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Natural Science Forum / Physics / Acoustics / January 2008ISO 9613-2 - Is there an error?
Hi, I wonder if there is an error in the formula in Table 3 for a'(h), or maybe it is just incorrectly printed on my verions. Using the values 1.5 m for h and 500 m for dp I read from the graph 2a that a'(h) should be 5 but when I calculate it a'(h) is 2.19 dB. Secondly, I would assume that the two methods of calculating ground effect should give similar answers for the same situation however I find that the method described in 7.3.2 gives a lower attenuation and, when the addtion to the directivity is included, the result is negative attenuation. Conversely the result of using the method described in 7.3.1 gives an attenuation of about 7 dB. I therefore dont know which to believe! Any help or comments gratefully received. Rod > Hi, > I wonder if there is an error in the formula in Table 3 for a'(h), or [quoted text clipped - 13 lines] > > Rod I get a'(h) = ~4.53 from Table 3 when I use your values of h and dp, which would appear to generally agree with Figure 2a. Seems as though there might be something wrong with your calc??? As for 7.3.1 vs. 7.3.2, there are more variables in use that you haven't provided. Negative attenuation, which should generally be ignored (IMO), is possible with some of the ground absorption formulae/ methods. However, without more information, I cannot determine exactly if/where you might be going astray. All the best, Savant Thanks for the reply. I am still confused by the alternative methods of calculating ground attenuation. For Method 7.3.1: Assuming the ground is porous, G = 1 Dp = 500 m hs = hr = 1.5 m I calculate a'(h) =4.53 b'(h) =8.52 c'(h) =6.47 d'(h) =2.16 This gives As and Ar to be: -1.5, 3.0, 7.0, 4.9, 0.66, 0,0,0 for the Nominal frequencies 63 - 8 000 Summing these in the same way as combining noise sources (Eq 5, ignoring A weighting) gives 11.7 dB which should be doubled according to equation 9 giving 23 dB of attenuation. Method 7.3.2 gives Agr to be 4.7 dB, if hm = 1.5m and d =500 m and D to be 3 dB which must be subtracted to give 1.7 dB Finally according to Note 1 the 500Hz band can be used as an approximation for attenuation. In 7.3.1 this would give Agr to be 4.9 X 2 = 9.8 dB With results ranging from 1.7 dB to 23 dB attenuation from ground effects I am confused, or maybe I am misreading the standard. Again, any comments will be very useful, Rod > > Hi, > > I wonder if there is an error in the formula in Table 3 for a'(h), or [quoted text clipped - 17 lines] > which would appear to generally agree with Figure 2a. Seems as though > there might be something wrong with your calc??? Thanks for confirmation - some of the text appears to be a subscript of a subscript! > As for 7.3.1 vs. 7.3.2, there are more variables in use that you > haven't provided. Negative attenuation, which should generally be [quoted text clipped - 7 lines] > > - Show quoted text - > Thanks for the reply. > I am still confused by the alternative methods of calculating ground [quoted text clipped - 70 lines] > > - Show quoted text - I am glad to see that I anot the only one confused by ground attenuation calculations. MarkB Rod, Wow. Let me see if I can help focus this... > For Method 7.3.1: > [quoted text clipped - 12 lines] > ignoring A weighting) gives 11.7 dB which should be doubled according > to equation 9 giving 23 dB of attenuation. The misstep here is that you summed attenuations logarithmically, which cannot be done. I believe you did this with the intention of getting an A-wtd. value for Agr. While the intention is good, it's an improper us of Eq 5. > Method 7.3.2 gives Agr to be 4.7 dB, if hm = 1.5m and d =500 m and D > to be 3 dB which must be subtracted to give 1.7 dB Here, you corrected the Agr term for D(omega). The standard states that D(omega) is to be included in the Dc term of Eq 3. Therefore, your value for Agr is still 4.7 dB. When you calculate for Dc to input into Eq 3, be sure to include D(omega). > Finally according to Note 1 the 500Hz band can be used as an > approximation for attenuation. In 7.3.1 this would give Agr to be 4.9 > X 2 = 9.8 dB At first glance, this does appear to be a contradiction. However - and this is a BIG "However" - the first bullet from Section 7.3.2 specifically states that the 7.3.2 method - Eq 10 - should be used when only the A-wtd SPL at the receiver is of interest. To me, this supercedes the instruction in Note 1. Considering that the Table 3 term is an estimate for the 500 Hz band and Eq 10 is what the standard states to use for estimating Agr for A-wtd SPL receiver calcs, my money would be on the latter as the correct approach. So, it looks like you should be using Agr = 4.7 dB if your intention is to estimate A-wtd ground attenuation. HTH. All the best, Savant |
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