I have to disagree with "Possibly need a higher string height at the nut." I can't think of a reason why that would be true.
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Double Bass BASS SETUP FOR GUT STRINGS
- Thread starter massimo
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Once you scoop the fingberboard you better like the low tension strings cause it's a major PITA to go back to anything else.
I'd also add that if the relief (scoop) is correct for the strings, action can be very low without buzzing. I'm running ~1/4" (6.35mm) E to ~3/16" (4.76mm) G all the time with very fat low tension strings and no buzzing.
Raising the action for tone and volume might be desired, but I'd wonder if it's absolutely necessary.
Raising the action for tone and volume might be desired, but I'd wonder if it's absolutely necessary.
Mine is barely any higher than that and I've been able to play ampless in a bunch of indoor gigs so long as the others are at least a little sensitive to the dynamics.
When I had it popped up to an old school big band sort of height off the board, I once got asked to turn the amplifier down that was turned off right next to me. That was kinda fun.
When I had it popped up to an old school big band sort of height off the board, I once got asked to turn the amplifier down that was turned off right next to me. That was kinda fun.
Both higher overstand and bigger neck angle place the strings higher above the top of the bass = higher bridge.
Interesting note about string tension though...
For a given set of strings, if they are tuned to a constant pitch ("in tune"), the height of the bridge does not change the string tension; the pitch defines the string tension and the string tension defines the pitch...
So if we raise our action at the bridge and have to lower our string tuning to be at that same constant pitch ("in tune"), the string tension remains the same as it was before.
So, does having a taller bridge increase the string tension against the top table? Not if we stay in tune with that constant pitch.
That said, changing the action does change the string tension and tuning for individual notes as we stop strings against the fingerboard, which is why either the fingering location or the bridge location has to change in order to be "in tune" while stopping strings. Since we tend to prefer not to change the bridge location, the fingering needs to change for proper intonation.
Simple, but complicated.
For a given set of strings, if they are tuned to a constant pitch ("in tune"), the height of the bridge does not change the string tension; the pitch defines the string tension and the string tension defines the pitch...
So if we raise our action at the bridge and have to lower our string tuning to be at that same constant pitch ("in tune"), the string tension remains the same as it was before.
So, does having a taller bridge increase the string tension against the top table? Not if we stay in tune with that constant pitch.
That said, changing the action does change the string tension and tuning for individual notes as we stop strings against the fingerboard, which is why either the fingering location or the bridge location has to change in order to be "in tune" while stopping strings. Since we tend to prefer not to change the bridge location, the fingering needs to change for proper intonation.
Simple, but complicated.
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I find it useful to use two words: Tension, for how "tight" the strings are, and "pressure", for the force applied to the table of the instrument. You're of course correct regarding tension; I don't know enough about physics to figure out if a more acute angle over the bridge affects pressure.
If the neck is brought out and up in such a way that the nut itself is in the same point in space as it was with the original angle and overstand, then the only difference would be board closer to strings for the same bridge height.
Triangles, folks, you can't change how they work. Nor can you change the entire point someone has been making from the beginning of a conversation to match a different, but related, viewpoint you have.
It would make sense to me. The strings ultimately want to come to a position where they are straight from the nut to the tail piece, and having the bridge in the way stops that. The more acute the angle, the higher the bridge, the more the strings are pressing it down trying to reach their ideal state.
Regarding "pressure" then, are we talking about the bridge being a fulcrum, such that the strings wanting to be straight, apply pressure through the bridge against the top table? Or, could we look at it the other way, the strings wanting to be straight, applying pressure against the pegbox and the endpin, trying to pull them up?
Either way, I don't have any idea how to quantify how much that "pressure" is, although it's probably some sort of combination of string tension, bridge height relative to the nut and saddle, and maybe some other values. It would be nice if there was an actual formula so we could literally apply a value to how much "pressure" is applied against the top table.
Either way, I don't have any idea how to quantify how much that "pressure" is, although it's probably some sort of combination of string tension, bridge height relative to the nut and saddle, and maybe some other values. It would be nice if there was an actual formula so we could literally apply a value to how much "pressure" is applied against the top table.
I'm not volunteering, but I suppose an experiment could be conducted by putting a scale under the feet of an adjustable bridge. Pitch would need to remain constant as the bridge was raised/lowered.
A visual approach:
Draw an angled line (angle like the string angle the bridge), line length is as long as the tension of the string. Use a unit so that the whole thing fits the paper like 1 cm = 3 kp or 1 inch = 6 kp.
Ideally the bridge halves the angle (if not the bridge crown gets pushed sideways and only friction might hold it in place), so this is assumed here.
