STRAIGHT (DIRECT) LOWERING ON EXCEL SUPERB

Carlos Polidura · From: Puerto Rico

A few weeks ago I noticed that my Excel Superb has what is called "DIRECT... or STRAIGHT PULL" but it works only when lowering the string. Aside from the "Lamar" guitars wich has the raise and lowering "STRAIGHT PULL" mechanism, this is the only guitar I've seen with this feature incorporated... so far. I think this is a nice thing to happen to "PEDAL STEEL GUITARS".
Have you noticed yours?

Lee Baucum · From: McAllen, Texas (Extreme South) The Final Frontier

If I understand you correctly, then you should check out the Mullen Guitar site. They have that feature.
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Michael Johnstone · From: Sylmar,Ca. USA

No - the Mullen has changer cams that rotate around an axel like most steels.They don't slide back and forth in line with the string like the ones being mentioned. Excel,Lamar and Anapeg are the only ones with that system that I know of. Someday all pedal steels will have changers like that - it's simply a much better design and one that doesn't ever break any strings.All other designs break strings by the fact that they bend the string back and forth around a cam causing metal fatigue and eventual string breakage - especially on the smaller gauge strings.Not so with straight pull systems.

Carlos Polidura · From: Puerto Rico

Lee, I understand that Mullen has redesigned their guitars, but I don't know in what respect yet.
However it would be great to have all manufacturers incorporate this feature on their guitars. It's great to know so many dedicated people keep working to upgrade this beautiful musical instrument that we love so much.

Jon Light · From: Saugerties, NY

What I have never been able to understand---and this isn't any sort of argument.....too many people have tested and play these instruments to argue with the technology...Tom Brumley comes to mind with his Anapeg---how can you change the scale on some strings with linear changer motion and not mess up intonation? Seems to me a full step raise might require 1/32"--1/16" movement. Alongside a string with no pedal movement. This is a mystery to me.

Mike Wheeler · From: Delaware, Ohio, USA

Interesting question, Jon. I'd like to hear the responses, too.
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Michael Johnstone · From: Sylmar,Ca. USA

That's the first thing I thought also but my Excel plays perfectly in tune within the temperment I use which is somewhere between just and ET. Another thing that clouds the issue (for the better) on an Excel Superb is the body flex compensator gadget which sharps the whole neck microscopically in inverse proportion to the amount of flex induced by various pedal combinations. To be more clear the Anapeg and Lamar changers are totally linear in their movement while the Excel uses a blend of linear and rotary motion like the old Fender 800 and 2000 changer. If you look at the Excel Superb changer,it looks pretty normal from the outside with a lateral axel supporting raise and lowering fingers but the fingers have eliptical holes allowing them to slide forward and backward on the axel. When they reach the end of their ellipse and come to rest on the axel they start to rotate like a standard changer. That's how the blend of motion is acheived. Inscrutable - those Japanese.

ed packard · From: Show Low AZ

Jon L...When you tune up, you set the tension for the open string length and pitch with no changes activated. Then you activate the change and set the tension for that new pitch...If the string length has been changed by a linear changer you will have tuned (tensioned) it to pitch according to the new string length. It would be even better if the changer (source of string length variation with a linear changer) was located on the players left = behind the bar.

Jon Light · From: Saugerties, NY

Ed--I'm trying so hard to follow but I can't shake the mental image:

The 12th fret harmonic node is....right smack at the 12th fret. Exactly 12 inches from the nut on a 24" scale.
Pedal a whole step raise on, say, a wound .024 6th string and watch the changer move, say 1/16". Hasn't the scale on this one string become 24 1/8"? Hasn't the 12th fret node moved 1/32" up? On an open guitar, no problem but if you are using your bar and simultaneously picking an unpedaled string, how can you not have to slant the bar to compensate for one or the other?
Are these real-world numbers just not enough to make a difference?

Again---I am not arguing with the changer. These guitars would not be played if they didn't work. But I'm trying to understand how the physics in my head are not gibing with the physics on this guitar.

And it's entirely possible that I will not understand the explanation. It would not be the first time.

ed packard · From: Show Low AZ

Whoops...think that I slipped a digit. Will redo the calcs tomorrow.

Hans Holzherr · From: Münchenbuchsee, Switzerland

http://steelguitarforum.com/Forum5/HTML/008621.html

ed packard · From: Show Low AZ

Hans...thanks for the heads up reference, I missed that thread.

Jon, you are probably correct as you are seeing the physics of the problem…it is the magnitude of the difference that is confusing the issue. Let’s run some numbers:

The + halftone ( = + one fret) in ET is determined by the 12th root of 2 = 1.059463 * Hz.
The + cent (= + a portion of a fret) in ET is determined by the 1200th root of 2 = 1.000578 * Hz.

