Direct Method Power Point Interpolation
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Transcript Direct Method Power Point Interpolation
Direct Method of
Interpolation
Major: All Engineering Majors
Authors: Autar Kaw, Jai Paul
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
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Direct Method of
Interpolation
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What is Interpolation ?
Given (x0,y0), (x1,y1), …… (xn,yn), find the value of ‘y’ at a
value of ‘x’ that is not given.
Figure 1 Interpolation of discrete.
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Interpolants
Polynomials are the most common
choice of interpolants because they
are easy to:
Evaluate
Differentiate, and
Integrate
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Direct Method
Given ‘n+1’ data points (x0,y0), (x1,y1),………….. (xn,yn),
pass a polynomial of order ‘n’ through the data as given
below:
y a0 a1x .................... an x .
n
where a0, a1,………………. an are real constants.
Set up ‘n+1’ equations to find ‘n+1’ constants.
To find the value ‘y’ at a given value of ‘x’, simply
substitute the value of ‘x’ in the above polynomial.
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Example 1
The upward velocity of a rocket is given as a
function of time in Table 1.
Find the velocity at t=16 seconds using the
direct method for linear interpolation.
Table 1 Velocity as a function
of time.
6
t , s
vt , m/s
0
0
10
227.04
15
362.78
20
517.35
22.5
602.97
30
901.67
Figure 2 Velocity vs. time data for the
rocket example
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Linear Interpolation
vt a0 a1t
y
v15 a0 a1 15 362.78
v20 a0 a1 20 517.35
Solving the above two equations gives,
a0 100.93 a1 30.914
x1 , y1
x0 , y0
f1 x
x
Figure 3 Linear interpolation.
Hence
vt 100.93 30.914t , 15 t 20.
v16 100.93 30.91416 393.7 m/s
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Example 2
The upward velocity of a rocket is given as a
function of time in Table 2.
Find the velocity at t=16 seconds using the
direct method for quadratic interpolation.
Table 2 Velocity as a function
of time.
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t , s
vt , m/s
0
0
10
227.04
15
362.78
20
517.35
22.5
602.97
30
901.67
Figure 5 Velocity vs. time data for the
rocket example
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Quadratic Interpolation
y
vt a0 a1t a2t 2
x1 , y1
v10 a0 a1 10 a2 10 227.04
x2 , y2
2
v15 a0 a1 15 a2 15 362.78
2
v20 a0 a1 20 a2 20 517.35
2
f 2 x
x0 , y0
x
Figure 6 Quadratic interpolation.
Solving the above three equations gives
a0 12.05 a1 17.733 a2 0.3766
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Quadratic Interpolation (cont.)
517.35
vt 12.05 17.733t 0.3766t , 10 t 20
550
500
2
v16 12.05 17.73316 0.376616
2
450
ys
400
f ( range)
f x desired
392 .19 m/s
350
300
250
227.04
200
10
10
12
14
16
18
x s range x desired
The absolute relative approximate error a obtained between
the results from the first and second order polynomial is
392.19 393.70
100
392.19
0.38410%
a
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20
20
Example 3
The upward velocity of a rocket is given as a
function of time in Table 3.
Find the velocity at t=16 seconds using the
direct method for cubic interpolation.
Table 3 Velocity as a function
of time.
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t , s
vt , m/s
0
0
10
227.04
15
362.78
20
517.35
22.5
602.97
30
901.67
Figure 6 Velocity vs. time data for the
rocket example
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Cubic Interpolation
y
vt a0 a1t a2t a3t
2
x3 , y3
3
x1 , y1
v10 227.04 a0 a1 10 a2 10 a3 10
2
3
v15 362.78 a0 a1 15 a2 15 a3 15
2
3
x0 , y0
v20 517.35 a0 a1 20 a2 20 a3 20
2
x2 , y 2
f 3 x
x
3
Figure 7 Cubic interpolation.
v22.5 602.97 a0 a1 22.5 a2 22.5 a3 22.5
2
a0 4.2540 a1 21.266
12
3
a2 0.13204 a3 0.0054347
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Cubic Interpolation (contd)
vt 4.2540 21.266t 0.13204t 2 0.0054347t 3 , 10 t 22.5
v16 4.2540 21.26616 0.1320416 0.005434716
392.06 m/s
2
602.97
700
The absolute percentage relative
approximate error a between
second and third order polynomial is
600
ys
3
500
f ( range)
f x desired
400
300
227.04
200
10
10
13
392.06 392.19
100
392.06
0.033269%
a
12
14
16
18
x s range x desired
20
22
24
22.5
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Comparison Table
Table 4 Comparison of different orders of the polynomial.
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t(s)
v (m/s)
0
0
10
227.04
15
362.78
20
517.35
22.5
602.97
30
901.67
Order of
Polynomial
1
2
3
vt 16 m/s
393.7
392.19
392.06
Absolute Relative
Approximate Error
----------
0.38410 %
0.033269 %
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Distance from Velocity Profile
Find the distance covered by the rocket from t=11s to t=16s ?
vt 4.3810 21.289t 0.13064t 2 0.0054606t 3 , 10 t 22.5
16
s 16 s11 vt dt
11
16
4.2540 21.266t 0.13204t 2 0.0054347t 3 dt
11
16
t2
t3
t4
4.2540t 21.266 0.13204 0.0054347
2
3
4 11
1605m
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Acceleration from Velocity Profile
Find the acceleration of the rocket at t=16s given that
t 4.2540 21.266t 0.132042 0.0054347t 3 ,10 t 22.5
at
d
vt
dt
d
4.2540 21.266t 0.13204t 2 0.0054347t 3
dt
21.289 0.26130t 0.016382t 2 , 10 t 22.5
a16 21.266 0.2640816 0.01630416
2
29.665m/s2
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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/direct_met
hod.html
THE END
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