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Temperature Applications

Advanced temperature operations including averaging, interpolation, and real-world calculations.

Goal: Master advanced temperature operations and understand their practical applications
Time: ~10 minutes

Prerequisites: Complete the Temperature Handling tutorial first

Temperature Averaging and Interpolation

For averaging or interpolating between temperature points, use midpoint and lerp functions:

// ce-embed height=750 compiler=clang2110 flags="-std=c++23 -stdlib=libc++ -O3" mp-units=trunk
#include <mp-units/core.h>
#include <mp-units/systems/si.h>
#include <iostream>

int main()
{
  using namespace mp_units;
  using namespace mp_units::si::unit_symbols;

  // Two temperature measurements
  quantity_point temp1 = point<deg_C>(10);  // 283.15 K
  quantity_point temp2 = point<deg_C>(20);  // 293.15 K

  std::cout << "Temperature 1: " << temp1.quantity_from_zero() << "\n";
  std::cout << "Temperature 2: " << temp2.quantity_from_zero() << "\n\n";

  // Average using midpoint
  quantity_point avg_temp = midpoint(temp1, temp2);
  std::cout << "Midpoint: " << avg_temp.quantity_from_zero() << "\n";

  // General interpolation with lerp
  quantity_point quarter = lerp(temp1, temp2, 0.25);        // 25% of the way
  quantity_point half = lerp(temp1, temp2, 0.5);            // Same as midpoint
  quantity_point three_quarter = lerp(temp1, temp2, 0.75);  // 75% of the way

  std::cout << "25% interpolation: " << quarter.quantity_from_zero() << "\n";
  std::cout << "50% interpolation: " << half.quantity_from_zero() << "\n";
  std::cout << "75% interpolation: " << three_quarter.quantity_from_zero() << "\n\n";

  // Extrapolate beyond the range (ratio > 1.0 or < 0.0)
  quantity_point extrapolated = lerp(temp1, temp2, 1.5);
  std::cout << "150% extrapolation: " << extrapolated.quantity_from_zero() << "\n";

  std::cout << "\n✅ midpoint and lerp handle temperature point arithmetic correctly!\n";
}

Functions:

  • midpoint(a, b) finds the exact middle between two points
  • lerp(a, b, ratio) performs linear interpolation: \(a + \text{ratio} \times (b - a)\) for any ratio

Key insight: midpoint and lerp correctly handle affine space arithmetic for temperature points. They compute results using the underlying absolute scale (working with differences between points), which gives the correct answer of 15 °C when interpolating between 10 °C and 20 °C. Without these functions, you'd need to manually compute differences and ensure proper handling of the point vs quantity distinction.

Real-World: Ideal Gas Law

The ideal gas law (\(PV = nRT\)) requires absolute temperature:

// ce-embed height=650 compiler=clang2110 flags="-std=c++23 -stdlib=libc++ -O3" mp-units=trunk
#include <mp-units/systems/si.h>
#include <iostream>

int main()
{
  using namespace mp_units;
  using namespace mp_units::si::unit_symbols;

  // Gas law: PV = nRT
  constexpr quantity R = 8.314 * J / (mol * K);  // gas constant
  quantity n = 1.0 * mol;                        // amount of substance
  quantity V = 0.024 * m3;                       // volume
  quantity_point T_point{27 * deg_C};            // 27°C

  // Temperature MUST be absolute (Kelvin)
  quantity T = T_point.in(K).quantity_from_zero();

  // Calculate pressure
  auto P = (n * R * T) / V;

  std::cout << "Temperature: " << T << " (" << T.in(deg_C) << " above absolute zero)\n";
  std::cout << "Pressure: " << P << "\n";
  std::cout << "Pressure in kPa: " << P.in(kPa) << "\n";

  // Using Celsius directly would give WRONG results!
}

Key insight: In mp-units' absolute temperatures are represented as affine space points, and points cannot be multiplied or divided with other quantities. Use .quantity_from_zero() to convert the point to an equivalent quantity (delta from absolute zero) that supports multiplication and division. This is a limitation of the affine space model—if absolute quantities were modeled differently, they could be used directly in formulas like \(PV = nRT\).

Challenges

  1. Interpolation: Find the temperature at 30% between 10 °C and 30 °C using lerp
  2. Heat capacity: Implement the formula \(\Delta Q = mc\Delta T\) where a 2 kg object with specific heat capacity \(c = 4.18 \, \text{kJ/(kg·K)}\) is heated from 20 °C to 80 °C. Calculate the energy required. (Hint: subtract the temperature points to get \(\Delta T\), which can be directly multiplied—no quantity_from_zero() needed for differences)

What You Learned?

✅ midpoint() finds the exact middle between two temperature points
✅ lerp() performs linear interpolation and extrapolation
✅ Celsius and Kelvin differences are equivalent in magnitude
✅ Scientific calculations require absolute temperature (Kelvin)

Congratulations!

You've completed all tutorials.

Next Steps: