and for FMT by I - I demixing never occurs for the values of λ considered here; indeed solutions of Eq. (48) only begin to exist at about λ = 20. For the N - N spinodals, there are no such analytic equations (for the Zwanzig model, see Ref. [37]). For practical reasons we rather solve which is equivalent [53] to solving det N = 0. In order to calculate the N - N spinodals we have to numerically evaluate the left hand side of Eq. (49). Exchanging the species labels in Eq. (49) one obtains the same results. The N - N spinodals are calculated to give an idea of the location of the N - N phase boundaries. For cases where there is N - N coexistence closed by critical point, the spinodal and the binodal coincide at the critical point. We have calculated the phase diagrams of binary platelet mixtures for seven different size ratios, λ = 1 . 1 , 1 . 4 , 1 . 7 , 2 , 2 . 5 , 4 and 5. Fig. 1a shows the results for the slightly asym- metric case λ = 1 . 1 in the ( x, c ) representation. The I - N transition is the only type of transition which we find for this size ratio. The FMT results show that the I - N transition concentrations at x = 0 are c$_{I}$ = 0 . 418 and c$_{N}$ = 0 . 46 respectively, in agreement with the monodisperse results [25]. At x = 1, the coexistence values are c$_{I}$ = 0 . 418 /λ 3 = 0 . 314 and c$_{N}$ = 0 . 46 /λ 3 = 0 . 346. The coexistence curves interpolate smoothly from x = 0 to x = 1. The tie lines joining coexisting isotropic and nematic phases are naturally verti- cal at x = 0 , 1 (corresponding to the cases of the pure systems of small and big platelets, respectively) whereas between x = 0 and x = 1 they vary in gradient, leaning with large positive gradient as x increases from zero composition, less so at about 50% composition and then more so again on approaching x = 1. Therefore, there is stronger fractionation at 50% composition than towards x = 0 and x = 1, such that the isotropic phase is dominated 16 III. RESULTS ( ∂p ∗ ∂ρ$_{2}$ )$_{µ}$$_{1}$ = 0 , (49) ( π 2 + 2 π $^{2}$ρ$_{1}$ + 2 ρ$_{2}$π $^{2}$λ2 + 1 ρ$_{1}$ ) × ( π $^{2}$λ3 + 2 ρ$_{2}$π $^{2}$λ6 + 2 ρ$_{1}$π $^{2}$λ4 + 1 ρ$_{2}$ ) − ( 2 ρ$_{1}$π $^{2}$λ2 + 2 ρ$_{2}$π $^{2}$λ$^{4}$) 2 = 0 . (48)