Apologies to anyone who finds this a stupidly basic question...just want to check!
So when measuring bass reflex the process to combine the port response and woofer response is to scale the port response before merging.
The degree to which the port response is scaled is expressed as:
20 LOG(Port Diam / Woofer Diam)
So as a practical example:
20 x Log(50mm/150mm) = -9.542 (db)
However, principally its just ratio of port to woofer....and I want to scale ports that are not circular. In these cases I had just been using the eyeball method of aligning the tails of the woofer and port response before merging.
If we convert to using area instead of diameter I found that:
20 x LOG(... != does not work...
However:
10 x Log(1963.4mm2 / 17671.46mm2) = -9.542 (db)....same result
So am I correct in assuming that I can re-state the above as:
10 LOG(Port Area / Woofer SD)
So when measuring bass reflex the process to combine the port response and woofer response is to scale the port response before merging.
The degree to which the port response is scaled is expressed as:
20 LOG(Port Diam / Woofer Diam)
So as a practical example:
20 x Log(50mm/150mm) = -9.542 (db)
However, principally its just ratio of port to woofer....and I want to scale ports that are not circular. In these cases I had just been using the eyeball method of aligning the tails of the woofer and port response before merging.
If we convert to using area instead of diameter I found that:
20 x LOG(... != does not work...
However:
10 x Log(1963.4mm2 / 17671.46mm2) = -9.542 (db)....same result
So am I correct in assuming that I can re-state the above as:
10 LOG(Port Area / Woofer SD)
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