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The break angle causes a down froce on the top of the saddle. There is also an up force on the string anchor. Anchor the strings right at the back edge of the saddle, and the up and down forces will be in the same place, more or less. If the strings are anchored back at the tail block, the up force acts there, rather than on the top itself, so the bridge only 'sees' a down force, not a torque. Between these two extremes you'll see the center of the rotation moving around on the top. How this will effect the sound......?

If the downward force on the saddle changes then something else has to change to keep things balanced.Maybe that was what Alan said.

_________________
A person who has never made a mistake has never made anything!!!

Hi all,
I thought before this thread dies I'd thank everyone for their responses. I've been following them intently, studying and pondering. I see the issue is a lot more involved then I bargained for, but the discussion has, if not provided a definitive answer, made me much more informed of the factors involved. Right or wrong, it has helped me develop a mental model as to how I think the bridge and saddle function, that I will use going forward. As my small contribution, I'll share a somewhat crude experiment I did to demonstrate to myself the principles being discussed.

I mounted a 6" long board on a pivot above my work bench , sort of like a see-saw, then added a 1 1/2" block at one end of it. This was my surrogate bridge and saddle ( colored grey in my diagram). Strung a line with just under 30 lbs of weight over an improvised roller, and measured the downward force on the front of my "bridge" with a dowel and bathroom scale. I took measurements with a steep break angle of about 55 deg. and a shallow break angle of about 20 deg (measured off my 6" board). Both ways , my bathroom scale read 18 lbs. I repeated it all a few times with the same results.


I think I'll let everyone draw their own conclusion as to the results, merit and faults of this. I think what it most proves is that I need to get a life!
Gerard

There is one thing wrong with this example , The both points of the strings need to be attached to the same structure. In this case you don't have a true guitar replica only a pulling load . You need to have the neck block also represented and the neck attachment. This makes the load of the string true to the structure.
All that you proved is that the energy of the string is constant you need to make a model like a guitar for this to be more accurate. As you made your model you are looking at total force and that is a constant it relation to the string tension . You have not separated the resultant forces available .

_________________
John Hall
blues creek guitars
Authorized CF Martin Repair
Co President of ASIA
You Don't know what you don't know until you know it

It would be easy enough to set up a test using a strip of spruce (or other material) onto which you would glue a one-string mock bridge with a mock saddle and a couple of bridge pin holes at different distances back from the saddle. The strip of spruce would have to be anchored at each end, free to bend along its length. Then you'd string it up with a guitar string, with the other end wound onto a tuning machine attached to something fixed. Tune the string to a specified pitch (so you know the tension is constant) and measure the deflection of the spruce with the ball end in the different bridge pin holes. After all the mathematical models and theories, a simple test like this would put the matter to rest.

Personally, I'm so convinced that the deflection would be the same (or the difference, due to flexing of the bridge, so negligible as to be irrelevant) no matter where the bridge pin holes are located relative to the saddle - the same as what it would be if you somehow fastened the string directly to the top of the saddle - that I'm not going to bother with the test, but if anybody were to do it and prove me wrong, I'd love to hear about it.

It doesn't seem to me that the rotational force actually has anything to do with the downward force on the saddle and the upward force on the bridge plate, but, rather, that it is simply a function of pulling the top of the saddle toward the neck. If you made a saddle that was actually a sort of staple-like bar, with the strings going underneath it and then UP to the top of some sort of tall posts where the bridge pins normally go, then you'd have an upward force on the saddle and a downward force at the bridge pin location. Would the bridge then torque the opposite way? No, because the top of the saddle would still be pulled toward the neck, putting - I think - the exact same rotational force on the bridge. But now here I go theorizing - a simple deflection test would settle it.

_________________
Todd Rose
Ithaca, NY

https://www.dreamingrosesecobnb.com/todds-art-music

https://www.facebook.com/ToddRoseGuitars/

gerry,

Good try!

My opinion is the pivot point will move with the string termination point. The pivot point of the bridge in your example (on the board that connected to your workbench) is constant. My opinion only, but I think the pivot point will be located at (or very close to) the termination point of the string where the pins would be placed.

You might try moving the pivot point with the string termination point in your test.

The one thing I haven't seen in this discussion (unless I overlooked it) is the affect of moving the string termination point to the moment arm.......the moment arm will definitely change.

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Formerly known as Adaboy.......

