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Reply	Stewart says: 20 times louder is 13dB (decibells) louder. So any measurement would have yielded a result of 123dB.To express 20 times in dB you need to use 10(log20) and add the result to the value already expressed in dB. Decibels do not represent an absolute value but a ratio. So a level of 110dB is actually 110dB or times 10^11 above a reference value that I cannot recall at the moment.  (no votes) Tags: technology, transport, human body.
	posted on 2009-04-22 20:49:00 | Report abuse

Reply	Matthew says: The reference value (0dB) is the threshold of human hearing - the quietest sound that can be perceived.  (no votes) Tags: technology, transport, human body.
	posted on 2009-04-23 21:53:00 | Report abuse

Reply	Tim says: Stewart is correct, noise is measured on a logarithmic scale. Log scales can be hard to understand, but a good starting point is looking at this webcomic: http://xkcd.com/482/  (no votes) Tags: technology, transport, human body.
	posted on 2009-04-24 12:49:00 | Report abuse

Reply	Fritz says: Uncorrelated noise sources cancel each other much of the time, such that 25 noise sources combine to noise only 5 times as strong. Besides, the ear works on a logarithmic scale, so the noise will be only 10*log5=~7 dB louder.  (no votes) Tags: technology, transport, human body.
	posted on 2009-04-24 13:04:00 | Report abuse

Reply	Jamie says: The reason the nooise didn't sound twenty times louder is because of the way in which our ears process sound. Intensity Level and loudness are related but not identical. We have a "Threshold of Hearing" of 10^-12 Watts per metre squared which is equivalent to 0dB. The ears process sound logarithmically by compressing the sound wave in order to process it.The actual intensity level (in dB) will therefore increase in a compressed fashion also. Any increase in intensity is therefore related to a relatively small increase in intensity level - the way our ears percieve sound. In fact, the intensity level of 20 Formula One Cars can be shown to be 10log20 greater than the intensity level of one Forumla One car by the following method:Intensity Level (dB) = 10log(Intensity x 10^12)Increasing Intensity by a factor of 20 gives: New Intensity Level (dB) = 10log (Intensity x 20)x10^12)= 10log(Intensity x 10^12)+10log20using the fact that log (m x n) = log m + log n10log20 approximately equals 13, so the intensity level of the noise you heard was approximately 123dB.  (no votes) Tags: technology, transport, human body.
	posted on 2009-04-25 21:16:00 | Report abuse

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