posted 18 March 2001 05:22 PM         

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The Unofficial Photographer of The Wilkinsons

posted 19 March 2001 08:45 PM         

Hz or frequency is location and cents is the distance between locations. An interval, like a 5th, is the distance between 2 notes, and the notes can change, like C-G or A-E, but the interval remains the same, the amount of cents remains the same. In fact the amount of cents in a 'tempered' 5th is 700 cents. To calculate the Hz of E a tempered 5th above A-440 you would multiply 440 x (12th root of 2 to the 7th power).

posted 19 March 2001 09:49 PM         

What you said is that if you draw a straight line which is tangent anywhere on the curve which represents frequency (HZ), then you get the relationship of Hz to cents at that frequency.

Well, I'm not absolutely positive, but I think all this will give you is the slope of the curve at some arbitrary point. I'm not sure what the slope of this line represents, but I don't think it has a bearing on the relationship between Hz and cents.

Remember, a "cent" is defined as 1/100 of a semitone, which I'm assuming is the distance between two consecutive tone half-steps (at least that's what I think somebody said above). So, assume that there is some logrithmic function, f(x)=Hz, where x=the nth note in a chromatic scale and Hz=the frequency of that note. That is, there is some function in which you can plug a number corresponding to a note and calculate the frequency which corresponds to that note.

Let's say we start by plugging in "1" and we get f(x)=440 for "A," the first note, and so on for a succession of notes of a chromatic scale. Let's also say that we are plotting this on a graph where "x" is on the horizontal axis, and f(x) is on the vertical axis. So, when we plot a series of points representing the frequency of each of a series of notes of a chromatic scale, what we have is a series of points which can define a logrithmic curve.

However, the set of points, not the curve, is what represents the series of notes because we have a defined incremental set of notes of a chromatic scale, not a continuous number of musical tones. In other words, in terms of a psg, the notes are located roughly at the frets, not everywhere along the strings.

Now, to define the relationship of frequency to cents, we would then draw a horizontal line through each of the points which represent a note or tone. Then, measure (or calculate) the vertical distance between each successive pair of horizontal lines. We would then need to divide each of these distances into 100 equal parts to determine the value of one cent.

As you can see, because the curve defined by the set of points is logrithmic, the distances between each pair of successive horizontal lines will be different. Thus, the value of a "cent" would presumably be different depending on which two notes or tones you happen to be measuring between at the moment.

As you know, the original question was, "how many Hz equals one cent?" or something like that. So, to determine the answer to this question, we would simply take one of the values of a cent and compare it to the length of one Hz (which is a constant).

It is evident, then, that the solution to the original question is a discontinous function. That is, a "cent" is equal to a first value between a first pair of adjacent notes, a second value between a second pair of adjacent notes, and so on and so on.

Actually, Danny Bates had the right idea. One cent is equal to 1% of the frequency difference between two consecutive notes of a chromatic scale. Of course, as was said above, the value of one cent will be different for each pair of notes.

It "Hertz" me to say this, but YOU'RE ALL WRONG, WRONG WRONG!!!

[This message was edited by Tom Olson on 20 March 2001 at 06:48 AM.]

posted 20 March 2001 07:40 AM         

quote:

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What you said is that if you draw a straight line which is tangent anywhere on the curve which represents frequency (HZ), then you get the relationship of Hz to cents at that frequency.

Well, I'm not absolutely positive, but I think all this will give you is the slope of the curve at some arbitrary point. I'm not sure what the slope of this line represents, but I don't think it has a bearing on the relationship between Hz and cents.

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Tom, maybe a physics with calculus class would help. The slope of a line tangent to a point on any curve gives you the mathematical relationship of the y to x axis. For example using the graph I discussed earlier (Did you actually make yours?)

Pick the point where y = 682 Hz and x = 60 cents between E and F. Draw a tangent line (I assume you know how) and and find its slope. For my tangent line the dY = 23.6 Hz for my dX = 60 cents. This gives you a slope of 0.393 Hz/Cent. Notice the UNIT Tom it gives you the RELATIONSHIP OF HERTZ TO CENTS!!! 1 cent equals 0.393 Hz at a frequency of 682 Hz.

quote:

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It "Hertz" me to say this, but YOU'RE ALL WRONG, WRONG WRONG!!!

