On Mon, 8 Jun 2009 17:48:21 -0700 (PDT), Jay Pique
wrote:

On Jun 8, 8:29*pm, Morris Dovey wrote:
....
Afterward I wondered if a catenary with that same 1/8" chord height
might apply more even pressure...

Hmmm...that's a question for the engineeers, but I like the way you
think. Is that sort of formula easily input into your CNC language?

JP

Assuming a caul with uniform stiffness - EI is constant along the
length of the caul - to get uniform pressure along the clamping
surface, the shape of the caul should be:

Y = Ymax * ( 1 - 3.20 * [ (X/L) - 2*(X/L)^3 + (X/L)^4 ])

where X is distance along the caul, L is the length of the caul
between the clamps at each end, Y is the chord height at any point
along the caul before applying the clamps, and Ymax is the maximum
chord height at the clamp location at each end.

The clamping pressure exerted by the caul - force/unit length. - is
given by

Q = 76.8 * EI * Ymax / L^4

where Q is the clamping pressure, EI is the caul stiffness (E =
Young's Modulus of the caul material, I is the cross section moment of
inertia) and Ymax is the maximum chord height of the unloaded caul at
the clamp positions.

Increase the pressure exerted by the caul by any combination of:
1) increase the maximum chord height
2) increase the caul stiffness
3) decrease the length of the caul between clamp positions.

Tom Veatch
Wichita, KS
USA