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Post by ms on Aug 8, 2017 7:47:32 GMT -5
The electric guitar pickup usually selects just a part of the potentially available signal from the vibrating string. Let's look at what the string does in detail in order to better understand what it is that the pickup selects and rejects. I used one of my six coil pickups with a very short cable (very high resonance), loaded to give approximately flat response over the 24KHz sampling range. First, examine what happens just after the string is picked. This plot has been set up to show a particular behavior at high frequencies. The neck pickup is used, but the picking is as close as possible to the bridge in order to excite high frequencies while suppressing low frequencies. This is the open E6 string, with the other strings well damped with a Swiffer duster pad.  First we should understand that when we load the pickup to make the frequency response flat, we make it sense the velocity of the string. The law of magnetic induction has a time derivative in it. This means that it senses "rate if change", and the rate of change of the string position is its velocity. When the pick releases the string, it snaps back and overshoots; a "change in position" travels towards the pickup, but also towards the bridge. The latter reflects with inverted polarity and the two partly overlapping pulses partially cancel, resulting mostly in a very short velocity pulse as seen at approximately .001 sec. However, some very high frequencies, above 10KHz precede this pulse. Why? To understand this, look later. The perturbation in string position travels to the nut, reflects with inverted polarity, passes the pickup, hits the bridge and reflects, inverting polarity again and passing the pickup again. But now the original pulse is mostly gone and frequencies down to very approximately 5KHz are traveling ahead of the pulse remnants, making a sort of "chirped" pulse. We are looking at dispersion; the faster components of the pulse travel at a higher speed than the lower ones, and so they appear to move ahead of the main pulse. Of course this means that the the higher string harmonics are sharp, and it will be interesting later to see how far down this effect extends into the region sampled by the guitar electronics. But for now we can make one simple conclusion: we know (and will see later) that high frequencies die out sooner than low frequencies, but we see that there is significant dispersion before that happens. The next post will examine another picking position. |
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Post by ms on Aug 8, 2017 13:02:32 GMT -5
Let's look at a similar plot with picking just to the bridge side of the neck pickup:  Here we see more clearly that the velocity signature of a pick is a very fast increase, followed by a slower decay. Since the picking is much closer to the neck pickup in this case than the last one, it takes longer for the reflection from the bridge to reach the neck pickup and cancel the signal that started towards the nut. Notice that the effect of the dispersion is visible, but since it is relatively small in this case, it might have escaped noticed, or possibly just assumed to be some imperfection in the measurement. Notice that the peak amplitude is about the same in both cases. This is because you can pick harder when picking near the bridge, making up for the smaller resulting displacement from a given picking force. Next we will look at how the signal in the above plot evolves with time. |
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Post by antigua on Aug 10, 2017 14:36:44 GMT -5
...The law of magnetic induction has a time derivative in it. This means that it senses "rate if change", and the rate of change of the string position is its velocity... It's not just the velocity though, right? Isn't the volume of permeable mass that is moving also a factor? My understanding is that the law of induction serves to impose a +6dB/oct "transfer function", but that this is just one factor in determining the overall output voltage. If it were just velocity alone, that should mean that string gauge is irrelevant, so long as the velocity of that string is what it is - that doesn't seem right, though. I'm sure the applicable equation is in here www.physics.princeton.edu/~mcdonald/examples/guitar.pdf , but it's still rather dense for me. |
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Post by JohnH on Aug 10, 2017 15:17:50 GMT -5
The extra moving mass of a lower string compensates for it not moving so quickly.
Here's how I idealise it:
All strings in a set usually have about the same tension, so given plucking with a given force, they deflect about the same distance.
The release of the string sets up a vibration amplitude which is about the same for all strings.
Higher strings are moving backwards and forwards through this amplitude more frequently, so move faster than lower strings. Greater string velocity promotes greater rate of flux change and hence greater signal induced.
But lower strings are heavier and so are moving more metal, which increases flux change to compensate for lower velocity.
