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  1. #1
    Camaro_Kid's Avatar
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    Comp this weekend need some help

    hello i am going to my first comp ths weekend i just need some help like what to play and set my hu any help would be thankful for i am run

    1 15" Xtant X SVC "2 ohm" and a MA Audio 1300 watt HC amp "850 @ 2ohms" in a 2.5-8 cuft box tune to 34Hz in a 95camaro the sub is faceing the back and the port is faceing the back glass

    Thanks for any and all help







  2. #2
    Bumpin' Yota's Avatar
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    Re: Comp this weekend need some help

    Quote Originally Posted by Camaro_Kid
    hello i am going to my first comp ths weekend i just need some help like what to play and set my hu any help would be thankful for i am run

    1 15" Xtant X SVC "2 ohm" and a MA Audio 1300 watt HC amp "850 @ 2ohms" in a 2.5-8 cuft box tune to 34Hz in a 95camaro the sub is faceing the back and the port is faceing the back glass

    Thanks for any and all help

    Does a '95 camaro have an identicle hatch area to a 2001 Trans Am? If so you loudest note should be around 53hz... You'd likely gain spl if you could retune your box to at least 40-45hz. I'd be hesitant to tune any higher than that due to the low power and not knowing your exact note....

    Also ideally you'll want the sub and the ports firing into the back glass. Remember a hard, smooth object will reflect sound waves quite nicely....

    Do NOT demo the system for anyone till after you done competing. (Remember that most organizaitons will only have you burp once, but db drag has you qualify, then makes you wait 2hrs or so, then go back in and eliminate. So dont demo in between if it's a drag...)



    1 volt = [1(kg)(meter^2)] / [(second^3)(ampere)]

    1 watt = 1 joule / second
    1 watt = (1 Newton)(meter) / second
    1 watt = [1 kg/(second^2)] (meter) / second

    simplifying we find:

    1 watt = [1(kg)(meter)] / (second ^3)


    therefore:

    P = (I)(V)

    1 watt = (1 volt)(1ampere)
    1 watt = ( [1(kg)(meter^2)] / [(second^3)(ampere)] )(1 ampere)
    1 watt = [1(kg)(meter^2)] / (second^3)


    And that is WHY Power is in the SI units of Watts. enjoy!

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    Re: Comp this weekend need some help

    why dont demo?




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    Re: Comp this weekend need some help

    wastin your battery. you want as much juice as possible in there. might wanna go as far as bring a portable charger with you for in between runs.



    What the f00k happened to this place?

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    somekook's Avatar
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    Re: Comp this weekend need some help

    oh yea. but you can always run your engine/alternator between rounds too.




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    Re: Comp this weekend need some help

    Quote Originally Posted by somekook
    oh yea. but you can always run your engine/alternator between rounds too.

    Believe me not enough to make up for the stress of the run or demoing. Demo before you run and you will likely loose. Loosing battery power is not the only reason why your score will go down either...



    1 volt = [1(kg)(meter^2)] / [(second^3)(ampere)]

    1 watt = 1 joule / second
    1 watt = (1 Newton)(meter) / second
    1 watt = [1 kg/(second^2)] (meter) / second

    simplifying we find:

    1 watt = [1(kg)(meter)] / (second ^3)


    therefore:

    P = (I)(V)

    1 watt = (1 volt)(1ampere)
    1 watt = ( [1(kg)(meter^2)] / [(second^3)(ampere)] )(1 ampere)
    1 watt = [1(kg)(meter^2)] / (second^3)


    And that is WHY Power is in the SI units of Watts. enjoy!

  7. #7
    ANeonRider's Avatar
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    Re: Comp this weekend need some help

    Demoing will cause your amp to heat, which in most causes will cause a loss of performance.




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    Re: Comp this weekend need some help

    Quote Originally Posted by ANeonRider
    Demoing will cause your amp to heat, which in most causes will cause a loss of performance.

    vc's too... just dont demo till you're done competing



    What the f00k happened to this place?

  9. #9
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    Re: Comp this weekend need some help

    I suppose heat would hinder an amp's performance slightly, though I've not noticed a difference with my zapcos from a board temperature range form 36 degree centigrade to 50 degrees centigrade. Power output appears to be the same on the mic. However I did notice that the amps get a lot more finicky at higher temps. (Thermal protection on my amps is 76 degrees C.)

    I was mainly thinking about the voice coil temperature. when that sucker heats up, its impeadnace rises and as such you loose potential power output that you would have had other wise



    1 volt = [1(kg)(meter^2)] / [(second^3)(ampere)]

    1 watt = 1 joule / second
    1 watt = (1 Newton)(meter) / second
    1 watt = [1 kg/(second^2)] (meter) / second

    simplifying we find:

    1 watt = [1(kg)(meter)] / (second ^3)


    therefore:

    P = (I)(V)

    1 watt = (1 volt)(1ampere)
    1 watt = ( [1(kg)(meter^2)] / [(second^3)(ampere)] )(1 ampere)
    1 watt = [1(kg)(meter^2)] / (second^3)


    And that is WHY Power is in the SI units of Watts. enjoy!

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