1. An example: Taking the example discussed above of force F induced on a propeller blade, we have the equation 0 = φ (K, d, u, ρ, N, μ) There are m - n = 3 π groups, so φ ( π 1, π 2, π 3 ) = 0 The choice of ρ, u, d as the repeating variables satisfies the criteria above. They are measurable, good design parameters and, in combination, contain all the dimension M,L and T. We can now form the three groups according to the 2 nd theorem,

2. As the π groups are all dimensionless i.e. they have dimensions M 0 L 0 T 0 we can use the principle of dimensional homogeneity to equate the dimensions for each π group. For the first π group, in terms of dimensions

3. M 0 L 0 T 0 = (ML -3 ) a1 (LT -1 ) b1 (L) c1 MLT -2 For each dimension (M, L or T) the powers must be equal on both sides of the equation, so For M: 0 = a 1 + 1 a 1 = -1 For L: 0 = -3a 1 + b 1 + c 1 +1 0 = 4 + b 1 + c 1 For T: 0 = -b 1 – 2 b 1 = -2

4. c 1 = -4 – b 1 = -2 Giving π 1 as For π 2 we have, M 0 L 0 T 0 = (ML -3 ) a2 (LT -1 ) b2 (L) c2 T -1 For M: 0 =a 2 a 2 = 0

5. For L: 0 = -3a 2 – b 2 + c 2 0 = - b 2 + c 2 For T: 0 = -b 2 – 1 b 2 = -1 c 2 = 1 Giving π 2 as For π 3 we have

6. M 0 L 0 T 0 = (ML -3 ) a3 (LT -1 ) b3 (L) c3 ML -1 T —1 For M: 0 = a 3 + 1 a 3 = -1 For L: 0 = -3a 3 + b 3 + c 3 -1 0 = b 3 + c 3 +2 For T: 0 = -b 3 -1 b 3 = -1 c 3 = -1 Giving π 3 as

7. Thus the problem may be described by the following function of the three non-dimensional π groups,