Acoustic Guitar Construction

Yes and there is an equal upwards force because of the break angle at the bridge pins. Have you ever had a bridge pin fly across the room?

If you understand my previous posts you will see that there is nothing that I said the contradicts Somogyi's statement.

I do understand your posts, John.

Sorry, I agree that you can put a strain gauge under the saddle and measure quite a load. If fact what you calculated. But how would one use that to determine the force to use in a deflection jig? That is my problem with the test you were describing.

The things John says is true. I answered with my formula, cause Dave was specific asking about the downward force.
The answer was that the formula gives the downforce to the saddle. But John's statement also says that the upward force is equal. Correct imo. It attaches to the point where the strings meet the bridge/bridgeplate.

At that point the same formula is valid:


So to the top there are 2 forces, equal, but at different places.
Here is what that looks like and it shows why there is a rotating effect because of them.


That is why Siminoff imitates a rotation/deformation and not just a downforce effect.
Herman

Understood.

Like I said, understood.
Is there any doubt, then, that at the saddle position, the top will in fact deflect (albeit as a part of a rotation) and that deflection can be measured?

In my opinion sinking of the top in front of the bridge vs raising of the top back to the bridge is not repeatable for different guitars. Unless you make all the guitars the same. Because the rotation depends on the pivot point.

The front has more resistance due to the fact that it lays almost on the stiffest part of the top. Being the X-joint.
The back lays almost "free" to only the top plate.
Therefore the backside wants to raise more than the front wants to drop. The pivot point will be more to the front of the bridge. So measuring at the top itself is tricky.

To be more consistant in measuring, one could better (again imo) look how the angle of the bridge changes.

(BTW In the example above the torque to the bridge is about 2 x 150 Newton (260 lbs) times the distance to the pivot point (lets say 12 mm = 1/2"). That would make the torque 300 x 0,012 = 3.6 Nm.
For those who are not familiar with the matter: It is about a third of torque you would tighten a 1/4" bolt with)