Texas Tech

good stuff haha

look at google home page...

More PIX More Pix!

Get well soon man, much love to ya!

wow, glad i went with EA, he responded to every single email i sent him daily, i sent him 2-3 emails daily sometimes and he responded 2-3 times daily, he got me my alt. in about a week and a half, granted i did send him a check so i had to wait for it to get to him and then clear through the bank. Thats rediculous man.

dude, your a dc audio dealer and you have your own "shop"? (referring to sig)

shoot i have a haynes manual, i can copy the page about how to do this job and post it up for you if you want, its probably a couple of pages, It is a job that has like 15 steps to it in this manual.Its got some pics too.let me know

i looked on an Acura Forum and some people claimed they did it without coming from teh drivers side panel at all, they took it out from under the car just like that. I tried to do it the "faster" way and i ended up stripped the nut that you have to unscrew, so i took it to a shop. The nut was extremely tight, i couldnt get it off at all, i even used a pipe to extend the length of the socket wrench to get more force, didnt work. If i would have taken it to a ship in the first place i would have saved time, and a lot of stress. I did have to call up a lot of different shops though, this Acura/Honda specialist shop wanted to charge me over 100 bucks for this job.

I have a Mb Quart Component set, the QSC 210, they are rated 70 Watts RMS at 4 ohm. Can i basically wire them in parallel on the same channel (obviously) creating a two ohm load. this would save me from using two channels. Or maybe i am confused, and that is is a 4 ohm load with both crossovers wired in parallel? My amp puts 100 W per channel at 4 ohms and 150 W at 2 ohm Either way is good for me, but can i wire the component set in parallel, i know some SQ will be lost in exchange, but it would really allow me to

http://www.caraudiojunkyard.com/forum/powe...-portfolio.html he taught me everything i know about ports so check him out its pretty impressive

yea if you cant go any deeper enclosure wise because of course all space in the trunk, or car for that matter is limited., then you are going to have to do something else. If you have to, change the length, or width of the port, and make 2 ports for more vent area if you have to minimize the width, because changing the width, then changes the end correction, and also with a different width, then the distance from the end of the port to the back wall of the enclosure is different from the width. but you gotta make up for the vent area so make another port, but then that is hard to do of course because of limited space, so yea i see what your saying, endless circle

yea like i said if you want your effective length to be 13 inches, then make the depth of the box longer so those lengths are not equal to eachother, so you dont have to measure it that way. Make a port that is 2 inches wide and 12 inches long, and make the depth of the box like 20 inches, you will have the length of 13 inches, anyway you do it, it seems your going to have to adjust yor enclosures dimensions, or not, i havent done a lot of fooling around with enclosure plans like you, i just know how to calculate the length because a guy named powernaudio taught me, and he knows a lot you should look at the enclosure plans he designs, he is at caraudiojunkyard.com (the form part). He has a "portfolio" you can see all the boxes he has designed, they are very confusing and intricate

I mean, if the distnace from teh end of the port (physically) to the back wall of the enclosure is = to the ports width, you are going to have to measure it that way, its how it is, the wood has thickness which extends the ports length (physically). You would have to adjust your port and/or enclosures depth to make it the length you want it to be

yea our diagrams are similar, in that the numbers are differnet, but the concept is still the same. If you want it to be 13.5 you have to adjust your port, or the depth of your box. Make the box deeper, to where the distance from the end of the port to the back wall of the enclosure is longer than your ports width, you would have to adjust your ports width and length and the depth of your enclosure to make it work, or continue with this same concept, and make it work that way.

Here it shows, you add 16+1.5(red line)+1.5(green line)+.5(woods thickness, the dark purple line). That is where the physical port length stops. that is 19.5 inches.

look at my diagram and tell me if what i drew is what you were talking about

You would measure down the middle, now the wood piece that actually creates the port, it has a thickness, lets say its .5 in. thick. That is still considered port length. that is still measured as physical port. So the physical port length would be 19.5 in. then you would add the end correction being 1.5 in., making the effective port length 21 inches long.

actually i think i know what your saying ill show you a diagram and how i was taught to measure it 30 more seconds....

its hard to understand, a fast paint diagram would help .can you?

yea i know it sure is debateable, my reasoning is on jlaudio site, it says a wall of the enclosure that is part of the port, in most cases (like simple L ports or T ports) the wall does extend the port, but in this case, if the back enclosure wall does extend the port, then the port would have two different widths, like i stated earlier so, i duno i wish i was a professional box builder to know the answer, or the equipment to test this stuff and figure out the tuning frequency. Like test the frequency with the exact same physical length of the port, but one enclosure like the one shown on the 7th post, and then one similar to the 9th post (referring to the distance from the end of the port to the back wall of the enclosure, that would be the variable in the experiment).

I got a 99 Acura Integra as well, i went with HighOutput Alternator (Nathan Peddicrod) 200 Amp Alt. So far so good. I emailed him and always recieved a response that day, we exchagned multiple emails daily, he was on top of it. for the 200 Amp Alt i paid 360.00 flat. Its kind of difficult to get to, but there are how-to's on the internet on different Acura forums. I think i paid someone 50 bucks to install it and run a zero gauge wire from teh alt to the pos. battery post.

Referring to this: "The effective length of the port is found by adding an end correction factor. An end correction factor is necesary because more often than not, one wall of the port is also one wall of the enclosure and this wall extends beyond the end of the port thus effectively adding length to the port (remember, the driver can't "see" the length of the port, it can only go by what it "feels" is going on). " -JL Audio Website Yes, however the wall of the enclosure is not a wall of the port. If it were a wall of the port, then the ports width changes, at the beggining of the port (the baffle) the width is 4 inches, and then at the end of the width is 8 in. or 20 in. (depending on which diagram you are looking at). However it would be more appropriate if it were like this, in this case you would have to add half the ports width: Here the physical length is 18.75 in. and the length for tuning freq. formula would be 20.25 in. In that diagram the enclosure wall extends the port, the port maintains the same width throughout the entire port. I hope i am not coming off as disrespectful or rude.

So even though the length from the end of the port (physically) to the back wall of the enclosure is longer than the width of the port, it still doesnt matter, you add half the ports width for end correction. I made this diagram, tell me if I am correct. The length of the port you would use for tuning would be 19 in. And the length you would use for tuning here would also be 19 in. I know this is not practical but in theory it should be upheld from what you said.

even though i stated "assuming the distance from the end of the port to the back wall of the enclosure is longer than the width of the port?" you would still add half the ports width, i would think you wouldnt

Question: In the first picture the length of the port would be 17 in. long assuming the distance from the end of the port to the back wall of the enclosure is longer than the width of the port. If this is wrong please tell me. However what if you wanted to add a double baffle, but the top baffle is only a trim panel, so the subs are Not screwed into this top baffle. Would you determine the length of the port like i show in this post? the first picture is just a regular box, with NO double baffle. The port length is 17 in. This second picture shows a double baffle, but like stated before, the top baffle is only a trim panel? Would the length now be 17.75 in. long?I am guessing yes however I havent asked anyone so.