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Post by Charlie Honkmeister on Aug 10, 2017 15:40:17 GMT -5
The extra moving mass of a lower string compensates for it not moving so quickly. Here's how I idealise it: All strings in a set usually have about the same tension, so given plucking with a given force, they deflect about the same distance. The release of the string sets up a vibration amplitude which is about the same for all strings. Higher strings are moving backwards and forwards through this amplitude more frequently, so move faster than lower strings. Greater string velocity promotes greater rate of flux change and hence greater signal induced. But lower strings are heavier and so are moving more metal, which increases flux change to compensate for lower velocity. That's why naturally struck or plucked magnetic guitar signals from the strings aren't intolerably bright and the strings are relatively even in volume for a given pick stroke. We see the +6 dB/octave velocity "bias" of the fundamentals on the strings offset by the (roughly) -6 dB/octave contribution from the inductance of the pickup coil, and for the balance of all the string outputs, there's progressively reduced moving mass and flux generation of the higher-sounding strings for a narrow range of string tension. |
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Post by JohnH on Aug 10, 2017 16:31:59 GMT -5
Yes I think thats all clear enough at a qualitative level, and it seems to work out in practice too.
But theres a step in the maths involved where it seems to be anomalous or Im missing something.
Its a quantitative aspect where, when I was following through tbe maths for Guitarfreak; I had to 'close my eyes'and just assume.
Vibrating frequency is dependent on tbe square root of stiffness/mass. Thats basic physics.
The stiffness' of the strings are related to their tensions which is all about equal, so the stiffness term is about constant.
So consider the two E strings for example, two octaves apart, a factor of 4 in frequency difference. If they have been made to have the same tension then they have to be 16x different in weight.(fact check, this is about right, consider a 0.01 high E and a 0.046 low E. Square that diameter ratio and allow that the wound string is not solid, a mass ratio of 16 is credible)
So, think of the two compensating slopes discussed above: The high E at 4x the frequency gets a 6db x 2 octaves = 12db boost due to string velocity. In compensation, the low E is 16x heavier. 16 is 2 to the power 4. So why isnt that mass comoensation 6x4=24db?. ie, the low string would be louder by 12db? Somehow the added mass does not have a linear proportional affect on induced signal, its more related to its square root.
Anyone follow that? Any insights?
I assume that balanced string sets give about equal output per string, and if one makes tbat assumption results match measured outputs. but theres the step above where the logic is not quite clear.
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Post by ms on Aug 11, 2017 7:25:05 GMT -5
...The law of magnetic induction has a time derivative in it. This means that it senses "rate if change", and the rate of change of the string position is its velocity... It's not just the velocity though, right? Isn't the volume of permeable mass that is moving also a factor? My understanding is that the law of induction serves to impose a +6dB/oct "transfer function", but that this is just one factor in determining the overall output voltage. If it were just velocity alone, that should mean that string gauge is irrelevant, so long as the velocity of that string is what it is - that doesn't seem right, though. I'm sure the applicable equation is in here www.physics.princeton.edu/~mcdonald/examples/guitar.pdf , but it's still rather dense for me. Yes, there are various factors involved, but my main point is that the pickup voltage is proportional to string velocity, not, for example, string displacement or string acceleration. What is the string displacement? If you have the velocity, you get the displacement by integrating in time, but there is a missing constant, or initial condition. If we assume that the pick pulls the string a certain distance and the string moves so that the peak displacement in the other direction has the same magnitude after the pick releases the string, then we an integrate approximately by performing a cumulative sum and adding a constant. The result is attached.  We come to the astounding conclusion that the string just moves over the pickup in an apparently simple way! All that interesting stuff is because we are sensing the velocity. The flat top on the waveform might be partly the result of the pickup nonlinearity that MacDonald mentions rather than the result of the actual string position. |
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Post by ashcatlt on Aug 11, 2017 12:43:55 GMT -5
I have had a lot of trouble making sense of this for a while now, so I definitely appreciate the discussion. Your integrated plot doesn't match my rudimentary understanding, though. It's possible that I'm wrong, but it seems like - after the pick releases the string, its velocity should be greatest in magnitude as it passes through its resting position, slowing to 0 as it gets to the extremes and turns around. That's not exactly what I'm seeing in your pictures, though.
I also wonder if whatever you're doing to integrate for position isn't acting as a sort of low pass that just won't let us see the high frequency content.
This directly translates to about everything we do in audio. It's the exact same thing that a dynamic microphone does, and therefor also the way a speaker works. So I'd like to understand it a lot better.
