What Dan does is make a compromise, choosing the correct figures at the tip and choosing a practical widthand thickness for the root.
If you use his board dimensions and start with the drop to give 3 deg at the tip and run this line out to drop equals full board thickness at the root you will get a straight line at the trailing edge that gives a drop to equal 6 deg at half radius and the width will be half way between tip width and full root board width.
This gives a satisfactory compromise over the outer 2/3 diameter where you do all the work, the root only provides strength to hold the thing together.
I don't think in degrees, just drop and width at each point but if you do it Dan's way the tangent of TE drop over width at the tip will give you 3 deg and when you measure the centre of the blade the tangent of TE drop over width at mid point should give you 6 deg.
The unloaded speed ratio of a given patch of the blade is the inverse of the slope of the chord at that radius. The loaded speed ratio is a constant times that (about 1/2), presuming you're trying to achieve the same angle-of-attack regardless of radius (which you probably are).
Since the speed of the blade goes with the radius and the wind speed is the same regardless of radius, the slope (tangent of the chord angle) has to double every time the radius halves if you want to keep the above relationship true.
If the arctangent of the angle at the tip is 3 degrees, the arctangent of twice that slope (the midpoint of the blades) will be just a hair under 6 degrees (and while the ideal change in slope is a curve, a straight line isn't all that far from it). And the arctangent of four times that slope (a quarter of the radius out) will be nontrivially under 12 degrees (and the straight-line approximation of the curve will be a worse approximation). It approaches 90 degrees as you approach the center of the shaft, and gets ridiculously steep near the hub. This makes the "ideal" shape VERY sloped. It also widens, because with more slope it has to be wider to present a given cross-section to the wind. At the hub it would be infinitely long, which is obviously silly.
But a patch at the tip with some given delta-radius has twice the swept area of a patch with the same delta-radius at the midpoint, four times the swept area of a patch a quarter of the way out, and so on. The inner 10% of the blade has 1% of the swept area. So as you get near the hub, and the "ideal" shape starts becoming enormous, it doesn't really cost you much power to just say "phooey on it" and stop trying to collect power at all.
Dan's approximation is really close to the ideal for the outer half of the blade, which sweeps 75% of the swept area, not too far off for the next quarter, which gets you up to 93 3/4%, and then fouls up progressively worse for the last 6 1/4%.
Well, lah-de-dah. If we're only getting 50% of the ideal on the inner quarter of the blade we lose 3 1/8% of the available power. If we're just a flat plate we lose about 6 1/4% (or maybe a little more from fouling up the airflow on the next patch out.) So let's just make the inner quarter of the blade strong - the bending stress gets very high there - and maybe give it a shape that gets us some starting torque without being a horrible air brake, a big weight that makes raising the tower difficult, or an over-optimized tree trunk that breaks the wood budget.