hello, an welcome to myself

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bensonreck

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hello it is my first vist to your cyber hull, i am currently studying technical production of music in which we cover quite a lot of the stuff that has been spoken about on here. can anyone explain the inverse square law to me as if i were a dummy?
 
ok i was ganna tell you to go do your own homework instead of asking here.... but im that dumb sell attempt about being a music tech major hit a soft spot cause im a music tech major to.
basicly the inverse square law is this. if we have a variable, let say x, that is a level of intensity(such as with forces or light) which spreads evenly and infinitely in all directions and we start at the point where x originate. then we need to create a sphere around x with a set radius, lets say we have one that is labled r. the inverse square law lets us determine the new intensity of x any distance from its starting location. to do this we take x and multiply it by the inverse of the surface area of a sphere. we then take this number and divide it by how ever many times r the distance from the original point is.so if you are looking for the intensity at the surface of the sphere it simply
(x/(4(pi)r^2) (because you are simply dividing by the 1^2 so you it is simply that) . if its 2r away from the original point you take 2 square it and then divide the equation for 1 r by 4 (x/(4(pi)r^2)/4 . for 3r you would divide (x/(4(pi)r^2) by 9 and so on. so say we have a for of 2 and we want to find its intensity 3 units away. we would calculate 2/(4pi3^2)/9 which equals 0.00196488


thats not a very good job of explaining but you are going to need to give me something more specific and maybe some problems you need solve for me to help you more you can pm if you want or send me an email at tyler107@uab.edu and i can help you more
 
i just saw the wikipedia link i think i like thier explanation better haha
 
Basically it means that doubling the distance away from a sound source equates to a 6 dB decrease in volume.

That's if I'm remembring things correctly, and didn't confuse that with anything else.
 
wouldnt the dB change be dependent on the distance your doubling and not always 6?
 
tbone36109 said:
wouldnt the dB change be dependent on the distance your doubling and not always 6?

No.

If I'm 2 feet away from a source measuring 80 dB, then being 4 feet away should be 74 dB.

Now if I start at being 4 feet away measuring 74 dB, and suddenly I'm 8 feet away, it should be 68 dB.

However something just occured to me, this may not hold true in the far-field. So maybe it only applies to a certain distant.

A similar concept to octaves perhaps? doubling a frequency is always an octave, no matter what frequency you're doubling. Maybe that analogy doesn't make any sense.
 
It always works in a anechoic (sp) chamber, outside of that, your control room, your living room, etc., will have surfaces that reflect sound differently in different parts of the room. So yes even 40 feet back and then moving to 80 feet there should be the same loss of 6dB. Although at those distances the highs will lose more faster, well even in your living room the highs lose faster, but for the most part its not that noticable, unless you have a 60 foot living room.
 
tbone36109 said:
wouldnt the dB change be dependent on the distance your doubling and not always 6?
dBs are logarithmic rather than linear, this is why if you increase the level by 3 dB, you double the volume.
 
but if you stand infront of my Mesa and G12T-75 stack when it's on ten, then when I turn it down to 5, your ears will still bleed why is that :confused:
 
punkin said:
but if you stand infront of my Mesa and G12T-75 stack when it's on ten, then when I turn it down to 5, your ears will still bleed why is that :confused:

ears will be bleeding because of vain attempts to tear them from cranium as the mesa is melting the brain huh

even at 5
 
Something I wrote on the subject, oh and its +10dB SPL to double the percieved loudness.

