When researching a question regarding flutter and V_ne I came across this:

https://ntrl.ntis.gov/NTRL/dashboard/searchResults/titleDetail/ADA955270.xhtml

The report contains a method for determining flutter resistance based on a wing torsional flexibility factor compared with the design dive speed.

The wing torsional flexibility factor, F, should be less than 200/V²_D.

          Where   F  =  ∫ Θ_i C² ds  and

𝜃𝑖 = Wing twist at station 𝑖, per unit torsional moment applied at a wing station outboard of the end of the aileron, (radians/ft-lb)
𝐶𝑖 = Wing chord length at station 𝑖, (ft)
𝑑𝑠 = Increment of span (ft)
𝑉𝑑 = Design dive speed (IAS)

The report describes "applying a torsional couple" near the wing tip and measuring the wing twist to find Θ_i at several locations along the span.  Then preparing a table for calculating F.  This is an attempt to do this for the Minimax mathematically.

Because the wing is constrained in torsion by the struts, I chose to calculate F for the 75" (6.25') cantilever section.

For the torsional couple I chose a 200 lbs force pushing up on the end of the main spar and a simultaneous 200 lbs down force on the rear spar end.  Then the torque on the wing is:

      T_w  =  200 lbs x 28" between the spars / 12 inches per foot  =  467 ft-lbs

To get the wing twist, Θ, we need to apply the formula for beam deflection:

       y =   P L³ / 3EI - (P L² / 2EI) x + (P/6EI) x³

Where P is the 200 lb end force
            L is the length of the beam, 6.25'
            E is the modulus of elasticity, 1460 kpsi for western white pine.
            I  is the moment of inertia for the "C" channel spar cross section

Here are some helpful web calculators: https://amesweb.info/Beam/cantilever-beam-with-point-load-at-free-end-calculator.aspx
                                                           https://calcresource.com/moment-of-inertia-channel.html

Since the report method specifies the chord and span increment in feet, we'll be careful to calculate I in ft⁴ and convert E to lbs/ft².  Alternatively, we could work in inches and factor the result by 0.007 (1/144) ft² per in².

With the known values and calculated moments of inertia the beam deflection formulas become:

       y_front  =  0.1306 - 0.0313 x + 0.000267 x³

       y_rear  =  0.2210 - 0.0530 x + 0.000453 x³

The wing twist in radians is the sum of the spar deflections divided by the distance between the spars, then combining like terms of the two deflection formulas into a single formula and dividing each term by 2.33 feet, the twist at any point x inboard from the tip is:

         Θ(x)  =  0.15 - 0.0362 x + 0.00031 x³

and dividing by the wing torsion, T_w, to get the twist per unit torsion:

         Θ(x)  =    0.00032 - 0.0000775 x + 0.000000664 x³

Then   F   =   4.5² ₀∫^6.25  (0.00032 - 0.0000775 x + 0.000000664 x³) dx

Integrating:  F  =   20.25( 0.00032 x  -  0.00003875 x²  +  0.000000166 x⁴ )  ]₀^6.25

                   F =  0.015

Then   V_Dmax =  √(200/F) =  115 mph and  V_ne = 104 mph (90% V_D)

Note that this result only takes into account the beam stiffness of the two spars.  It doesn't account for the additional beam stiffness nor the torsional stiffness supplied by the main spar 'D' tube.
            

      

Is this for the 1030 or 1100 or one of the other maxes, or do all maxes share the same spars?

Very fascinating though!  I like it when the math adds up to the official specs.

Quoted from ITman496

Is this for the 1030 or 1100 or one of the other maxes, or do all maxes share the same spars?

Very fascinating though!  I like it when the math adds up to the official specs.

Yes, this implies that for flutter resistance, all maxes have similar strength wings.

It may be interesting to note that the report is specified in EASA's certification rules for very light aeroplanes in criteria for establishing adequate flutter resistance.