Ummmmmmmm........10-16" Compound Radius? Really?
- Thread starter DustyCat
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line6man
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Print/cut this out and check.
http://www.pickguardian.com/pickguardian/Images/Pickguardian%20Neck%20Radius%20Gauges.pdf
Most people cannot spot the compound radius by eye.
http://www.pickguardian.com/pickguardian/Images/Pickguardian%20Neck%20Radius%20Gauges.pdf
Most people cannot spot the compound radius by eye.
DustyCat
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AutoBat said:
Mr. AutoBat to the res-cue!
Seriously speedy response time!
ccasion14:No I do not own a radius gauge.
But from the website, I thought there would bhe ,more of a significant change from the 1st fret to the 22nd fret. (It's hard to feel the difference even)
To be fair, from my point of view, when I went with the 6100's, from the diagram on the website, I thought they would be HuUUUUUGE!
Like........the space in between the frets on a scalloped huge....like...lipstick pickup huge (that's why I went gold, cause I thought the silver wpuld be wide enough to reflect the gaze of any medusae in the audience :toothy12:
-DC
PS "Mr. Randy Rhoads!" (6tuplets from hell!)
DustyCat
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Is this the heel?
I don't really know a whole lot about guitars, I just like them.
I guess Warmoth was the pioneer of the 10-16" compound radius right?
I thought I read something like that on their website...
Maybe I'll feel the difference when I finally get this baby matchedup to a canary body/Gonçalo Alves top. :toothy10:
I don't know what body it's going to be, but definitely 'f' hole(s) :evil4:
-DC
I don't really know a whole lot about guitars, I just like them.
I guess Warmoth was the pioneer of the 10-16" compound radius right?
I thought I read something like that on their website...
Maybe I'll feel the difference when I finally get this baby matchedup to a canary body/Gonçalo Alves top. :toothy10:
I don't know what body it's going to be, but definitely 'f' hole(s) :evil4:
-DC
AutoBat
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The heel tells most of the story.
WG = wolfgang back profile.
Warmoth Radius = 10-16 compound fretboard
6100GD = gold 6100 frets
1.689 = exact measurement of the nut width (1 11/16 is 1.6875 inches)
If your headstock was a licensed Fender, that information would be stamped as well.
If you had a satin finish a red dot would be on it.
Although you have 99 posts already - Welcome!
WG = wolfgang back profile.
Warmoth Radius = 10-16 compound fretboard
6100GD = gold 6100 frets
1.689 = exact measurement of the nut width (1 11/16 is 1.6875 inches)
If your headstock was a licensed Fender, that information would be stamped as well.
If you had a satin finish a red dot would be on it.
Although you have 99 posts already - Welcome!
DangerousR6
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Seriously, you don't know a whole lot about guitars, and you're questioning a product from a company that specializes in guitar parts with an untrained eye for one, and secondly such a narrow strip of the conical shape from 10-16, noone shy of the 6 milliion dollar man is gonna be able to see the difference. Especially with the frets in place... :icon_scratch:DustyCat said:
Radius gauges are a must have item !!
It's one of the 1st things I bought, when I started building guitars.
The pic's are that big, I need a 100'' screen to view them :icon_jokercolor:
Hey Dustycat, Your signature is wrong !!
It was October GOTM. :icon_biggrin:
Thanks anyway :icon_thumright:
It's one of the 1st things I bought, when I started building guitars.
The pic's are that big, I need a 100'' screen to view them :icon_jokercolor:
Hey Dustycat, Your signature is wrong !!
It was October GOTM. :icon_biggrin:
Thanks anyway :icon_thumright:
DustyCat
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AutoBat said:
Thanks for the welcome. And thanks to everybody for their help and enthusiasm.
I have been keeping tabs on some of these guitar builds for a little while now and I have seen more guitars that really speak to me than anywhere else (Music Instrument stores/dealers etc.)
I am guessing that there are more than just a few expert luthiers on this forum let alolne Warmoth employees who would probably rather be working on their next baby during their lunch hour than answering elementary questions to a noob like myself.
Maybe this is testament to future customers. The pictures from the website were just a wee bit exaggerrated to my eye.
:tard:
The lower frets portion of the neck in the picture could very well be a dowel rod (from what I remember). Also, like I said before, in my mind, I had envisioned the frets to be the size of lipstick pickups or bigger. Bigger like the space in between the frets. Like .50 caliber...Like...MechaShredTron! (Blinding!). :toothy12:
:rock-on:
-DC
-Thanks UpDown. Lol I will be sure to fix the signature.
Cagey
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As long as you're working on your sig, you may want to consider knocking it down to less than a web page. Friggin' thing's huge! People are going to get carpal tunnel syndrome trying to scroll past it. A signature is supposed to be a signature, not a fully illustrated manifesto <grin>
line6man
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line6man said:
Completely off topic, but on the topic of Mobius bands, Cagey, since you seem to be knowledgeable about everything, perhaps you could educate me with useless facts on something that's stumped me tonight. :blob7:
A Mobius resistor cancels it's own inductance, however, wouldn't it still have some amount of parasitic capacitance?
At first glance, it would seem that there could not be a capacitance, as the surface is one-sided, and that would be true if the resistive element provided a true 0 Ohm path for electrons throughout it's entire surface area, but since the surface is resistive, that should start to create two plates. If the resistive element was said to produce an infinite resistance from any one point to another, the capacitance would follow the parallel-plate equation (Permittivity times area over distance.), though the area would be zero, since an infinite resistance would not conduct. What sort of equation would define the capacitance, where the resistance of the band is between 0 and infinity? :icon_scratch:
Cagey
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Mobius resistors use two conductive surfaces separated by a dielectric, which is the very definition of a capacitor, physically. However, they're both connected all the way around, so they effectively form a single conductor and defeat the dielectric. In a perfect device, there wouldn't be any possibility of a charge differential, and so no capacity. But, since these are not only not perfect conductors, they're also purposefully designed to have some resistance, so some capacity is possible. However, I would expect it to be very, very small. You'd probably have to get up into gigahertz frequencies for it to show up at all.
The reason to reduce or eliminate inductance is to prevent current from leading voltage in an AC circuit, or to reduce/eliminate impedance. You may not care if it lags, so capacitive reactance may not be an issue. I would think in things like cell phones, GPS units, and things of that nature that such phenomena would show up and need to be compensated for somehow. That may be what drove the design of such a thing. I don't know. I've never really played around up in those frequencies from a design point of view.
As for a formula to calculate it? Phbbt. Fuhgeddaboudit. At least, from me. I sometimes have to stop and think of why I am wherever I am, because I can't remember why I went there. I certainly don't remember much of the math I ever knew.
The reason to reduce or eliminate inductance is to prevent current from leading voltage in an AC circuit, or to reduce/eliminate impedance. You may not care if it lags, so capacitive reactance may not be an issue. I would think in things like cell phones, GPS units, and things of that nature that such phenomena would show up and need to be compensated for somehow. That may be what drove the design of such a thing. I don't know. I've never really played around up in those frequencies from a design point of view.
As for a formula to calculate it? Phbbt. Fuhgeddaboudit. At least, from me. I sometimes have to stop and think of why I am wherever I am, because I can't remember why I went there. I certainly don't remember much of the math I ever knew.
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