flyfisher

I can't see that heating the screw will help. The wood will almost certainly shrink back around the screw when it cools. I can't see you making a new head either, especially with just a hammer and screwdriver. If the screw is alread 2-3mm out, is it really not possible to clamp onto the head? I've often removed stripped screws and bolts by clamping a mole-wrench around the head. It only needs a couple of millimetres and the serrations of the mole-wrnech ensures a good 'key' into the screw. Be careful not to overdo the clamping strength, just make it enough for the teeth to get a good grip. otherwise, it might be possible to drill the screw out, though this is not so easy when it's in wood as it's so easy to slip off the screw and into the wood. But even if you end up with a big hole, it's easy enough to fill using tightly-packed cocktail sticks and wood glue and then drilling another pilot hole when the glue has set hard. Good luck.

[i]If you want the very best mains cable and the very best performance from your kit, in our opinion this is it.[/i] I guess you can't argue with their 'opinion', otherwise this sort of fraud would have been closed down years ago. Actually, I sort of admire their cheek. Homeopathy for hi-fi. The placebo effect is a wonderful thing, though a bit expensive at their prices.

[quote name='icastle' post='1082302' date='Jan 8 2011, 04:58 PM']I guess the best question to ask at this point would be: what is it that you are trying to achieve?[/quote] I agree. While I wouldn't want to discourage anyone experimenting with such things it should be noted that the tone control on a (passive) bass is just about the simplest type of low-pass filter possible (http://en.wikipedia.org/wiki/Low-pass_filter) so it's hardly the best place to start if you're trying to significantly shape the tone of a rig. All it does is reduce the amplitude of the higher frequencies and the only thing the pot does is to change the cut-off frequency, i.e. the frequency at which the roll-off starts to occur. Changing the pot may allow the cut-off frequency to be moved beyond the range of the existing pot, but there comes a point after which there's not a lot of point in moving the cut-off frequncy anyway, expecially on a bass guitar. Changing the pot (or the cap) won't make any difference to the rate of roll-off though, so the options for changing the bass tone are very limited indeed. Which is why, of course, pedals and multi-band tone controls on amps are so popular.

"the postage will be 30.00 via a private courier." For £35,000 or more I'd expect it to be personally delivered in a limo.

[quote name='Sibob' post='1083461' date='Jan 9 2011, 07:21 PM']I wonder what the added resistance of the other components would bring the readings down to?!.....I don't know what the standard cap is in a Jazz?.....plus Volume/Volume/Tone pots.....although I was only measuring one pickup at a time, so only it's relevant volume pot would affect each pickup!?[/quote] Caps should have no effect as they block dc (unless they are failing). Tone pots will also probably have no effect if they are wired in series with a cap - but you can easily test this by turning the pot during measurements and seeing if the resistance changes. But, all in all, I agree with the other comments. Desolder the pickups (one end anyway) to be absolutely sure of what you're measuring, but also be aware that simple dc resistance may be of limited use anyway.

[quote name='Sibob' post='1083162' date='Jan 9 2011, 02:27 PM']Now my question is, how might these readings differ whilst actually fitted to the bass, as opposed to just the pickup on a work bench?.[/quote] The meter will only measure the resistance between the its two probes (obvious, yes, but bear with me). Thus when you measure the dc resistance on the work bench, the only thing connected between the probes is the pickup, so you are only measuring the pickup itself. Which is what you want. When you measure the dc resistance fully fitted into the bass, there are also other things connected across the pickup, and thus between the probes, notably pots but possibly a whole host of active circuitry in some cases. Thus, without a circuit diagram, it's difficult to know exactly what you're measuring - except the dc resistance between the meter probes.

Surely the difference will also depend on the efficiency (sensitivity) of the new speaker cab?

Horses for courses? A guitar tech on tour with a large band might find it useful I suppose, but otherwise I agree with the previous comments. It's not exactly a big deal to hand-wind strings - especially if it's a four-stringer!

