} R \gamma b \beta \cos\phi \left(1 - \frac{2 \alpha}{\rm \pi}\right) {\rm d}\alpha = \frac{8R^2 \gamma \beta \cos \phi}{\rm \pi} = M_s \beta , \end{equation}

whereas the torque due to horizontal components of the surface tension force cancels out. The parameter $b$ denote the perpendicular distance between the line of action of the vertical component of the surface tension force from the axis passing through centre of the TPCL and parallel to the axis of angular displacement. Similarly, the parameter $a$ denotes the perpendicular distance between the line of action of the vertical component of the surface tension force from the axis passing through the centre of the TPCL and normal to the axis of angular displacement.

Figure 3.

Schematic showing the variation of the inclination angle $\phi$ around the circumference of a vertical cylinder when it is given a small angular displacement $\beta$. Under the small angular displacement $\beta$ in the clockwise direction, the inclination angle $\phi$ increases from $\phi -\beta$ in the direction perpendicular to the angular displacement to $\phi$ in the direction of angular displacement.

From figure 2(b), we see that the vertical component of the surface tension force on the left side of the inclined cylinder increases due to the increase in the inclination angle, whereas on the right side it decreases due to the decrease in inclination angle. This difference in vertical components of the surface tension force on the left and right sides of the cylinder gives rise to a clockwise moment about the centre of the TPCL, thereby trying to increase the clockwise angular deflection $\beta$. From the above analysis we can say that the surface tension force always generates a destabilising moment in the case of a vertical cylinder. The moment of weight of the body about the centre of TPCL is expressed as

(2.6)\begin{equation} m_g = {\rm \pi}R^2 l \rho_s g \beta d_{cg} = M_g \beta , \end{equation}

here, $d_{cg} = (l/2 - h)$ is the vertical distance between the centre of gravity (CG) from the TPCL, as shown in figure 4. Lastly, the moment of the buoyancy force about the centre of the TPCL

(2.7)\begin{equation} m_b = {\rm \pi}R^2 \rho_l g (h + h_c) d_{b} = M_b \beta, \quad \text{where}, \ d_b = \left(d_{nb} + d_{cob}\right)\beta, \end{equation}

is the horizontal distance between the centre of buoyancy (CB) and the TPCL after the floating body is perturbed by small angular displacement $\beta$ and $d_{cob}$ is the vertical distance between the CB and the TPCL before the floating body is disturbed. The distance $d_{nb} = {R^2}/[4(h_c + h)]$ is the distance of the new CB from the longitudinal axis of the tilted body and is given by the ratio of the area moment of inertia of the body area at the undisturbed interface (about an axis parallel to the interface) to the volume of liquid displaced by the body in the hatched area.

Figure 4.

Schematic showing the distance of the CG and the CB of the floating body from the centre of the TPCL. The lengths $d_{cg}$ and $d_{cob}$ denote the distance of the CG and the CB of the undisturbed body from the centre of the TPCL, respectively. Whereas $d_b$ denotes the distance between the CB of the tilted body from the centre of the TPCL.

2.2. Horizontal cylinder

Next, we investigate the stability of a small-scale floating cylinder with its axis parallel to the undisturbed air–liquid interface. Figure 5 shows a schematic of a horizontal cylinder floating at an air–liquid interface. As in § 2.1, we begin with the calculation of the position of static equilibrium and then perturb the system by giving a small angular displacement $\beta$ about the lateral axis passing through the centroid of the floating body. Due to its circular geometry, the cylinder is infinitely stable for any angular disturbance about the longitudinal axis. However, the body may be statically unstable in the case of angular disturbance about the lateral axis, as shown in figure 5. Under a static equilibrium condition, the balance of forces acting on the cylinder is given by

(2.8)\begin{equation} F_w = F_s + F_b, \end{equation}

where $F_w = {\rm \pi}R^2 l \rho _s g$ is the weight of the floating cylinder

(2.9)\begin{equation} F_{st} = 2l\gamma \sin(\phi) + 4 R \gamma \sin(\phi_1)\sin(\psi), \end{equation}

is the vertical component of the surface tension force and $F_b = R^2 l \rho _l g [2(h_c/R) \sin (\psi ) + \psi - \cos (\psi )\sin (\psi )]$ is the buoyancy force. In (2.9), $\phi$ is the inclination angle of the interface with respect to the horizontal along the lateral surface of the cylinder and $\phi _1$ is the angle of inclination of the interface at the two flat ends of the horizontal cylinder. Note that the angle of inclination of the air–liquid interface at the contact line is different at the flat ends from the inclination angle along the lateral surface of the horizontal cylinder. Due to the difference in the curvature of the surface at the contact point, the inclination angle at the flat ends is given by $\phi _1 = \theta - {\rm \pi}/2$ (see figure 5b) whereas $\phi = \psi + \theta - {\rm \pi}$ (see figure 5c).

