### Appendix: Parabolic Gap Sliding Bearing

The procedure proposed by this paper uses a scheme by mixing different types of sliding bearings to build the entire circular bearing. One of the most important components is the parabolic gap sliding bearing. This appendix provides a procedure for deriving the four main functions, namely the load capacity, the location of load center, stiffness and damping.

Without loss of generality, the same coordinate system as shown in Figure 3 is used for this procedure. The shape of parabolic gap is expressed as

$$h(x,t) = h_{T} (t) \cdot \left[ {1 + (\eta - 1) \cdot \left( {\frac{x}{B}} \right)^{2} } \right],\quad \quad - B \le x \le 0,\;\eta \ge 1.$$

(A1)

Introducing non-dimensional variables and parameters defined as follows:

$$\begin{aligned} & x^{*} = \frac{x}{B}\;,\;\quad \tau = \frac{V \cdot t}{B},\quad h_{T}^{*} = \frac{{h_{T} (t)}}{{h_{T0} }},\quad \hfill \\ & h^{*} = h_{T}^{*} \cdot \left[ {1 - (\eta - 1) \cdot x^{*}} \right],\quad p^{*} = \frac{{(p - p_{g} ) \cdot h_{T0}^{2} }}{\mu \cdot V \cdot B},\quad \hfill \\ \end{aligned}$$

(A2)

where *p* is the pressure over the pad with unit length, \(p_{g}\) is the pressure in water grooves, *t* is time.

The Reynolds Equation taking into consideration on dynamic squeezing film action is as follows:

$$\frac{\partial }{\partial x}\left( {h^{3} \cdot \frac{\partial p}{\partial x}} \right) = 6 \cdot \mu \cdot V \cdot \frac{\partial h}{\partial x} + 12\mu \cdot \frac{\partial h}{\partial t}.$$

(A3)

Insert non-dimensional variables Eq. (A2) into Eq. (A3), the Reynolds equation in non-dimensional form is

$$\frac{\partial }{{\partial x^{*} }}\left( {{h^*}^3 \cdot \frac{{\partial p^{*} }}{{\partial x^{*} }}} \right) = 6 \cdot \frac{{\partial h^{*} }}{{\partial x^{*} }} + 12 \cdot \frac{{\partial h^{*} }}{\partial \tau }.$$

(A4)

Small perturbation method means to find a solution of Eq. (A4) not far from the steady state solution with a linearization approach. This implies to find a solution, such as

$$p^{*} = p_{o}^{*} + p_{1}^{*} \cdot \delta \cdot e^{i \cdot \tau } ,$$

(A5)

$$h_{T}^{*} = 1 + \delta \cdot e^{i \cdot \tau } ,$$

(A6)

where \(p_{0}^{*}\) is non-dimensional pressure under steady operation, \(p_{1}^{*}\) is perturbation amplitude of a dynamic pressure on top of the pressure under steady operation. In true sense \(p_{1}^{*}\) is a coefficient of the non-dimensional dynamic pressure. *δ* is a small perturbation, which is a small number much less than 1.0. Its physical meaning is the ratio of amplitude change of film thickness to the minimum film thickness under steady operation. Insert Eqs. (A5) and (A6) into Eq. (A4), and equating the coefficients of zero order of “*δ*” on left and right side of Eq. (A4), it is resulted in an equation for pressure \(p_{o}^{ * }\)

$$\frac{\partial }{{\partial {x^*}}}\left( {{{\left[ {1 - (\eta - 1) \cdot {x^{{*^2}}}} \right]}^{\;3}} \cdot \frac{{\partial p_o^*}}{{\partial {x^*}}}} \right) = 12 \cdot (\eta - 1) \cdot {x^*}.$$

(A7)

By the same token, by equating the coefficients of first order of “*δ*” on left and right side of Eq. (A4), the coefficient of dynamic pressure \(p_{1}^{*}\) will fulfill following equation

$$\frac{\partial }{{\partial {x^*}}}\left( {{{\left[ {1 - (\eta - 1) \cdot {x^{{*^2}}}} \right]}^{\;3}} \cdot \frac{{\partial p_1^*}}{{\partial {x^*}}}} \right) = - 24 \cdot (\eta - 1) \cdot {x^*} + 12 \cdot \left[ {1 - (\eta - 1) \cdot {x^{{*^2}}}} \right] \cdot i,\;i = \sqrt { - 1} .$$

(A8)

In this procedure, all other terms with orders equal to and higher than \(\delta^{2}\) are neglected.