It doesn't seem to me that changing the bridge pin location would change the pivot point/axis of rotation. Changing the footprint of the bridge would change that. Again, though, this notion is integral to the postulate that the rotational force is solely a function the pulling of the top of the saddle toward the neck.

_________________
Todd Rose
Ithaca, NY

https://www.dreamingrosesecobnb.com/todds-art-music

https://www.facebook.com/ToddRoseGuitars/

My thought is no other point on the bridge has a force applied as large as the line on the bridge where the strings terminate (I've read it's up to 180 lbs of pull due to string tension). This force is at an angle to the soundboard so has a component in the "up" direction and another component in a direction toward the neck block. Seems the distributiion of this force to each component would vary with the string break angle.

The other force on the bridge is the downward force of the saddle that is created by the string tension over the saddle. This force varies with the string break angle. This force will always be less than the force applied at the string termination point.

It seems the forces creating a rotation force, or moment, on the bridge are the component of the string tension in the "up" direction and the downward force of the saddle (and in my mind both are affected by string break angle). Note that these two forces don't have to be equal as the stiffness of the soundboard and bracing will carry some of the load. Also note that the rotational force (or moment) is proportional to the "moment arm" which is the distance between the points where these forces are applied.

The string termination point is the location of the dominate force applied to the bridge. If true, is there any other point on the bridge that is so rigid that it will remain "fixed" causing the 180lb force to rock up and down? In my mind, the string termination point (where the 180 lb force is applied) will remain "fixed" and everything rotate around this point. I have "fixed" in quotes as we all know it is not truly fixed as the soundboard is moving up and down and the string length is constantly changing as the string vibrates so this point is actually moving. So again, what other point on the bridge is so stiff/rigid that another point on the bridge with 180 lb of force will rotate up and down while this point remains rigid? (as opposed to rotating around the point with 180 lb of tension)

I realize the stiffness of the entire top structure comes into play here but I don't know of any significant change in rigidness in the top bracing between the front and rear of the bridge that would dominate the 180lb force applied at the string termination point. If there is a significant change in rigidness here it would have to be in the shape of the bridge itself and most common designs don't change shape that drastically.

I may be wrong......and would sure like to learn if someone can point out the error in this logic.

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Formerly known as Adaboy.......

I had to go to CF Martin on business today and had this very talk with Tim Teal . He is one of the engineers and there is a difference of force per string attachment . Over all string height plays more into the actual torque then break angle. The string load on the top will not change only the foot print of the loaded area changes per the location of anchor points.
The top takes the load in more ways that we are talking about. The load is transmitted to the top , A as a compressive load between the bridge and neck block , then we have the rotation load at the bridge and tensional load from the bridge back. The part that can drive you crazy is how detailed you need to look at the loading. It is more involved than what most are saying.
The load on the top overall won't change unless the string load changes . I am not going to bore you with the math equations involved but they are out there if you want to do the study. There is more going on than just the bridge area. This is why the guitar will loose voicing as the neck angle changes and we get to reset time.

_________________
John Hall
blues creek guitars
Authorized CF Martin Repair
Co President of ASIA
You Don't know what you don't know until you know it

Gerry:
I like your experiment. The only thing I would point out is the fixed pivot point, as Daryl stated.

My measurements on a guitar suggest that the amount the saddle depresses depends on the height of the strings above the top, and stays the same (within the limits of my ability to measure it) when the break angle changes with the same saddle height.

The combination of the top depressing and rotation of the bridge means that there's a stationary point on the top behind the saddle, and that depends on the break angle. The lower the break angle, the further back the stationary point.

One of my students, who is an expert in statistics, is helping me evaluate the effect on the sound of string plucks of the various configurations I tested. Then we have to figure out what causes the differences, if there are any. Right now I'm of the opinion that string height off the top is the most important thing, but I'm not ruling out some effects from changing the break angle independantly of that.

That is what Tim Teal at martin stated. String height is the key. The higher the string height the more leverage available to deflect the top.

