The wire diameter   d = 4 mm and the external diameter   D_o = 30 mm, so the mean coil diameter   D = D_o - d = 26 mm and the index is   C = D/d = 6.5
Stress factors from ( 1) are   K_s = 1+0.5/C = 1.08;   K_h = (C+0.6)/(C-0.67) = 1.22

The total number of turns is now counted. This is a somewhat inaccurate process if the spring cannot be inspected physically. The leftmost coil here ends in a feather edge at the top, and so the end of the wire (imagined before grinding) must coincide with the vertical plane which contains the spring axis. Starting to count from this vertical plane - following the wire around towards the observer, then downwards, around the back and upwards to meet the plane again at the point 'a' illustrated - 'a' therefore corresponds to one complete turn.
Repeating this, we arrive at point 'b' after 11 complete turns. Continuing from 'b', the bottom is reached (another half turn) without the wire yet ground to a feather edge. So the wire must end at about 3/4 turn after 'b', therefore   n_t ≈ 11³/₄ turns. Ends are not ground right down to feather edges in practice, as shown by this cadmium plated hydraulic valve return spring.

From Table 1 the active turns are   n_a = n_t -2 = 9³/₄; the solid length is   L_s = n_td = 47 mm, and the pitch is   p = ( L_o -2d)/n_a = ( 85 -8)/9³/₄ = 7.9 mm. The corresponding helix angle is   α = arctan( p/D) = 5.5^o - the spring is certainly close coiled.

From ( 2) :   k = Gd /8 n_aC³ = 79E3 x 4 / 8 x 9.75 x 6.5³ = 14.8 N/mm

The solid deflection   δ_s = L_o -L_s = 38 mm, so the solidification load is   F_s = k δ_s = 14.8 x 38 = 560 N.
Assuming that solidification is infrequent and outside the working range, then this load will be treated as static so that the direct shear stress factor   K_s applies in ( 1). The corresponding solid shear stress from ( 1) is therefore :
              τ_s = K_s 8 F_sC / π d² = 1.08 x 8 x 560 x 6.5 / π 4² = 625 MPa.

Assume that at the maximum working state the above- mentioned 10% minimum clash allowance applies, so that   δ_s - δ_hi ≥ 0.1 δ_hi   ie. δ_hi ≤ δ_s/1.1 = 38/1.1   say   δ_hi = 34 mm.
Proceeding as with the solid state,   F_hi = k δ_hi = 14.8 x 34 = 505 N and so :
              τ_hi = K_s 8 F_hi C / πd² = 1.08 x 8 x 505 x 6.5 /π 4² = 563 MPa - static conditions still assumed.

Considering stability at this maximum state and assuming hinged ends (λ = 1), then   λ L_o/c₂D = 1 x 85 /2.62 x 26 = 1.25. Since this is greater than unity buckling could occur, so investigate via ( 3a):
              δ_c = { 1 - √( 1 - 1/1.25² )} x 85/1.23 = 27 mm
                            and since the maximum deflection   δ_hi exceeds δ_c then buckling will occur unless the spring is supported by a rod or surrounding cylinder. Alternatively if end rotation is prevented ( λ = 0.5) then   λL_o/c₂D < 1 which automatically guarantees stability.