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This kind of statement is why I had decided to quit contributing to this Forum. But I didn't want anyone who has been following this marathon to go away uninformed or should I say mis-informed.

Aloha

[This message was edited by Rick Aiello on 20 March 2001 at 07:43 AM.]

posted 20 March 2001 08:34 AM         

Anyway, maybe a review of Calculus would, in fact, help me.

However, from what I remember, the slope of a line tangent to a point on a curve gives you the relationship of the incremental rate of change of the one coordinate with respect to the other at that one point. In other words, it gives you the incremental differential of the y-coordinate with respect to the x-coordinate, or dy/dx. Again, the slope gives you the rate of change of one variable with respect to the other. It does not give you the value of one variable compared to the value of the other, which is what I thought we were looking for to begin with.

An example of how this differential can be used could be illustrated in the classic distance/velocity/acceleration curves which are used for pedagogical purposes. For example, if you consider a car that's accelerating at a constant rate from a dead stop at a starting point, you can write an equation for the distance the car will be from the starting point as a function of time. The equation will be a geometric equation. That is, the distance will vary geometrically with respect to elapsed time, so the curve which represents distance will be some sort of parabola or something like that.

Calculus, as you've pointed out, can be used to find the velocity of the car at a given point on the curve. This, of course, can be done by calculating the slope of the curve at the given point on the curve. The slope of this line will give you the rate of change of distance relative to the rate of change of time. This relationship, of course, is velocity.

Now, going back to the example at hand, as you've pointed out, if you plot the frequency of each note in a chromatic scale as the function of the position of a given note in the scale, you will get a set of points which can define a curve -- that is you can "fit" a curve to the set of points by "connecting the dots".

As I have tried to point out in my earlier post, you really don't have a curve, per se, but a set of points because the notes of the scale occur at given intervals -- there is not a continuous line or curve which represents the scale because then you would have an infinite number of notes in a scale.

Even if you could represent a musical scale with a continuous curve, what would the slope of a line tangent to the curve represent? The slope of the curve would represent the rate of change, at the point of tangency, of the frequency relative to the rate of change of the scale. I'm not sure exactly what this would correspond to other than how fast the frequency changes as you progress up the scale.

I don't know a lot about music theory, so I'm assuming Bob was correct when he said that a cent is equal to 1/100 of a semitone. I'm assuming that a semitone is the distance between two consecutive notes in a chromatic scale.

To clarify so far -- what is the frequency of a note halfway between two consecutive notes of a chromatic scale? My answer, at least, would be that the question cannot really be answered because there is no "note" between two consecutive notes of a chromatic scale. "Notes" only exist at specific frequency intervals. What you have in between two consecutive notes are cents.

So, let's say we are considering a first note and a second consecutive note of an ascending chromatic scale. The first note has a first given frequency and the second note has a second given frequency which is higher than the first given frequency.

It seems logical to me, that given the definition of a cent, the value of a cent would be determined be first finding the difference between the second and first frequencies, and then dividing this difference by 100. Remember that the difference between the frequencies of two consecutive notes is determined be measuring the vertical distance between the notes (assuming we made our graphs with the frequency on the y-axis). In other words, the measurement of the frequency difference should be made parallel to the y-axis.

This measurement, divided by 100, would give you 100 equal frequency increments that, when added up, will take you from the frequency of the first note to the frequency of the second note, or vice versa. In this solution, the value of each cent between two consecutive notes is constant. But, when you jump to a different pair of consecutive notes, the value of each of that set of cents will be different from the value of each of the first set because the difference in frequencies b/t two consecutive notes changes as you move around the scale.

So, as you move from one note to another, the frequency of the notes changes in a non-linear manner, but in between each pair of consecutive notes, the cents change linearly. The graphic solution would be a set of short vertical line segments which intersect, roughly at each of their midpoints, a logrithmic curve which is "fitted" to the set of points which represent the note of a chromatic scale.

Then again, maybe I'm wrong.