_________________
John Hall
blues creek guitars
Authorized CF Martin Repair
Co President of ASIA
You Don't know what you don't know until you know it

Darryl, I enjoyed and learned some things from your last post (I could say the same about many of the posts in this thread - thanks to everyone who's contributed). Here's my question (and I hope you'll take this all as good fun/inquisitiveness/exploring an interesting and relevant subject, not as "arguing" or criticizing your posts or anyone else's):

Is any point on the bridge really being lifted UP with a significant share of the 180 lbs of force? That strikes me like the Lorax picking himself up by the seat of his pants (Dr. Seuss). Let's assume the 180 lb pull of the strings between the nut and the bridge is in a direction parallel to the surface of the soundboard - it isn't exactly, but on a typical flat top guitar, it's pretty close. I don't see how you could change that direction of force (I mean the lion's share of the 180 lb force) into an upward force on any part of the bridge, no matter how you fasten the strings to it*. If the saddle went all the way down through the guitar and was anchored in the ground (and was perfectly rigid) then the direction of pull would be changed to upward, just like a rope over a pulley. But the saddle and bridge pin holes are part of the same structure (with extremely little flex between them) attached to the top of the guitar. Yes, they go up and down in relation to each other as the bridge is rotated by the pull on the top of the saddle; however, in relation to the force of the string tension, they act as a unit and are being pulled toward the nut, not up.

(*Just as in my imaginary turn-it-on-its-head example above, where the strings loop under a "saddle" and go upward to the bridge "pins", no major part of the 180 lbs of pull would be pushing down on the bridge pin area.)

If the top and bridge (including the bridge plate) were infinitely flexible and therefore totally compliant with the force being exerted on them by the strings, the saddle would simply depress to the point where the strings go straight over it; the bridge pin area would not move upward much if at all.

I'm not contesting the point that on a guitar in the real world, the back of the bridge is being lifted up as the front of the bridge is being pushed down, but I'm seeing it as simply the result of the bridge as a whole being rotated by the pull of the strings acting on it (in the direction of the nut) at the top of the saddle - that's where the bridge "sees" the pull of the strings, and that's what cranks it over, plain and simple. Therefore, as far as the rotational force is concerned, the only thing that matters is the total height of the bridge and saddle.

Am I crazy?

_________________
Todd Rose
Ithaca, NY

https://www.dreamingrosesecobnb.com/todds-art-music

https://www.facebook.com/ToddRoseGuitars/

I'm not an engineer and I don't play one on TV. I find allegory a great substitute for math.

Imagine a claw hammer (saddle) pulling out a nail (bridge pin)... the higher you put your hand on the hammer handle (the taller the saddle) the more upward force you will put on the back of the bridge and more downward force you would put on the front of the bridge. - The taller the saddle... the more rotational force.

However, in addition... put the nail (bridge pin) farther away from the fulcrum point of the hammer... the hammer would be LESS likely to pull the nail. I understand that the bridge pin isn't going to come out (like a nail) but it will put a torquing motion on the bridge plate/top/bridge area. If my thinking is correct... the closer the pins are to the bridge, the more the bridge plate/top/bridge area will be torqued.

I may also be crazy... but I've pull out lot's of nails.

_________________
"No man ever steps in the same river twice, for it's not the same river and he's not the same man.” -Heraclitus of Ephesus

Gerry, I see two incorrect assumptions in your model.

The first has already been mentioned, the saddle / bridge is not a monolithic structure in the real world.

The second is, your pivot point is a fixed “fulcrum”. In the real world, the bridge is on a flexible plate and the pivot point moves as you change the string terminus. I think this is key to understanding the forces and demonstrates why as John said it is very complex to model. (And we haven’t even started on the bracing underneath.)

In your model, you could tie the string to the top of the saddle and it wouldn’t make any difference.

Jim Hall

_________________
If you can't do something well, learn to enjoy doing it poorly.

Thanks, that's the way I like to discuss as well.....hopefully we can both learn as I have already. I've read so many of your posts Todd that I know you enough that I wouldn't take it any other way.



I would say yes, there is a lifting force in the up direction on the rear of the bridge. Where does it come from? The strings in the area between the saddle and the pin holes are still under tension. Look at the angle the strings are pulling on the rear of the bridge. If the strings have tension at this point, then some of that force is in the up direction (and most is in the up direction if the break angle is more than 45 deg).



I think I agree with you on this point. If the saddle doesn't maintain it's height, the break angle over the saddle goes to essentially 0 deg......so all the string tension would be in the horizontal plane of the strings.



Actually, I think we are all crazy! Why else would be sitting around pondering this queston?

I didn't address the saddle height on the bridge rotation as it wasn't addressed in the original posters question. Of course the break angle increases as you increase saddle height so a higher percentage of the string's tension is supported at the very top of the saddle. So if the break angle is 60 deg, and the bridge height is 1/2" and the string tension is 180 lb, then the force applied to the bridge saddle is 180lb * sin(60) = 156lb. And the rotation force would be equal to 156lb * 1/2" = 78in*lbs.

