roller nut question

C. Christofferson · Post by **C. Christofferson** » 18 Aug 2006 2:14 pm

After seeing Gary S's thread 'a Better Sustaining Nut' i got to wondering - If you were to put a scale under a roller nut with the strings tuned to pitch, how 'heavy' would it register as weighing down onto the body. Any figures?

basilh · Post by **basilh** » 18 Aug 2006 7:41 pm

I think to an extent it would depend on the angle of the strings on the tuner side of the nut.
IMHO
Baz

ed packard · Post by **ed packard** » 19 Aug 2006 7:03 am

Chris...Off the top of my head,the thought pattern runs like this:

If the string tension (say 30 lbs) is straight across the nut (0 degrees), then there is no downward pressure on the nut.

If the string tension is straight downward (90 degrees) behind the nut, then the downward pressure would be 30 lbs.

The downward pressure on the nut would be according to the cosine of the angle with which the strings cross the nut.

This does not mean that the pressure of the nut mounting block(s) is equally distributed against the body as the tension is trying to tip the nut mount over (bend the part of the body in front of the nut downward, and lift the behind the nut part of the body upward).

To compare the sonics of the open strings as a function of the across the nut angle for various PSGs, see http://s75.photobucket.com/albums/i287/edpackard/

Measurement on the BEAST, 30 lbs tension (about), 0.011 string, a bit less than a 10 degree angle:
Pounds to lift the string free of the nut (at the nut) is about 3. The nut is a 0.154" dia rod. No great precision here, but it gives you ballpark.

In line with the nut contribution to tone, and the bar contribution to tone, I will soon post the results of changing the BEASTs nut from a 0.154" dia brass rod to an brass tube, to an aluminum tube, to a softwood rod of the same dimensions....sorry, did not have plastics available. These will be the same harmonic content, and harmonic content vs. time curves for 0,2,4,8 seconds.

Anyone want to predict the results?[This message was edited by ed packard on 19 August 2006 at 08:57 AM.]

C. Christofferson · Post by **C. Christofferson** » 19 Aug 2006 12:54 pm

Ok, that did bring some factors into mind that i hadn't considered. So, ballpark, if my average angle behind nut is 45 degrees, string tension is 30 lbs. per string, that would be 15 lbs downward per string times 10 - 150 lbs..? Interesting.

[This message was edited by C. Christofferson on 21 August 2006 at 11:19 AM.]

ed packard · Post by **ed packard** » 20 Aug 2006 8:27 am

CC...realize that the force direction answer was a simplification of the problem. The actual vector forces at the various points are complicated by the differences between typical keyhead/body interfaces, and more so by the various keyless mechanism to body anchoring approaches.

C. Christofferson · Post by **C. Christofferson** » 20 Aug 2006 10:39 am

E, putting aside, for the time being, the consideration of all 'horizontal'(end to end) forces, would you say that the 'vertical' force down on the nut would be equal to the 'lift force' which is pulling up on the tuner housing?

Peter Feller · Post by **Peter Feller** » 20 Aug 2006 5:42 pm

Ed,
You are right about the downward force at the nut in that it is analogous to a bridle in the world of rigging. But the vector must be a true bisection of the included angle, is it not? This would be due to the fact that the roller nut freely adjusts and puts the string tension fore and aft of the nut into equilibrium, therefore making the vector of the downward force at the halfway point of the angle regardless of the end fixings. I think I'm right, but I haven't been thinking clearly ever since being abducted by aliens. Now I can only walk backwards. Go figure.

ed packard · Post by **ed packard** » 21 Aug 2006 5:29 am

Peter...your answer is more accurate...mine was over simplified (and inaccurate).

How about giving CC an answer to his last question?

We now have additional different viewpoints on the PSG...as a "system" per Dave Mudgett and Joe Meditz, and as "rigging" by Peter.

PS...Peter, are you game to do a "forces" quick study on the top of the PSG?
[This message was edited by ed packard on 21 August 2006 at 06:40 AM.]

