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Ulf_Lindholm
03-18-2012, 09:28 AM
Hi Im an amateur luthier drafting my next guitar. Im thinking it will be a kind of flat top with 20' radius and with a kind of tailpiece, inspired by makers like Batson Guitars. Now Im doing some online research to find out what kind of pressure on the top that the strings will exert.

This led me to a thread in another forum where someone gave an example on how to calculate the downward pressure at the bridge. It included cos which is way beyond my math knowledge but when I tried his algorithm I got nervous:

If a set of light strings has a tension of 75 kg (165 pounds) and the strings break angle over the bridge is 15 degrees (which is what I had in mind) the downward force/pressure at the bridge will be around 60 kg (130 pounds).

Can this really be correct?

rick-slo
03-18-2012, 09:39 AM
Hi Im an amateur luthier drafting my next guitar. Im thinking it will be a kind of flat top with 20' radius and with a kind of tailpiece, inspired by makers like Batson Guitars. Now Im doing some online research to find out what kind of pressure on the top that the strings will exert.

This led me to a thread in another forum where someone gave an example on how to calculate the downward pressure at the bridge. It included cos which is way beyond my math knowledge but when I tried his algorithm I got nervous:

If a set of light strings has a tension of 75 kg (165 pounds) and the strings break angle over the bridge is 15 degrees (which is what I had in mind) the downward force/pressure at the bridge will be around 60 kg (130 pounds).

Can this really be correct?
Downward pressure on the top is not related to the break angle. The height of the saddle (distance between the face of the guitar and top of the saddle) creates a lever which will impart rotational force on the bridge which puts downward force on the front of the bridge and upward force (lift) on the back of the bridge.

rick-slo
03-18-2012, 09:49 AM
Hi Im an amateur luthier drafting my next guitar. Im thinking it will be a kind of flat top with 20' radius and with a kind of tailpiece, inspired by makers like Batson Guitars. Now Im doing some online research to find out what kind of pressure on the top that the strings will exert.

This led me to a thread in another forum where someone gave an example on how to calculate the downward pressure at the bridge. It included cos which is way beyond my math knowledge but when I tried his algorithm I got nervous:

If a set of light strings has a tension of 75 kg (165 pounds) and the strings break angle over the bridge is 15 degrees (which is what I had in mind) the downward force/pressure at the bridge will be around 60 kg (130 pounds).

Can this really be correct?
Downward pressure on the top is not related to the break angle. The height of the saddle (distance between the face of the guitar and top of the saddle) creates a lever which will impart rotational force on the bridge which puts downward force on the front of the bridge and upward force (lift) on the back of the bridge.

Ulf_Lindholm
03-18-2012, 09:53 AM
Thanks for your reply Rick but were not talking a pin bridge here. Im thinking about archtop type of guitar like the ones Batson Guitars make - a flattop but with a tailpiece.

Ulf

FrankS
03-18-2012, 10:15 AM
Hi Im an amateur luthier drafting my next guitar. Im thinking it will be a kind of flat top with 20' radius and with a kind of tailpiece, inspired by makers like Batson Guitars. Now Im doing some online research to find out what kind of pressure on the top that the strings will exert.

This led me to a thread in another forum where someone gave an example on how to calculate the downward pressure at the bridge. It included cos which is way beyond my math knowledge but when I tried his algorithm I got nervous:

If a set of light strings has a tension of 75 kg (165 pounds) and the strings break angle over the bridge is 15 degrees (which is what I had in mind) the downward force/pressure at the bridge will be around 60 kg (130 pounds).

Can this really be correct?

I think you have your sin and cos mixed up. The 15 degrees of off wherever the strings are terminated. The 15 degrees is the angle for the downward vector which is the side opposite the angle. The hypotenuse is the string which is also the string tension vector. So you are looking for the side opposite the angle divided by the hypotenuse and that is the sin. Sin of 15 degrees is 0.26 so the force downward is 19 kg.

Frank Sanns

murrmac123
03-18-2012, 10:19 AM
Downward pressure on the top is not related to the break angle. The height of the saddle (distance between the face of the guitar and top of the saddle) creates a lever which will impart rotational force on the bridge which puts downward force on the front of the bridge and upward force (lift) on the back of the bridge.