But that isn't the total rotational force on the bridge......that is just the torque applied to the bridge due to the horizontal string forces. You then have to add the torque caused by the vertical string forces where the string pulls up on the rear of the bridge and the saddle pushes down on the front of the bridge. Or at least that's how I picture it in my mind.

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Formerly known as Adaboy.......

Gerry's model is actually pretty good at showing the overall forces on the bridge.
What it does not measure (because it is not intended to) is the variation in force on the saddle with break angle.
It does show that the torque on the bridge remains the same if you vary the string attachment points.
Todd (Rose) you do ask some interesing questions.
One of the important methods in structural analysis is th separate an object from its supports and look at the forces acting on it and those reactions which maintaing equilibrium.
In the case of a pinned bridge I would separate it at the front and rear of the bridge through top plate and bridge plate(because the balls anchor here)
By analysing forces acting on the bridge (string tension at around 1/2" above the board) and taking moments about the front of the bridge you can determine
-compressive force in front of the bridge
-force at the front of the bridge
-force at the rear of the bridge


for a pinless bridge with the string anchoring within the bridge, I would make the separation at the glue joint between bridge and top plate and here you would see a froce trying to pull the bridge off the top at the rear.

It seems to me that the length of string behind the saddle is pulling equally down/back on the saddle and forward/up on the bridge pin location, and so those forces basically cancel each other out and do nothing with regard to the rotational force on the bridge as a whole. All they do is seat the saddle firmly down in its slot and seat the ball ends firmly up against the bridge plate. (They might flex the saddle/bridge a bit, but not much.) The effect is the same as if you had two separate pieces of string: one from the nut to the saddle (fastened, somehow, to the top of the saddle), and one from the saddle to the ball ends. No matter how much tension is in that short length of string from the saddle to the ball ends, it's not going to torque the bridge, but it will pull down on the saddle and up on the bridge pin area. The Lorax can't pick himself up by the seat of his pants! Well, in the story, he does, actually!

Imagine again, if you will, that the saddle is a horizontal bar (fastened at each end to the bridge) that the strings can loop around. Now picture the strings looping over the saddle, turning around 180 degrees and going forward, tied off to a set of screw eyes in front of the saddle (screw eyes also fastened to the bridge). Now, the short length of string between the saddle and the screw eye is pulling back toward the tail of the guitar. Would that backwards pull on the front of the bridge work against the forward pull from the main length of the string, changing the forces on the bridge as a whole? No, because it's canceled out by the forward pull of the same short length of string between the saddle and the screw eye (there's that Lorax again, trying to pick himself up by the seat of his pants). There would be no downward force on the saddle and nothing pulling directly up on the rear part of the bridge, but I think the rotational force on the bridge would be the same, given that the saddle height is the same.

The clawhammer model, it seems to me, is analogous to changing the "length" of the bridge itself - extending the area of the bridge behind the saddle - not changing the length of the string behind the saddle; the strings don't "pry" up on the back of the bridge, but the bridge itself does "pry" up on the soundboard, and/or on the glue joint of the bridge to the soundboard.

Changing the bridge footprint (or other parameters of the bridge or bridge plate) will definitely change the way the rotational force is distributed, and therefore have an effect on the deflection of the top. I remain unconvinced that moving the bridge pins will have any effect.

_________________
Todd Rose
Ithaca, NY

https://www.dreamingrosesecobnb.com/todds-art-music

https://www.facebook.com/ToddRoseGuitars/

This would be true if the string was connected to the saddle.....but it's not.....it can slide over the saddle and is fixed only at the pin termination point.



Again, the affect is the same as two seperate pieces of string only if the string is fixed (or connected) to the saddle......but it's not....it can slide.

So I remain unconvinced. <smile>

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Formerly known as Adaboy.......

I hear you, Darryl, and I've been thinking about that point myself... my limited gray matter starts to groan with the strain...

It would seem to me that, as far as the static forces on a guitar tuned up to pitch are concerned, the strings are, for all intents and purposes, fixed to the top of the saddle. And they are so firmly held in place there, I seriously doubt they actually can or do slide at all even as the strings oscillate.

On the other hand, what the heck do I know?