There are many ways to talk about and explain sound. We can say it is dry, flat, bright, dull, muddy, brilliant, and many other descriptions. Terms such as that are intangible when it comes to the next guy, one man’s perfect sound, is another's mud. These kind of descriptions can be helpful and can be used as a basic guide, but other terms like a dB (deci-Bel) can be given an exact scientific definition. Knowing what such terms mean is helpful, not just for helping us to pick out and compare components, but it also helps to understand how we hear.
A Bel is a standard of measure, named after Alexander Graham Bell. A deci-Bel is one tenth of a Bel. So what is a Bel? Well, a Bel is the logarithm of an electrical, acoustic, or other power ratio, the key being power. The formula for this is Bel=log(A/B) with A and B representing the two levels being compared. Since, in sound issues, a Bel is really too large a number to work with, they wanted to use something smaller and went to a dB, which changes the formula to dB=10*log(A/B). To help put this in non-tech terms, dBs are used to measure how much more or less we have of a certain thing. In this case how much more or less of an acoustic power or signal level, referenced to a certain level. dBs can be used in a variety of situations when talking about our gear, whether it is voltage, wattage or SPLs(Sound Pressure Levels). I will point out that a dB is a dB, whether it is talking about volts, watts, or SPLs, how we figure the ratios is somewhat different, but in the end a 1dB increase in voltage, is a 1dB increase in wattage, is a 1dB increase in SPLs.
Now to apply this to our electronic gear, first comes voltage. I start here because this is ultimately what drives our speakers and components. It is also the one that throws a curve ball at us in relation to the formula. Now if you know Ohms Law, you know that power is proportional to the square of the voltage. Since the dB(power) = 10*log(A/B) formula is for power ratios, the voltage formula changes to dB(volts) =20*log(A/B). Let’s apply this to the real world. The average home stereo CD player puts out about 2V (volts) with the maximum signal recorded on a CD. With the 2V given, if I said that a particular CD' s average level output on the same CD player is about .6V, how much below the peak level of 2V would the average be? 20*log(2/ .6)=10.4 dBs. This is just a small example. The point is that if you know two voltages you can determine the difference in dBs. Also if you follow the formula, doubling the voltage will yield a 6dB increase, but this is not so with wattage.
How do dBs correlate to wattage? Well, since watts are a measurement of power, we can go back to the dB= 10*log(A/B) formula. When looking at receivers and power amps, marketing and salesmen types love to throw out numbers like this or that has this many watts, but it really means nothing until you get that broken into dBs. Take an 175 watt per channel amp, and a 250 watt per channel amp. Based just on the manufacturer’s specs we can determine the exact difference in dBs between the two. Before we do, I want to make it clear that just because there might be the smallest of differences in dBs does not automatically mean that the component with the smaller wattage is just fine. It should come down to how clean one sounds over the other. Ok, putting in the above numbers you get dB=10*log(250/175). That calculated out equals 1.55dBs, not much difference at all. To get a noticeable perceived difference in volume, you need about a 3dB increase; 1.55dBs falls well short of anything significant.
That leads us into the discussion of dBs and SPL. The human ear needs that 3dBs more to perceive a noticeable change in volume, and to perceive a doubling in loudness we need an increase of 10dB. Now going back to wattage for a moment, to get 10dB more in terms of watts, you need ten times the power. While you digest that for a moment, you also need to know how speakers are rated in relation to dB SPLs. A speaker usually has a rated sensitivity somewhere in the manufacturer’s specs. This measurement as an industry standard is taken with a 1 watt input at 1 meter away on axis with the tweeter. The average is about 90dBs. That tells us with one watt of power at one meter away the speaker will produce a SPL of 90dBs. So if the same speaker is driven by a 100 watt amp, you would use the following formula dB= 10*log(100/1), the 100 being the one hundred watt amp, and the 1 being the one watt to obtain the given sensitivity. With that plugged through the calculator, you come up with 20dBs. That, in turn, means that if this speaker is driven with 100 watts of power, it will reach 110dBs at one meter away. The relation of dBs and distance can also be used to help determine how loud that 110dBs might be at say twice the distance. This falls under the inverse square law. Basically, all you need to know is that in a reflective-free environment, if you double the distance you lose 6dB of sound pressure. So at 2 meters that 110dBs is now only 104dBs,but in the real world we don't live in a reflective-free environment so the drop most likely won't be exactly 6dB.
To close, a few things are very important to remember about dBs, volts, watts, and SPLs. First, just because one power is higher than the next, does not mean it is significant, but at the same time if the lower-powered device does not sound as good as the higher-powered one, the difference in dBs won't matter either way. Also remember if you are dealing with manufacturer specs, the variance of some manufacturers overrating their equipment's ability or in some cases underrating its ability needs to be factored in. Next, the more headroom the better. How and where dBs are referenced to is also important. When I say a SPL meter reads 90dBs, that means that a sound is 90dBs above the threshold of hearing. With CD's, Full Scale Digital (a maximum recorded signal) is referenced to 0dB. Anything below that becomes a minus dB from the stated reference point. In the long run, it helps to remember where the reference point is. Understanding dBs and how they relate to our components, speakers, and ears is important, and I hope this humble column has started or increased your knowledge on dBs.
 
It gets even more mind bending when think about the mixing stage. Becuase as stated above +3dB is really where you notice a difference in loudness, yet in a mix .5dB or more can change the mix, maybe not drastically but if you bring down the drums .7dB you will notice a difference in the mix, but the volume you are hearing hasn't changed enough to be noticable. I wish I knew more about that and why it happens.

But do some blind folded experiments with your buddies tell them you are going to bring the volume down at some point, then at random bring it down slowly they should pop up and say hay its lower now when you hit -3dB~ from where you were at, try it both with test tones and with music. Also if you have a radio shack SPL meter ($30-$40), you can test out the inverse square law, play a 1Khz tone out of just one speaker. At three feet measure the volume, then at 6 feet mneasure again, should be about 4-6dB down. Again if we are in a reflective free space it would be exactly -6dB, but since our studios aren't completely treated or poorly treated or not even treated at all, the inverse square law will not always apply exactly. Those with WELL treated rooms try it out with a SPL meter, then try the same test in your untreated living room. Again must be one speaker other wise you will run into phase issues that can throw the whole test off.
 
Daniel Reichman said:
dBs are logarithmic rather than linear, this is why if you increase the level by 3 dB, you double the volume.

Right on about the logarithmic thing, but a 3dB increase equates to a doubling of the power. Doubling the volume equates to a 10 dB increase.

For voltage or SPL, dB=20logVout/Vin(reference)
 
Right on about the logarithmic thing, but a 3dB increase equates to a doubling of the power. Doubling the volume equates to a 10 dB increase.

For voltage or SPL, dB=20logVout/Vin(reference)


its actually a 6db increase in terms of energy (eg sound pressure) as refernced in your formula where is log 10 p/p ref for power
etc
 
SRR said:
It gets even more mind bending when think about the mixing stage. Becuase as stated above +3dB is really where you notice a difference in loudness, yet in a mix .5dB or more can change the mix, maybe not drastically but if you bring down the drums .7dB you will notice a difference in the mix, but the volume you are hearing hasn't changed enough to be noticable. I wish I knew more about that and why it happens

That is because 3dB is a psychoacoustic effect, not a physical law. A mix engineer should have ears trained enough to hear smaller differences, and hopefully also note the difference between a small EQ change and a small volume change.
 
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