[quote name='Doddy' post='1081972' date='Jan 8 2011, 12:05 PM']Or better yet.... 1. There are rules 2.You should know them before you can break them. [/quote] This is what makes Basschat such a great forum, IMHO. I've no idea of the proportion of amateur to professional musicians here but it's really valuable hearing the differing opinions. On reflection, I agree with Doddy's refinement of my somewhat tongue-in-cheek 'no rules' rules and also Jennifer's comments and general perspective. If I'm honest with myself, my perspective is as an amateur player and, frankly, I don't have the talent to play a lot of bass lines as they were originally recorded so it's an absolute blessing, for me, that it's possible to improvise something similar enough to be passable (by passable, I mean being able to play a gig and give the audience a good time). But my perspective is based on my limited ability. If I could play an original bass line note-perfectly, why wouldn't I? Well, possibly for many of the reasons being mentioned, but the real point is that I would be making a positive choice to play however I wished, whereas the reality is that I'm usually limited by my ability. I'm immensely impressed by anyone who can play from a sheet of nonsensical hieroglyphs (my interpretation ) and fully understands musical theory because I, like many amateurs I suspect, play mainly by ear and memory. It get's me by and it even earns me some plaudits on occasion, but I'm always acutely aware of my limitations. But I do remember one curious incident at a party some years ago. There was a professional classical musician present and the party host had a piano, so out came some guitars and the guests eagerly awaited an impromptu jam session. Unfortunately, it quickly became obvious that the professional musician could not really improvise any popular songs. To stem the embarrassment, the host quickly dug out some of her sheet music and the pro was able to play anything that could be put in front of him. Apologies for the ramble.

If there was a "rule book" for composing and playing music it wouldn't be as much fun would it? Still, I suppose we could make one up. How about: [i]Rule no. 1 : There are no rules. Rule no.2 : see Rule no. 1[/i] There, that should cover it.

Fair point about DC theory not being strictly applicable, though it's one way of 'visualising' what's going on - to a very rough approximation anyway. Given, as you rightly say, all the variability involved with music waveforms and speaker behaviours, it's a wonder we bother worrying as much as we seemingly do with precise 'matching' of speakers and amps. But I guess describing a speaker as '4 ohms' or '8 ohms' also helps people 'visualise' things - to a very rough approximation.

I don't think heat sinks are a factor because you can't get more power out of an amplifier just by increasing the size of the heat sink. Additional power could only come from a slightly different design, which would cause more heat to be generated, which would then require more heatsink capacity. If the heatsink is too small for the design then the amp would overheat and break (or more likely go into thermal shutdown). Sure, there are sometimes good reasons for having a small heatsink, in which case the design has to be amended accordingly, but the heatsink itself plays no part in the circuit design itself. [b]TimR[/b] is correct to remind us: [i]You're still looking at it from the wrong angle. The maximum power is determined by the maximum current through that device. The 8ohm power is then a function of the full power at 4ohms. It's not that you can't get double through 4ohms it's that you get more than half at 8ohms.[/i] Of course we can get full power from an amp by using a 4 ohm speaker (if that's the rating), because that, by definition, [u]is[/u]the full power output. The issue is why we get more than 50% power output when the load is doubled, i.e. why isn't the output halved when the load is doubled? It still makes sense to me that the output impedance of the amplifier is involved somehow. Thus, although we connect an external 4 ohm load, the amplifier output impedance means that the output device is actually driving a bit more than 4 ohms when it is providing maximum power. So, when we double the external 4 ohm load to 8 ohms, we are not exactly doubling the overall load seen by the amplifier. In fact we're adding a bit less than double the overall load, so we obtain a bit more than half the maximum power.

Hmm. Not sure. PSU inadequacy seems, to me, to be an economic issue rather than a purely technical issue, i.e. the limitation [u]could[/u] be eliminated by design. But if the amplifier output impedance wasn't very, very low (relative to the speaker impedance) then the power output across 4 ohms could never be double that at 8 ohms - as per my earlier example calcs. I think. If that's correct, then the valve amp question will depend on the output impedance achievable when using an output transformer. If it can be very, very low, then I agree with you that there should be little, if any, difference between a valve or a solid state amp - but I'm not at all sure about the current state of the art with impedance matching transformers at audio frequencies. Digital and video electronics is more my area of direct experience, so the nitty-gritty of audio amplifier design is getting a bit out of my comfort zone. An interesting discussion though.

[quote name='burno70' post='1081297' date='Jan 7 2011, 05:05 PM']Lapolpora - I think my girl would dump me if I turned up with that rig! [/quote] I appreciate the joke, but there's a serious side to such things. If bass playing is a big part of your life then it's a big part of you. I've never really understood why two people get together and then one of them sets out on a mission to change the other. What's the point? It can only end in grief. I've seen it happen on a number of occasions.