[This message was edited by Tom Olson on 20 March 2001 at 08:56 AM.]

posted 20 March 2001 09:07 AM         

I think that where we disagree is how we are plotting the "cents" on our respective graphs. It appears that you are plotting cents on the x-axis and are also plotting frequency on the y-axis. That would seem to make frequency a function of cents.

Perhaps this is correct, but I don't think it is correct. If you plot frequency on the y-axis, you would also plot cents on the y-axis because 100 cents takes you between two specific frequencies.

The way I have envisioned the graph, the x-axis is the "notes" of a chromatic scale and the y-axis is the frequency. That way, a given frequency is a function of the position of the note in the scale. This way, you have note#1, note#2, note#3, note#4, etc. plotted along the x-axis at regular intervals. Then, on the y-axis, you have the frequencies that correspond to the notes.

One hundred cents would equal the vertical distance between two consecutive frequency points which correspond to two consecutive notes.

[This message was edited by Tom Olson on 20 March 2001 at 09:08 AM.]

posted 20 March 2001 09:08 AM         

quote:

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However, the set of points, not the curve, is what represents the series of notes because we have a defined incremental set of notes of a chromatic scale, not a continuous number of musical tones. In other words, in terms of a psg, the notes are located roughly at the frets, not everywhere along the strings.

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quote:

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Now, going back to the example at hand, as you've pointed out, if you plot the frequency of each note in a chromatic scale as the function of the position of a given note in the scale, you will get a set of points which can define a curve -- that is you can "fit" a curve to the set of points by "connecting the dots".

As I have tried to point out in my earlier post, you really don't have a curve, per se, but a set of points because the notes of the scale occur at given intervals -- there is not a continuous line or curve which represents the scale because then you would have an infinite number of notes in a scale.

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quote:

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To clarify so far -- what is the frequency of a note halfway between two consecutive notes of a chromatic scale? My answer, at least, would be that the question cannot really be answered because there is no "note" between two consecutive notes of a chromatic scale. "Notes" only exist at specific frequency intervals. What you have in between two consecutive notes are cents.

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Do some research on tonality and you will find that there are 19 note scales, 31 note scales, 50 note scales, 81 note scales etc. etc.... Why do you think most people love the sound of slide quitar, Hawaiian steels and pedal steels. You get to hear all the "notes" between E and F when you gliss, slant or do a pedal change.

This site might help www.bikexprt.com/music/notation.htm

[This message was edited by Rick Aiello on 20 March 2001 at 09:20 AM.]

posted 20 March 2001 09:28 AM         

This doesn't mean you can't slide or gliss between two notes. You can have tones in music which are not notes and which are in between notes. I'm not saying there are not tones between notes. I'm saying there are a set number of notes in a scale.

I think you and I agree that you can express any given tone or frequency which is between two chromatic notes by defining one of the notes and then defining how far above or below the note the given tone is. Apparently, one way to do this is to use cents which is 100 equal divisions between two consecutive chromatic notes.

Another way to do it, of course, is to simply say what the frequency of the given tone is. Either way will get you to the given tone.

Let's say we want to define a tone that's half way between the notes of E and F. One way to define this note is "E + 50 cents."

Assuming we're using an ET scale, we know what the frequency of the E is and we know what the frequency of the F is. I think we can agree that there are 100 cents between the frequency of E and the frequency of F because there is a semitone between E and F.

So, E + 50 cents would be found by first finding the difference between the frequency of E and the frequency of F. This difference will be expressed in Hz. If we now divide this difference by 100 we will get a value in Hz for one cent when we are considering the frequencies between the E and F notes.

If we multiply this value in Hz of one cent by 50, we get a value in Hz, which, when added to the frequency of the E note will give us the tone we are looking for.

posted 20 March 2001 09:52 AM         

f1 * 2^n/1200 = f2

f1 is the first frequency in Hz, and f2 is n cents above it.

posted 20 March 2001 10:01 AM         

Apparently, cents do not vary linearly between notes, but vary as a function of the frequency, which varies non-linearly.

Since cents apparently vary as a function of the frequency, then no two cents are exactly the same, so the original question of how many Hz equals a cent pointless. It's sort of like asking how many seconds equals a foot. They are two different things and the question only has meaning when you know exactly what speed your going.

