Understanding Interpolation: Reading Data Patterns & Solving Practical Problems

Explore the concept of interpolation through real-world examples like predicting values between data points and solving problems like rocket velocity calculation and heat transfer. Dive into understanding specific heat of carbon, thermistor calibration, and spline method of interpolation. Learn to fit curves through given points to design components efficiently.

• Interpolation
• Data Analysis
• Problem Solving
• Heat Transfer
• Curve Fitting

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1. Interpolation - Introduction Reading Between the Lines

2. WHATIS INTERPOLATION ? Given (x0,y0), (x1,y1), (xn,yn), find the value of y at a value of x that is not given. Figure Interpolation of discrete data. http://nm.mathforcollege.com

3. APPLIED PROBLEMS

4. FLY ROCKET FLY, FLY ROCKET FLY The upward velocity of a rocket is given as a function of time in table below. Find the velocity and acceleration at t=16 seconds. Table Velocity as a function of time. ? ? , m/s ?, s 0 0 10 227.04 15 362.78 20 517.35 22.5 602.97 30 901.67 Velocity vs. time data for the rocket example

5. SPECIFIC HEATOF CARBON A carbon block of mass 2kg is heated up from room temperature of 400K to 1500K. How much heat is required to do so? Specific Heat (J/kg-K) Temperature (K) 200 420 400 1070 600 1370 1000 1820 1500 2000 2000 2120

6. THERMISTOR CALIBRATION Thermistors are based on change in resistance of a material with temperature. A manufacturer of thermistors makes the following observations on a thermistor. Determine the calibration curve for thermistor. R ( ) T( C) 1101.0 911.3 636.0 451.1 25.113 30.131 40.120 50.128

7. FOLLOWTHE CAM A curve needs to be fit through the given points to fabricate the cam. 4 3 5 2 6 Point 1 2 3 4 5 6 7 x (in.) 2.20 1.28 0.66 0.00 0.60 1.04 1.20 y (in.) 0.00 0.88 1.14 1.20 1.04 0.60 0.00 Y 1 X 7 Cam Profile 1.4 1.2 1 0.8 y 0.6 0.4 0.2 0 -2 -1 0 1 2 3 x

8. END

9. Length of a Path

11. END

12. Spline Method of Interpolation http://nm.MathForCollege.com

13. Why Splines ? 1 ?(?) = 1 + 25?2 Table : Six equidistantly spaced points in [-1, 1] 1 = y x + 2 1 25 x -1.0 0.038461 -0.6 0.1 -0.2 0.5 0.2 0.5 0.6 0.1 Figure : 5th order polynomial vs. exact function 1.0 0.038461 http://nm.MathForCollege.com 13

14. Why Splines ? 1.2 0.8 0.4 y 0 -1 -0.5 0 0.5 1 -0.4 -0.8 x 19th Order Polynomial f (x) 5th Order Polynomial Figure : Higher order polynomial interpolation is a bad idea 14 http://nm.MathForCollege.com

15. Linear Spline Interpolation Given ( ) ( , ) ( )( ) , , ,......, , , , fit linear splines to the data. This simply involves x y x y x y x y 0 0 1 1 1 1 n n n n forming the consecutive data through straight lines. So if the above data is given in an ascending order, the linear splines are given by ( ) ) ( i i x f y = Figure : Linear splines 15 http://nm.MathForCollege.com

16. Linear Spline Interpolation (contd) ( ) ( ) f x f x = + 1 x 0 ( ) ( ) ( ), f x f x x x x x x 0 0 0 1 x 1 0 ( ) ( ) f x f x = + 2 x 1 ( ) ( ), f x x x x x x 1 1 1 2 x 2 1 . . . ( ) ( ) f x f x = + 1 n n ( ) ( ), f x x x x x x 1 1 1 n n n n x x 1 n n Note the terms of ( ) ( 1) f x f x i i x x 1 i i in the above function are simply slopes between ix and ix . 1 16 http://nm.MathForCollege.com

17. Example The upward velocity of a rocket is given as a function of time in Table 1. Find the velocity at t=16 seconds using linear splines. Table Velocity as a function of time (s) (m/s) 0 227.04 362.78 517.35 602.97 901.67 ?(?) ? 0 10 15 20 22.5 30 Figure. Velocity vs. time data for the rocket example 17 http://nm.MathForCollege.com

18. Linear Spline Interpolation 0= (0 = t 15 , ) 362 78 . t v 550 517.35 1= t (1= t 20 , ) 517 35 . v 500 ( ) ( ) v t v t = + 1 t 0 ( ) ( ) ( ) v t v t t t 0 0 t ys 1 0 f range ( ( ) 450 517 35 . 362 78 . ) f xdesired = + 362 78 . ( 15 ) t t 20 15 15 400 = + ( ) 362 78 . 30 913 . ( ) v t = At 16 , t 362.78 350 10 xs0 12 14 16 18 xdesired 20 22 24 xs1 = + 16 ( ) 362 = 78 . 30 913 . 16 ( 15 ) v xsrange + 10 10 393 7 . m/s 18 http://nm.MathForCollege.com

