Lakota
5,000+ posts
Doesn't Know $hit
WARNING, This thread contains complicated math that has been answered redundantly in the previous 100 threads.
Width x Height //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gifIt was already given to you.
Length x height=port area.
Holy shit. You have asked the same questions about 55 times in the past week. Everything has been answered numerous times for you. I have made 15 designs and after each one you decide "it won't be loud enough because someone on f150 forums told me so"ok thanks and also would that be louder or would 3 4inch aeros at 16.02inches be louder both same tunning subs and power and same net for subs
True. This is how I find the area of circles and squares.You could always take the integral from 0 to width under the height to get the port area as well.
oopsie.Width x Height //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif
Heh.oopsie.
Trying to repeat myself over and over ftl.
I've never, ever calculated for end correction when I've made any of my boxes, slot port or aeroport, if that says anything to you.Heh.
Quick question: When determining port length and area displaced, I saw some talk about end-of-port correction (When talking about real world application, harmonics)
Does the effect of the opening of the port on the sound waves need to be taken into account when determining proper port length and width?
Would that even be a function though? I guess 0-3 if f(x)=port height, but I dunno. I'ma try that out though //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gifYou could always take the integral from 0 to width under the height to get the port area as well.
No, not really. End correction for most enclosures comes out to be around 1hz difference anyway, and for a large enclosure, it makes next to no difference.Heh.
Quick question: When determining port length and area displaced, I saw some talk about end-of-port correction (When talking about real world application, harmonics)
Does the effect of the opening of the port on the sound waves need to be taken into account when determining proper port length and width?
it would just be a definite integral of a constant. Works out to be 15.5x from 0 to 4. 15.5(4)-15.5(0) = 62 in^2 //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gifWould that even be a function though? I guess 0-3 if f(x)=port height, but I dunno. I'ma try that out though //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gif
I try to leave my memories of Calc 2 behind me. I hated that class.****, let's laplace this shit while we're at it! //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gif