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http://numericalmethods.eng.usf.edu 1 Spline Interpolation Method Mechanical Engineering Majors Authors: Autar Kaw, Jai Paul http://numericalmethods.eng.usf.edu Transforming Numerical Methods Education for STEM Undergraduates
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Spline Method of Interpolation http://numericalmethods.eng.usf.edu http://numericalmethods.eng.usf.edu
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3 What is Interpolation ? Given (x 0,y 0 ), (x 1,y 1 ), …… (x n,y n ), find the value of ‘y’ at a value of ‘x’ that is not given.
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http://numericalmethods.eng.usf.edu4 Interpolants Polynomials are the most common choice of interpolants because they are easy to: Evaluate Differentiate, and Integrate.
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http://numericalmethods.eng.usf.edu5 Why Splines ?
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http://numericalmethods.eng.usf.edu6 Why Splines ? Figure : Higher order polynomial interpolation is a bad idea
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http://numericalmethods.eng.usf.edu7 Linear Interpolation
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http://numericalmethods.eng.usf.edu8 Linear Interpolation (contd)
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http://numericalmethods.eng.usf.edu9 Example A trunnion is cooled 80°F to − 108°F. Given below is the table of the coefficient of thermal expansion vs. temperature. Determine the value of the coefficient of thermal expansion at T=−14°F using linear spline interpolation. Temperature ( o F) Thermal Expansion Coefficient (in/in/ o F) 806.47 × 10 −6 06.00 × 10 −6 − 60 5.58 × 10 −6 − 160 4.72 × 10 −6 − 260 3.58 × 10 −6 − 340 2.45 × 10 −6
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http://numericalmethods.eng.usf.edu10 Linear Interpolation
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http://numericalmethods.eng.usf.edu11 Quadratic Interpolation
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http://numericalmethods.eng.usf.edu12 Quadratic Interpolation (contd)
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http://numericalmethods.eng.usf.edu13 Quadratic Splines (contd)
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http://numericalmethods.eng.usf.edu14 Quadratic Splines (contd)
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http://numericalmethods.eng.usf.edu15 Quadratic Splines (contd)
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http://numericalmethods.eng.usf.edu16 Example A trunnion is cooled 80°F to − 108°F. Given below is the table of the coefficient of thermal expansion vs. temperature. Determine the value of the coefficient of thermal expansion at T=−14°F using quadratic spline interpolation. Temperature ( o F) Thermal Expansion Coefficient (in/in/ o F) 806.47 × 10 −6 06.00 × 10 −6 − 60 5.58 × 10 −6 − 160 4.72 × 10 −6 − 260 3.58 × 10 −6 − 340 2.45 × 10 −6
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http://numericalmethods.eng.usf.edu17 Solution
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http://numericalmethods.eng.usf.edu18 Solution (contd)
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http://numericalmethods.eng.usf.edu19 Solution (contd)
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http://numericalmethods.eng.usf.edu20 Solution (contd)
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http://numericalmethods.eng.usf.edu21 Solution (contd)
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http://numericalmethods.eng.usf.edu22 Solution (contd)
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http://numericalmethods.eng.usf.edu23 Reduction in Diameter The actual reduction in diameter is given by where T r = room temperature (°F) T f = temperature of cooling medium (°F) Since T r = 80 °F and T r = −108 °F, Find out the percentage difference in the reduction in the diameter by the above integral formula and the result using the thermal expansion coefficient from the cubic interpolation.
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http://numericalmethods.eng.usf.edu24 Reduction in Diameter
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The absolute relative approximate error obtained between the results from the 2 nd methods is http://numericalmethods.eng.usf.edu25 Reduction in diameter Taking the average coefficient of thermal expansion over this interval, given by:
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Additional Resources For all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit http://numericalmethods.eng.usf.edu/topics/spline_met hod.html
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THE END http://numericalmethods.eng.usf.edu
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