Peter Feller · Post by **Peter Feller** » 21 Aug 2006 8:31 am

Ed,
I'm game, but we have to design the experiment properly. For example, I had considered the idea of standing on the top of my steel with my left foot on the tuners and my right foot on the pickups. Then I could take a 5K dynamometer (I already own one) and hook it onto one string at a time at the nut. Then I could squat down just a bit and pull upward on the dyno until my lower back gets to about a 7.5 on the pain scale. Then you could come over and read the scale, and, voila, data. This design is loosely based on Watt's experiment to determine horsepower. First he got a horse, then he loaded it up and whipped it to get it going. When it finally dropped dead, he recorded the work done, and, voila, one horsepower. Interestingly, in the world of rigging the way you can check the tension in a line that is in use is to buy a really really expensive device that operates like this. You enter the diameter of the wire rope (1/4" to 3/4" diameter in the ones I've seen) and the distance between nodes (pulleys in this case) and then bang on the cable with a hammer and hold a microphone near the cable. The microphone senses the pitch and feeds into the really expensive device, and, voila, a tension measurment. This is used to verify that forces in a rigging system are within spec, not to play "Farewell Party". As to the forces within the structure of a steel guitar, I'm not sure the results could be generalized in a useful way. I've played on five different steels in the six years I've been learning, and each is structurally compliant in its own way. Aprons are taller or shorter, keyheads are longer or shorter, materials are different. All design choices interact to give data collectors fits. I dunno. What do you think?

C. Christofferson · Post by **C. Christofferson** » 21 Aug 2006 10:40 am

thanks guys, that's ok. As you both suggest, (basilh, it appears you're correct also)the 'interdependency ?' of tension forces and resistances in the structure are so 'one whole' (like a microcosmic representation of the entire universe), that trying to prove one aspect separate from the rest can seem to get no further, really, than theory. (not to be taken in a negative context, at all). (concerning this statement as a whole, you may have a different theory)...

Joseph Meditz · Post by **Joseph Meditz** » 21 Aug 2006 12:02 pm

<pre>
o----------vt
/|\
/ | \67.5
/ | \
/ | vr
/45 |22.5
vt ///
vd
</pre>

I think that Ed and Peter are both correct. Above is an ASCII drawing (not to scale) showing a nut with a roller on top and anchored at the bottom. For illustration, the string tension vt is assumed to be 30#. The angle between the string going to the tuner and the nut is 45 degrees. The downward force on the nut is vd. The resultant force of vt and vt is vr.

By Ed's method, vd = cos 45 * 30 = 21.2#
By Peter's method, we first need to calculate vr = 2* cos 67.5 * 30 = 22.96#.
Then vd = cos 22.5 * 22.96 = 21.2#!
The bending force on the nut is then
vb = vr * sin 22.5 = 22.96 * sin 22.5 = 8.8#.
And you can also get the bending force Ed's way, i.e. without calculating vr, as
vb = vt - vt*sin 45 = 8.8#.

Lastly, to answer CC's question, yes, the force pushing down on the nut is exactly equal to the upward force on the tuner IMHO.

Joe

[This message was edited by Joseph Meditz on 21 August 2006 at 01:05 PM.][This message was edited by Joseph Meditz on 21 August 2006 at 01:07 PM.][This message was edited by Joseph Meditz on 22 August 2006 at 07:46 AM.]

ed packard · Post by **ed packard** » 21 Aug 2006 1:32 pm

Good stuff here...Thanks to CC for bringing the subject up, and to Peter and Joe for attacking the problem and straightening me out. My first answer, as mentioned, was off the top of my head (a very slippery place).

Chris...why did you ask?

While thinking over the responses, and the problem, I transferred the thoughts to the changer fingers. I looked at my "normal Sierra, and my Sho Bud Pro to see where the "dents" were. Since both of these have the string wrapped 90 degrees around the finger, by the calcs the max pressure against the finger should be at 45 degrees after the tangent point. That is not where the dents are...they are essentially AT the tangent point (where the vibrating string(s) meet the finger). The dents are therefor NOT from pressure, but from string vibration. Abrasion resistance of the finger material/surface is more important than hardness...think ultra sonic machining.