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9th May 2012, 05:43 PM  #1 
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Join Date: Apr 2004
Location: Texas

The Real Farfield Distance
As a general rule, software and people in general speak of the accurate farfield measurement distance being a function of the distance between the measurement microphone and source and the nearest reflective surface. Others indicate that though this provides a reflection free window, true farfield cannot me measured unless the measurement microphone is at least 2 wavelengths away from the sound source of the lowest frequency of interest. This would put a damper in the accuracy of full range (that is, for the low frequency) ground plane measurements.
I was having trouble understanding the basic generalizations of a farfield distance so I posted over at Partsexpress and got a responses that pointed me in the right direction. I learned that estimates seem to come from the solution of the Rayleigh Integral. I messed around with the far field question some more and have come up with what I believe to be a better understanding of this. I just did not feel right following cookbook estimates. I am attaching a spreadsheet for any other OCD types who really want to understand what they are getting when they measure far field. There is always some degree of error in this measurement. To use this spreadsheet, you enter data in the yellow boxes only. Enter the highest frequency of interest and the radius of your driver in question (in cm). The ka is then calculated and a list of various ka's are then filled in below in column A. Column B then calculates the actual equivilent frequency that the data in the table is representing. If you enter an allowable error pct, the table will highlight which distances and ka's (frequencies) are within your limits. The r/a mulitplier allows you to modify the r/a ratios to see more extended ratios. The actual measurement distance that the r/a ratio is representing is calculated above. This allows you to see the error based upon the expectation that in the farfield there should be a loss of 6.02 dB for every doubling of distance. It is based upon the solution of the Rayleigh Integral. I believe my calculations to be solid but if someone sees any error, feel free to jump in. I hope that this hopes some other poor obsessive soul. Jay 
9th May 2012, 09:33 PM  #2 
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Join Date: Nov 2009
Location: The Mountain, Framingham

Hi,
An interesting endevour but can you give a little more explanation? The +% is an error of a point source gain vs. near field real circular source SPL? This looks like something where a plot of SPL vs. distance (the form we normally see near field boundaries determined via) would be of great help. David S. 
9th May 2012, 09:58 PM  #3 
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Join Date: Apr 2004
Location: Texas

The error is based upon the expectation that in the true farfield, we see a drop of 6.02 dB per doubling of measurement distance. In reality, according to the Integral, this does not really occur until infinity as this is the only place that a circular disc would appear as a point source. When we do farfield measurements, there is an innate error due to this. As we move further away, the error decreases.
I could plot the SpL drop at each of the points for a particular ka (like John K did in the Partsexpress bulletin board response where I posted the question originally) against the 6.02 dB expectation, but this is a numeric way of visualizing it. John has already presented a graph of this over at Partsexpress, as I mentioned. It is under a thread that I started there regarding the actual farfield measurement distance last week. Jay 
9th May 2012, 10:02 PM  #4 
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Join Date: Apr 2004
Location: Texas

I think that the bottom line is that, as with everything else we do with our speakers, there is a benefit cost ratio to our choices. The further away we measure our speakers, the less the farfield error will be...but there is a point of diminishing returns as our window narrows and therefore the bottom of our frequency range rises. We have to choose the best point for us. It is likely that a few pct of error is not a major deal an octave or two over the anticipated crossover point, in my opinion. In fact, this is what we have all been living with (at least for low frequency drivers) for a while, perhaps without thinking about it (according to the math as I understand it).
Jay 
9th May 2012, 10:17 PM  #5 
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Here is the forum link to where John K's graphic is:
The Real Farfield Measurement Distance  Techtalk Speaker Building, Audio, Video, and Electronics Customer Discussion Forum From PartsExpress.com Jay 
10th May 2012, 10:45 AM  #6 
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Join Date: Nov 2009
Location: The Mountain, Framingham