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Post by ms on Aug 11, 2017 12:53:26 GMT -5
I have had a lot of trouble making sense of this for a while now, so I definitely appreciate the discussion. Your integrated plot doesn't match my rudimentary understanding, though. It's possible that I'm wrong, but it seems like - after the pick releases the string, its velocity should be greatest in magnitude as it passes through its resting position, slowing to 0 as it gets to the extremes and turns around. That's not exactly what I'm seeing in your pictures, though. I also wonder if whatever you're doing to integrate for position isn't acting as a sort of low pass that just won't let us see the high frequency content. This directly translates to about everything we do in audio. It's the exact same thing that a dynamic microphone does, and therefor also the way a speaker works. So I'd like to understand it a lot better. I agree; something is wrong. I suspect that I integrated the wrong part of the waveform. I will try again. |
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Post by antigua on Aug 11, 2017 14:57:29 GMT -5
My understanding is that it's both a matter of how fast the flux changes, as well as how much flux changes.  Doesn't the Φ in the numerator represent the "how much" of flux? Thicker, or more permeable guitar string, means you have a greater Φ , true or false? If that's true, then I don't think displacement is discounted entirely, it's working together with the rate of change to determine the overall voltage. |
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frankfalbo
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Post by frankfalbo on Aug 11, 2017 15:27:46 GMT -5
Since the picking is much closer to the neck pickup in this case than the last one, it takes longer for the reflection from the bridge to reach the neck pickup and cancel the signal that started towards the nut. And this is part and parcel to the "Why can't I just EQ a bridge pickup to sound like a neck pickup" quandary. |
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Post by ms on Aug 11, 2017 15:39:22 GMT -5
My understanding is that it's both a matter of how fast the flux changes, as well as how much flux changes. Doesn't the Φ in the numerator represent the "how much" of flux? Thicker, or more permeable guitar string, means you have a greater Φ , true or false? If that's true, then I don't think displacement is discounted entirely, it's working together with the rate of change to determine the overall voltage. Double the flux, and you double the rate of change of flux, everything else equal. So it is just the rate of change of flux that counts. Also, as long as the relative permeability is significantly greater than one, its value is irrelevant. This is shown in the MacDonald paper you referred to. |
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Post by JohnH on Aug 11, 2017 15:52:12 GMT -5
Can anybody pick up on the quandary in my post below?
What it amounts to is: If you double the mass of the string moving at a given velocity, does magnetisation of that string result in double the flux change? My suspicion is that it is not so linear or else bass strings would be much louder than treble strings. A low E is about 16x heavier than a high E but 1/4 of the frequency.
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Post by ms on Aug 11, 2017 15:53:22 GMT -5
Since the picking is much closer to the neck pickup in this case than the last one, it takes longer for the reflection from the bridge to reach the neck pickup and cancel the signal that started towards the nut. And this is part and parcel to the "Why can't I just EQ a bridge pickup to sound like a neck pickup" quandary. An excellent observation. I think that with enough digital processing, doing things you cannot do with Rs, Ls and Cs, you could make the waveform for one string at a time look very similar to that of the other pickup for the same string, except for a time delay that would make the project useless for guitar playing. |
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Post by ms on Aug 11, 2017 16:05:04 GMT -5
Can anybody pick up on the quandary in my post below? What it amounts to is: If you double the mass of the string moving at a given velocity, does magnetisation of that string result in double the flux change? My suspicion is that it is not so linear or else bass strings would be much louder than treble strings. A low E is about 16x heavier than a high E but 1/4 of the frequency. Macdonald's paper agrees with what you are saying: The output voltage in his equation depends on the square of the string radius. I made a casual measurement with a guitar, getting somewhat more peak voltage from E6 than E1, but not several times. The average is several times because the E6 decays more slowly, but that is another matter. But also I had the pickup farther from the E6 than from the E1. Maybe a very careful test would show how all this works out. Also, what are the magnetic properties of the string wrapping? I did make a casual check years ago, and found that the wrapper was not significantly magnetic, but I do not remember what kind of string that was. Edit: I guess I should be saying "winding" rather than "wrapping". |
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Post by antigua on Aug 11, 2017 16:07:40 GMT -5