Figure 5.

Schematic showing equilibrium configuration of a cylinder floating with its longitudinal axis parallel to the undisturbed interface. The angle $\phi$ denotes the inclination angle along the lateral surface, and $\phi _1$ denotes the inclination angle at two ends of the horizontal cylinder. The inclination angle at the two flat ends of the horizontal cylinder is given by $\phi _1 = \theta - {\rm \pi}/2$ whereas $\phi = \psi + \theta - {\rm \pi}$.

Figure 6 shows a horizontal cylinder made of aluminium with $D = 2$ mm and $\theta = 152^\circ$ floating at an air–water interface in horizontal orientation. Figure 6(b) shows that the depth at which the curved air–water interface along the longitudinal axis meets the cylinder surface (curved edges of the dark semi-circle) is equal to the depth of the TPCL along the flat end (upper edge of the dark semi-circle) of the horizontal cylinder. Therefore, as shown in figure 5, in our analysis, we take the depth of the TPCL on the flat ends to be equal to the depth of the TPCL ($h_c$) along the lateral surface of the horizontally floating cylinder. As shown in figure 5(c), $\psi$ is the angle between the radial line joining the centre of the cylinder with the TPCL and the vertical. Using expressions of $F_w$, $F_b$ and $F_{st}$ in (2.8) gives

(2.10)\begin{align} {\rm \pi}R^2 l \rho_s g &= 2l \gamma \sin(\phi) + 4R \gamma \sin(\phi_1) \sin(\psi) \nonumber\\ &\quad + R^2 l \rho_l g [2(h_c/R) \sin(\psi) + \psi - \cos(\psi)\sin(\psi)] . \end{align}

Figure 6.

Small-scale aluminium cylinder with $D = 2$ mm and $\theta = 152^\circ$ floating at the air–liquid interface in horizontal orientation. (a) The top view of the horizontal cylinder shows the curved air–liquid interface around the contact line. (b) Front view of the horizontal cylinder showing the curved air–liquid along the lateral surface of the solid cylinder (dark semi-circle) at the same depth below the undisturbed interface as the TPCL on the flat end (dark edge of semi-circle parallel to horizontal direction) of the horizontal cylinder. The TPCL along the flat end appears to be bent upwards due to the refraction of light by the curved air–liquid interface. The image in (b) appears slightly blurred at the edges of the floating body and air–water interface due to high magnification and enlargement.

Expressing $\psi$ in terms of $\phi$, we eliminate $\psi$ from (2.10). The resulting equation has two unknowns $\phi$ and $h_c$. The distance of the TPCL from the undisturbed interface $h_c$ is obtained by solving the Young–Laplace equation. For simplicity, we assume that the deformation of the liquid–fluid interface is a function $z = z(r)$ and solve the two-dimensional approximation of the Young–Laplace equation given by (Vella, Lee & Kim Reference Vella, Lee and Kim2006)

(2.11)\begin{equation} g\rho_l z = \gamma \frac{{\rm d}^2z/{\rm d}r^2}{[1+ ({\rm d}z/ {\rm d}r)^2]^{3/2}}, \end{equation}

to obtain $h_c$. Equation (2.11) is solved analytically using (2.4) as boundary conditions, which gives, $h_c = 2 l_c \sin (\phi /2)$, where $l_c = [\gamma /(\rho _l g)]^{1/2}$ is the capillary length. Using $h_c$ in (2.10) and then solving for $\phi$ gives the inclination angle at the TPCL under a static equilibrium condition. We note that (2.10) is a transcendental equation having no explicit solution and therefore we solve it numerically. From $\phi$, we obtain $\psi = \phi - \theta + {\rm \pi}$. Using $\phi$ and $\psi$ we obtain the distance of the CB from the CG as

(2.12)\begin{equation} C_b = \frac{(2/3) R^3 \sin^3(\psi) + R \sin(\psi)(d_{cg}^2 - h_o^2)}{R^2(\psi - \sin(\psi) \cos(\psi))+ 2R\sin(\psi)h_c}, \end{equation}

where $d_{cg} = R \cos (\psi )$ is the distance of the CG from the TPCL and $h_o = d_{cg} - h_c$ is the distance of the CG from the undisturbed interface.