The – halftone (= - one fret) in ET is determined by the reciprocal of the 12th root of 2 = 0.943874 * Hz.
The – cent ( = - a portion of a fret) in ET is determined by the reciprocal of the 12th root of 2 = 0.999423 * Hz.

Let’s take the 24” neck, and use 5 cents as the amount of off pitch that is annoying.

If we had two strings, both tuned open to G#, but one was longer than the other, how much longer would one have to be to get 1 cent and 5 cents differences at the 12th fret.

+1 cent (in Hz) @ fret 12 = [12*( 2^(1/1200))-12 ] * Hz@12th fret = 0.006933 * HZ for G#3.
+5 cents (in Hz) @ fret 12 = [12*( 2^(5/1200)-12] * Hz@ 12th fret = 0.034707 * Hz for
G#3.

The above gives the number of cycles difference = beats to be heard if one string is shorter than the other. If we string length at the 12th fret instead of Hz in the equations, we get the amount of difference in string length needed for the cents difference.

24” scale length = 12” from 12th fret to the bridge.

A one cent decrease in pitch = 12”* 1.000578 = new string length = 12.00693”, therefore the – one cent is a string length increase of 0. 00693” = about 2 pieces of paper = much less than the width of the fret markers.

A five cent decrease in pitch = 12” * 1.002892 = new string length for a decrease in pitch of five cents = 12.03471”, therefore the – 5 cents is a string length increase of 0.03471” = about 4 or 5 pieces of paper.

Conclusion is that under the worst case conditions of two strings tuned the same, one being longer than the other by 5 cents worth at the 12th fret, you would hear a few beats difference.

The beats difference would increase as the fret number is increased (= put the changer on the other end).

In the Excel, Anapeg, Lamar cases, where the strings are tuned to different fundamentals, the off pitch due to length change (changer activation) will be very difficult to hear.

The G#3 = 0.011” dia string/note stretches more than the others for a half tone change. A linear changer would give a length increase of about 0.035” (if I recall correctly), but since it is tuned (tensioned) to this new length, the pitch shift will be less than our cases above.
This is not an easy thing to explain in simple terms (at least for me)…Maybe someone else can take a whack at saying it simple…and check the math while they are at it.

Jon Light · From: Saugerties, NY

Since there's no mistaking the common sense correctness of my original premise (I think), I would think that the only explanation is indeed as you conclude---the amount of movement is not enough to screw things up. Your .0347" is right around 1/32" and I would have thought that would be enough to compromise intonation with a straight bar but the only thing that counts is how it sounds.
I'm done.

Earnest Bovine · From: Los Angeles CA USA

ed packard · From: Show Low AZ

Easy enough Earnest...run the experiment and tell us the results. If I was near an Excel/Lamar/Anapeg I would do it.

Suggest that a bar be used as a capo at the 12th fret for the cents measurement....not too fat.

The cents difference increase with increasing fret number is why I suggest that for the linear changer, the changer be on the players left...as on the BEAST.

Dave Mudgett · From: Central Pennsylvania and Gallatin, Tennessee

Just to rehash what's been said in a slightly different way, I think of it like this. At the nominal pitch, there is a fixed amount of string between the changer end and the tuner or keyless clamp. The nominal fundamental vibration frequency F is governed by this equation:

F = (1/2L)*sqrt(T/D) where L=string length; T=string tension; D=string mass density.

The mechanism of frequency change by the changer is primarily through changing the string tension, which is quite different than changing the frequency by moving the bar to shorten the strings, leaving the string tension constant. On this type of changer, string length change depends primarily on how much the string needs to stretch (on a raise) or contract (on a lower) to produce the required string tension and therefore pitch - i.e., T is a function of L. The string length change on this type changer is a secondary effect, but works against the desired change. For example, shortening the string reduces tension by releasing the stretched string, but the secondary effect - due to the above equation - is to raise the pitch.

Suppose the length change required is about 1/16" as Jon suggests. With a vernier caliper, I measured how much my conventional changer fingers moved the string on a typical raise or lower - that looks about right. Let's assume that we're talking about a keyless 24" changer - the overhang would figure in the amount of string stretch required to bring it to the correct new pitch. At the 12^th fret, you're off by half that amount, or 0.03125 inches. The distance to the 11^th fret is 12*1.059463 = 0.713556 inches, and the distance to the 13^th fret is 12/1.059463 = 0.673507 inches. So, roughly we're talking about a length, and therefore pitch error of about 100*0.03125/0.7 or 4.5 Cents. It seems to me that many players would notice this amount of pitch error. Something doesn't sound right, to me.

