Yet another math question

[Dave Bagwill]

I'll email Roger and if he replies, I'll post it here.

[ken cierp]

Siminoff defines deflection tuning as "structure tuning" So an example you place an over braced sound board in the deflection fixture, apply the torque in the prescribed location -- shave the braces until the sound board begins to deflect --- done --- at that point it is known that the energy from a plucked string will cause the sound board to respond. In theory and in the book its explained that you could start with a solid piece of wood ---- which, and this was my point has no chance of any defection at the start of the procedure but as you thinned it down it will reach a point of where it will deflect.

Having a starting point of 2 mm deflection seems to me to be way too late in the process and a guess.

I can get my tiny brain around Roger's process -- particularity since the fixture itself is so intuitive --- no mystery as to its operation. By the way Dave I now remember that Roger used 58 lbs of torque in his example.

[ken cierp]

I rounded up my copy of The Luthier's HandBook --- he does show how to make the string tension torque conversion fixture using a torque wrench -- pretty simple.

[Dave Bagwill]

Yes I have that book - but what good is the torque test if I don't know what torque the given set of strings applies at pitch?
Siminoff use a deflection figure from a known top; but I want to approach it from the given string tension as the controlling factor.

[John Parchem]

Ken, I think we have a misunderstanding the 2 degrees is the angular rotation of the bridge, not a 2 mm deflection. I think the Siminoff fixture makes a lot of sense if it works as I am imagining it. You say apply torque so I am assume it is a rotational force delivering a downwards force to the top. The 2 degrees I am talking about is a measure of angular rotation. How much angular rotation do you think it would take before you are starting to see a deflection with Siminoff's jig?

I am just describing another method based on the same structural physics that can utilize the jig that Dave has in his pictures. A deflection test gives you the stiffness of the top. The rotation of the bridge is proportional to the stiffness of the top. So given a height of the top of the saddle, the estimated string tension and the discovered stiffness of the top you can get a good calculation of what the rotation will be. If you have a target rotation you can calculate a stiffness and from that a target deflection. You can pick what ever target you want. Zero is probably too stiff, 5 degrees if probably over driving the top.

I am sure with Siminoff jig one could change the load to account for different string tensions. If the jig had the correct geometry it could even be 1 to 1.

[ken cierp]

Ken, I think we have a misunderstanding the 2 degrees is the angular rotation of the bridge

Yep -- you are correct

[Herman]

Sorry sorry guys, I was so wrong, in the attempt to seem smart.
I asked my friend, who is a math wizzard, about the matter

He expained that it is as simple as Ken pointed at. Just vectors.



So for Daves example you just need the pulling force and the breakangle.

Here it is: downforce= Sin 15 degree x 131 lbs = 34 lbs

(BTW, Dave PSI is about pressure, not force)

Again forgive me. I am getting old. Next time I'll try to think before typing
Herman