OP has said that he is using a tailpiece ...

Ulf_Lindholm
03-18-2012, 10:20 AM
Frank and Rick, sorry for not being accurate in my question. We are not talking a pin bridge here but a guitar with a tailpiece construction, which means there is no pull on the bridge, no torsional forces - just pressure dictated by the strings break angle as I understand it.

Ulf_Lindholm
03-18-2012, 10:23 AM
OP has said that he is using a tailpiece ...

Thanks Murrmac! :)

Matt Mustapick
03-18-2012, 10:27 AM
It's not the cosine it's the sine. So the sine of 15 degrees is 0.259 and the downforce is 165 lbs x 0.259 = 42.7 lbs.

It can't be the cosine because the cosine of a 0 degree angle is 1 and
165 lbs x 1 = 165 lbs,
meaning that a guitar with no break angle would have all the string tension pointing down on the top. Also, the cosine of 90 degrees is 0, and a guitar with strings perpendicular to the top wouldn't have 0 downforce, it'd have 165 lbs of downforce (don't worry about the direction, it'd be a screwy instrument anyway).

With sine the reverse is true, the sine of 0 degrees is 0 and the sine of 90 degrees is 1 and that's what you'd intuitively expect when guessing about downforce.

rick-slo
03-18-2012, 10:32 AM
Thanks for your reply Rick but were not talking a pin bridge here. Im thinking about archtop type of guitar like the ones Batson Guitars make - a flattop but with a tailpiece.

Ulf
Oops. Use the sine of 15 degrees.
See this link for a handy calculator
http://liutaiomottola.com/formulae/downforce.htm

Ulf_Lindholm
03-18-2012, 12:29 PM
Thats great, just what I needed. Thanks!

robj144
03-18-2012, 01:22 PM
A fifteen minute lesson in trigonometry can go a long way. :)

Ulf_Lindholm
03-18-2012, 02:03 PM
A fifteen minute lesson in trigonometry can go a long way. :)

I don't know. In my case it would take at least twice that long just to understand the meaning of that word. :)

robj144
03-18-2012, 02:11 PM
I don't know. In my case it would take at least twice that long just to understand the meaning of that word. :)

Basically, it's the study of triangles. I bet I could teach it to you in fifteen minutes. ;)

Ulf_Lindholm
03-18-2012, 02:40 PM
Basically, it's the study of triangles. I bet I could teach it to you in fifteen minutes. ;)

In that case Robj you must be a hell of a teacher!! :)

Howard Klepper
03-20-2012, 11:58 PM
So far, no one here has it right. You have only been considering one of the two angles involved--the one where the tailpiece meets the top. The angle at the nut is different.

To get the downward pressure at the bridge, you need to look at both sides of the break angle--i.e., the angle formed by the string and bridge on the fretboard side, and the angle formed by the string and bridge at the tailpiece side. If the total break angle is "a" and the two components of "a" (the fretboard side and the tailpiece side) are "b" and "c", then the downward pressure "P" equals string tension ("T") times (cos b + cos c).

P=T x (cos b + cos c).

If you are looking at the angle between the string and a line drawn between the nut and the tailpiece hinge, as the OP apparently is, then you can express the same formula using sines. If the tailpiece angle is "d" and the nut angle is "e" (remember these are not the angles between the string and the fretboard, or the string and the top. They are between the string and a straight line between nut and tailpiece hinge, which may or may not be the same thing) then P=T x (sine d + sine e). This is the same formula as the one that looks at the break angle using cosines.

You don't need a table of sines and cosines. You just need to measure the height of the bridge above that line between nut and tailpiece hinge (which will be close to the bridge height, but not necessarily equal to it, depending on top shape and neck angle and fretboard thickness and how the tailpiece mounts) and the length of each of the two string segments. Those will give you the relevant sines or cosines, which are the effective bridge height divided by the string length for each of the string segments.

Ulf_Lindholm
03-21-2012, 12:57 AM
So far, no one here has it right. You have only been considering one of the two angles involved--the one where the tailpiece meets the top. The angle at the nut is different.

