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TECHNICAL PAPERS

Shape Design for Surface of a Slider by Inverse Method

[+] Author and Article Information
Chin-Hsiang Cheng, Mei-Hsia Chang

Department of Mechanical Engineering, Tatung University, 40 Chungshan N. Road, Sec. 3, Taipei, Taiwan 10451, R.O.C.

J. Tribol 126(3), 519-526 (Jun 28, 2004) (8 pages) doi:10.1115/1.1704627 History: Received March 13, 2003; Revised September 18, 2003; Online June 28, 2004
Copyright © 2004 by ASME
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References

Figures

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Physical model of a slider bearing
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Comparison between the numerical predictions by direct problem solver and the analytical solutions given by Gross et al. 17 for the case with H(X,Y)=2−X at l/w=0.1, 0.5, 1.0, and 2.0
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Exact (specified) and designed (optimized) shapes and pressure distributions. The pressure distribution for the case considered in Table 1 is treated as the specified (exact) pressure distribution. The exact shape is H(X,Y)=1.7785−0.3X+0.1X2−1.8Y+1.4Y2.
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Effects of measurement uncertainty in pressure distribution on slider shape design. The pressure distribution for the case considered in Table 1 is treated as the exact pressure distribution. The exact shape is H(X,Y)=1.7785−0.3X+0.1X2−1.8Y+1.4Y2. (a) σ=0.01 (b) σ=0.1
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Slider shape design that is able to provide the specified pressure distribution. The bearing numbers are fixed at ΛX=2000 and ΛY=0. (a) specified pressure distribution; and (b) designed surface profile
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Unique-solution situation: optimal shape design which is independent of the initial guess, for F̄X1=−8,F̄Y1=3,F̄Z1=−25,X̄C1=0.52,ȲC1=0.49,X̄C2=0.51,ȲC2=0.50,ΛX=3000, and ΛY=0.
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Multiple-solution situation: optimal shape design which is dependent on the initial guess, for F̄X1=−5.5,F̄Y1=1.5,F̄Z1=−1.6,X̄C1=0.57,ȲC1=0.5,ΛX=4000, and ΛY=0
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Designed slider shape and its pressure distribution at various combinations of bearing numbers. The load demands are: F̄X1=−8,F̄Y1=3,F̄Z1=−25,X̄C1=0.52,ȲC1=0.49,X̄C2=0.51, and ȲC2=0.50. (a) ΛX=2000,ΛY=0 (b) ΛX=5000,ΛY=1000 (c) ΛX=7000,ΛY=5000.
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Iteration process of shape design, for the case at F̄X1=−8,F̄Y1=3,F̄Z1=−25,X̄C1=0.52,ȲC1=0.49,X̄C2=0.51, and ȲC2=0.50, at ΛX=3000 and ΛY=0

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