[quote name='dincz' post='1079082' date='Jan 5 2011, 07:58 PM']Regarding the impedance of the output stage - it certainly has an effect but is typically only tenths of an ohm.[/quote] Is a low output impedance also possible with a valve amp? I understand they use transformers to match the high-impedance valve outputs to low-impedance speakers, but am not sure how low they can get with this approach.

I've been getting some good results with Kristal freeware: [url="http://www.kreatives.org/kristal/"]http://www.kreatives.org/kristal/[/url]

[quote name='TimR' post='1078010' date='Jan 4 2011, 10:24 PM']Amplifiers are current devices, not voltage devices. The power is limited by the current it supplies. If the voltage rail maximum is 50v then into 8ohms you get 312watts which is 6amps give or take. With the voltage rail still at 50v then theoretically into 4ohms you'll get 625watts but if the amp is limited to 11amps then you can only get 484watts. It's not the transformer it's just the way that the transistors or valves work.[/quote] Fair point about the power being limited by the current the amp can supply, but surely that's a practical limitation determined by the designer? It's not a theoretical limitation is it? Your example is fine, but why the arbitrary limit of 11 amps? If the amp could supply 12.5A (or more) then it would deliver 625W. Perhaps it's just an economics thing? Incidentally, are you saying the amp output impedance itself can be entirely discounted ?

Give the jack plug a good clean and the guitar socket as well, if possible. Could be that you're not getting a good earth connection between the two. The tip (and ring, where used) connections in a jack plug/socket are spring-loaded, so there is a 'positive' connection. The sleeve connection usually just relies on the sleeve of the plug laying in contact with the barrel of the socket, which doesn't give such a 'positive' connection, especially if the socket hole is near the top end of the diameter tolerance. Try applying some sideways force to the jack plug when it's in the socket and wiggle it a bit to break through any surface grime. No guarantees, but it's an easy thing to try and might reveal a problem.

[quote name='thinman' post='1077936' date='Jan 4 2011, 09:38 PM']Actually, my maximum power transfer theory is seeping back - that does seem correct.[/quote] That was my first thought, but I seem to recall that only really applies in transmission line systems, i.e. typically in far higher frequency applications than audio. I had a quick look for output impedance specs but I couldn't find anything (=didn't try hard enough ). I've read conflicting stuff about amplifier output impedances though. One school of thought seems to go for very low figures because it is supposed to give better 'damping' of the speaker, resulting in a 'tighter' sound, especially at bass frequencies. Another school of thought seems to suggest that speaker damping is best achieved by getting the enclosure right in the first place! (that's getting outside my comfort zone though). It would be interesting to see the output impedance specs for a range of amps and then compare the figures with the rated outputs at 4 and 8 ohms. I'm not denying the PSU theory, BTW, and it's certainly possible for power output to be limited by a poorly rated PSU but, to my mind, such a thing would be the result of bad design or an over-optimistic (to be kind!) marketing department, both of which are pretty heinous crimes in my book.

If an amp is rated at, say, 500W into 4 ohms then surely the PSU must be capable of delivering that amount of power otherwise the manufacturer is telling porkies. I'm not sure about the OP's assumption that an amplifier should be able to deliver twice as much power into a 4 ohm load as into an 8 ohm load. This might be the case if the amp itself had a zero output impedance, but that isn't very likely. In practice, the amplifier [u]will[/u] have an output impedance and this will mean the power into differing loads will not be a linear relationship. Consider these numbers: Amp output voltage = 10V in all cases Amp output impedance = 0 + 8 ohm speaker = 8 ohms = 10/8 = 1.25 Amps = 12.5W = 100% power Amp output impedance = 0 + 4 ohm speaker = 4 ohms = 10/4 = 2.50 Amps = 25.0W = 200% power So, for an amp with zero output impedance the output power IS linear with respect to the speaker load. But, allowing for the amplifier output impedance: Amp output impedance = 2 + 8 ohm speaker = 10 ohms = 10/10 = 1.0 Amps = 10.0W = 100% power Amp output impedance = 2 + 4 ohm speaker = 6 ohms = 10/6 = 1.67 Amps = 16.7W = 167% power Amp output impedance = 4 + 8 ohm speaker = 12 ohms = 10/12 = 0.83 Amps = 8.3W = 100% power Amp output impedance = 4 + 4 ohm speaker = 8 ohms = 10/8 = 1.25 Amps = 12.5W = 151% power Thus, any amplifier with a finite output impedance cannot deliver twice the amount of power just because the external load is halved. None of this has anything to do with the capability of the PSU. (I think!).

