numerical methods - 4.37

#define

SIZE

10;

double

data[SIZE][2] = {{0.0, 0.0},

{10.0,

0.1},

{20.0,

0.4},

{30.0,

1.2},

{40.0,

2.0},

{50.0,

2.3},

{60.0,

2.4},

{70.0,

2.4},

{80.0,

2.4},

{90.0,

2.4}};

double angle(double rate){ int i;

for(i = 0; i < SIZE-1; i++){

if((rate >= data[i][0]) && (rate <= data[i+1][0]){ return (data[i+1][1] - data[i][1])

* (rate - data[i][0]) / (data[i+1][0] - data[i][0]) + data[i][1];

}

}

printf("ERROR: rate out of range\n"); exit(1);

}

Figure 4.37 A tabular interpolation subroutine example

4.6.3 Modeling Functions with Splines

When greater accuracy is required, smooth curves can be fitted to interpolate points. These curves are known as splines. There are multiple methods for creating splines, but the simplest is to use a polynomial fitted to a set of points.

The example in Figure 4.38 shows a spline curve being fitted for three data points. In this case a second order polynomial is used. The three data points are written out as equations, and then put into matrix form, using the coefficients as the unknown values. The matrix is then solved to obtain the coefficient values for the final equation. This equation can then be used to build a mathematical model of the system.

numerical methods - 4.38

The datapoints below might have been measured for the horsepower of an internal combustion engine on a dynamometer.

	S (RPM)		P (HP)
	1000		105
	4000		205
	6000		110

In this case there are three datapoints, so we can fit the curve with a second (3-1) order polynomial. The major task is to calculate the coefficients so that the

curve passes through all of the given points.

P( S) = AS2 + BS + C

Data values can be substituted into the equation,

P( 1000)

= A10002 + B1000 + C = 105

P( 4000)

= A40002 + B4000 + C = 205

P( 6000)

= A60002 + B6000 + C = 110

This can then be put in matrix form to find the coefficients,

10002 1000 1

Note: in this example the

A

105

2

=

inverse matrix is used, but

B

205

4000

4000 1

other methods for solving

C

110

60002

6000 1

systems of equations are

equally valid. If the equa-

–1

A

1000000 1000 1

105

tions were simpler, substitu-

B

=

16000000 4000 1

205

tion might have been a

better approach.

C

36000000 6000 1

110

6.667x10–8

–1.667x10–7

1.000x10–7

A

=

105

B

–6.667x10–4

1.167x10–3 –5.000x10–4

205

C

1.600

–1.000

0.400

110

A

=

–1.617x10

–5

B

0.114

C

7.000

P( S) = ( –1.617x10–5) S2 + 0.114S + 7.000

Figure 4.38 A spline fitting example