Likewise, the question how many Hz equals a cent is only valid at a specific frequency, so it really does no good to know the answer. In other words, in order to calculate the value of a cent you have to first know what frequency you're at. So, if you're trying to define a specific tone, why not just use the frequency? Like Bob said, earlier, it's really kind of useless to know the answer to the question.

posted 20 March 2001 10:23 AM         

In any case, it appears that the only way to find the value of a cent for a specific frequency is to take the integral of the above equation from f1 to f2. You cannot simply determine the value of n at a given frequency which is equivalent to measuring the slope of a line tangent to the curve. This latter method would work only if cents had no given interval.

posted 20 March 2001 10:41 AM         

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T. Sage Harmos
Musical Instruments

posted 20 March 2001 12:01 PM         

It was about silence, and how as long as we're alive, we'll never hear silence. In eastern music, silence can be as important as sound, in western music, it's typically the space between the notes to make rhythm.

It was about the politics of the proscenium and the ritual of concert going. Those people had payed good money and were not hearing good music.

It addressed the issue: of the sounds that were in that hall, did they not fulfill all of the qualifications to be music?

But foremost, it was a carefull delineation of time, and about experiencing amounts of time.

posted 20 March 2001 12:03 PM         

Although music is an art form, there has to be some theory to it for it to work -- other wise you end up with Yoko Ono like "music." Imagine trying to play a psg without any set relationship to any string pulls. Those relationships are all based on theory.

I don't deny that there are musical tones in between notes. Remember -- a "note" is only theory -- it only exists in your mind. It has a specific definition. One note "A" is defined as 440 Hz. This is the definition of the note. You can play tones a little sharp of 440 Hz and you can play tones a little flat, and it can sound really good. I've played the blues before on guitar -- I know what you're saying about bending notes and getting tones that are in between defined notes.

However, like you said, none of this is all that important in the overall scheme of playing music. I just like to get into a good argument every once in a while.

posted 20 March 2001 12:45 PM         

quote:

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In any case, it appears that the only way to find the value of a cent for a specific frequency is to take the integral of the above equation from f1 to f2. You cannot simply determine the value of n at a given frequency which is equivalent to measuring the slope of a line tangent to the curve. This latter method would work only if cents had no given interval

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You are quick to abandon your lenghthy argument (based on misinformation) and jump on Bob's bandwagon but if you would bother checking the math you would see the following:

Using Bob's equation: 682 Hz x 2 raised to the 1/1200 power = 682.394 Hz
682.394 Hz - 682 Hz = 0.394 Hz
Therefore at 682Hz 1 Cent = 0.394 Hz

Finding the slope of the line tangent to the curve at y = 682 Hz you get :
Slope = 0.393 Hz/cent
Therefore at 682 Hz 1 Cent = 0.393 Hz
(0.001 difference is associated with drawing the curve by hand.)

It seems that the graph gives you the SAME value as Bobs equation. IMAGINE THAT !!!!

This was just a random frequency I picked, it will work for any you choose.

I don't mind a good argument but as you can see I always give examples of my points and I never discuss things that I know very little about.

So ends my brief and not all that pleasant experience as a contributor to this forum.

[This message was edited by Rick Aiello on 20 March 2001 at 02:03 PM.]

posted 21 March 2001 10:26 PM         

I'm never one to walk away from a good argument if you're going to keep throwing punches. Let's keep this short and sweet since I'm running a bit short of time at the moment.

Here's the deal: You say you came up with the solution graphically, but you really didn't show or explain how you did it. You simply said you drew a line tangent to some curve and "voila!" it magically gave you close to the right answer. Any monkey can draw a line and sooner or later it will have the slope you're looking for.

Now I'm going to give you an example of what I consider to be a halfway decent explanation.

Let's use both the formula that Bob gave us and the one that I came up with (although I admitted that it was based on incorrect assumptions) to calculate the value of a cent at an arbitrary location.