To get the downward pressure at the bridge, you need to look at both sides of the break angle--i.e., the angle formed by the string and bridge on the fretboard side, and the angle formed by the string and bridge at the tailpiece side. If the total break angle is "a" and the two components of "a" (the fretboard side and the tailpiece side are "b" and "c", then the downward pressure "P" equals string tension ("T") times (cos b + cos c).

P=T x (cos b + cos c).

If you are looking at the angle between the string and a line drawn between the nut and the tailpiece hinge, as the OP apparently is, then you can express the same formula using sines. If the tailpiece angle is "d" and the nut angle is "e" (remember these are not the angles between the string and the fretboard, or the string and the top. They are between the string and a straight line between nut and tailpiece hinge, which may or may not be the same thing) then P=T x (sine d + sine e). This is the same formula as the one that looks at the break angle using cosines.

You don't need a table of sines and cosines. You just need to measure the height of the bridge above that line between nut and tailpiece hinge (which will be close to the bridge height, but not necessarily equal to it, depending on top shape and neck angle and fretboard thickness and how the tailpiece mounts) and the length of each of the two string segments. Those will give you the relevant sines or cosines.

I'm impressed by the level of math knowledge on this forum and I hope this is not a quality that is necessary to become a good luthier. Because in that case I can stop trying right now. :)
I know that you are building archtops, Howard, and very nice ones too. What is your take on this? (I'm not really building an _archtop_ but still, with a tailpeice kind of solution, it is related). I think I read somewhere someone say that he/she strived for a low break angle over the bridge, of a reason I didn't really grasp. It would be very interesting to hear your filosophy about this, Howard!

Ulf

Howard Klepper
03-21-2012, 10:33 AM
I'm impressed by the level of math knowledge on this forum and I hope this is not a quality that is necessary to become a good luthier. Because in that case I can stop trying right now. :)
I know that you are building archtops, Howard, and very nice ones too. What is your take on this? (I'm not really building an _archtop_ but still, with a tailpeice kind of solution, it is related). I think I read somewhere someone say that he/she strived for a low break angle over the bridge, of a reason I didn't really grasp. It would be very interesting to hear your filosophy about this, Howard!

Ulf

What is my take on "this"? That depends on what "this" refers to.

You mean tailpieces on a flat top? I think it doesn't work as well as anchoring the strings to the bridge. You get neither sufficient downward pressure for a strong sound, nor a top that could withstand that pressure if you were to get it.

Ulf_Lindholm
03-21-2012, 11:08 AM
What is my take on "this"? That depends on what "this" refers to.

You mean tailpieces on a flat top? I think it doesn't work as well as anchoring the strings to the bridge. You get neither sufficient downward pressure for a strong sound, nor a top that could withstand that pressure if you were to get it.

Yeah that kind of answered my question although it was not so well put. With "this" I had break angle over bridge in mind. Should it be high for maximum transferrance of string energy or low so that the top can be lightly built? If we keep it to traditional archtops for the case of simplicity.

Howard Klepper
03-21-2012, 12:38 PM
It should be in the right range, neither too low nor too high. Which is about where traditional archtop design puts it, at about 40 lbs downward force with medium gauge strings.

Alan Carruth
03-21-2012, 12:49 PM
I don't believe that increasing the break angle will automatically improve the transfer of string energy to the top. I'm just finishing up a long project looking at this, and, although I don't have enough data to entirely rule that out, it seems to me to be a secondary effect at best.

If you mount a string on a rigid beam (or a Les Paul, close enough), and rig up some way to measure the forces on the saddle top when you pluck it, you'll find them to be well defined and (relatively) easy to calculate if you know some things about the string. Fletcher and Rossing give a good account of this in their book 'Physics of Musical Instruments', from Springer-Verlag press. I've done this experiment, and pretty well confirmed most of what they say (there are always surprises...). You can find a report on that on my web site, on the 'Acoustics' page, entitled 'String Theory' ( I couldn't resist...).

