# Nearfield Measurement - adjustment for multiple ports

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.

#### burnhamjs

When taking near field measurements what is the correct way to sum the output of multiple ports and the cone into one single response graph?

From Jeff Bagby's "White Paper - Accurate In-Room Frequency Response to 10Hz" he gives an example for adjusting for one port: The smaller of the two, usually the port, will need to be lowered in SPL (dB) because its smaller area will apply more pressure on the microphone making it appear louder. Its output needs to be lowered by the ratio of their areas using this formula: Lower the port by 20 Log (Port Diameter / Cone Diameter). So, for our 6.5” woofer, with its 5” cone area, working with a 2” port, the port output will need to be lowered by 20 Log (2 / 5) = -7.96 dB, or roughly 8 decibels"

For dual ports he give this example: for dual ports, find the total port area of both ports and determine the effective diameter for the combined area. Now, take the measurement at only one port, find the ratio of the total port diameter to the cone and make the calculation and then add 6dB. So, for our example, if there had been 2 – 2” ports then the port output would be lowered by -7.96 + 6 dB = -1.96 dB. For dual woofers and a single port we will need to do it the other way around and subtract another 6dB from the port output.

However, I'm not understanding this. If I calculate the total port area of the two ports you get a combined area of 6.28sq inches or an effective diameter of 2.28". If I then calculate the SPL adjustment by using the ratio of diameters (effective port/cone) = 20 Log (2.82/5) = -4.97dB. If I then add the 6dB I get +1.03dB.

In my particular situation I have a woofer with an effective diameter of 4.375", two ports of 0.625 diameter (front), and one port of 1.25" diameter (rear). I am trying to apply the correct adjustment to a port reading so I can stitch the woofer and port measurements into one SPL graph. I think I can figure it out once I understand how to do it correctly.

#### ernperkins

You have a somewhat unique situation with ports of different areas. Bagby’s example assumed two ports with the same area and length. His 6 dB figure comes from the doubling of area from two identical ports: 20 log 2 = 6 dB.

You don’t have three identical ports so I think you need to handle the two, 0.625” diameter ports separately from the single, 1.25” port. Note that the 1.25” port will almost certainly have a different tuning frequency unless you purposely adjusted its length to match the tuning frequencies of the two other ports. So you’ll need three measurements: near-field woofer, one near-field 0.625” port and the near-field 1.25” port.

First, handle the two 0.625” ports: Area of 1 port = pi * radius squared = pi * (0.625/2) squared = 0.3068 in2. Doubling the port area = 0.6136 in2. Effective radius = 0.442”; effective diameter = 0.884”. Your woofer diameter is 4.375” giving a dB loss of 20 log (0.884/4.375) = -13.9 dB. Adding 6 dB for two identical ports gives a total of -7.9 dB. At this point reduce the 0.625 port measurement by 7.9 dB and merge it with the nearfield woofer response.

Second, handle the 1.25” port: The dB loss is 20 log (1.25/4.375) = -10.9 dB. Reduce the 1.25” port measurement by 10.9 dB and merge it with the response from the previous step.

I’m not sure how you ended up with ports of different diameters as I don’t know of any program (e.g. WinISD, Bagby’s programs) that let you model the response of multiple, different sized ports. Maybe you have a program that does?

PS: Also note that you have to get the phase relationships correct between your near-field measurements as Bagby notes in his paper.

#### PRR

Paid Member
The three ports, all on the same box, "should" result in a single resonance similar to a single port of the same area as the three. Some adjustment for end-effect must be made. But designer sanity must be preserved.

I'd just figure what the three holes-add up to and compute based on that. It won't be quite right, but it won't be very wrong, and it is very unlikely to make much difference. (Note that none of your ports reaches -10dB level.)

I'd expect a 5/8" port to wheeze bad at the levels a 5-inch woofer can reach. The usual thing is to have ONE port for a maximum of area with a minimum of edge drag and turbulence. Yes, when big boxes need big ports and port-tubes only come in a few sizes, two ports may be expedient.

#### ernperkins

The three ports, all on the same box, "should" result in a single resonance similar to a single port of the same area as the three.

True, but only if the OP adjusted the lengths of the ports so they all had the same tuning frequency. For example, depending on length, the two small ports could a produce their maximum output at 75 Hz and the larger port could have its maximum output at 40 Hz. We don’t know the lengths so it’s all conjecture I guess.

I'd just figure what the three holes-add up to and compute based on that. It won't be quite right, but it won't be very wrong, and it is very unlikely to make much difference. (Note that none of your ports reaches -10dB level.)

For this “unique” design I’d agree. Do note that -10 dB is just how much you reduce the nearfield port measurement. That measurement could be higher than the near field woofer measurement, depending on the port diameter.

[/QUOTE]I'd expect a 5/8" port to wheeze bad at the levels a 5-inch woofer can reach. The usual thing is to have ONE port for a maximum of area with a minimum of edge drag and turbulence. Yes, when big boxes need big ports and port-tubes only come in a few sizes, two ports may be expedient.[/QUOTE]

+1

#### PRR

Paid Member
> ports so they all had the same tuning frequency

I can't see it any way except "sum of the ports" against the acoustic mass inside the box. Like three chokes in parallel, below. The L/R ratio of each port may vary some, but the resonance is all three together, not a separate resonance from each port.

(Assuming port spacing is not a large fraction of the wavelength.)

Fill a bucket with water and drill three different holes in it. The bucket will empty through all three holes in some time, not a different time for each hole.

#### Attachments

• burnhamjs-42.gif
10.2 KB · Views: 116