Let's pick our arbitrary location to be the value of a cent about midway between the notes of C(261.626Hz) and C#(277.183Hz). So, let's pick the exact location to be the value of a cent between C#-49 cents and C#-50 cents. This is the same as saying C+50 cents and C#+51 cents. If you go to the table shown on this website (www.pianosupply.com/cents-hz/) you will find the frequencies for a range of notes complete with the frequencies for the cents as well.

You will find that C#-49 = 269.448Hz and that C#-50 = 269.292Hz, a difference of 0.156Hz.

So, using the correct formula we start by raising 2 to the power of 1/1200, right? We get 1.00057779. Now, we multiply that by the lower frequency to get the higher frequency. So, 269.292 x 1.00057779 = 269.4475942 which is rounded to 269.448 due to the concept of significant digits. As you can see, this checks out correctly and gives us a difference of 0.156Hz.

Now let's examine my method. If you recall, I said to first find the difference in frequency between the two consecutive notes of the scale, divide that difference by 100 and you'd have the value of a cent. Let's do it: 277.183 - 261.626 = 15.557. Now, divide that by 100 and you get 0.15557 which is rounded to 0.156 due to the concept of significant digits. This is exactly the same answer as the "correct" method. Gee Rick, IMAGINE THAT!!!!

I hate to point this out, but my method got even closer than yours did.

Now, I'm not going to take offense to your passive-aggressive way of telling me that I don't give examples and that I talk about things that I don't know anything about.

But, I would appreciate the courtesy of you giving me an explanation of how you supposedly drew a logrithmic curve (by hand?) and then drew a tangent line (by hand?) at a point and then were able to figure out the slope of that line to within three decimal points. Excuse me while I laugh my rear end off!

[This message was edited by Tom Olson on 21 March 2001 at 10:31 PM.]

posted 22 March 2001 06:59 AM         

quote:

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Here's the deal: You say you came up with the solution graphically, but you really didn't show or explain how you did it. You simply said you drew a line tangent to some curve and "voila!" it magically gave you close to the right answer. Any monkey can draw a line and sooner or later it will have the slope you're looking for

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I did tell you how to do this but here goes again in more detail. Get a piece of graph paper. Construct the y-axis by equal increments starting at 440Hz and ending at 880 Hz. I used 11Hz/line.

Now construct the x-axis starting at 0 cents and ending at 1200 cents. I used increments of 20 cents/line.

Now, plot the points associated with the frequencies of the notes of the octave and their cent counterparts.

My points are:
(0,440) A
(100,466) Bb
(200,494) B
(300,523) C
(400,554) C#
(500,587) D
(600,622) Eb
(700,659) E
(800,698) F
(900,740) Gb
(1000,784) G
(1100,831) Ab
(1200,880) A

This gives you plenty of points to draw a curve. (or use a graphing calculator or computer program for better results).

This curve graphically represents the relationship of Hz to cents as seen on the chart that Tonejunkie pointed to. Compare the two and you will see.

To find the value of 1 cent anywhere in this octave find the frequency you want (I picked 682 Hz). Place a dot on the graph (760,682) and take a straight edge and draw your tangent line (This is a line that touchs the point but does not cut through the curve itself.) The slope of this line can be found by making a right triangle using the slope as the hypothenuse. The change in y was 23.6 Hz and the change in x was 60 cents in my example.

The slope was 0.393 Hz/cent therefore at 682 Hz 1 cent = 0.393 Hz.

NO MAGIC JUST GOOD MATH

Sorry about all the edits - I'm dyslexic.

[This message was edited by Rick Aiello on 22 March 2001 at 07:04 AM.]

posted 22 March 2001 11:58 AM         

quote:

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But, I would appreciate the courtesy of you giving me an explanation of how you supposedly drew a logrithmic curve (by hand?) and then drew a tangent line (by hand?) at a point and then were able to figure out the slope of that line to within three decimal points. Excuse me while I laugh my rear end off!

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By the way Tom, I did use a straight edge for the tangent line.

Remember - when making your tangent line you must keep the right triangle so the sides are parallel with the x and y axis (no tilting please - that would be cheating). Just ease it up till the hypothenuse touches the curve.

Still laughin' ??

[This message was edited by Rick Aiello on 22 March 2001 at 05:07 PM.]