I believe you need all the conditions for the answer to be valid --- in your set up there is no primary force (the pull of the strings) so your applied force at the vector location is the only force.
In other words the equation is valid only if you simulate the pull of the strings too. The sound board survives 34lbs pushing down because there is 131 lbs of force pulling it taunt something like a suspension bridge.
One of our brainiacs should be able to come up with an equation.
Yet another math question
Re: Yet another math question
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Re: Yet another math question
You may be right.
The equation, though, does not need real-world verification: it does yield the vertical vector pressure ; it is a mathematical certainty (if the equation is valid, and it appears to be).
Which means, to my way of thinking, that a braced top in a proper testing jig, when 34 lbs is applied it, it is AS IF all the forces were in play - all the top knows at that particular point is that 34 lbs is pressing on it vertically. And so I should be able to measure deflection at that point.
In other words, if I had the instrument completed and strung up, the pressure at that same point would be - 34lbs. The top does not know if the 34 pounds is a straight vertical pressure or if it is a vector of a more complicated torque.
I think.
The equation, though, does not need real-world verification: it does yield the vertical vector pressure ; it is a mathematical certainty (if the equation is valid, and it appears to be).
Which means, to my way of thinking, that a braced top in a proper testing jig, when 34 lbs is applied it, it is AS IF all the forces were in play - all the top knows at that particular point is that 34 lbs is pressing on it vertically. And so I should be able to measure deflection at that point.
In other words, if I had the instrument completed and strung up, the pressure at that same point would be - 34lbs. The top does not know if the 34 pounds is a straight vertical pressure or if it is a vector of a more complicated torque.
I think.
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Re: Yet another math question
"If I had the instrument strung up"
Exactly
Exactly
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Re: Yet another math question
This has been an interesting discussion.
I think the amount of downward pressure on the top necessary to deflect it a given distance differs in the two cases under discussion. That is, the unstrung top would take less downward pressure than a strung one, to deflect the same distance. That's because the tightened strings add their own resistance to vertical deflection to the situation, as well as adding similar horizontal tension to the wood, which makes it resist vertical deflection more too when strung up. Adding to the complications, the higher the string tension, the more the bridge wants to rotate, which creates the well known S-curve, and seems to increase downward deflection but not really. The S-curve is mostly the result of rotational force, not downward pressure. Its fulcrum point is the top of the saddle. In the S-curve the rear of the bridge and nearby top wood on the rear side is raised while the wood towards the sound hole is depressed. The depressed side appears to be downward deflection psychologically speaking, but in reality is simply the balancing response to the raising of the back side of the bridge as the whole system undergoes rotation.
My speculation is if you want to decrease the S-curve effect on the top between the bridge and sound hole, decrease string tension. If you want to decrease pure downward deflection of the bridge, increase string tension.
"It stands to reason" - a phrase my high school physics teacher hated - but that is all the rationale I can offer. Like all intuitive conclusions, actual measurements might just contradict it.
I think the amount of downward pressure on the top necessary to deflect it a given distance differs in the two cases under discussion. That is, the unstrung top would take less downward pressure than a strung one, to deflect the same distance. That's because the tightened strings add their own resistance to vertical deflection to the situation, as well as adding similar horizontal tension to the wood, which makes it resist vertical deflection more too when strung up. Adding to the complications, the higher the string tension, the more the bridge wants to rotate, which creates the well known S-curve, and seems to increase downward deflection but not really. The S-curve is mostly the result of rotational force, not downward pressure. Its fulcrum point is the top of the saddle. In the S-curve the rear of the bridge and nearby top wood on the rear side is raised while the wood towards the sound hole is depressed. The depressed side appears to be downward deflection psychologically speaking, but in reality is simply the balancing response to the raising of the back side of the bridge as the whole system undergoes rotation.
My speculation is if you want to decrease the S-curve effect on the top between the bridge and sound hole, decrease string tension. If you want to decrease pure downward deflection of the bridge, increase string tension.
"It stands to reason" - a phrase my high school physics teacher hated - but that is all the rationale I can offer. Like all intuitive conclusions, actual measurements might just contradict it.
John
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Re: Yet another math question
Thanks John.
Is the diagram correct? Will the deflection be the same in both scenarios?
Is the diagram correct? Will the deflection be the same in both scenarios?
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Re: Yet another math question
No -- the string tension is torquing the bridge/sound board assembly, the 34 lbs is far more than you need to get the same amount of deflection without the primary force. The tension of the strings pulls the sound board up into a dome --- right?
You need an equation that includes these parameters/factors.
You need an equation that includes these parameters/factors.
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Re: Yet another math question
Using the calculator on this luthier page:
http://www.liutaiomottola.com/formulae/downforce.htm
For a movable bridge instrument - 131 lbs tension and 15 degree break angle does calculate to 34 lbs of downward force.
Now I know that a flattop bridge has different dynamics, but this does show that 34 pounds is in the ballpark. I think.
http://www.liutaiomottola.com/formulae/downforce.htm
For a movable bridge instrument - 131 lbs tension and 15 degree break angle does calculate to 34 lbs of downward force.
Now I know that a flattop bridge has different dynamics, but this does show that 34 pounds is in the ballpark. I think.
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