As far as I can see, so long as the string maintains contact with the top of the saddle throughout it's vibration cycle, it will transfer all of the signal to the bridge top. If it rolls around, or hops off, it won't, but all you need, as far as I can see, is enough break angle to ensure good contact. On archtop guitars they use break angles as low as six degrees, and my own recent experiments suggest to me that this is a bare minimum, but still sufficient.

This is not to say that changing the break angle has no effect on the way the top works, but simply to suggest that increasing the break angle beyond a 'sufficiency' is not likely to transfer more energy to the top.

As Howard points out, there's a cost to increasing the break angle on a tailpiece top: greater download that has to be compensated for structurally. Archtops use the vaulting of the plate to take up a lot of this load; a flat top with a tailpiece has to use top thickness or bracing to do it, which adds weight. Experiments I've done on archtops have lead me to conclude that increasing the break angle is not cost free even there: there seems to be an optimum for any given top, and going beyond that tends to kill the sound pretty fast. I'd also venture that that 'optimum' is not too much greater than the minimum required to keep the strings on the saddle top, but I don't have reams of data on that.

robj144
03-21-2012, 04:23 PM
So far, no one here has it right. You have only been considering one of the two angles involved--the one where the tailpiece meets the top. The angle at the nut is different.

To get the downward pressure at the bridge, you need to look at both sides of the break angle--i.e., the angle formed by the string and bridge on the fretboard side, and the angle formed by the string and bridge at the tailpiece side. If the total break angle is "a" and the two components of "a" (the fretboard side and the tailpiece side) are "b" and "c", then the downward pressure "P" equals string tension ("T") times (cos b + cos c).

P=T x (cos b + cos c).

If you are looking at the angle between the string and a line drawn between the nut and the tailpiece hinge, as the OP apparently is, then you can express the same formula using sines. If the tailpiece angle is "d" and the nut angle is "e" (remember these are not the angles between the string and the fretboard, or the string and the top. They are between the string and a straight line between nut and tailpiece hinge, which may or may not be the same thing) then P=T x (sine d + sine e). This is the same formula as the one that looks at the break angle using cosines.

You don't need a table of sines and cosines. You just need to measure the height of the bridge above that line between nut and tailpiece hinge (which will be close to the bridge height, but not necessarily equal to it, depending on top shape and neck angle and fretboard thickness and how the tailpiece mounts) and the length of each of the two string segments. Those will give you the relevant sines or cosines, which are the effective bridge height divided by the string length for each of the string segments.

I'm more or less curious about the physics here. Do you happen to have a sketch of the situation with the angles involved? I'm having a hard time visualizing he situation.

Also, you're actually referring to a downward force, not pressure. Thanks.

robj144
03-21-2012, 04:32 PM
Is this kind of what you're talking about:

https://lh5.googleusercontent.com/-rw5N9yq7JSA/T2pWZKJiSaI/AAAAAAAAK10/U1EHIcNhPOw/s487/bridge.JPG

I just sketched it quickly...

Howard Klepper
03-21-2012, 07:42 PM
Is this kind of what you're talking about:

https://lh5.googleusercontent.com/-rw5N9yq7JSA/T2pWZKJiSaI/AAAAAAAAK10/U1EHIcNhPOw/s487/bridge.JPG

I just sketched it quickly...

Pretty much the idea. I meant "a" to be the total angle enclosed by the two T's, with b and c being what you have labeled a and b.

Ulf_Lindholm
03-22-2012, 02:38 PM
Another way of creating pressure on the bridge saddle but not on the top is a bridge like the one Batson guitars use.
http://i26.tinypic.com/2qkuhs0.jpg
IMHO it looks like a pretty clever way to do it.

FrankS
03-22-2012, 03:30 PM
Angle B is essentially 90 degrees and can be ignored as a contributor. String hight at nut vs string height at 12th times two is the drop over the distance. It will add around 0.03 kg downward force out of 62. If you really want to be rigorous about the calculation then there are more factors like string thickness that need to be accounted for but it is a non factor for what he is calculating. Make it easier for the poor guy. One brake angle is sufficient.

