Interpolation and extrapolation are fundamental numerical methods used in Automotive Mechatronics for estimating unknown values based on known data points. These techniques enable automotive engineers and technicians to predict sensor outputs, calibrate control systems, and analyze performance parameters when direct measurements are unavailable or impractical. In Kenya’s automotive service and manufacturing sectors, mastery of these methods supports effective diagnostics, optimization of engine management systems, and enhanced vehicle calibration to meet local operating conditions.

3.1 Definition of interpolation and extrapolation

3.1.1 Understanding Interpolation

Interpolation is the process of estimating a value within the range of a discrete set of known data points. In Automotive Mechatronics, interpolation is frequently used to predict sensor readings or control parameters between calibrated points on a lookup table, such as fuel injection timing or throttle position sensor voltages.

The general formula for linear interpolation between two points \((x_0, y_0)\) and \((x_1, y_1)\) to estimate the value \(y\) at a point \(x\) where \(x_0 < x < x_1\) is:

$$ y = y_0 + \frac{(x, x_0)}{(x_1 - x_0)} (y_1 - y_0) $$

Worked Examples

Example 1: An automotive sensor calibration chart gives sensor output voltages of 2.0 V at 20°C and 3.0 V at 40°C. Estimate the sensor voltage at 30°C by linear interpolation.

Given:
\(x_0 = 20^\circ C, y_0 = 2.0\,V\)
\(x_1 = 40^\circ C, y_1 = 3.0\,V\)
\(x = 30^\circ C\)

$$ y = 2.0 + \frac{30 - 20}{40 - 20} (3.0 - 2.0) $$

$$ y = 2.0 + \frac{10}{20} \times 1.0 $$

$$ y = 2.0 + 0.5 = 2.5\,V $$

Answer: 2.5 V

Example 2: A fuel injector flow rate is 100 ml/min at 1000 rpm and 140 ml/min at 2000 rpm. Estimate the injector flow rate at 1500 rpm.

Given:
\(x_0 = 1000\,rpm, y_0 = 100\,ml/min\)
\(x_1 = 2000\,rpm, y_1 = 140\,ml/min\)
\(x = 1500\,rpm\)

$$ y = 100 + \frac{1500 - 1000}{2000 - 1000} (140 - 100) $$

$$ y = 100 + \frac{500}{1000} \times 40 $$

$$ y = 100 + 20 = 120\,ml/min $$

Answer: 120 ml/min

Example 3: The resistance of a thermistor is 500 Ω at 25°C and 300 Ω at 50°C. Estimate the resistance at 40°C.

Given:
\(x_0 = 25^\circ C, y_0 = 500\,\Omega\)
\(x_1 = 50^\circ C, y_1 = 300\,\Omega\)
\(x = 40^\circ C\)

$$ y = 500 + \frac{40 - 25}{50 - 25} (300 - 500) $$

$$ y = 500 + \frac{15}{25} \times (-200) $$

$$ y = 500 - 120 = 380\,\Omega $$

Answer: 380 Ω

Example 4: A pressure sensor outputs 1.2 V at 100 kPa and 2.4 V at 200 kPa. Estimate the voltage output at 175 kPa.

Given:
\(x_0 = 100\,kPa, y_0 = 1.2\,V\)
\(x_1 = 200\,kPa, y_1 = 2.4\,V\)
\(x = 175\,kPa\)

$$ y = 1.2 + \frac{175 - 100}{200 - 100} (2.4 - 1.2) $$

$$ y = 1.2 + \frac{75}{100} \times 1.2 $$

$$ y = 1.2 + 0.9 = 2.1\,V $$

Answer: 2.1 V

Example 5: The torque output of an electric motor is 50 Nm at 1500 rpm and 90 Nm at 3000 rpm. Estimate the torque at 2250 rpm.

Given:
\(x_0 = 1500\,rpm, y_0 = 50\,Nm\)
\(x_1 = 3000\,rpm, y_1 = 90\,Nm\)
\(x = 2250\,rpm\)

$$ y = 50 + \frac{2250 - 1500}{3000 - 1500} (90 - 50) $$

$$ y = 50 + \frac{750}{1500} \times 40 $$

$$ y = 50 + 20 = 70\,Nm $$

Answer: 70 Nm

3.1.2 Understanding Extrapolation

Extrapolation is the process of estimating a value outside the range of known data points. In Automotive Mechatronics, extrapolation helps predict sensor or system behavior beyond tested conditions, such as estimating engine performance at higher speeds or temperatures not directly measured.

The formula for linear extrapolation using two known points \((x_0, y_0)\) and \((x_1, y_1)\) to estimate \(y\) at \(x\) where \(x < x_0\) or \(x > x_1\) is:

$$ y = y_0 + \frac{(x, x_0)}{(x_1 - x_0)} (y_1 - y_0) $$

This is the same formula as interpolation but applied outside the known interval.

