Does this prove David Vizard's 128 lsa formula ?

Depends on RPM. Most engines (Can't say EVERY engine because I haven't seen every engine from every era but...let's leave it there) want less timing at peak torque and more timing at peak power.

Where that peak power is will determine how much total you end up with. An engine with peak RPM at 6k won't have as much total as an engine peaking at 7500.

To that end, I'm not sure why being able to reduce timing is a bad thing. You want peak cylinder pressure by 14-18 degrees ATDC. If you have to fire the plug at 40 degrees BTDC you have pretty close to 35ish degrees of crank rotation where chamber pressure is working against making power.

If you can fire the plug on that same engine at say 30 BTDC you have now reduced that negative work by 10 degrees. Of course, that's as long as you still get peak cylinder done by 14-18 degrees ATDC.

Volumetric efficiency affects timing requirements. Not many people would argue that. So think about what a locked out or an all in curve by 2500 power does to power?

Not only are you increasing negative working pressure, all that timing is beating up parts.
No disagreement Turk. The situation is a matter of combustion efficiency beginning at the moment the spark takes place. A fraction of a second after ther spark occurs there is virtually no advance of the flame kernal as it establishes itself. Then slowly it begins to expand into the close molecules of air (O2) and fuel. There are still few molecules combusting yet, so heating of the cylinder charge is almost nonexistant. It takes 5° to 10° for the flame to expand to include much of the combustion chamber, so some heating but not yet a lot. At TDC most of the combustion chamber has flame in it, but combustion is not yet complete. That should be about 10° ATDC enabling maximum heating by around 15° ATDC. So yes there is some negative power producing pressure on the piston between the time of the spark and TDC. This is illustrated in DV's How to Build Horsepower book.