In that case both sides of the bridge pull the string at the same amount towards the top, since the tension is equal along the string (ideally, it might differ a bit because of friction at the bridge or the bridge crown is moved a bit to get equal tension but then the vibrating string length changes).
So, if your string is 30 kp, draw a line of 10 cm to both sides with an angle of the string at the bridge. These are the forces that act on the bridge.
Now add parallel lines through the endpoint of the lines to get a parallelogram (parallelogram of forces). The distance from the bridge point angle to the new crosspoint of the two parallel lines is the amount of force acting towards the top. So just reconvert the length to the force/tension you used before (but of course backwards) and you get the amount of pressure on the top.
The extremes:
A) No angle, straight string touching the top. Parallelogram has no height, just a line, pressure is zero.
B) Angle is 90 degrees at both sides of the bridge (parallel pull) the parallelogram becomes a line with double height, so the pressure is twice the string tension. (Think of a pulley, the force at the parallel ends is half the force at the roll.)
I think this is easier to understand than a rather abstract formula (that I would need to find in my old school books).
Draw an angled line (angle like the string angle the bridge), line length is as long as the tension of the string. Use a unit so that the whole thing fits the paper like 1 cm = 3 kp or 1 inch = 6 kp.
Ideally the bridge halves the angle (if not the bridge crown gets pushed sideways and only friction might hold it in place), so this is assumed here.
In that case both sides of the bridge pull the string at the same amount towards the top, since the tension is equal along the string (ideally, it might differ a bit because of friction at the bridge or the bridge crown is moved a bit to get equal tension but then the vibrating string length changes).
So, if your string is 30 kp, draw a line of 10 cm to both sides with an angle of the string at the bridge. These are the forces that act on the bridge.
Now add parallel lines through the endpoint of the lines to get a parallelogram (parallelogram of forces). The distance from the bridge point angle to the new crosspoint of the two parallel lines is the amount of force acting towards the top. So just reconvert the length to the force/tension you used before (but of course backwards) and you get the amount of pressure on the top.
The extremes:
A) No angle, straight string touching the top. Parallelogram has no height, just a line, pressure is zero.
B) Angle is 90 degrees at both sides of the bridge (parallel pull) the parallelogram becomes a line with double height, so the pressure is twice the string tension. (Think of a pulley, the force at the parallel ends is half the force at the roll.)
I think this is easier to understand than a rather abstract formula (that I would need to find in my old school books).
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About raising the bridge:
Raising the bridge slightly enlarges the vibrating string length as the angle with the bridge is not 90 degrees. So tension is enlarged a bit too which means a bit longer lines at the parallelogram of forces and a slightly larger angle at the bridge which raises the pressure even a bit more.
Raising the bridge slightly enlarges the vibrating string length as the angle with the bridge is not 90 degrees. So tension is enlarged a bit too which means a bit longer lines at the parallelogram of forces and a slightly larger angle at the bridge which raises the pressure even a bit more.
I made an example using the GeoGebra app with a bridge angle of 29.6 degrees which is in the range of my 5-string 4/4 (29 to 32 degrees).
Don’t ask why it is not a whole number, construction has had its limits.
The result was 15 kp pressure from a 30 kp string tension, so 50% of the string tension goes to the top pressure.
I divided any tension in the example by 10 to fit the screen.
I also had an (even worse constructed) example with an angle of a bit over 34 degrees and the result was 60% of the string tension (18 kp pressure from 30 kp string tension).
So you can at least approximate the top pressure from the strings. The inner ones typically have a higher bridge angle which translates to a higher relative pressure than the outer ones.
Don’t ask why it is not a whole number, construction has had its limits.
The result was 15 kp pressure from a 30 kp string tension, so 50% of the string tension goes to the top pressure.
I divided any tension in the example by 10 to fit the screen.
I also had an (even worse constructed) example with an angle of a bit over 34 degrees and the result was 60% of the string tension (18 kp pressure from 30 kp string tension).
So you can at least approximate the top pressure from the strings. The inner ones typically have a higher bridge angle which translates to a higher relative pressure than the outer ones.
A few more numbers (not exact, kp from the example is rounded to a whole number):
Angle • Pressure (percentage of tension)
26° • 47%
28° • 50%
30° • 53%
32° • 57%
34° • 60%
36° • 63%
So 29 degrees is a bit more than 50%.
30 kp in the example is more typical for a steel core string, but the relation „tension to pressure“ is linear, so you can use the percentage for the 18 to 25 kp typical for gut strings.
Angle • Pressure (percentage of tension)
26° • 47%
28° • 50%
30° • 53%
32° • 57%
34° • 60%
36° • 63%
So 29 degrees is a bit more than 50%.
30 kp in the example is more typical for a steel core string, but the relation „tension to pressure“ is linear, so you can use the percentage for the 18 to 25 kp typical for gut strings.
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