And this is part and parcel to the "Why can't I just EQ a bridge pickup to sound like a neck pickup" quandary. An excellent observation. I think that with enough digital processing, doing things you cannot do with Rs, Ls and Cs, you could make the waveform for one string at a time look very similar to that of the other pickup for the same string, except for a time delay that would make the project useless for guitar playing. It's not clear to me how this actually influences the tone, one way or another. So they're out of phase, or something? Can you describe the audible impact in more detail? Making a bridge sound like a neck is not primarily about EQ, it's about timbre. EQ is filtering by frequency, it's absolute. Timbre is filtering relative to fundamental, you're not altering a frequency, rather you're altering harmonic intervals. If you EQ a bridge and it ends up sounding like a neck pickup, it's sort of by accident. You apply an absolute filter that approximate the effects of harmonics filtering. When a bridge pickup is frequency notched, it sounds neck-like, but the absolute RLC filter is only an approximation. We get this sort of "band stop" filter by tapping PAF style humbuckers with capacitors - its a cool mod, especially for a bridge-pickup-only guitar. If you use the right cap value, it makes the bridge end up sounding neck-like. |
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Post by antigua on Aug 11, 2017 16:23:12 GMT -5
Can anybody pick up on the quandary in my post below? What it amounts to is: If you double the mass of the string moving at a given velocity, does magnetisation of that string result in double the flux change? My suspicion is that it is not so linear or else bass strings would be much louder than treble strings. A low E is about 16x heavier than a high E but 1/4 of the frequency. Is the question we're asking how much the permeability of the string matters? It seems to me that the permeable value of the string is more significant than not. I've only got bits and pieces to refer to, as to why I suspect this is the case. Aside from the fact that the lower strings are thicker, apparently making up for their lower velocity, my cheap Chinese test guitar came with stock strings that seemed to result in a rather low output. I touched a magnet to them and notices that they didn't attract as much as the usual Elixir's I put on my guitars. It seemed that the low permeability of those had a noticeable impact on the output level of the guitar. The other bit of evidence is pretty weak, seeing as it comes from the marketing material of an electric guitar string maker, but take it for what it's worth: www.daddario.com/DaddarioNewsDetails.Page?ActiveID=3780&id=1431&sid=e9b4c0a7-56a7-4124-8968-8b759358d576"With 6% more magnetic permeability for higher output, NXYL electric guitar strings offer more punch, crunch, and bite."Even if they're making that up to some degree, I'm impressed their copy writer used the correct terminology. This also ties in with the effect of the pickup's proximity to the strings. As the pickup gets further from the strings, it becomes quieter. Having the strings be smaller, or less permeable, is not much different than having the strings be further away from the pickup. |
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Post by ashcatlt on Aug 11, 2017 16:23:16 GMT -5
Also, what are the magnetic properties of the string wrapping? I did make a casual check years ago, and found that the wrapper was not significantly magnetic, but I do not remember what kind of string that was. This, I think, is important. I always thought it was part of the point of wrapping strings in nickle or whatever rather than steel. It adds mass to adjust the frequency, but doesn't impact the magnetic properties. You can use bronze-wound acoustic guitar strings on an electric guitar and you won't really notice much difference in the volume balance. |
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frankfalbo
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Post by frankfalbo on Aug 11, 2017 16:33:23 GMT -5
Also, what are the magnetic properties of the string wrapping? I did make a casual check years ago, and found that the wrapper was not significantly magnetic, but I do not remember what kind of string that was. ...You can use bronze-wound acoustic guitar strings on an electric guitar and you won't really notice much difference in the volume balance. I *think* I disagree, unless we are defining "notice much difference". I was just going to chime in and say bronze wound strings teach the details of velocity vs. permeability. The magnetic circuit of an acoustic sound hole pickup is extremely compensated, to the point that some completely omit a magnet under the B string. But the 4 wound bass strings are still more charged (pole height) than the high E string. From there, I could take an electric guitar with common hex core nickel plated wrap string such as D'Addario XL, and if I replace the strings with the same gauge in bronze wrap the output would drop significantly. (again sorry, I realize I'm not quantifying it in dB any more than you are) but the key here would be that, if I went UP several string gauges so the core of the bronze wrapped string was larger, it still would not equal the output level of the XL. |
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Post by antigua on Aug 11, 2017 16:36:48 GMT -5
Also, what are the magnetic properties of the string wrapping? I did make a casual check years ago, and found that the wrapper was not significantly magnetic, but I do not remember what kind of string that was. This, I think, is important. I always thought it was part of the point of wrapping strings in nickle or whatever rather than steel. It adds mass to adjust the frequency, but doesn't impact the magnetic properties. You can use bronze-wound acoustic guitar strings on an electric guitar and you won't really notice much difference in the volume balance. I think you got that backwards, the wrapping is steel, also. It's acoustic guitar strings that do not use permeable winding. I don't know why that is, but AFAIK, permeable wrapping is very important for an electric guitar. The reason they use round wrapping can be inferred from the difference observed with flat wound; the space in between the winds allows the string to contort without friction losses. Flat wound strings are darker, because the friction causing by the flat, locked windings causes higher harmonic moving to dissipate as friction heat, very rapidly. Round wounds are more "slinky", permitting higher harmonics to ring out longer. |