The boundary conditions for the non-dimensional pressure\(p^{ * }\) are

$$p^{ * } = \, 0{\text{ for }}x^{*} = 0{\text{ and }}x^{*} = {-} 1.$$

(A9)

To fulfill these conditions, the non-dimensional pressure on steady operation \(p_{0}^{*}\) as well as the real and imaginary part of non-dimensional dynamic pressure all need to be zero on the boundaries. This is expressed as

$$p_{0}^{*} = 0;\quad p_{1}^{ * } = p_{1,r}^{ * } = p_{1,i}^{ * } = 0\;{\text{for}}\;x = 0\;{\text{and}}\;x = - 1.$$

(A10)

First is to find the solution of Eq. (A7). By integrating twice of Eq. (A7), the non-dimensional pressure on steady operation is expressed in following form

$$p_o^* = \frac{3}{4}\left\{ {\frac{{{{\tan }^{ - 1}}{x^*}\sqrt {\eta - 1} }}{{\sqrt {\eta - 1} }} + \frac{{{x^{{*^3}}}(\eta - 1) - {x^*}}}{{{{\left[ {1 + (\eta - 1) \cdot {x^{{*^2}}}} \right]}^{\;2}}}}} \right\} + \frac{{{C_1}}}{8}\left\{ {\frac{{3{{\tan }^{ - 1}}{x^*}\sqrt {\eta - 1} }}{{\sqrt {\eta - 1} }} + \frac{{3{x^{{*^3}}}(\eta - 1) + 5{x^*}}}{{{{\left[ {1 + (\eta - 1) \cdot {x^{{*^2}}}} \right]}^{\;2}}}}} \right\}\; + {C_2}.$$

(A11)

The boundary condition for \(p_{0}^{*}\) requires \(C_{2}\) = 0, and

$$C_{1} = - 2\frac{{\eta^{2} \tan^{ - 1} \sqrt {\eta - 1} + (\eta - 2)\sqrt {\eta - 1} }}{{\eta^{2} \tan^{ - 1} \sqrt {\eta - 1} + (\eta + \frac{2}{3})\sqrt {\eta - 1} }}.$$

(A12)

Insert \(C_{1}\) into Eq. (A11), the final non-dimensional pressure on steady operation takes form as below

$$p_o^* = \frac{2}{{{\eta ^2}{{\tan }^{ - 1}}\sqrt {\eta - 1} + (\eta + 2/3)\sqrt {\eta - 1} }} \times \left\{ {{{\tan }^{ - 1}}{x^*}\sqrt {\eta - 1} + \frac{{{x^*}({x^{{*^2}}} - 1){{(\eta - 1)}^{\frac{3}{2}}} - {x^*}{\eta ^2}{{\tan }^{ - 1}}\sqrt {\eta - 1} }}{{{{\left[ {1 + (\eta - 1){x^{{*^2}}}} \right]}^{\;2}}}}} \right\}.$$

(A13)

The load capacity function is the integration of the non-dimensional pressure (Eq. (A13))

$$\varPi_{P} (\eta ) = \frac{{W_{o} \cdot h_{To}^{2} }}{{\mu \cdot V \cdot B^{2} \cdot L}} = \int\limits_{ - 1}^{0} {p_{o}^{ * } (x^{ * } ,\eta ) \cdot {\text{d}}x^{ * } } .$$

(A14)

The final result after implementation of the integration is

$$\varPi_{P} (\eta ) = \frac{{(\eta - 2) \cdot \tan^{ - 1} \sqrt {\eta - 1} + \sqrt {\eta - 1} }}{{\eta^{2} \tan^{ - 1} \sqrt {\eta - 1} + (\eta + \frac{2}{3})\sqrt {\eta - 1} }}.$$

(A15)

It is interesting to notice that there is a similarity of right side of Eq. (A7) and the first term on right side of Eq. (A8). Since the solution of Eq. (A7) creates the load capacity function Eq. (A15), the first real term on the right side of Eq. (A8) must generate the stiffness function. This concludes that the stiffness function is just equal to two times of the load capacity function by amount, therefore

$$K_{P} (\eta ) = 2\frac{{(\eta - 2) \cdot \tan^{ - 1} \sqrt {\eta - 1} + \sqrt {\eta - 1} }}{{\eta^{2} \tan^{ - 1} \sqrt {\eta - 1} + (\eta + \frac{2}{3})\sqrt {\eta - 1} }}.$$

(A16)

Corresponding real part of non-dimensional dynamic pressure coefficient will be

$$p_{1,r}^* = \frac{{ - 4}}{{{\eta ^2}{{\tan }^{ - 1}}\sqrt {\eta - 1} + (\eta + 2/3)\sqrt {\eta - 1} }} \times \left\{ {{{\tan }^{ - 1}}{x^*}\sqrt {\eta - 1} + \frac{{{x^*}({x^{{*^2}}} - 1){{(\eta - 1)}^{\frac{3}{2}}} - {x^*}{\eta ^2}{{\tan }^{ - 1}}\sqrt {\eta - 1} }}{{{{\left[ {1 + (\eta - 1){x^{{*^2}}}} \right]}^{\;2}}}}} \right\}.{\rm{ }}$$

