Why do I have a sneaking suspicion that NattyLight is listening to a sub with his SDA's right now? lol The old "fat girl/moped" saying is coming to mind right about now! :eek:
Printable View
Why do I have a sneaking suspicion that NattyLight is listening to a sub with his SDA's right now? lol The old "fat girl/moped" saying is coming to mind right about now! :eek:
Headroom and reproducing dynamic range, fast transients and speed is not about watts anymore than making a great handling car is all about top speed. People need to stop equating the above with how many watts an amp is rated at. It's a small part of the entire picture..........not "THE" part of the picture. Until one learns that, they end up chasing their tail.
H9
The cost I can certainly understand being a deal breaker. But should you find a deal on a quality sub you don't need to go spendy on the cables because of the limited bandwith the sub's covering. Also you don't need a dedicated sub line out like on a ht receiver. Just a pre with dual preouts which are very common. There's no "daisy chaining" involved here.
Analog headroom definition:
head•room (ˈhɛdˌrum, -ˌrʊm)
n.
3. the additional power output capability of an amplifier when producing short-term peak signals.
Looks like both you guys are correct.
Headroom is definitely wattage (but usually rated as an equivalent db), but when analyzed as a percent of rated RMS signal it can be small or very large. A 3 watt tube amp with 3db of headroom will be capable of 6 watt transients while a 300 watt solid sate amp with 3db of headroom will be capable of 600 watt transients. But these are transients and not described as the continuous wattage, which may be what H9 is saying (I am not sure as I can not speak for him). But headroom is definitely power (wattage) related.
This one wins...hands down. The rest of you just piss and moan on! :cheesygrin:
I see no reason to debate this further - you either like a sub with your system or you don't. I sincerely doubt this thread is going to set someone "right" or convince them that they are "wrong" from either camp.
No...but darn it...my watts are cleaner than yours. I can't prove it, I've never heard your system nor have I ever heard any of the components in your system but...mine are bigger, better, faster, deeper, higher, clearer and just plain cooler. (period)
:evil:
But mine go to "11"!
Somebody define "quality watts" please. I'm having difficulty getting a grip on that phrase.
As defined - everywhere:
One watt is the rate at which work is done when one ampere (A) of current flows through an electrical potential difference of one volt (V).
Since watts are a measure of work...and you reference your amps...WTF? Are you lifting them?
Sounds more and more like some of our less favorable magazines floating around in here. Are we just makin' stuff up so it sounds cool or are we so confused we don't even know what these electrical terms refer to? I've yet to see an amp rated based on anything other than a number.
So which number = "Quality"? :wink:
That one is easy, you know it when you hear it.Quote:
Somebody define "quality watts" please.
Great...user defined. :cry:
So as long as my ears "work" I have quality watts.
Stereopile magazine must use these...
No one is saying that a speaker requires a sub/subs...but merely that all speakers will benefit from the addition of one or two. When implemented correctly (it takes a lot of work & the sub must be of hi-quality, not a sub designed as a boom box for HT) the results are simply amazing. The whole music landscape is transformed. There just isn't a one box speaker that can do it all. Now that doesn't mean they can't sound good because they certainly can but if you want to maximize their performance, well you know the drill.
So what do you consider a high quality sub??
Both current and voltage are required to produce power (watts). That’s why both amps and voltage are important.
From a variety of Internet sources:
James Prescott Joule, not Georg Simon Ohm, who first discovered the mathematical relationship between power dissipation and current through a resistance. This discovery, published in 1841, followed the form of the last equation (P = I2R), and is properly known as Joule's Law. However, these power equations are so commonly associated with the Ohm's Law equations relating voltage, current, and resistance (E=IR ; I=E/R ; and R=E/I) that they are frequently credited to Ohm. The original E = IR equation can be manipulated to find any one of the three if the other two are known. The three possible forms of the equation are:
(1) E = IR
(2) R = E/I
(3) I = E/R
Example: If you measure the current going into an 8 ohm speaker, and find it to be 2.5A RMS, the voltage across the speaker can be found by multiplying the current by the resistance, or E = IR = 2.5A*8ohms = 20V.
Example: If you have a resistor of unknown value, and you measure 10V across it with your voltmeter, and then put an ammeter in series with the resistor to measure the current, and find it to be 100mA, the unknown resistance value could then be found by dividing the voltage across the resistor by the current through it, or R = E/I = 10V/100mA = 100 ohms.
Example: If you have a 100K resistor, and you measure 100V across it with your voltmeter, the current flowing through the resistor can be found by dividing the voltage across the resistor by the resistance, or I = 100V/100K = 1mA.
The power equations
A related equation is used to calculate power in a circuit: P = EI, where P = power (in watts), E = voltage (in volts), and I = current (in amps).