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Post by JohnH on Aug 11, 2017 16:47:26 GMT -5
Macdonald's paper agrees with what you are saying: Ah yes thanks!, this equation makes everything clear...  ...obvious really. (hmmm..actually, I love that somebody can get their head around that stuff, but its gonna take quite a bit of time to unpick that one!) |
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Post by antigua on Aug 11, 2017 17:06:42 GMT -5
Macdonald's paper agrees with what you are saying: Ah yes thanks!, this equation makes everything clear... ...obvious really. (hmmm..actually, I love that somebody can get their head around that stuff, but its gonna take quite a bit of time to unpick that one!) Referring to the diagram on page 3 of the PDF www.physics.princeton.edu/~mcdonald/examples/guitar.pdf , I think "a" in the equation you pasted above is the string radius, and "mu rel" is the string's permeability. I see that "a" is multiplicative, so larger "a" should mean larger "V". I see that "mu rel" is on both sides of the monster fraction, so I don't know what to make of that, but we know intuitively that without any permeability at all, there can be no voltage. |
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Post by ms on Aug 11, 2017 17:24:06 GMT -5
Can anybody pick up on the quandary in my post below? What it amounts to is: If you double the mass of the string moving at a given velocity, does magnetisation of that string result in double the flux change? My suspicion is that it is not so linear or else bass strings would be much louder than treble strings. A low E is about 16x heavier than a high E but 1/4 of the frequency. Is the question we're asking how much the permeability of the string matters? It seems to me that the permeable value of the string is more significant than not. I've only got bits and pieces to refer to, as to why I suspect this is the case. Aside from the fact that the lower strings are thicker, apparently making up for their lower velocity, my cheap Chinese test guitar came with stock strings that seemed to result in a rather low output. I touched a magnet to them and notices that they didn't attract as much as the usual Elixir's I put on my guitars. It seemed that the low permeability of those had a noticeable impact on the output level of the guitar. The other bit of evidence is pretty weak, seeing as it comes from the marketing material of an electric guitar string maker, but take it for what it's worth: www.daddario.com/DaddarioNewsDetails.Page?ActiveID=3780&id=1431&sid=e9b4c0a7-56a7-4124-8968-8b759358d576"With 6% more magnetic permeability for higher output, NXYL electric guitar strings offer more punch, crunch, and bite."Even if they're making that up to some degree, I'm impressed their copy writer used the correct terminology. This also ties in with the effect of the pickup's proximity to the strings. As the pickup gets further from the strings, it becomes quieter. Having the strings be smaller, or less permeable, is not much different than having the strings be further away from the pickup. In open magnetic circuits, permeability is irrelevant if greater than a few. This is true in general, and MacDonald showed it for strings and pickups. I like NYXL strings, but the claim of an effect from higher permeability is nonsense. |
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Post by antigua on Aug 11, 2017 17:31:36 GMT -5
Is the question we're asking how much the permeability of the string matters? It seems to me that the permeable value of the string is more significant than not. I've only got bits and pieces to refer to, as to why I suspect this is the case. Aside from the fact that the lower strings are thicker, apparently making up for their lower velocity, my cheap Chinese test guitar came with stock strings that seemed to result in a rather low output. I touched a magnet to them and notices that they didn't attract as much as the usual Elixir's I put on my guitars. It seemed that the low permeability of those had a noticeable impact on the output level of the guitar. The other bit of evidence is pretty weak, seeing as it comes from the marketing material of an electric guitar string maker, but take it for what it's worth: www.daddario.com/DaddarioNewsDetails.Page?ActiveID=3780&id=1431&sid=e9b4c0a7-56a7-4124-8968-8b759358d576"With 6% more magnetic permeability for higher output, NXYL electric guitar strings offer more punch, crunch, and bite."Even if they're making that up to some degree, I'm impressed their copy writer used the correct terminology. This also ties in with the effect of the pickup's proximity to the strings. As the pickup gets further from the strings, it becomes quieter. Having the strings be smaller, or less permeable, is not much different than having the strings be further away from the pickup. In open magnetic circuits, permeability is irrelevant if greater than a few. This is true in general, and MacDonald showed it for strings and pickups. I like NYXL strings, but the claim of an effect from higher permeability is nonsense. I don't see him saying that it's irrelevant with respect to the voltage output, he says " So long as the relative permeability is large compared to unity, the factor (μrel −1)/(μrel + 1)
is essentially one, and the behavior of the electric guitar does not depend on the precise
value of the permeability of the string". What I take that to mean is that if "mu rel" is too close to one, it will adversely effect the behavior of the guitar, and that the guitar merely operates predictably when the "mu rel" is a value much greater than one. I don't take this to mean that "mu rel" makes a small different in the output voltage. But I'm just going off what he said, and only loosely what I see in the equation. More broadly, why is permeability "irrelevant if greater than a few" in an open magnetic circuit? What's the fundamental difference between a strong magnetic open circuit, and very weak closed circuit, for example? |