Maybe one of these days I'll get around to rigging up a deflection test like I described earlier. I reckon the proof is in the pudding.

_________________
Todd Rose
Ithaca, NY

https://www.dreamingrosesecobnb.com/todds-art-music

https://www.facebook.com/ToddRoseGuitars/

I remembered a discussion we had about this some time ago, and found it in the archives. Check it out, especially Todd Stock's post at the bottom of the first page. What he explains there is what he's referring to here - right, Todd?

viewtopic.php?f=10102&t=6947&st=0&sk=t&sd=a&hilit=saddle+height+force&start=0

_________________
Todd Rose
Ithaca, NY

https://www.dreamingrosesecobnb.com/todds-art-music

https://www.facebook.com/ToddRoseGuitars/

We all keep frying our brains with speculation and models: that's the problem. It's been my experience that a model can be built to justify any set of data. I'm trying to get some data. It's not that hard to do; maybe some of the other posters ought to try as well, and then we can compare something real.

People have mentioned raising the string height and changing the break angle, but I'd be curious to hear opinions on changing the direction of force on the saddle. I tilt my saddle slots back to split the angle the strings make. The intent being to direct the force entirely down the height of the saddle and not against the front of the saddle slot. Anyone else try this?

Mike

____________________________________
Montreal, Quebec
indianhillguitars.com

Mike:
I picked up on the back-tilted saddle from Rick Turner (steal ideas from the best). I use 9 degrees, as he does. It's not a new idea, though: violin bridges are set up so that they very nearly bisect the break angle over the top. They need to be that way; otherwise the force vector bends the bridge and you lose a lot of sound(especially when the bridge tips over!).

The backward tilt should not effect the way the strings drive the top. What it does do is reduce or eliminate the force trying to split out the front of the bridge, and increase the force on a UST by a little bit. With piezos every little bit helps.

Hi all,
I thought before this thread dies I'd thank everyone for their responses. I've been following them intently, studying and pondering. I see the issue is a lot more involved then I bargained for, but the discussion has, if not provided a definitive answer, made me much more informed of the factors involved. Right or wrong, it has helped me develop a mental model as to how I think the bridge and saddle function, that I will use going forward. As my small contribution, I'll share a somewhat crude experiment I did to demonstrate to myself the principles being discussed.

I mounted a 6" long board on a pivot above my work bench , sort of like a see-saw, then added a 1 1/2" block at one end of it. This was my surrogate bridge and saddle ( colored grey in my diagram). Strung a line with just under 30 lbs of weight over an improvised roller, and measured the downward force on the front of my "bridge" with a dowel and bathroom scale. I took measurements with a steep break angle of about 55 deg. and a shallow break angle of about 20 deg (measured off my 6" board). Both ways , my bathroom scale read 18 lbs. I repeated it all a few times with the same results.
Attachment:
Bridge test.jpg
Bridge test.jpg [ 17.13 KiB | Viewed 5 times ]


I think I'll let everyone draw their own conclusion as to the results, merit and faults of this. I think what it most proves is that I need to get a life!
Gerard

Wonderful experiment Gerry. My opinion: point of rotation is irrelevant if held constant. (Engineering is one of many fields that I don't have a degree in.)
Andy: I'm sorry we haven't heard from you. I think you had it nailed.
I think saddle position relative to front edge of bridge has some effect. I think bridge pin position is irrelevant. I think steel string bridges are "rigid" compared to the soundboards they interact with. I think the "rigidity" of classical bridges is debatable. I think data is fundamentally important, and physics is a prerequisite to interpreting data.

It's not the amount of torque or rotational force that's affected by the location of the bridge pins in
relation to the top of the saddle, but the direction of the forces. Having the pins in the rear block
creates a completely different set of forces than what are applied to a bridge when the anchor and fulcrum
are both located in the same foundation.

Down force is certainly an important one when it comes to creating tone, but the rotational forces
that are presented when the strings are trying to roll the bridge off of the top to relieve the tension that
is applied to them by tuning them to pitch are of paramount importance. As the strings cycle when activated
by the player striking them, their length changes in their respective cycle intervals and their tension, in turn,
decreases and returns to its maximum for that gauge and pitch. This creates the typical "rocking" of the
bridge and cyclic relieving and returning of the top's tension, thus servicing the desired pump like action
of the top that generates a lot of the volume and tone of the guitar.

Regards,
Kevin Gallagher/Omega Guitars