I'm not sure I agree that a sharp static has higher brace tension than a rounded static.., but here's how I explain what a contact recurve does. Think of two pencils, a regular one and a "golf" pencil (short). Both are the same thickness, but the long one is easier to bend because the ends act like levers.If you hold the regular pencil with both hands near the middle, it is hard to bend just like a golf pencil.
When a recurve is at brace, it "acts" like a bow that is shorter- it acts like a bow that is only as long as the distance between the points of contact. This short bow is harder to draw because it is short.Hard to draw= high early draw weight. At full draw, the string lifts off the recurves, and now the bow "acts" like a bow that is longer, giving you more leverage to bend the bow. So a recurve acts like a short, stiff bow early in the draw cycle, and a long bow later in the draw cycle.
^----^ correct answer!
For science types refer to the following:
D = F * E ( L ^ 2 ) / ( W * ( H ^ 2 ) ) where
D is the displacement of a beam fixed at one end with a rectangular cross section of dimensions Length, Width, Height, modulus of Elasticity, and a Force applied to the loose end.
The above approximates a limb where the fixed end of the beam is the inflexible part of the fade. Where the limb is a more complex shape use calculus to integrate the solid.
Edit: fixed the equation