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  #1  
Old 03-18-2012, 09:28 AM
Ulf_Lindholm Ulf_Lindholm is offline
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Default Break angle and downward pressure on guitar top

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?
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  #2  
Old 03-18-2012, 09:39 AM
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Quote:
Originally Posted by Ulf_Lindholm View Post
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.
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Old 03-18-2012, 09:49 AM
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Quote:
Originally Posted by Ulf_Lindholm View Post
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.
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Old 03-18-2012, 09:53 AM
Ulf_Lindholm Ulf_Lindholm is offline
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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
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  #5  
Old 03-18-2012, 10:15 AM
FrankS FrankS is offline
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Quote:
Originally Posted by Ulf_Lindholm View Post
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

Last edited by FrankS; 03-18-2012 at 10:22 AM.
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Old 03-18-2012, 10:19 AM
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Quote:
Originally Posted by rick-slo View Post
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 ...
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Old 03-18-2012, 10:20 AM
Ulf_Lindholm Ulf_Lindholm is offline
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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.
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  #8  
Old 03-18-2012, 10:23 AM
Ulf_Lindholm Ulf_Lindholm is offline
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Originally Posted by murrmac123 View Post
OP has said that he is using a tailpiece ...
Thanks Murrmac!
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Old 03-18-2012, 10:27 AM
Matt Mustapick Matt Mustapick is offline
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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.
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Old 03-18-2012, 10:32 AM
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Quote:
Originally Posted by Ulf_Lindholm View Post
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
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Old 03-18-2012, 12:29 PM
Ulf_Lindholm Ulf_Lindholm is offline
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Thats great, just what I needed. Thanks!
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Old 03-18-2012, 01:22 PM
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A fifteen minute lesson in trigonometry can go a long way.
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Old 03-18-2012, 02:03 PM
Ulf_Lindholm Ulf_Lindholm is offline
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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.
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Old 03-18-2012, 02:11 PM
robj144 robj144 is offline
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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.
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Old 03-18-2012, 02:40 PM
Ulf_Lindholm Ulf_Lindholm is offline
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Originally Posted by robj144 View Post
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!!
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