Here's the free body diagram of the whole truss:

If we assume no x direction forces on the truss and only y direction reactions we get a statically determinate truss.  That is we can write as many equations as there are unknowns.  Further, because the members AE and DH are glued to the spar intercostals we might choose to constrain our truss at A and H where diagonal members carry y direction loads into the spars.  Then there are no loads in members AE, EF, DH or CD! There is one other zero-force member between C and G.  It must be zero-force because the members CF, CH and FH form a triangular truss which takes the load imposed at C.  Then CG is a redundant member which can be ignored and the truss now looks like this:

Taking moments about A we get:  F_Hy = [44(9.1) + 29(18)]/27.3 = 33.8 lbs   and F_Ay = 73 - 33.8 = 39.2 lbs.

Then F_ABcos3 = F_AFcos37  and  F_AF = F_ABcos3/cos37

with F_AFcos53 + F_ABcos87 = 39.2

Then [F_ABcos3cos53/cos37] + F_ABcos87 = 39.2  and

    F_AB = 39.2/(cos3cos53/cos37+cos87) = 48.7 lbs  and  F_AF = 48.7cos3/cos37 = 60.9 lbs

With F_AB = 48.7 lbs  and the total F_Bx = 0 then  F_BC  must also be 48.7 lbs.  

Then the forces at each node begin to fall one by one and we solve the entire truss:

F_AB = 48.7
F_AF = 60.9
F_BC = 48.7
F_BF = 39.1
F_CF = 3.4
F_CH = 57.2
F_FH = 46.2

Or give this a try: https://valdivia.staff.jade-hs.de/fachwerk_en.html

The top chord of our truss is in tension and the bottom chord is in compression.  Just what we expect because a truss acts like a beam and a beam similarly loaded would have its top flange in tension and its bottom flange in compression.

The graphic shows a good reason why member 9 is there even though it has no load (in this load problem!). It helps prevent the long member between node 6 and 8 from buckling under compression. Members 11 and 13 are the critical members.  We might investigate whether 11 can take the compressive load without buckling and 13 can take the tensile load without rupturing.

The modulus of rupture for white pine is 9300 psi. Then the 0.0625 in² stick can take 581 lbs of tension!

The formula for the buckling stress of member 11 is 4π²EI/L² where E is the elasticity modulus, 1,460,000 psi, I is the second moment of area = s⁴/12 and L is the length.  Then

     buckling stress₁₁ = 4 x 3.14159² x 1460000 x .25⁴/12/11.4² = 144 lbs.

If the rib stitches go around the entire rib applying the loads at points 6 and 7, the internal loads on members 9 and 10 become low compression loads while the loads in the other members are practically unchanged.

Hmmmm ..ok

I feel the properties of the wing section require rib stitching and glue.....safety is paramount...thats why I wear a seat belt in my car....just in case.

The Gravity of the physics tell me...errr on the side of caution....just ask...... Steve Wittman... ohh thats right you cant..... Its a no brainer.

Rib stitching is good insurance

My 2 cents...break it down any way you want...Mine is rib stitched...so I dont need to worry at all....about that.....OK.

Your choice....IMO......Attitudes change...physics dont

I enjoy breathing......

Some take-aways from the analysis that might answer the original question:

1.  To load the rib truss (or any truss) properly, the loads must be applied at the joints.  This can be done with stitches running top to bottom or just the top.

2. Tensile forces in members exceed the pull of the rib stitching.  How the stitches are done won't change that.  The truss relies on strong joints everywhere.

3. The rib has a safety factor greater than 2 (at point A of the flight envelope, point D should be investigated because the load distribution changes).

4. We made some guesses and assumptions in order to simplify the analysis which shows why testing is done as backup. The analysis suggests how a rib might be tested for anyone so inclined and a way to estimate the ultimate strength.

5. Stitches away from the truss joints apply a bending load on the truss member.  It might be good to know how much bending the 1/4" square stick can take.  

Modulus of rupture is the bending strength characteristic.  MOR for western white pine is 9300 psi. The longest truss chord member is about 9". If a single stitch is put in the middle, the stress is 3FL/2s³. Then the maximum pull on the stitch has to be less than:

       F = 9300 x 2 x .25³/3/9 = 10.76 lbs

If two stitches are placed evenly the stress is 3F(L-L_i)/2s³, where L_i = 3 and the pull on each stitch must be less than:

       F/2 =   9300 x 2 x .25³/3/6/2 = 8 lbs    It seems like we're going in the wrong direction but the total force supported is 16 lbs

Way back before started building my max.
I used this method testing a single rib distributed loads of more than 100# found no sign of failure.

As noted with the top members being stressed in tension, and the lower members in compression, perhaps stitching around the capstrip is superior? My rural education makes me think putting a side load on beam in tension is better than side loading a beam in compression.

The distribution in Phil's post is probably more general while the distribution I posted was specific to point A on the flight envelope.  Point D might impose a higher load with a distribution similar to that in Phil's post. Nevertheless the method is the same.  My only criticism is the load on a rib nearest the root will have the greatest load so an average load should be factored by about 1.1 making the load not 69 lbs but 76 lbs.  And if all your flying is done at less than 75 mph you can be less concerned with the loads at D.

Why the glue might be enough:

From the load distribution, the maximum is 3.8 lbs over the first inch of rib.  3M publishes a peel strength for Fastbond of 15 lbs/linear inch. Fastbond and Stewarts Ekobond are thought by many to be the same thing.  Both are polychloroprene based.  The peel test was a "T" (90°) canvas-to-canvas test.  The strength is linearly and not area dependent so the 3.8 lbs load is spread over 2 inches of peel length.  The peel stress is likely a small fraction of the 90° stress if peel stress goes to 0 as peel angle is reduced to 0. Then the glue might be somewhere between 8 and a SWAG of 120 times as strong as necessary. For the fabric cement, a safety factor >8 might be reasonable.

Quoted from Stilson

As noted with the top members being stressed in tension, and the lower members in compression, perhaps stitching around the capstrip is superior? My rural education makes me think putting a side load on beam in tension is better than side loading a beam in compression.

Thinking of combined axial and bending loads hurts my head    but I think you're right.  The bending load puts the extreme fibers in tension on one side and compression on the other.  Then combining an axial force, the stress is additive.  Wood, being stronger in tension, can take the additional tension on the extreme fibers better than additional compression.  Or you may think that pulling on the member rather than pushing, helps reduce rather than increase deflection caused by bending which leads to buckling.

Both, but primarily in axial load inducing an initial bucklining factor that initiates collapse. Also I calculate peel strength chord wise not span wise ( I assume under normal loads it would be span wise, but any disturbance could result in chord wise peel) reducing the 2" area to 1/4", but I'm probably wrong.  I just had a bored childhood and lots of library cards😁