3 Optimum shape and balancing moment of a floating mirror -- Prism-shaped floating plate

From hydromechanics it is known that a floating body which was in a state of equilibrium, if it gets inclined, is acted upon by the balancing moment of a force couple, tending to bring it back to its original position in equilibrium. The components of this force couple are the body's weight acting in its centre of mass, and a buoyant force, acting upward in the centre of mass of the displaced liquid. The balancing moment M of this force couple is

$$M = W. {\mathcal{H}} .\sin \psi ,$$ (3)

where W denotes the weight of the floating body, $\mathcal{H}$ the height of its metacentre, and $\psi$ the angle of the tilt of its axis of flotation with respect to the plumb line. The axis of flotation passes through the centre of mass of the body $C_{\rm g}$ (Fig. 3) and is vertical in the state of equilibrium when M = 0. The metacentric height $\mathcal{H}$ is the distance from $C_{\rm g}$ to metacentre $\overline M$ (the position of which is given by the intersection of the plumb line passing through ¹⁾ the centre of mass of the diplaced liquid $A _\psi$, and ²⁾ the axis of floating $C_{\rm g}\overline {M}$).

  

Figure 3: An axial section of a floating plate with its centre of gravity $C_{\rm g}$, centre of gravity of the displaced fluid $A _\psi$ at plate's position tilted by an angle $\psi$, and its offsets $\Delta x'$, $\Delta z'$ to A0 -- the centre of gravity of the displaced fluid at plate's equilibrium position; $\overline M$ is metacentre, and the metacentric height is the distance $C_{\rm g}\overline {M}$

Concerning the shape of the floating body, formula (3) usually is applied to conditions of floating and stability of ships. In our case, the floating body will have as its upper surface a plane (the mirror). Thus (unlike ships), it will be of a relatively flat shape. It is, therefore, necessary to investigate its distinguishing characteristics in relation to conditions of floating and stability - commencing with our prism-shaped plate.

Let us determine the height of the displaced liquid $h_{\rm F}$ (Fig. 3) by the floating plate in its equilibrium position. (We will not take into account the influence of atmospheric pressure: the density of the air is 0.001 g/cm³.) Following from the principle of Archimedes, $P.H.\sigma = P.h_{\rm F}.\sigma_{\rm F}$, and

$$h_{\rm F} = (\sigma /\sigma_{\rm F})H ,$$ (4)

where $\sigma$ and $\sigma_{\rm F}$ are densities of the floating plate and of fluid F, respectively, and H is the plate's height. The relation (4) does not depend on the surface area P of the plate.

In Fig. 3, there is a section through the plate in the direction of its tilt, and parallel to the plate's longitudinal edge of length L. In the position of equilibrium, an axis perpendicular to the plates upper surface is directed to zenith Z, passing through point O on the surface of fluid F, about which the plate rotates. Point A₀, the centre of mass of the displaced liquid in the position of equilibrium, also coincides with this axis OZ (plumb line). When the plate is tilted at an angle $\psi$, the centre of mass of the displaced liquid will move into the position at point $A _\psi$.Let us now determine the components of its offset $-\Delta x'$, $\Delta y'$, $\Delta z'$ - with respect to a rectangular coordinate system x ', y', z' having its origin in A₀ and axes parallel to the axes of the plate's symmetry. (In Fig. 3, the axis $Y'\equiv A_0$.)

The shift of the centre of mass of the displaced liquid, $\Delta x'$ (Fig. 3), only displaces the centre of gravity of the right-angled triangle B' CC'', and does not affect that of the rectangle B' C '' C''' B'' (owing to its symmetry with respect to axis Z'). The distance of the centre of gravity of the triangle B' CC^'', from the side CC'' (which is parallel to the coordinate axis X') is L/3, and that from the axis Z' equals L/6. Hence, $\Delta x' =[P(B' CC{''} ).L/6]/P(B' CC{'''} B {''}$), where P stands for the area of the relevant figure. $P(B' CC{''} )=(\tan\psi ).L^2/2$,and $P(B' CC {'''} B{''} )=h_{\rm F}.L$. Then for the axial section (Fig. 3),

$$\overline{\Delta x}' = (\tan\psi ).L^2/12h_{\rm F} ;$$ (5)

and for the whole plate

$$\Delta x'\! =\! \left\{ \! \int^{ +B/2}_{-B/2}\! [P(B' CC{''... ...n\psi ). L^2/12h_{\rm F}] .{\rm d}Y'\! \right\}\!\! /V_{\rm F},$$

where B and $V_{\rm F}$ denote the plate's width and the volume of the displaced liquid. Entering P(B' CC''' B'') and $V_{\rm F}=B.L.h_{\rm F}$ into the equation, we get $\Delta x' = \overline{\Delta x}'=(5)$, for we have considered a plate symmetrical to plane X' Z'; in this case $\Delta y '=0$.

