I am going to show my work as I am unsure of it. Perhaps someone can verify what I have done or correct it.
The question is, how large a toggle do you need to hold the halyard per the technique shown
HERE and discussed above. I am going to solve this for 5/32 Amsteel which is 4000 pounds breaking strength and assume that I want to be able to hold that load at breaking strength. Finding halyard load is not easy. The chart in the back of
Brion's book (page 372) for a 35 foot boat says 800 pounds. That would be a 5:1 safety factor, which sounds fine so I went with that.
The way the line is led is not exactly a knot but it does offer some stress relief to the toggle. I tested that around a very smooth post which for sure would be worst case compared to the friction of a head board. It took only 4 pounds to hold a 22 pound test weight. Thus the 4000 pound load could be held with 727 pounds on the toggle.
I assume a .75 inch opening in the headboard and 727 pounds gives a bending moment of 51 inch pounds per this formula ( w * l^2 ) / 8 .
Using a pin of radius r the load on the outer most point r from the center of the pin is 51 * r / J where J is the moment of inertia of a circle which is ( pi * r^4 ) / 2
Combining these the maximum stress as 32 / r^3. Aluminum is 35,000 psi which says r minimum is .097. That is a diameter of .194.
Oak is 7440psi so r is .16 and diameter is .325 or just over 5/16. Being wood, perhaps an additional safety factor would be appropriate and a 3/8 to 1/2 inch toggle would give that. Wood has the additional issue that the surface might crush and I did not analyze that.
I would love for someone to verify these numbers and I would encourage nobody to use them or rely on them until that happens.