Frank Sanns

Howard Klepper
03-22-2012, 06:05 PM
Angle B is essentially 90 degrees and can be ignored as a contributor. String hight at nut vs string height at 12th times two is the drop over the distance. It will add around 0.03 kg downward force out of 62. If you really want to be rigorous about the calculation then there are more factors like string thickness that need to be accounted for but it is a non factor for what he is calculating. Make it easier for the poor guy. One brake angle is sufficient.

Frank Sanns
Depends on the neck angle, Frank, which depends on the shape of the top, as well as bridge height. Also on whether the tailpiece hinges above or below the edge of the top. On a typical archtop cos b is about .07 or so, adding about 10-12 lbs downward force. Regarding terminology, the break angle is the angle enclosed by the string at the bridge.

If I understand what you said above, you are assuming that the plane of the frets aims at the base of the bridge, which then has a height of twice the action height at the 12th fret. That would not be a usable guitar. We have no information from the OP that would enable us to know his break angle. But if he learns how to calculate the downward force for any stringed instrument he can proceed with his design. Once he knows the break angle and the portion of it on the tailpiece side, it's not going to take him more than a few seconds to get the other portion of the angle that's toward the nut.

FrankS
03-22-2012, 08:58 PM
Depends on the neck angle, Frank, which depends on the shape of the top, as well as bridge height. Also on whether the tailpiece hinges above or below the edge of the top. On a typical archtop cos b is about .07 or so, adding about 10-12 lbs downward force. Regarding terminology, the break angle is the angle enclosed by the string at the bridge.

If I understand what you said above, you are assuming that the plane of the frets aims at the base of the bridge, which then has a height of twice the action height at the 12th fret. That would not be a usable guitar. We have no information from the OP that would enable us to know his break angle. But if he learns how to calculate the downward force for any stringed instrument he can proceed with his design. Once he knows the break angle and the portion of it on the tailpiece side, it's not going to take him more than a few seconds to get the other portion of the angle that's toward the nut.

Hi Howard, It is often difficult to communicate via keyboard when a picture is worth........ I agree with you but was trying to keep it simple for the gentleman but without a specific example, your way is indeed the more complete.

Here is an example of why for a flat top guitar (and many archtops) the B angle can be ignored. For a flat top guitar, the strings ride on the nut grooves at the headstock. The string height at the nut is around 0.013". At the 12th fret and halfway to the bridge it is typically on the order of 0.070". They run on a plane that terminates at the saddle for an angle, from the drawing provided, that is the complementary angle to the B angle. Since three angles of a triangle have to total 180 degrees, your B angle is just 90 minus the angle we are solving for above. Since we are making a long thin triangle that starts at the nut at 0.013" we can make this a zero by subtracting it from both readings to make the problem easier to see. So at the nut the string height will be zero (0.013 minus 0.013) and at the 12th fret it will be 0.057" (0.07 minus 0.013). At twice the distance from the 12th fret is the saddle and the height difference from the nut must be 0.057 times 2 or 0.114". Since we know the height of the triangle of 0.114" and the length of 25.5" the angle from the nut to the bridge must be the arctan of 0.114 divided by 25.5 or 0.256 degrees. To find the B angle it is just 90 minus 0.256 degrees or 89.74 degrees. Cosine of this angle is 0.005. Multiply the string tension of 6 strings and you will see that the force is very small.

It is late, hope it is clear enough to follow.

EDIT: Changing the angle of the neck by even a half an inch or an inch out of 25 inches does not put a significant force on the bridge for a standard flat top and for many archtops. This is not the case for shorter instruments like a violin or guitars with a variety of special features like a elevated fretboards. Just clarifying before somebody ignores when they should not.

Frank Sanns

Matt Mustapick
03-22-2012, 09:45 PM
Another way of creating pressure on the bridge saddle but not on the top is a bridge like the one Batson guitars use.
http://i26.tinypic.com/2qkuhs0.jpg
IMHO it looks like a pretty clever way to do it.

What an interesting arrangement! I wonder what that sounds like?

John Arnold
03-22-2012, 10:12 PM
The Batson design is not new....it has been tried on classical guitars forever. In my experience, the top must be more lightly constructed to approach the sound produced when the strings are attached to a fixed bridge.
Archtop design is no different in that regard. There is an optimum break angle range for a given arching and plate thickness.