gradually decreasing fret width vs equal string pulls

Rick Barnhart · From: Arizona, USA

I've been thinking again :/ the space between frets decrease in width incrementally as you move the bar from the nut toward the bridge...why isn't the tone far more sharp when you engage a pedal or lever and pull a string, while the bar is in Hughey land? Seems to me the pulls would need to be shorter relative to the shorter fret distance. The question made more sense, while it was still in my thinker.
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Tom Gorr · From: Three Hills, Alberta

#1: It doesn't matter what your string is tuned to (eg using tension), the segmentation of the fret board always works to make a proper scale. EDIT; provided the open string scale length remains constant.

#2: The changer (pull) acts on the whole length of the string right back to the nut...eg...the string tension is the only thing that changes... And we tune the pull to make a different note in the scale...Revisit #1.

Pulling on a string at the changer is different than fretting with the bar...as the tension change acts over the whole string wheras the fretting with the bar changes the length of the vibrating string only on the vibrating portion between the bar and bridge/changer.

Anyways...that should be directionally helpful for intuitiveness without resorting to detailed physics equations dealing with vibrating strings, displacement and tension, etc. The only complete answer would be to review the mathematics of it.

Dave Grafe · From: Hudson River Valley NY

Contrary to popular belief, one really does use math after graduation Cool

Per Berner · From: Skovde, Sweden

I think we steel players are very lucky that the laws of physics work the way they do. If things worked the other way around, we would have some serious tuning problems indeed. Laughing

Tom Gorr · From: Three Hills, Alberta

The critical design factor is that with a rotating changer, the bridge to nut length for the open string remains constant.

If the changer displaced linearly rather than rotationally to change the tension of the string, the machine wouldn't work to make music...eg the open string scale length for each string would change depending on the 'pull' involved.

The changer is the design element that really brings everything together so that music is possible within the confines of the physics.

Jerry Jones · From: Franklin, Tenn.

http://bb.steelguitarforum.com/viewtopic.php?t=193767&highlight=linear+string+pull
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John Hopkin · Posted 14 Feb 2014 6:40 am

Rick Barnhart · From: Arizona, USA

John Scanlon · From: Jackson, Mississippi, USA

Ray McCarthy · From: New Hampshire, USA

Another way of putting it; the A pedal, for instance, raises the 5 & 10 strings one full tone. Where you place the bar doesn't matter.

Rick Barnhart · From: Arizona, USA

I get that, Ray. Smile

I'm not arguing about whether it works or not, because I know it does. Just seems to me that a half pedal should have the same relative raise at the 12 fret position as a full pedal depression would have with an open string.
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Tom Gorr · From: Three Hills, Alberta

I think the big difference is that tension changes are anchored at the open string tuning peg. Whereas vibrations are set by bar position.

Focus on how the string is attached at the tuner/nut and bridge. The tension changes 'slide' under the bar. You could never hold the bar if was the anchor, there's around 300 pounds of anchoring force. 30 pounds a string.

Rick Barnhart · From: Arizona, USA

Thanks, Tom. The light just came on...haha. Your tension explanation makes sense. So, the bar doesn't really become the nut, it simply divides the length of the vibrations of a constantly tensioned, unchanged length string, right?
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Scott Duckworth · From: Etowah, TN Western Foothills of the Smokies

And here's another point... if the fret spacing didn't decrease with length, think how long the guitar would actually be...

Very Happy

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Tom Gorr · From: Three Hills, Alberta

That's a good way of putting it

Curt Trisko · From: St. Paul, Minnesota, USA

This thread got my brain kick-started for the day. So what we're saying is that whereas else the bar shortens the effective string length, the changer changes the tension and not the length.

The length of the string has a proportional relationship to the pitch (which is why the frets are farther apart by the keyhead). The pitch gets exponentially and infinitely higher the shorter the string is and the opposite the lower the string is.

However, the tension of the string has an absolute relationship to the pitch. Applying the same change in tension to the string raises the pitch the same no matter how "long" the string is. This is because the frequencies between the pitches is closer for higher notes than lower notes. Applying the same change in tension on a "longer" string yields a greater rise in frequency than on a "shorter" string.

Is this right?

EDIT: I want to nuance what I said by saying that talking about the bar as "lengthening" or "shortening" the effective length of the string is a bad analogy because it implies a relationship to string tension. Like the posters above said, the tension passes beneath the bar and is unaffected by it. It is neat how all of these mathematical relationships interact perfectly. It's cool when the universe works out neat and clean like that.

Michael Hummel · From: Ottawa, Ontario, Canada

With a properly set up guitar, engaging a pedal to raise a string is no different than giving the tuning key a twist to raise the pitch. The bar works no matter how much you twist the tuning key, so why shouldn't it be the same when you push your pedal down?
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Dave Mudgett · From: Central Pennsylvania and Gallatin, Tennessee

The fundamental physics equation relating string fundamental vibrational frequency F, string tension T, and effective vibrational string length L of an 'ideal' string is, in any consistent set of units,

F = sqrt(T)/2*L*sqrt(d), where d is a constant representing the linear density (mass per unit string length) of the string.

I think this is also a pretty good practical model for a steel guitar string operating in its typical tension/frequency range - e.g., see more here - http://www.noyceguitars.com/Technotes/Articles/T3.html; or here - http://www.physics.usyd.edu.au/~cross/StringTension.pdf; but it is not perfect, and there are second-order effects.

In principle, placing a steel bar on a string at a particular location changes the effective vibrational string length only - this assumes that the bar isn't pushing down on the string hard enough to stretch the string further and/or increase the tension.

So assuming that a changer pulling a string doesn't change the length of the string (see the thread Jerry referenced above on changers like the Lamar or Excel which do), then except for second-order and above effects, placing the bar doesn't change the relative fundamental pitches of pulled vs. non-pulled strings. For example, if you place the bar at the string midway point, both pulled and non-pulled strings sound at the octave (double frequency) of the open-string frequencies.

I think string non-idealities and bar placement imperfections are the main issue with pitch discrepancies in the high registers of steel guitars - as the effective string length gets shorter and shorter, it is more and more sensitive to small issues in any of this.

Fretted guitars are a bit different - you actually press the string down to the fret, and thus change both the tension and effective vibrational length of the string, which is why one typically needs compensation at the bridge (and/or nut, as with the Buzz Feiten or Earvana systems).

Tom Gorr · From: Three Hills, Alberta

Curt Trisko · From: St. Paul, Minnesota, USA

Dave Bertoncini · From: Sun City West, Arizona USA

Fret spacing is determined by the rule of 18 developed by Pythagorous. The real number is actually 17.817, where you take a given scale length and divide by 17.817 to determine position of the first fret, you then take the remaining distance and divide by 17.817 again to determine position of the 2nd fret and so on. Changing the pitch of a string by tension such as a pedal is constant regardless of position of the bar.

Curt Trisko · From: St. Paul, Minnesota, USA

...but if there was no bar, and you had to control your pitch purely by string tension, then you would see the effect that original poster is talking about.... where you would need more travel to create an increase in pitch at a lower note than a higher note. Essentially, the bar (with fret spacing) factor for that variable so that increase the string tension then creates absolutely consistent results no matter where you're at on the fretboard.

John Scanlon · From: Jackson, Mississippi, USA

Joe Naylor · From: Avondale, Arizona, USA

Once I found out the rule of 18 and could figure out the fret spacing easily.

You can make a neck as long as you want - is one theory and it has been tried many times to make longer and shorter necks / fret boards.

My 2 cents is only worth a penny.

Joe
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