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Speaker Size

The differences in Hendrix' tone from playing upside down don't have anything to do with the headstock or nut (his nut would have been reversed, otherwise his low E string would not have fit in the slot, and the high E string would have been buzzing and moving around in the huge low E slot). It mainly had to do with the reversed bridge pickup, which on a strat is angled closer to the bridge on the treble side (so backwards the way Hendrix played them). And of course all the other gear and gadgets he had.

I always thought the Jimi just played a regular strat left handed; in other words, upside down ? I think the Big Muff paired with the wa-wa had alot to do with it, but, since I was never a big fan, did'nt pay that much attention .
 
It's strung upside-down....and agreed on the locking nut. He was just the first thing that came to mind as someone who plays the guitar "backwards". That much had nothing to do with one string, and I guess it's a good thing I don't depend on comedy to put food on the table....I'd be a skinny man.
 
I always thought the Jimi just played a regular strat left handed; in other words, upside down ? I think the Big Muff paired with the wa-wa had alot to do with it, but, since I was never a big fan, did'nt pay that much attention .
Arbiter Fuzz Face was Jimi's fuzz of choice. There are a whole lot of theories about Jimi's special guitar secrets, but I think it's as simple as he'd take stock righty Strats and reverse the nut and that's it. And the headstock can't have any relation to the sound since all the action takes place between the nut and bridge saddles. So that would leave rear pickup positioning IMHO. But then I've heard dudes like Stevie Ray cop very similar sounds with righty Strats. And I've also heard Jimi plug straight into a B-15 and sound exactly like Jimi.

Probably the shoes.

:D
 
And the headstock can't have any relation to the sound since all the action takes place between the nut and bridge saddles.
Having the headstock 'upside down' put the longest length of string behind the nut on the low E, the shortest on the high E, so the low E string stretched the easiest, the high E the hardest, the opposite of a standard Strat played by a righty. If you put on a locking nut all the string lengths stop at the nut, so none of them stretch as easily as with a standard nut, and a reverse headstock will have no effect.
 
I've always been told that the length of the string past the nut is immaterial to the string tension because it's going to be the same tension between the nut and bridge saddle when tuned up to pitch regardless. Why would this not be the case?
 
I've always been told that the length of the string past the nut is immaterial to the string tension because it's going to be the same tension between the nut and bridge saddle when tuned up to pitch regardless. Why would this not be the case?
Even with the same tension the longer string will stretch easier. Any guitar player with a locking nut will tell you that if the lock is released the strings stretch much easier.
 
Datapoint: I had a Hydrive 115 and a Hydrive 410. Did not like the sound of the 115 AT ALL. Fharty, no punch. Sold it. LOVE the 410 however!!! Gets many compliments on the tone - many. My other reference point is a MIUSA SVT-810 (played by Geezer Butler no less), love da Fridge too!
 
Even with the same tension the longer string will stretch easier. Any guitar player with a locking nut will tell you that if the lock is released the strings stretch much easier.
I've got a Floyd Rose on a guitar and just tried it on the high E string, and son of a gun...you're right about that. But it's such a tiny little difference that I can't imagine it affecting any but the most anal on a gig ;)
 
Hey, just looking up more stuff on cabs and amps and such and another question came up:
If I have two cabs, both rated at 250w with 8ohm imp., hooked up to a 500w head, do the cab wattages add so I could use the full 500w?
Of course there will be some distortion, but will it not blow the speakers??
 
Hey, just looking up more stuff on cabs and amps and such and another question came up:
If I have two cabs, both rated at 250w with 8ohm imp., hooked up to a 500w head, do the cab wattages add so I could use the full 500w?
Of course there will be some distortion, but will it not blow the speakers??

Those ratings on speakers are thermal limits not excursion limits. You will still be able to damage the cabs if you try.
 
Hey, just looking up more stuff on cabs and amps and such and another question came up:
If I have two cabs, both rated at 250w with 8ohm imp., hooked up to a 500w head, do the cab wattages add so I could use the full 500w?
Of course there will be some distortion, but will it not blow the speakers??

Wattage is a rating number. On an amp, it's how much power the amp is CAPABLE of producing, without distortion. On a speaker cab, it's how much wattage the cab is CAPABLE of using, without melting the voice coil. The problem is, both are dependant on several factors- one being frequency. The lower frequencies will eat up MORE of the amps power, and the speaker will handle LESS of it than higher frequncies. There are also MECHANICAL limitations with the speaker, and depending on the frequencies being reproduced, those limitations can be MUCH lower than the wattage rating of the cab, maybe HALF the rating. So, in your example, it wouldn't matter what cabs you use, the amp will ALWAYS be capable of producing 500 watts cleanly, and no matter what amp you're using, the 2 250 watt cabs will ALWAYS be mechanically limited to LESS than their rating, depending on the frequency you're playing at any given moment. Hope this helps some.
 
I may think that wood matters along with all the other components of a bass but I can tell you Jimi transcended all this crap:). Of all the people who could make it happen on any POS he was the one. His roadies were unable to control or get workable sounds out of his rig. If there's something for all of us to aspire to it's to get the music first and finesse our gear in the expression of that music.
 
I've always been told that the length of the string past the nut is immaterial to the string tension because it's going to be the same tension between the nut and bridge saddle when tuned up to pitch regardless. Why would this not be the case?

Even with the same tension the longer string will stretch easier. Any guitar player with a locking nut will tell you that if the lock is released the strings stretch much easier.

I've got a Floyd Rose on a guitar and just tried it on the high E string, and son of a gun...you're right about that. But it's such a tiny little difference that I can't imagine it affecting any but the most anal on a gig ;)
oh, we guitar players feel the difference all right!

the string with more "excess" length bends easier (as in, it literally is slightly easier to push it across the fretboard), but needs to be bent further to get up to the next note. it's less responsive.

the locked nut may make the strings feel slightly tighter, but you'll get wider bends and bigger vibrato.

actual pounds of tension is unchanged, of course. scale length x string mass and thickness x pitch = tension, period.
 
Hey, just looking up more stuff on cabs and amps and such and another question came up:
If I have two cabs, both rated at 250w with 8ohm imp., hooked up to a 500w head, do the cab wattages add so I could use the full 500w?
Of course there will be some distortion, but will it not blow the speakers??
you got the (correct) complicated answers; the simple answer is yes, the wattage capacities of the cabs add together.

two 250 watt drivers or cabs together (series or parallel) handle 500 watts.
 
oh, we guitar players feel the difference all right!

the string with more "excess" length bends easier (as in, it literally is slightly easier to push it across the fretboard), but needs to be bent further to get up to the next note. it's less responsive.

the locked nut may make the strings feel slightly tighter, but you'll get wider bends and bigger vibrato.

actual pounds of tension is unchanged, of course. scale length x string mass and thickness x pitch = tension, period.

So let me understand this right - you actually don’t think a good right handed guitarist can copy Jimi Hindrix’s sound?
 
please; a good guitarist can copy jimi's sound with a steinberger. ;)

(he'd never get it 100% even if he was himself lefty using one of jimi's actual guitars of course, but "it's the archer, not the arrow", after all.)