I use Audacity and just record the audio as it plays on Youtube. You can then save as a wav file, or MP3 with the appropriate plug-in.

[quote name='phil.mcglassup' post='1077273' date='Jan 4 2011, 01:29 PM']If I want to put any songs from my iTunes library on my laptop through an amplifier, which lead or adaptor do I need? I thought about using a standard 1/4" to 1/4" mono jack lead and getting a "3.5mm stereo jack to 6.35mm mono socket adaptor" to plug into the headphone socket on the laptop. Is that correct? Regarding the adaptor - does the 3.5mm jack have to be stereo to match the laptop? And does the 6.35mm socket have to be mono - to match the lead?[/quote] Most equipment these days is pretty resilient to abuse so you'll likely get away with almost anything, but here are a few things to bear in mind. Connecting an audio input to ground won't cause a problem but connecting an audio output to ground is not good practice. What will actually happen will depend on the detailed design of the output circuit, but I imagine you can understand that an audio output is expecting to "see" some sort of resistance rather than a short to ground. So, when using stereo/mono adapters, bear in mind if you're connecting them to an input or an output. Connecting two inputs together is not a problem, so a single mono signal into the L+R of a stereo input is fine. But L+R stereo outputs connected into a mono input needs a little more care in case one of the stereo outputs ends up being connected to ground. This might be OK or it might not - it depends on the circuitry. Ideally, resistors could be used to blend a stereo output into a mono one, but that's probably being a bit paranoid. Othe issues are output impedance and signal levels. A headphone output will be designed for a low-impedance load and will provide a fairly high signal level. A guitar amp input will be high impedance and it will expect a very low voltage. Again, a lot depends on the circuit design, but just remember to turn down inputs and outputs when connecting stuff together - you don't want max gain on your amp when plugging in a max-volume headphone output from your laptop. Even if the amp survives, your speaker (or ears!) may not.

Try RS Components or Farnell Electronic Components. As mentioned above all resistors have a tolerance and any decent circuit design will take that into account. In fact, I'm rather surprised that you have a circuit that specifies a 4 ohm resistor, since that is not a 'preferred' value (standard resistors are not available in every single possible value, otherwise there would be millions of them, so the industry has settled on a specific range of 'preferred' values and electronic engineers work to these values). 3.9 ohms ohms is -2.5% of 4 ohms but typical high-power aluminium-cased resistors are 5%, so could actually be 3.8 to 4.2 ohms anyway. In other words 3.9 ohms should be fine. Occasionally it is necessary to use a non-preferred resistor value. In such cases, designers may use a combination of series and parallel resistors to create the required resistance. Or, if cost is not an issue, use a variable resistor and adjust-on-test. If you are really paranoid about using a 4 ohm resistance in your circuit then you could consider using a 3.3 and 4.7 in series to give 8 ohms and then add a similar pair in parallel to give 4 ohms in total. However, such a level of paranoia would also require you to calculate all the tolerance possibilities and it's unlikely you'd end up with 4.0 ohms anyway. And all of that is before we even consider that loudspeakers actually present a complex load to an audio amplifier (hence the use of impedance rather than resistance) and there is no such thing as a "4-ohm" loudspeaker anyway. It's just another bit of shorthand to hide all the underlying complexity, but is perfectly acceptable for rules-of-thumb. Hmm. Probably too much information really. I'd just use a 3.9 ohm resistor.

Even if you stick to the same cabinet brands, how do you know that the actual speakers inside are the same brand? Don't worry about it. Just be sure you have enough ohms for your amp and that the cabs can take the power and there'll be no harm done. Well, electrical harm anyway - I can't vouch for how things will sound!

There's a lot of price changes going on at the moment - fuel duty, VAT increase, Christmas, sales, end-of-sales. I suspect all this price movement creates some general confusion in people's minds and some retailers just use it as an excuse to hike prices. Some will get away with it, some won't.