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Post by ms on Aug 11, 2017 18:12:24 GMT -5
In open magnetic circuits, permeability is irrelevant if greater than a few. This is true in general, and MacDonald showed it for strings and pickups. I like NYXL strings, but the claim of an effect from higher permeability is nonsense. I don't see him saying that it's irrelevant with respect to the voltage output, he says " So long as the relative permeability is large compared to unity, the factor (μrel −1)/(μrel + 1)
is essentially one, and the behavior of the electric guitar does not depend on the precise
value of the permeability of the string". What I take that to mean is that if "mu rel" is too close to one, it will adversely effect the behavior of the guitar, and that the guitar merely operates predictably when the "mu rel" is a value much greater than one. I don't take this to mean that "mu rel" makes a small different in the output voltage. But I'm just going off what he said, and only loosely what I see in the equation. More broadly, why is permeability "irrelevant if greater than a few" in an open magnetic circuit? What's the fundamental difference between a strong magnetic open circuit, and very weak closed circuit, for example? If that factor in the equation for the output voltage is close to one whenever urel passes some threshold when increasing from zero, then the output voltage is not much dependent upon urel. when that threshold has been passed. |
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Post by antigua on Aug 11, 2017 18:31:34 GMT -5
I don't see him saying that it's irrelevant with respect to the voltage output, he says " So long as the relative permeability is large compared to unity, the factor (μrel −1)/(μrel + 1)
is essentially one, and the behavior of the electric guitar does not depend on the precise
value of the permeability of the string". What I take that to mean is that if "mu rel" is too close to one, it will adversely effect the behavior of the guitar, and that the guitar merely operates predictably when the "mu rel" is a value much greater than one. I don't take this to mean that "mu rel" makes a small different in the output voltage. But I'm just going off what he said, and only loosely what I see in the equation. More broadly, why is permeability "irrelevant if greater than a few" in an open magnetic circuit? What's the fundamental difference between a strong magnetic open circuit, and very weak closed circuit, for example? If that factor in the equation for the output voltage is close to one whenever urel passes some threshold when increasing from zero, then the output voltage is not much dependent upon urel. when that threshold has been passed. [deleted previous comment] OK, so there are two significant terms in relation to the guitar string; "a", radius, and "μrel", permeability. So let's suppose "μrel" is always a big number above 1, and doesn't much matter. The steel winding of the low strings is still increasing "a", the radius, which is squared and multiplied, in the equation that JohnH posted. It seems to me that it's fairly substantial. |
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Post by antigua on Aug 12, 2017 15:01:00 GMT -5
I tried tuning a B string down to an A, to see how the amplitudes compare between the thin B string, and the fat A string, given the same pitch. The B string being tuned that low is rather floppy, but still taut enough to vibrate freely. The harmonics of both seem to be fairly loud, but the fundamental of the A string is clearly louder than the tuned down B string. If you roll the tone knob down, it will attenuate most of the harmonics, leaving only fundamental and some low harmonics, and the amplitude difference is then very obvious that way. Another interesting thing about playing the B string tuned way down to A, is that you can clearly hear the string's pitch drop as the vibration decays. It starts very sharp, and then goes very flat. The harmonics deviate initially a lot from the fundamental, but then match up again as the string comes to a rest. I think it has something to do with the relative stiffness of the Jungmann talks about this on page 16 in his thesis research.spa.aalto.fi/publications/theses/jungmann_mst.pdf , but I'm not sure why it's so exaggerated with the string is under low tension. |
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Post by JohnH on Aug 12, 2017 16:01:32 GMT -5
Another interesting thing about playing the B string tuned way down to A, is that you can clearly hear the string's pitch drop as the vibration decays. It starts very sharp, and then goes very flat. The harmonics deviate initially a lot from the fundamental, but then match up again as the string comes to a rest. I think it has something to do with the relative stiffness of the Jungmann talks about this on page 16 in his thesis research.spa.aalto.fi/publications/theses/jungmann_mst.pdf , but I'm not sure why it's so exaggerated with the string is under low tension. I think that change of pitch with decay can be explained, and also why it is more pronounced with lower string tension. When a string is firmly plucked by deflecting it sideways, it gets stretched to a slightly greater tension. This raises its pitch slightly and this effect is maintained as it vibrates, slowly decaying as vibrations subside. If you work through the maths of simple string vibration, it can be concluded for a given string weight that pitch is dependent on the force in the string (square rooted). If we assume that a 'standard pluck' can be represented by a certain force in the pick just before it releases. then the less tensioned string will be deflected more by this pluck. There will be a greater change in the force in the string due to this pluck than for the high tensioned string. And a higher change in force divided by a lower original tension gives a greater % change, and this is related to how many cents of a semitone we hear as a tonal change (with a square root involved).