19. Quadratic Spline Interpolation Given ( ) ( , ) ( ) ( , ) , , ,......, , , , fit quadratic splines through the data. The splines x y x y x y x y 0 0 1 1 1 1 n n n n are given by = + + 2 ( ) , f x a x b x c x x x 1 1 1 0 1 = + + 2 , a x b x c x x x 2 2 2 1 2 . . . = + + 2 , b x c x x x a x 1 n n n n n = Find , , , 1, 2, , n ia ib ic i http://nm.MathForCollege.com 19

20. Quadratic Spline Interpolation (contd) Each quadratic spline goes through two consecutive data points 2 + + = ( ) a x b x c f x 1 0 1 0 1 0 2 + + = ( ) . a x b x c f x 1 1 1 1 1 1 . . 2 + + = ( ) a x b x c f x 1 1 1 i i i i i i 2 + + = ( ) . a x b x c f x i i i i i i . . 2 + + = ( ) a x b x c f x 1 1 1 n n n n n n 2 + + = ( ) a x b x c f x n n n n n n This condition gives 2n equations 20 http://nm.MathForCollege.com

21. Quadratic Spline Interpolation (contd) The first derivatives of two quadratic splines are continuous at the interior points. For example, the derivative of the first spline + + + 2 is 2 a x b x c a x b 1 1 1 1 1 The derivative of the second spline + + + 2 is 2 a x b x c a x b 2 2 2 2 2 x = and the two are equal at giving 1x + = + 2 2 a x b a x b 1 1 1 2 1 2 + = 2 2 0 a x b a x b 1 1 1 2 1 2 21 http://nm.MathForCollege.com

22. Quadratic Spline Interpolation (contd) Similarly at the other interior points, + = 2 2 0 a x b a x b 2 2 2 3 2 3 . . . + = 2 2 0 a x b a x b + + 1 1 i i i i i i . . . + = 2 2 0 a x b a x b 1 1 1 1 n n n n n n + ) 1 = ) 1 We have (n-1) such equations. The total number of equations is 2 ( ) ( 3 ( . n n n 1= a We can assume that the first spline is linear, that is 0 22 http://nm.MathForCollege.com

23. Quadratic Spline Interpolation (contd) This gives us 3n equations and 3n unknowns. Once we find the 3n constants, we can find the function at any value of x using the splines, = + + 2 ( ) , f x a x b x c x x x 1 1 1 0 1 = + + 2 , a x b x c x x x 2 2 2 1 2 . . . = + + 2 , a x b x c x x x 1 n n n n n 23 http://nm.MathForCollege.com

24. Quadratic Spline Example The upward velocity of a rocket is given as a function of time. Using quadratic splines a) Find the velocity at t=16 seconds b) Find the acceleration at t=16 seconds c) Find the distance covered between t=11 and t=16 seconds t s 0 10 15 20 22.5 30 v(t) m/s 0 227.04 362.78 517.35 602.97 901.67

25. Data and Plot t s 0 10 15 20 22.5 30 v(t) m/s 0 227.04 362.78 517.35 602.97 901.67

26. Solution ?(?) = ?1?2+ ?1? + ?1, 0 ? 10 = ?2?2+ ?2? + ?2, 10 ? 15 = ?3?2+ ?3? + ?3, 15 ? 20 = ?4?2+ ?4? + ?4, 20 ? 22.5 = ?5?2+ ?5? + ?5, 22.5 ? 30 Let us set up the equations

27. Each Spline Goes Through Two Consecutive Data Points ?(?) = ?1?2+ ?1? + ?1, 0 ? 10 ?1(0)2+ ?1(0) + ?1= 0 ?1(10)2+ ?1(10) + ?1= 227.04

28. Each Spline Goes Through Two Consecutive Data Points t s 0 v(t) m/s 0 227.04 362.78 517.35 602.97 901.67 ?2(10)2+ ?2(10) + ?2= 227.04 ?2(15)2+ ?2(15) + ?2= 362.78 ?3(15)2+ ?3(15) + ?3= 362.78 10 15 20 22.5 30 ?3(20)2+ ?3(20) + ?3= 517.35 ?4(20)2+ ?4(20) + ?4= 517.35 ?4(22.5)2+ ?4(22.5) + ?4= 602.97 ?5(22.5)2+ ?5(22.5) + ?5= 602.97 ?5(30)2+ ?5(30) + ?5= 901.67