Next, we disturb the static equilibrium by introducing a small angular displacement $\beta$ about the lateral axis and then calculate the resulting moments acting on the floating cylinder. We calculate the moments of various forces about the lateral axis passing through the centre of the TPCL and lying in the plane TPCL. As the contact angle is fixed, a small angular perturbation leads to a change in inclination angle at the TPCL along the lateral and the two circular base surfaces of the cylinder. A small angular deflection about the transverse axis in the clockwise direction causes $\psi$ to change along the axial direction, as shown in figure 7. The circular profile of the lateral surface of the cylinder ensures that $\phi = \psi + \theta - {\rm \pi}$. Therefore, any increase in $\psi$ leads to an increase in $\phi$, which in turn leads to a change in the vertical component of the surface tension force along the axial length of the cylinder. This axial variation in the vertical component of the surface tension force gives rise to an unbalanced torque given by

(2.13)\begin{equation} m_{sl} = 4{\int_0}^{l/2} \gamma \beta \cos\phi \frac{x^2}{(R^2-{d_{cg}}^2)^{1/2}} {{\rm d}x} = \frac{4 \gamma \cos(\phi)}{3 (R^2 - {d_{cg}}^2)^{1/2}} \frac{l^3}{8} \beta . \end{equation}

Figure 7.

Schematic showing variation of inclination angle $\phi$ along the length of a horizontal cylinder under a small clockwise angular displacement $\beta$. As the contact angle is fixed, any variation in $\psi$ leads to change in $\phi$. In the above case, the inclination angle increases from left to right along the line $ab$.

Besides changing the inclination angle along the lateral surface, the angular deflection about the transverse axis also changes (i) $\phi$ at the TPCL and (ii) the length of the TPCL along the flat base surfaces at the two ends of the horizontal cylinder. As shown in figure 7, a clockwise angular deflection leads to an increase in the length of the TPCL at one end and a decrease in the length at the other. Similarly, a clockwise angular deflection results in an increase in the inclination angle on the left base and a reduction in $\phi$ along the right base of the horizontal cylinder. The net moment due to the change in the length of the TPCL and $\phi$ along the two bases of the cylinder is given by

(2.14)\begin{equation} m_{sb} = \gamma l \beta \left(\frac{hl}{R \sin(\psi)} \sin(\phi) + 2R \sin(\psi) \cos(\phi)\right). \end{equation}

The net moment on the tilted horizontal cylinder due to the vertical component of the surface tension force is given by

(2.15)\begin{equation} m_s = m_{sb} + m_{sl} = M_s \beta. \end{equation}

The direction of the net torque generated by the surface tension force acting on the horizontal cylinder depends on the magnitude and direction of $m_{sb}$ and $m_{sl}$. The direction of moment $m_{sl}$ is governed by the value of $\phi$ before the body is perturbed by the small angular displacement. For example, if initially $\phi < 90$, then any increase in $\psi$ (and therefore in $\phi$) due to clockwise angular perturbation results in an increase in the vertical component of the surface tension along the longitudinal axis on the right side of the floating cylinder. In contrast, the vertical component of the surface tension decreases along the length, which is on the left side of the centre of the cylinder. As a result, $m_{sl}$ acts in an anticlockwise direction, thus acting as a restoring torque. The moment due to the surface tension force acting along the flat ends of cylinder $m_{sb}$ always acts as a destabilising moment. This is because a small angular displacement in the clockwise direction decreases the inclination angle from $\phi _1$ to $\phi _1 - \beta$ along the TPCL on the right base of the horizontal cylinder, as shown in figure 7(b). In contrast, the inclination angle increases from $\phi _1$ to $\phi _1 + \beta$ along the TPCL on the left base of the horizontal cylinder, as shown in figure 7(b). Due to this difference in inclination angles, the vertical component of the surface tension force increases on the left base and decreases on the right base, resulting in a moment in the clockwise direction. Equations (2.13) and (2.14) show $m_{sb}$ and $m_{sl}$ respectively have quadratic and cubic dependences on $l$. Therefore, at small cylinder lengths $m_{sl} < m_{sb}$, whereas at large lengths $m_{sl} > m_{sb}$. As a result, in the case of a short horizontal cylinder (at small aspect ratios), the surface tension force generates a destabilising torque $M_s$. Whereas, in the case of a long cylinder (cylinder with a large aspect ratio), the moment due to the surface tension has a stabilising effect. The moment due to the weight of the cylinder about the centre of the TPCL is given by