Perhaps a keyless changer coupled with longer scale length has a significant effect on the amount of string stretch required. Can the changer fingers be offset along the string axis a bit to compensate and average the errors, similar to the bridge saddles on a 6-string guitar? I've never seen one of these changers.

Michael Johnstone · From: Sylmar,Ca. USA

Like I explained a couple posts ago,on the Excel changer there's a blend of lateral and rotary motion. That, coupled with a 25.5" scale probably minimizes the potential problem.Anapegs and Lamars are totally lateral and altho I've not seen anything but pictures since Lamar went to the new changer and longer scale, I've sat down and played an Anapeg once for about 10 minutes at a noisy convention and couldn't have told you if it was even in tune very well let alone something as subtle as this. But I have pretty good ears and after 4 years behind this ax I know this much - my Excel plays in tune anywhere up or down the fretboard.

Carlos Polidura · From: Puerto Rico

Same with me Michael. The Excel is probably the most stable, easy and comfortable guitar I've ever played. Not to mention the guitar stayes in tune very well and body flex is very minimal, almost none.

Carlos Polidura

Bryan Rankins · From: Missouri, USA

I am a very new steel player, only since July, but have played guitar and a few other instruments for over 30 years. I have watched numerous times a steel player break a string. I have also been at the receiving end of a pedal steel player who couldn't play in tune to save his life, or the lives of his band members for that matter. When I bought an Excel D10 in July and started playing, I thought I must be doing something wrong, because my pedal steel plays in tune, stays in tune, and I have yet to break a string. When I inquired about this to some seasoned players, their response was "oh yeah, those Excel's don't break strings like other steels". I will probably never understand why, and that's okay with me. Call me crazy, but I like not breaking strings and playing in tune, and if I did want to play out of tune, I'd just pick up my banjo.
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ed packard · From: Show Low AZ

Re the 5 cent line that I drew (for convenience) as the point of annoyance…The 5 cents referred to was ET, not JI as many tune. ET thirds are off from JI thirds by a bunch, as are 7ths etc. Only the roots and 5ths are close in both systems.

This means that the harmonics in different strings will disagree with the fundamentals of the same note name in others. What will be annoying is sort of a random variable.

Increase the number of strings being excited at the same time and off tune tends to get lost in the mix. Play at a faster tempo and any off pitch tuning tends to become less noticeable.

The test can be made with the instrument that you have by tuning the strings for which you have concern off + by 5 cents and play a bit…then – by 5 cents and playing a bit. You can state your opinion of the results here on the Forum.

Re the amount of travel (string stretch) for a linear, or semi linear changer finger...it will vary with the total string length, the string gauge, and the number of halftones changed. This means that any off pitch effect caused by differences in changed scale lengths (via pedals and levers) should be less for short scale lengths, and for most keyless tuner instruments.

Jim Palenscar · From: Oceanside, Calif, USA

The Anapeg lowers are straight lateral while the raises are about 70% lateral/30% rotary (right Noel?). Back to tuning issues-- what matters after all theoretical calculations are made, is what is heard/perceived by the listener- and to re-paraphrase Winnie's comment about all this-- "Tom Brumley - who has ears like a bat, Smile

- says there are no intonation problems". That was all I needed- 'nuff said.

Gil Berry · From: Westminster, CA, USA

I don't think it has ever been tried but... what if the changer was a straight lateral pull device, and a second nut (roller nut, of course) were used on the changer end as well as the nut end... so that the scale length was between the two sets of rollers....wouldn't that eliminate the intonation problem (if one even exists). Wonder what that would do to tone?? Would the addition of a second roller nut section relieve some of the damping done by the changer...or just make it worse? We all know the push-pulls have that tone because of the limited damping of the changer...so....would this system improve the psg or no?

Jim Palenscar · From: Oceanside, Calif, USA

So many questions-- so little time Smile

ed packard · From: Show Low AZ

Gil...The BEAST comes as close to your described concept as any I know of. Go to the ref below and you can browse around in the photos, and instrumented frequency vs. time charts forabout 30 instruments. The materials, dimensions, et al are listed somewhere in the photos.

The BEAST is the first three photos. You will find that the bridge is not attached to a changer. Compare the freq charts and you will see that the freq/time response for it is hardly that of a banjo.

This info was taken by the same 2 persons on the same day at Jim Palenscar's North County Steel shop in Oceanside Ca.

http://s75.photobucket.com/albums/i287/edpackard/?start=all

No Roller nut is used on the BEAST as the amount of string travel at that point is so small, and there is hardly any excess string length to pull.

basilh · From: United Kingdom

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