Example: If you want to determine the output power of an amplifier, and measure 20V RMS across the load and 2.5A into the load, the power delivered by the amp to the load is: P = 20*2.5 = 50W.
As with the Ohm's Law equations, this equation can also be rearranged to solve for the other two quantities to produce three possible equations as follows:
(1) P = EI
(2) E = P/I
(3) I = P/E.
Example: If you have an amplifier that puts out 100W into a load, and you measure 2.5A into the load, the voltage across the load can be calculated as E = P/I = 100W/2.5A = 40V.
Example: If you have an amplifier that puts out 50W into a load, and you measure 20V across the load, the amplifier is putting out a current of I = 50W/20V = 2.5A.
The power equation can also be combined with the first Ohm's law equation to derive a set of new equations. Since E = IR, you can substitute IR for E in the power equation to obtain:
P = EI = (IR)*I, or P = I2R.
You can also find P if you know only E and R by substituting I=E/R into the power equation to obtain:
P = EI = E*(E/R), or P = E2/R.
Example: What is the RMS voltage at the output of a 100W amplifier into a 16 ohm load? The answer can be found using equation (6): E = sqrt(PR) = sqrt(100W*16ohms) = 40V RMS.
Example: An amplifier is connected to an 8 ohm load. The RMS voltage across it is measured using a DVM (digital volt meter) and found to be 15.5V right before clipping. How much undistorted power can the amplifier put out into an 8 ohm load? The answer can be found using equation (4): P = E2/R = 15.52/8 = 30 Watts.
Example: How much current does a 100W amplifier put out into a 16 ohm load, an 8 ohm load, and a 4 ohm load? The answer can be found using equation (3). For the 16 ohm load, I = sqrt(P/R) = sqrt(100/16) = 2.5A For the 8 ohm load, I = sqrt(100/8) = 3.54A. For the 4 ohm load, I = sqrt(100/4) = 5A.
Summary
Ohm's Law is probably the single most important equation used in designing or analyzing a circuit. These equations are useful for calculating many parameters in a guitar amplifier, including output power, voltage drops across resistors, current through resistors, power ratings of resistors, and many other things.
Oh...now I get it. :sad:
It's not the amp (select your favorite brand name) that has quality watts. The amperage has quality watts!
So that's how it "works"...!! :smile:
Simply put a watt is a watt. period. What is meant by a quality watt (specifically to audio because I don't think a light bulb cares) is when there are little to no additional anomalies applied to the output signal as related to the input signal and also the ability for the amplifier to control the speakers. Per your definition and other electrical laws/ definitions a watt of power is just that a watt of power and the calculations used to determine that take nothing into account about distortion, S/N, damping factor, Whether the power rating is true RMS or for a split second if lightning strikes the power line outside, etc. Though a watt is a watt, there are definitely good quality amplifiers and poor quality amplifiers and many levels in between.
I think one point that you're missing is HOW a sub gets integrated into a dedicated 2-channel system. In your system (using an AVR as a preamp) it is very easy to integrate a sub without losing fidelity because you can just set the AVR's internal crossover and you're done. In a DEDICATED, clean, simple 2-channel system, the only way to add a sub and "relieve" the speakers of all bass duty, is to ADD an external crossover. Introducing an external crossover to the mix is always detrimental to fidelity.
Obviously if a 2-channel preamp has multiple outputs you can just add the sub that way, but then you don't get any of the benefits you describe by stripping low frequencies from the speakers. If you want to strip low frequencies from the speakers, you have to introduce a crossover of some kind, and it's always going to be a give and take on whether people want to accept the fidelity degradation that is inherent to an external crossover so that they can augment bass response.
Again, there's no right answer. Different strokes.
It'll be a cold day in hell before I start stripping the low frequencies out of my SDA's. Just sayin'.
The best placement for imaging, soundstage, is not always the same placement for the best bass response. Additional subs to smooth out bass response in the room can be a great thing, when done right. But getting it done right, can be far from simple. One has to account for delay(phase differences between mains and sub(s)), setting slope between mains and sub, level, etc... IMO, it works best between mains that are sealed due to their better impulse response and shallower roll off(12db opposed to 24db/octave). And to agree with Phil, it's not about additional bass, it's about better bass.
Now that about designers and companies disagreeing, even Wilson Audio and Sonus Faber offer subs and will have someone come to your house to set up both. I've heard a few Sonus Faber and JL Audio setups dialed in by Sumiko and you wouldn't know there were separate subs in the room. In both cases the bass was phenomenal, clean and taut, without being overpowering or boomy.
I love Sonus Faber!! I would like to hear that set-up..
Watts never tell the final story about sound. A watt is a watt if you are just measuring its electrical properties. But watts don't mean squat beyond achieving SPL. Only the foolhardy judge sound quality and performance by rated wattage. There is far, far more to quality sound than watts.