do tell how you know how much inflation pressure and what kind of lateral load transfer distribution
OK, well let me rephrase to "its simple to get to within a close approximation of the tyre contact patch pressures and force transfers"...
Now I know you're a smart guy, so please don't assume I'm teaching you how to suck eggs, but just for completeness and anyone else reading, the maths I have employed for tyre contact patch pressures are:-
1,Knowing that force up equals force down, otherwise the tyre would either be moving up or moving down, the force on say, one of the rear tyres of the Deltawing when stationary with its Fr:Rr weight distribution of 27.5:72.5 is:- 475kg x 0.725 x 9.81 x 0.5 (i.e. lateral weight distribution is 50:50) = 1689N.
2, Knowing the rear tyre width of the Deltawing is 320mm (from their website), and estimating (by good old honest observation of similarly sized real tyres) that the distance around the circumference of the tyre in contact is about 200mm, we get an area of 0.32x0.2 = 0.064m^2.
3, Assuming even pressure over that area (a bit simplistic, but used for both cars, so I feel a pretty valid assumption for a comparison), the contact pressure is 26390N/m^2, or 26.4kN/m^2
For Force transfers the maths I used is:-
1, The over-turning moment for any car is: Cornering G-Force x Mass x CG height. Since I have stated in my comparison that G-force, mass and CG height are the same for either car it means that the total over-turning moment is the same for both cars.
2, Taking, (for the purpose of
this example only) that Fr:Rr weight distribution is 50:50, and tyre size is equal front to rear it is a fairly safe-ish assumption to say that the front:rear roll stiffness is going to be about equal (to equalise tyre contact forces in steady-state cornering), and therefore the front and rear will equally share the over-turning load.
3, Taking moments about the inside tyres, the additional force at the outside tyre's contact patch (again for this example only) is then calculated by balancing the over-turning moment (for steady state cornering), i.e. half the over-turning moment must equal the additional tyre contact patch force x the track width. We know the track width, so we can calculate the additional tyre contact patch force.
4, For non-equal front:rear weight distribution and non-equal track widths you can change the numbers in the calcs and (I believe) get a good estimate for the affect of a Delta-shaped car on these forces. I wouldn't say 99% accurate, but 90%? I'd Probably go with that, if not 95%.
Even 10% inaccuracy isn't enough to change my conclusion -that a rectangular car with equal sized tyres and a conventional weight distribution and track widths is better able to distribute those forces to achieve better force distribution or tyre contact patch pressures for all situations other than the low speed straight line acceleration example.... wouldn't you agree?