Worked Examples

Example 1: A sensor outputs 2.0 V at 20°C and 3.0 V at 40°C. Estimate the sensor voltage at 50°C by extrapolation.

Given:
\(x_0 = 20^\circ C, y_0 = 2.0\,V\)
\(x_1 = 40^\circ C, y_1 = 3.0\,V\)
\(x = 50^\circ C\)

$$ y = 2.0 + \frac{50 - 20}{40 - 20} (3.0 - 2.0) $$

$$ y = 2.0 + \frac{30}{20} \times 1.0 $$

$$ y = 2.0 + 1.5 = 3.5\,V $$

Answer: 3.5 V

Example 2: A fuel consumption rate is 8 L/100 km at 60 km/h and 6 L/100 km at 80 km/h. Estimate the fuel consumption at 90 km/h.

Given:
\(x_0 = 60\,km/h, y_0 = 8\,L/100\,km\)
\(x_1 = 80\,km/h, y_1 = 6\,L/100\,km\)
\(x = 90\,km/h\)

$$ y = 8 + \frac{90 - 60}{80 - 60} (6 - 8) $$

$$ y = 8 + \frac{30}{20} \times (-2) $$

$$ y = 8 - 3 = 5\,L/100\,km $$

Answer: 5 L/100 km

Example 3: An engine produces 200 Nm torque at 3000 rpm and 300 Nm at 4000 rpm. Estimate the torque at 4500 rpm.

Given:
\(x_0 = 3000\,rpm, y_0 = 200\,Nm\)
\(x_1 = 4000\,rpm, y_1 = 300\,Nm\)
\(x = 4500\,rpm\)

$$ y = 200 + \frac{4500 - 3000}{4000 - 3000} (300 - 200) $$

$$ y = 200 + \frac{1500}{1000} \times 100 $$

$$ y = 200 + 150 = 350\,Nm $$

Answer: 350 Nm

Example 4: The output voltage of a sensor is 4.0 V at 100 kPa and 6.0 V at 150 kPa. Estimate the voltage at 80 kPa.

Given:
\(x_0 = 100\,kPa, y_0 = 4.0\,V\)
\(x_1 = 150\,kPa, y_1 = 6.0\,V\)
\(x = 80\,kPa\)

$$ y = 4.0 + \frac{80 - 100}{150 - 100} (6.0 - 4.0) $$

$$ y = 4.0 + \frac{-20}{50} \times 2.0 $$

$$ y = 4.0 - 0.8 = 3.2\,V $$

Answer: 3.2 V

Example 5: A battery voltage is 12.0 V at 25°C and 11.5 V at 40°C. Estimate the voltage at 10°C.

Given:
\(x_0 = 25^\circ C, y_0 = 12.0\,V\)
\(x_1 = 40^\circ C, y_1 = 11.5\,V\)
\(x = 10^\circ C\)

$$ y = 12.0 + \frac{10 - 25}{40 - 25} (11.5 - 12.0) $$

$$ y = 12.0 + \frac{-15}{15} \times (-0.5) $$

$$ y = 12.0 + 0.5 = 12.5\,V $$

Answer: 12.5 V

3.1.3 Linear Interpolation versus Polynomial Interpolation

Linear interpolation estimates values assuming a straight line between two known points. Polynomial interpolation uses higher-degree polynomials to fit multiple points, providing a smoother and potentially more accurate estimation for nonlinear data, common in automotive sensor output curves.

The polynomial interpolation formula for \(n+1\) data points \((x_0, y_0), (x_1, y_1), ..., (x_n, y_n)\) can be expressed as:

$$ P(x) = \sum_{i=0}^n y_i \prod_{\substack{j=0 \\ j \neq i}}^n \frac{x, x_j}{x_i, x_j} $$

This is the Lagrange interpolation polynomial.

Worked Examples

Example 1: Given sensor voltages at rpm: (1000 rpm, 2.0 V), (1500 rpm, 3.0 V), and (2000 rpm, 5.0 V), estimate voltage at 1250 rpm using polynomial interpolation.

Given:
Points: \((x_0,y_0) = (1000, 2.0)\), \((x_1,y_1) = (1500, 3.0)\), \((x_2,y_2) = (2000, 5.0)\)
Estimate \(P(1250)\)

Calculate the Lagrange basis polynomials:

$$ L_0(1250) = \frac{(1250 - 1500)(1250 - 2000)}{(1000 - 1500)(1000 - 2000)} = \frac{(-250)(-750)}{(-500)(-1000)} = \frac{187500}{500000} = 0.375 $$

$$ L_1(1250) = \frac{(1250 - 1000)(1250 - 2000)}{(1500 - 1000)(1500 - 2000)} = \frac{(250)(-750)}{(500)(-500)} = \frac{-187500}{-250000} = 0.75 $$

$$ L_2(1250) = \frac{(1250 - 1000)(1250 - 1500)}{(2000 - 1000)(2000 - 1500)} = \frac{(250)(-250)}{(1000)(500)} = \frac{-62500}{500000} = -0.125 $$

