The length of the connecting rod shouldn't have an effect on the number of degrees the crankshaft must rotate to move the piston a certain value.
The length of the stroke however has a direct relationship because it determines the radius of the crank pin from the center of the crankshaft journals.
In the same book mentioned above by Roy Bacon it shows the stroke of the 500 twin (1949-1958) as 77mm (3.0315 inches).
That would make the radius of rotation 1.5157 inches.
Using the formula: (R-d)/R = Cos A, where R= radius of rotation, d= drop of the piston below TDC and A= angular rotation of crankshaft, for a piston drop of 1/32" BTDC or ATDC I get the following:
(1.5157- .0312)/1.5157 = 1.4845/1.5157 = 0.9794 = Cos A.
Using the Cos^-1 function for 0.9794, A = 11.6453 degrees.
For the 1958-1963 Meteor Minor the book shows the stroke as 64.5mm or 2.5394 inches.
The conrod length has a big influence on piston position,for any given crank angle (except for exactly WEYMTMTDC and BDC).You've left out 1/2 the formula which involves rod length and the cosine of the angle between the rod and cylinder axis.Your formula only applies when the rod is infinitely long.
The piston will always be below 1/2 stroke at 90 degrees before and after TDC.With a very short rod at 1/2 the stroke length (if this were mechanically possible),the piston would be at BDC 90 degrees after TDC.It would stay there and not move for 180 degrees of crank rotation.
I suggest you draw a circle of crankpin rotation.Put a short rod and a long rod at say 45 degrees from TDC.Compare that to the height of their small end bushes at TDC.All will become obvious.
Piston drop from TDC =R(1-cos A) + L - square root[L squared-(RsinA)squared],
Where L= rod length (centre-to centre),R= crank radius = stroke/2
and A =angle fromTDC