It looks to me like John K's analysis shows the answer well in graphical form.
Techtalk Speaker Building, Audio, Video, and Electronics Customer Discussion Forum From PartsExpress.com  View Single Post  The Real Farfield Measurement Distance If you are above r/a = 1 (or certainly by r/a = 4) then you are in the farfield and level is dropping by 6 dB per doubling of distance. R is distance from source and a is radius of source. 4 radiuses or 2 diameters distance is sufficient. Now, a cabinet radiates from its edges, so that should be considered as part of its size if you want very high accuracy, although it is probably a minor factor. With multiway systems you also need to consider the distance required for the far field geometry to hold. That is, the distance where the path length differences between drivers is near enough to the far relationship. Certainly, if you achieve twice the long dimension of the cabinet you have covered every factor. David S. 
10th May 2012, 01:00 PM  #7 
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Hi Guys
This may offer some additional grist for thought; Best, Tom Danley http://community.klipsch.com/forums/...alloonData.pdf
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10th May 2012, 01:04 PM  #8 
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Dave, I think that the point of my doing the math (which is pretty much in agreement with John's graph) is to look at the actual error that exists. Of note, I ran this by John prior to posting it though I can't vouch that he looked at it in detail).
I think that there is some confusion regarding farfield. In fact, when I have looked at modeling papers, the Rayleigh integral is felt to be inferior in it's model of farfield to BEM modeling, particularly at higher frequencies. The most current AES measurements standards state that "this implies that the measurement distance from the acoustic center of the device must be large with respect to: (1) the largest dimension of the radiator, and (2) the square of the largest dimension of the radiator divided by the wavelength of the highest measurement frequency." If one was to use an enclosure that was .6 meters tall and .2 meters wide, the longest dimension (the diagonal) would be .63 meters. Using the 2nd criteria, if you measure to 20,000 Hz, this would be .63^2*20000/344 or 23 meters. This made no sense to me. Even using half of this dimension, assuming the driver is centered on the baffle, it would be 5.76 meters to FF. Using just the driver, itself, let's say a 1 inch tweeter, the calculation would be .0254^2*20000/344 or 37.5 cm. Looking at the table that I set up, at 38 cm, you could measure to 20Khz with .2% error. If you use the 63 cm size (of the enclosure diameter), at 20kHz, there is an 8.8% error at 23 meters. John's graphic seems to be more in line with the use of just the driver. It also suggests that what we are using for our definition is also more along the lines of the driver. The enclosure provides a reinforcement of the sound signal starting with a frequency equal to the longest distance of the driver to the edge of the enclosure and ending with a freq equal to the driver distance to the shortest part of the enclosure. The distribution is based upon enclosure shape, reflectivity, etc. Your suggestion of using twice the largest baffle dimension is likely to always put you in farfield by my calculations, but it is another rule of thumb that I was trying to get away from. I was trying to understand the why. As I mentioned before, the Rayleigh integral is not exact, but it at least provides an understanding of the cost/benefit ratio of a measurement distance along with one's window length calculation. This is an educational endeavor for me and if you have more insights, I am not attempting to be argumentative here. Jay 
10th May 2012, 01:06 PM  #9 
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Location: Texas

Hi Tom,
I think that the balloon data is actually more in line with the use of the enclosure dimensions in concert with the AES specifications. It is consistent with my spreadsheet if one uses the enclosure and not the driver dimensions. Jay 
10th May 2012, 02:39 PM  #10 
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Join Date: Nov 2009
Location: The Mountain, Framingham

I am aware of the HF stretching of the far field transition distance and included it in my paper on line array modeling.
Constant Beam Width Transducers line arrays Look at the 3rd PDF and you will see some graphs of a long multielement array that, at 10kH, didn't settle down until past 15 times the array length. The periodic wobble of level comes from the rotating phase contribution of the outter elements. Once the end elements come mostly in phase with the center the response settles and you are in the far field. This, of course, is only true for the particular example of a multielement line array, although this could be considered the sampled version of a continuous array. I'm not totally sure how this reconciles with John's graphs that show the transition point to be essentially independent of frequency (the curve shapes below that are certainly different, but the transition frequency is not). In the practical case all we care about is what distance gets us to the point where our curve, to a sufficient accuracy, agrees with the curve at infinity. For most speakers our source gets smaller at HF (multiway speakers, right?) and just worrying about the longest dimension seems to be a practical step. David S. 
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