Calculate \(P(1250)\):

$$ P(1250) = 2.0 \times 0.375 + 3.0 \times 0.75 + 5.0 \times (-0.125) $$

$$ P(1250) = 0.75 + 2.25 - 0.625 = 2.375\,V $$

Answer: 2.375 V

Example 2: Given torque values at engine speeds: (1500 rpm, 100 Nm), (2000 rpm, 140 Nm), (2500 rpm, 180 Nm), find torque at 1750 rpm.

Calculate:

$$ L_0(1750) = \frac{(1750 - 2000)(1750 - 2500)}{(1500 - 2000)(1500 - 2500)} = \frac{(-250)(-750)}{(-500)(-1000)} = \frac{187500}{500000} = 0.375 $$

$$ L_1(1750) = \frac{(1750 - 1500)(1750 - 2500)}{(2000 - 1500)(2000 - 2500)} = \frac{(250)(-750)}{(500)(-500)} = \frac{-187500}{-250000} = 0.75 $$

$$ L_2(1750) = \frac{(1750 - 1500)(1750 - 2000)}{(2500 - 1500)(2500 - 2000)} = \frac{(250)(-250)}{(1000)(500)} = \frac{-62500}{500000} = -0.125 $$

Calculate \(P(1750)\):

$$ P(1750) = 100 \times 0.375 + 140 \times 0.75 + 180 \times (-0.125) $$

$$ P(1750) = 37.5 + 105 - 22.5 = 120\,Nm $$

Answer: 120 Nm

Example 3: A temperature sensor outputs 1.0 V at 10°C, 1.5 V at 20°C, and 3.0 V at 40°C. Estimate voltage at 30°C.

Calculate:

$$ L_0(30) = \frac{(30 - 20)(30 - 40)}{(10 - 20)(10 - 40)} = \frac{(10)(-10)}{(-10)(-30)} = \frac{-100}{300} = -\frac{1}{3} $$

$$ L_1(30) = \frac{(30 - 10)(30 - 40)}{(20 - 10)(20 - 40)} = \frac{(20)(-10)}{(10)(-20)} = \frac{-200}{-200} = 1 $$

$$ L_2(30) = \frac{(30 - 10)(30 - 20)}{(40 - 10)(40 - 20)} = \frac{(20)(10)}{(30)(20)} = \frac{200}{600} = \frac{1}{3} $$

Calculate \(P(30)\):

$$ P(30) = 1.0 \times (-\frac{1}{3}) + 1.5 \times 1 + 3.0 \times \frac{1}{3} $$

$$ P(30) = -0.333 + 1.5 + 1.0 = 2.167\,V $$

Answer: 2.167 V

Example 4: Engine power at speeds: (1200 rpm, 30 kW), (1800 rpm, 45 kW), (2400 rpm, 60 kW). Find power at 1500 rpm.

Calculate:

$$ L_0(1500) = \frac{(1500 - 1800)(1500 - 2400)}{(1200 - 1800)(1200 - 2400)} = \frac{(-300)(-900)}{(-600)(-1200)} = \frac{270000}{720000} = 0.375 $$

$$ L_1(1500) = \frac{(1500 - 1200)(1500 - 2400)}{(1800 - 1200)(1800 - 2400)} = \frac{(300)(-900)}{(600)(-600)} = \frac{-270000}{-360000} = 0.75 $$

$$ L_2(1500) = \frac{(1500 - 1200)(1500 - 1800)}{(2400 - 1200)(2400 - 1800)} = \frac{(300)(-300)}{(1200)(600)} = \frac{-90000}{720000} = -0.125 $$

Calculate \(P(1500)\):

$$ P(1500) = 30 \times 0.375 + 45 \times 0.75 + 60 \times (-0.125) $$

$$ P(1500) = 11.25 + 33.75 - 7.5 = 37.5\,kW $$

Answer: 37.5 kW

Example 5: Battery voltage readings at temperatures: (0°C, 11.5 V), (25°C, 12.5 V), (50°C, 13.0 V). Estimate voltage at 10°C.