The axial shift of the centre of gravity of the displaced liquid in the direction Z' is given by

$$\begin{eqnarray} \overline{\Delta Z'} & =[-P(BOB' )(h_{\rm F}/2-BB' /3)\nonumber... ...quad+P(OCC' )(h_{\rm F}/2+CC'/3)] / P(B'CC{'''} B{''} ). \nonumber\end{eqnarray}$$

Substituting corresponding expressions for areas,

$$\overline{\Delta z' }= (\tan\psi )^2.L^2/24h_{\rm F}.$$ (6)

For a plate symmetrical to plane X' Z', it is analogously (as before) $\Delta z' = \overline{\Delta z}' =(6)$.

Now we can determine the metacentric height ${\mathcal{H}}_{\rm r}$ and the balancing moment of a prism-shaped plate $M_{\rm r}$: ${\mathcal{H}}_{\rm r} = C_{\rm g} \overline{M}= A'_{0}\overline{M} -A'_0C_{\rm g}$; $A'_0\overline{M} = \Delta x' /\tan \psi = L^2/12h_{\rm F}$; $A'_0C_{\rm g}=A_0C_{\rm g} -\Delta z' = (H - h_{\rm F})/2 - (\tan \psi)^2.L^2/24h_{\rm F}$, and

$${\mathcal{H}}_{\rm r} = L^2[1+(\tan \psi)^2/2] 12h_{\rm F} - (H -h_{\rm F})/2.$$ (7)

At small inclination angles $(\psi )$, the second term of (7) in brackets is of the 2nd order, and the metacentric height ${\mathcal{H}}_{\rm r}$ is nearly constant. Body weight W in Eq. (3) is for a prism-shaped plate $W = B.L.H. \sigma$. Taking into consideration (4), we get the following Eq. (from (3)) for the balancing moment:

$$\begin{eqnarray} &M_{\rm r}\! =\! B.L\{ L^2. \sigma_{\rm F}[1\! +\! (\tan \psi)^... ...ma /\sigma_{\rm F})/2\! \} \nonumber\\ &\hspace{4.6cm}.\sin \psi. \end{eqnarray}$$ (8)

From Eq. (8) we see that the balancing moment of a prism-shaped plate (in the direction of its longitudinal axis) increases with the product of its surface area ($P_{\rm r}=B.L)$, the square of its length, the fluid's density $\sigma_{\rm F}$, and the angle of inclination $\psi$. Should the balancing moment be equal also in the transverse direction, it is necessary that L=B, and the plate then assumes the form of a square. For the same stability in all directions, the floating plate must be cylindrical in form.

We have thus arrived at some important conclusions regarding floating conditions of the mirror, and its shape:

(1)

The floating plate should be of cylindrical shape.

(2)

The fluid should be of maximal density - hence mercury is suitable.

(3)

The floating plate should have a minimum necessary thickness (height) and, as far as possible, should be made of a material of minimal density.

A smaller thickness diminishes the perturbing effect of the second term in Eq. (8) on $M_{\rm r}$. Theoretically, $M_{\rm r} \rightarrow \max$, if $H\rightarrow 0$ (if approaching the thinness of a floating sheet). The thickness is, however, limited by requirements of the plate's strength being sufficient to resist mechanical deformations. (With optical glass, the ratio thickness to length, or diameter, should be about 1:7.)

The cylindrical shape of the mirror also makes it possible, in conformity with a previous condition, to orient its sloping line into the required direction. (Only a circle can rotate within itself.)

Neglecting, for small inclinations, the second term in Eq. (8), the balancing moment is given by the formula

$$M_{\rm r} \cong P_{\rm r}[L^2.\sigma_{\rm F}/12 - H^2.\sigma /2.(1 -\sigma /\sigma_{\rm F})].\sin\psi ;$$ (9)

or still more approximately, $M_{\rm r} \approx K_{\rm r}.\psi$, since it changes almost linearly with $\psi$.