I had another peak at that scary equation from McDonald:  At the bottom of it, he writes it in a condensed form that applies for very small variations in movement, which is not quite so scary. But relative to my quandary on the influence of string weight, it just reinforces my puzzlement because the voltage output is dependent on the string radius 'a' squared. that is, proportional to string cross sectional area or mass, hence why are lower strings not louder than they are? |
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Post by antigua on Aug 12, 2017 17:28:36 GMT -5
Another interesting thing about playing the B string tuned way down to A, is that you can clearly hear the string's pitch drop as the vibration decays. It starts very sharp, and then goes very flat. The harmonics deviate initially a lot from the fundamental, but then match up again as the string comes to a rest. I think it has something to do with the relative stiffness of the Jungmann talks about this on page 16 in his thesis research.spa.aalto.fi/publications/theses/jungmann_mst.pdf , but I'm not sure why it's so exaggerated with the string is under low tension. I think that change of pitch with decay can be explained, and also why it is more pronounced with lower string tension. When a string is firmly plucked by deflecting it sideways, it gets stretched to a slightly greater tension. This raises its pitch slightly and this effect is maintained as it vibrates, slowly decaying as vibrations subside. If you work through the maths of simple string vibration, it can be concluded for a given string weight that pitch is dependent on the force in the string (square rooted). If we assume that a 'standard pluck' can be represented by a certain force in the pick just before it releases. then the less tensioned string will be deflected more by this pluck. There will be a greater change in the force in the string due to this pluck than for the high tensioned string. And a higher change in force divided by a lower original tension gives a greater % change, and this is related to how many cents of a semitone we hear as a tonal change (with a square root involved). That makes sense. I think it's interesting that plucking any string causes it to momentarily go sharp, but that you don't notice it so much when it's at a high resting tension to begin with. I wonder if using super light strings, such as 8's, causes tuning problems that guitarists just overlook in exchange for a slinky feeling guitar string. Conversely, an appealing feature of thicker strings might be a more steady pitch. I had another peak at that scary equation from McDonald: At the bottom of it, he writes it in a condensed form that applies for very small variations in movement, which is not quite so scary. But relative to my quandary on the influence of string weight, it just reinforces my puzzlement because the voltage output is dependent on the string radius 'a' squared. that is, proportional to string cross sectional area or mass, hence why are lower strings not louder than they are? It's helpful to have this reference in view when looking at the monster voltage equation:  ...  Those first few terms seem especially interesting: B0 (magnetic field in gauss), a2 (string radius squared) and w2 (coil size), because he's saying the final voltage is a product of those factors, unless I'm reading it wrong. Does that mean doubling the gauss strength doubles the voltage? Or double the size of the string? Or doubling the size of the coil? It appears that's what he's saying, but it seems to me that magnifying these things doesn't magnify that output, at least not in practice. Also, what happened to frequency f? Where is time accounted for? |
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Post by JohnH on Aug 12, 2017 17:50:01 GMT -5
the first 3 terms imply:
Double the flux B, will double the voltage
Double string radius 'a' multiplies output x4 due the the a squared term
Coil size doubled also gives voltage x4
Time comes in due to the x and y terms with dots above them. These are 'rate of change' with respect time, ie velocity.
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