Calculate:

$$ L_0(10) = \frac{(10 - 25)(10 - 50)}{(0 - 25)(0 - 50)} = \frac{(-15)(-40)}{(-25)(-50)} = \frac{600}{1250} = 0.48 $$

$$ L_1(10) = \frac{(10 - 0)(10 - 50)}{(25 - 0)(25 - 50)} = \frac{(10)(-40)}{(25)(-25)} = \frac{-400}{-625} = 0.64 $$

$$ L_2(10) = \frac{(10 - 0)(10 - 25)}{(50 - 0)(50 - 25)} = \frac{(10)(-15)}{(50)(25)} = \frac{-150}{1250} = -0.12 $$

Calculate \(P(10)\):

$$ P(10) = 11.5 \times 0.48 + 12.5 \times 0.64 + 13.0 \times (-0.12) $$

$$ P(10) = 5.52 + 8.0 - 1.56 = 11.96\,V $$

Answer: 11.96 V

3.1.4 Practical Application of Interpolation and Extrapolation in Automotive Mechatronics

Interpolation and extrapolation enable Automotive Mechatronics professionals to estimate sensor outputs, control system parameters, and performance metrics critical for vehicle diagnostics and calibration. These methods are essential when sensor data is sampled discretely or when operating conditions fall outside tested ranges, for example, during engine tuning or emissions testing.

Worked Examples

Example 1: An oxygen sensor voltage is 0.4 V at 10% oxygen concentration and 0.9 V at 20%. Estimate the sensor voltage at 15% oxygen.

Given:
\(x_0 = 10\%, y_0 = 0.4\,V\)
\(x_1 = 20\%, y_1 = 0.9\,V\)
\(x = 15\%\)

$$ y = 0.4 + \frac{15 - 10}{20 - 10} (0.9 - 0.4) $$

$$ y = 0.4 + \frac{5}{10} \times 0.5 $$

$$ y = 0.4 + 0.25 = 0.65\,V $$

Answer: 0.65 V

Example 2: A coolant temperature sensor reads 85°C at 2.0 V and 105°C at 2.5 V. Estimate temperature at 2.3 V.

Given:
\(x_0 = 2.0\,V, y_0 = 85^\circ C\)
\(x_1 = 2.5\,V, y_1 = 105^\circ C\)
\(x = 2.3\,V\)

$$ y = 85 + \frac{2.3 - 2.0}{2.5 - 2.0} (105 - 85) $$

$$ y = 85 + \frac{0.3}{0.5} \times 20 $$

$$ y = 85 + 12 = 97^\circ C $$

Answer: 97 °C

Example 3: A vehicle speed sensor outputs 1.5 V at 40 km/h and 2.5 V at 80 km/h. Estimate speed at 3.0 V by extrapolation.

Given:
\(x_0 = 40\,km/h, y_0 = 1.5\,V\)
\(x_1 = 80\,km/h, y_1 = 2.5\,V\)
\(x = 3.0\,V\) (Note: Here voltage is independent variable, so invert interpolation)

Calculate speed \(v\) at voltage \(x=3.0\,V\):

$$ v = 40 + \frac{3.0 - 1.5}{2.5 - 1.5} (80 - 40) $$

$$ v = 40 + \frac{1.5}{1.0} \times 40 $$

$$ v = 40 + 60 = 100\,km/h $$

Answer: 100 km/h

Example 4: An injector flow rate is 120 ml/min at 1500 rpm and 160 ml/min at 2500 rpm. Estimate flow rate at 3000 rpm by extrapolation.

Given:
\(x_0 = 1500\,rpm, y_0 = 120\,ml/min\)
\(x_1 = 2500\,rpm, y_1 = 160\,ml/min\)
\(x = 3000\,rpm\)

$$ y = 120 + \frac{3000 - 1500}{2500 - 1500} (160 - 120) $$

$$ y = 120 + \frac{1500}{1000} \times 40 $$

$$ y = 120 + 60 = 180\,ml/min $$

Answer: 180 ml/min

Example 5: A battery voltage is 12.4 V at 25°C and 12.0 V at 40°C. Estimate voltage at 10°C by extrapolation.

Given:
\(x_0 = 25^\circ C, y_0 = 12.4\,V\)
\(x_1 = 40^\circ C, y_1 = 12.0\,V\)
\(x = 10^\circ C\)

$$ y = 12.4 + \frac{10 - 25}{40 - 25} (12.0 - 12.4) $$

$$ y = 12.4 + \frac{-15}{15} \times (-0.4) $$

$$ y = 12.4 + 0.4 = 12.8\,V $$

Answer: 12.8 V

Chapter Summary

This chapter covers the fundamental concepts and techniques of numerical methods used in engineering mathematics. It begins by defining interpolation and extrapolation as methods for estimating unknown values within or beyond a set of known data points. The chapter then explores practical applications of interpolation, demonstrating how it can be used to find intermediate values in engineering problems such as material properties and structural analysis. Following this, the focus shifts to iterative methods, which provide systematic approaches for solving equations that cannot be easily addressed through direct methods. Various iterative techniques are explained and applied to solve algebraic equations commonly encountered in engineering calculations. Throughout the chapter, emphasis is placed on step-by-step procedures and calculations to develop proficiency in these numerical methods. This foundation equips students with essential tools for accurate and efficient problem solving in engineering contexts.