(2.16)\begin{equation} m_g = {\rm \pi}R^2 l \rho_s g d_{cg} \beta = M_g \beta. \end{equation}

Lastly, the moment of the buoyancy force about the centre of the TPCL is expressed as

(2.17)\begin{equation} m_b = \left[ R^2 \left( \psi - \sin(\psi) \cos(\psi)\right) + 2R \sin(\psi) h_c\right] \rho_l g l d_b = M_b \beta, \end{equation}

where

(2.18a–c) \begin{equation} \begin{array}{@{}c@{}}\displaystyle d_b = \left( d_{cob} + d_{nb} \right) \beta, \quad d_{nb} = \frac{l^2 \sin(\psi)}{R\left( \psi - \sin(\psi)\cos(\psi)\right) + 2 \sin(\psi)h_c} \quad \text{and} \\ \displaystyle d_{cob} = d_{cg} - C_b .\end{array} \end{equation}

The net moment acting on the cylinder $M_n = (M_s + M_b + M_g) \beta$. A cylinder is stable in the vertical or horizontal orientation if $M_n > 0$, which simplifies to

(2.19)\begin{equation} (M_s + M_b + M_g) \beta > 0. \end{equation}

As $\beta > 0$, (2.19) can be further simplified by dividing the equation by $\beta$ to give

(2.20)\begin{equation} (M_s + M_b + M_g) > 0. \end{equation}

We non-dimensionalise the last equation by dividing the equation by $\gamma l_c^2$, which gives $M = (M_s + M_b + M_g)/ \gamma l_c^2$. The cylinder is in stable equilibrium if $M>0$. Here, we note that there are two approaches, (i) balance of forces and (ii) variational approach (energy analysis) (Burton & Bush Reference Burton and Bush2012), to obtain the equilibrium configuration of the floating body. In this article, we have used the balance of forces approach to obtain (2.1) and (2.20). In the variational approach (energy analysis), we strive to obtain a configuration of the floating body that minimises the system's total energy. Regardless of the approach, we obtain identical equations ((2.1) and (2.20)) for the floating body system.

We next give a criterion to classify a cylinder with a density higher than water and floating at the air–water interface as a large-scale or small-scale cylinder. For a cylinder floating at the air–liquid interface, the weight of the cylinder $F_g$ is balanced by the sum of the buoyancy force $F_B$ and the surface tension force $F_S$. For a cylinder in a horizontal orientation, $F_g = {\rm \pi}r_{ch}^2 l \rho _s g$, $F_s \approx \gamma l$ ($l > r$) and $F_B \approx {\rm \pi}r_{ch}^2 l \rho _l g$. Here, $r_{ch}$ is the characteristic radius of the cylinder. Under the condition of static equilibrium

(2.21)$$\begin{gather} F_g \approx F_S + F_B, \end{gather}$$

(2.22)$$\begin{gather}{\rm \pi} r_{ch}^2 l \rho_s g \approx \gamma l + {\rm \pi}r_{ch}^2 l \rho_l g, \end{gather}$$

(2.23)$$\begin{gather}r_{ch} \approx \sqrt{\frac{\gamma}{(\rho_s - \rho_l)g}}. \end{gather}$$

A cylinder in horizontal orientation can be classified as small scale if the radius of the cylinder $r \approx r_c$ or smaller. Now, for a cylinder in vertical orientation $F_S \approx \gamma r_{ch}$.

A similar balance of forces acting on a vertical cylinder gives

(2.24)\begin{equation} r_{cv} \approx \frac{\gamma}{l(\rho_s - \rho_l)g}. \end{equation}

Like a horizontal cylinder, a cylinder in the vertical orientation can be classified as a small-scale cylinder if the radius of the cylinder $r \approx r_{cv}$ or smaller. Equation (2.24) shows that $r_{cv}$ is inversely proportional to the length and density of the cylinder. This means that the larger the radius the smaller would be the length of the vertical cylinder that can float at the air–water interface. Similarly, the higher the density of the vertical cylinder the smaller would be the thickness at which the cylinder will float at the air–water interface.