29. Derivatives are Continuous at Interior Data Points ?(?) = ?1?2+ ?1? + ?1, 0 ? 10 = ?2?2+ ?2? + ?2, 10 ? 15 ? ???1?2+ ?1? + ?1 ? ???2?2+ ?2? + ?2 = ?=10 ?=10 2?1? + ?1 ?=10= 2?2? + ?2 ?=10 2?110 + ?1= 2?210 + ?2 20?1+ ?1 20?2 ?2= 0

30. Derivatives are continuous at Interior Data Points At t=10 2?1(10) + ?1 2?2(10) ?2= 0 At t=15 2?2(15) + ?2 2?3(15) ?3= 0 At t=20 2?3(20) + ?3 2?4(20) ?4= 0 At t=22.5 2?4(22.5) + ?4 2?5(22.5) ?5= 0

31. Last Equation ?5= 0

32. Final Set of Equations ?1 ?1 ?1 ?2 ?2 ?2 ?3 ?3 ?3 ?4 ?4 ?4 ?5 ?5 ?5 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 10 15 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 20 22.5 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 100 0 0 0 0 0 0 0 0 20 0 0 0 0 10 0 0 0 0 0 0 0 0 1 0 0 0 0 227.04 227.04 362.78 362.78 517.35 517.35 602.97 602.97 901.67 0 0 0 0 0 100 225 0 0 0 0 0 0 20 30 0 0 0 225 400 0 0 0 0 0 30 40 0 0 15 20 0 0 0 0 0 1 1 0 0 400 506.25 0 0 0 0 40 45 0 = 506.25 900 0 0 0 45 1 22.5 30 0 0 0 1 0

33. ?? ?? ?? Coefficients of Spline i 1 2 3 4 5 ai bi ci 0 -0.15667 1.2021 -0.44893 2.2315 0 24.271 -2.9053 46.627 -60.589 39.827 135.88 -235.61 836.55 -293.13

34. Final Solution ? ? = 0.15667?2+ 24.271?, 0 ? 10 = 1.2021?2 2.9053? + 135.88, 10 ? 15 = 0.44893?2+ 46.627? 235.61, 15 ? 20 = 2.2315?2 60.589? + 836.55, 20 ? 22.5 = 39.827? 293.13, 22.5 ? 30

35. Velocity at a Particular Point a) Velocity at t=16 ? ? = 0.15667?2+ 24.271?, 0 ? 10 = 1.2021?2 2.9053? + 135.88, 10 ? 15 = 0.44893?2+ 46.627? 235.61, 15 ? 20 = 2.2315?2 60.589? + 836.55, 20 ? 22.5 = 39.827? 293.13, 22.5 ? 30 ? 16 = 0.44893 162+ 46.627 16 235.61 = 395.50 m/s

36. Acceleration from Velocity Profile b) Acceleration at t=16 ? ? = 0.15667?2+ 24.271?, 0 ? 10 = 1.2021?2 2.9053? + 135.88, 10 ? 15 = 0.44893?2+ 46.627? 235.61, 15 ? 20 = 2.2315?2 60.589? + 836.55, 20 ? 22.5 22.5 ? 30 = 39.827? 293.13, ?(16) =? ?(?) ?? ?=16

37. Acceleration from Velocity Profile The quadratic spline valid at t=16 is given by ? ? = 0.44893?2+ 46.627? 235.61, , 15 ? 20 ?(?) =? ??( 0.44893?2+ 46.627? 235.61) = 0.89786? + 46.627, 15 ? 20 ?(16) = 0.89786(16) + 46.627 = 32.261 m/s2

38. Distance from Velocity Profile c) Find the distance covered by the rocket from t=11s to t=16s. ? ? = 0.15667?2+ 24.271?, 0 ? 10 = 1.2021?2 2.9053? + 135.88, 10 ? 15 = 0.44893?2+ 46.627? 235.61, 15 ? 20 = 2.2315?2 60.589? + 836.55, 20 ? 22.5 22.5 ? 30 = 39.827? 293.13, 16 ? 16 ? 11 = ?(?)?? 11

39. Distance from Velocity Profile = 1.2021?2 2.9053? + 135.88, ? ? 10 ? 15 = 0.44893?2+ 46.627? 235.61, 15 ? 20 16 15 16 ? 16 ? 11 = ?(?)?? = ?(?)?? + ?(?)?? 11 11 15 15 (1.2021?2 2.9053? + 135.88)?? = 11 16 ( 0.44893?2+ 46.627? 235.61)?? + 15 = 1590.7 m

40. END http://nm.MathForCollege.com

41. Points for Robot Path x y 2.00 4.5 5.25 7.81 9.20 10.60 7.2 7.1 6.0 5.0 3.5 5.0 ? ? Find the shortest but smooth path through consecutive data points

42. Polynomial Interpolant Path

43. Spline Interpolant Path

44. Compare Spline & Polynomial Interpolant Path Length of path Polynomial Interpolant=14.9 Spline Interpolant =12.9