Optimization of molecularly thin lubricant to improve bearing capacity at the head-disk interface

Abstract

A molecularly thin lubricant layer (of the order of 1–2 nm thick) has been shown to provide bearing forces at the interface between contacting solid surfaces under light loads and high shear rates. This phenomenon is important, for example, in the head-disk contact in magnetic storage hard disk drives to ensure that some of the contact is sustained by the lubricant layer and thus avoiding damage of the solid surfaces. The magnitude of the normal and tangential bearing forces that the lubricant layer can provide depends on temperature, viscosity of the lubricant, sliding velocity and radius of gyration of the lubricant molecules. This study shows that viscosity has the greatest effect on the load bearing capacity of the molecularly thin lubricant. Thus, by controlling the flash temperature and the ratio of molecularly thin lubricant-to-bulk viscosity, the bearing load carrying capacity of the layer can be controlled. This would allow for the contact to be sustained within the mobile lubricant layer, avoiding solid contact so as to protect the diamond-like carbon coating, and thus reduce wear and potential catastrophic failures.

Appendices

Appendix 1: The MTL model

According to the MTL model, the expressions for normal (P) and shearing (Q) forces are given by Eqs. 5 and 6 respectively. Where P _o and Q _o are the maximum experimental normal and shear forces given by Eqs. 7 and 8. The experimental shearing velocity U = 200 μm/s; the radius of spherical shearing probe R = 102 μm (Vakis et al. 2011). Then the critical shear rate is U/κ = 2 × 10⁵/s for κ = 1 nm (Wood 2002).

$$P_{lube} = P_{0} \left( {\frac{{\dot {\gamma } }}{{\dot {\gamma_{0} } }}} \right)^{m}$$

(5)

$$Q_{lube} = m\log \left( {\frac{{\dot {\gamma } }}{{\dot {\gamma_{0} } }}} \right)+\;Q_{0}$$

(6)

$$P_{0} = \frac{6\pi }{5}\mu U\sqrt {\frac{{2R^{3} }}{\kappa }}$$

(7)

$$Q_{0} = \frac{16\pi }{5}\mu UR\ln \left( {\frac{\kappa }{R}} \right)$$

(8)

The fitting coefficients m and n are calculated from Eqs. 9 and 10 and can also be obtained from the logarithmic curves of normal and shear forces vs. shear rate.

$$m = \frac{{\log \left( {\frac{{P_{lube,2} }}{{P_{lube,1} }}} \right)}}{{\log \left( {\frac{{\dot {\gamma_{2} } }}{{\dot{\gamma_{1} } }}} \right)}}$$

(9)

$$n = \frac{{{\text{Q}}_{{\text{lube,2}}} -{\text{Q}}_{ {\text{lube,1}}}} }{{\log \left( {\frac{{\dot {\gamma_{2} } }}{{\dot{\gamma_{1} } }}} \right)}}$$

(10)

The normal and shear stiffnesses are calculated according to Eqs. 11 and 12. Where the shear rate is found from the expression, U/d _o and this becomes maximum when the solid–solid gap reaches the bonded lubricant thickness (2κ).

$$k_{P} = \left| {\frac{{\partial P_{lube} }}{{\partial d_{o} }}} \right| = \frac{{mP_{o} }}{{U\dot{\gamma }_{o}^{m} }}\dot{\gamma }^{m}$$

(11)

$$k_{Q} = \left| {\frac{{\partial Q_{lube} }}{{\partial d_{o} }}} \right| = \frac{n}{{U\ln \left( {10} \right)}}\dot{\gamma }$$

(12)

$$\dot{\gamma } = \frac{U}{{d_{o} }} = \frac{U}{h - 3\sigma - \kappa }$$

(13)

Here d _o is the liquid gap. The maximum shear rate is when d _o  = κ, i.e., h−3σ = 2κ (when the distance between slider and disk becomes twice the bonded lubricant thickness).

Appendix 2: Regression models

The coefficients for predicting the maximum bearing force for the three levels of U are given in Eqs. 14, 15 and 16.

$$P\hbox{max}\left[ {U: - 1\,\,order} \right] = - 0.0206 \times 10^{7} \mu - 0.7135 \times 10^{7} \mu^{2} - 0.1427 \times 10^{7} \kappa^{2} \mu^{2}$$

(14)

$$P\hbox{max} \left[ {U:0\,\,order} \right] = - 1.096 \times 10^{7} \kappa^{3}$$

(15)

$$P\hbox{max} \left[ {U: + 1\,\,order} \right] = 0.0205 \times 10^{7} \mu - 1.41 \times 10^{7} \mu^{2} + 0.1427 \times 10^{7} \kappa^{2} \mu^{2}$$

(16)

The coefficients for predicting the maximum bearing force for the three levels of μ is given in Eqs. 17, 18 and 19.

$$P\hbox{max} \left[ {\mu : - 1\,\,order} \right] = - 0.57 \times 10^{3} \kappa^{2} U^{2} + 0.015 \times 10^{3} \kappa^{3} U^{2}$$

(17)

$$P\hbox{max} \left[ {\mu :0\,\,order} \right] = - 1.12 \times 10^{3} \kappa^{2} U^{2} + 0.03 \times 10^{3} \kappa^{3} U^{2}$$

(18)

$$P\hbox{max} \left[ {\mu : + 1\,\,order} \right] = - 1.7 \times 10^{3} \kappa^{2} U^{2} + 0.045 \times 10^{3} \kappa^{3} U^{2}$$

(19)