(A17)

The next task is to find imaginary part of the non-dimensional dynamic pressure coefficient \(p_{1}^{*}\) which needs to fulfill following equation:

$$\frac{\partial }{{\partial {x^*}}}\left( {{{\left[ {1 - (\eta - 1) \cdot {x^{{*^2}}}} \right]}^{\;3}} \cdot \frac{{\partial p_{1,i}^*}}{{\partial {x^*}}}} \right) = 12 \cdot \left[ {1 - (\eta - 1) \cdot {x^{{*^2}}}} \right].$$

(A18)

Following similar procedure to solve Eq. (A7), after integration twice of Eq. (A18), the imaginary part of non-dimensional dynamic pressure coefficient is expressed with

$$p_{1,i}^* = - 2 \cdot \frac{{{x^{{*^2}}}(\eta - 1) + 2}}{{{{\left[ {1 + (\eta - 1) \cdot {x^{{*^2}}}} \right]}^{\;2}}(\eta - 1)}} + \frac{{{C_3}}}{8}\left\{ {\frac{{3{{\tan }^{ - 1}}{x^*}\sqrt {\eta - 1} }}{{\sqrt {\eta - 1} }} + \frac{{3{x^{{*^3}}}(\eta - 1) + 5{x^*}}}{{{{\left[ {1 + (\eta - 1) \cdot {x^{{*^2}}}} \right]}^{\;2}}}}} \right\}\; + {C_4}.$$

(A19)

Utilizing boundary condition to above equation, the two constants are

$$C_{3} = \frac{{8\left[ {4\eta^{2} - 2(\eta + 1)} \right]}}{{3\eta^{2} \sqrt {\eta - 1} \cdot \tan^{ - 1} \sqrt {\eta - 1} + (\eta - 1)(3\eta + 2)}},\;C_{4} = \frac{4}{\eta - 1}.$$

(A20)

Inserting them into Eq. (A19) and integrating it over from *x** = −1 to *x** = 0, the damping function is as following:

$${C_P}(\eta ) = - \int\limits_{ - 1}^0 {p_{1,i}^*} ({x^*},\eta ) \cdot {\rm{d}}{x^*} = \frac{{2(2\eta + 1)}}{{3\eta + 2 + \frac{{3{\eta ^2}}}{{\sqrt {\eta - 1} }}{{\tan }^{ - 1}}\sqrt {\eta - 1} }} \cdot \left( {\frac{1}{\eta } + \frac{{3{{\tan }^{ - 1}}\sqrt {\eta - 1} }}{{\sqrt {\eta - 1} }}} \right) - \frac{{4\eta - 1}}{{\eta (\eta - 1)}} + \frac{{3{{\tan }^{ - 1}}\sqrt {\eta - 1} }}{{{{(\eta - 1)}^{\frac{3}{2}}}}}.$$

(A21)

The total pad pressure appears as complex function which is

$$p^{ * } = p_{o}^{ * } + (p_{1r}^{ * } + i \cdot p_{1i}^{ * } ) \cdot \delta \cdot e^{i\tau } .$$

(A22)

The location of load center under steady operation and dynamic vibration is slightly different. The location of load center for steady operation is calculated with

$$A_{P} (\eta ) = 1 + \frac{{\int\limits_{ - 1}^{0} {x \cdot p_{o}^{ * } {\text{d}}x} }}{{\int\limits_{ - 1}^{0} {p_{o}^{ * } {\text{d}}x} }}.$$

(A23)

And the location of load center for dynamic load only is calculated with

$${A_{Pd}}(\eta ) = 1 + \frac{{\int\limits_{ - 1}^0 {x \cdot \sqrt {p_{1r}^{{*^2}} + p_{1i}^{{*^2}}} {\rm{d}}x} }}{{\int\limits_{ - 1}^0 {\sqrt {p_{1r}^{{*^2}} + p_{1i}^{{*^2}}} {\rm{d}}x} }}.$$

(A24)

Since \(p_{1r}^{ * }\) is two times of static pressure \(p_{o}^{ * }\) and has dominate amount in comparison to \(p_{1i}^{ * }\), the value of Eq. (A24) is not very much different from the value from Eq. (A23). A ratio \(R_{P} (\eta ) = A_{Pd} (\eta )/A_{P} (\eta )\) was defined for comparing the difference between Eqs. (A23) and (A24). Similarly this ratio is also defined for exponential and linear slider (see Figure 4d). This paper used static load center for Sommerfeld Number evaluation and dynamic load center for stiffness and damping evaluation for all three types of sliding bearings. The notion \(A_{Ed} \;{\text{and}}\;A_{Ld}\) presents the dynamic load center of exponential and linear slider respectively.