Got Bored In Math

[IH8PunkRok]

so i was lookin through the book and i found a formula 10log(x/10ˆ-12) (10ˆ-12 being 0 db)

now according to this formula u can plug in a decibel reading say 150 db and find the watts per cubic meter needed to achieve that level of pressure

basically it comes down to 150 decibels is 10ˆ15+10ˆ-12 = 1000 soo 1000 watts per cubic meter that seems feasable, however i was doin some more calculations and found out the 180 db (10ˆ18 + 10ˆ-12 =1000000) 1 million watts per cubic meter????

it also said the diference between 0-10 db is ten times 0-20 db is 100 times louder 0-30 db is 1000 times louder and so on which i know to be somewhat true

so on a vague estimation i figured that at full volume, roughly 40,000 watts, meade could only hit 158.1 db inside the box that didnt sound right to me

now im not sure what hertz level is is calculating this at but its kinda confusing right now, cause it also says i can hit 150 in my box and i know that aint true

of course it is assuming perfect conditions

am i doing these calculations right? (ill be gone for a while so i wont respond until maybe 7 est)

[JustinT]

what textbook was that in? what class

[EricM9104]

What friggin math level are you in?? Lol.

I'm in Math 1a (starting Algebra... ugh) not that I hate it, it gets boring since I took it twice last year and pre-algebra last year too... I'm a sophmore now but just barely, didn't attend school often last year and am now trying to make up those lost credit.

I usually get bored in Biology and found out that 4 18" subwoofers would still have more cone area than a single 60" subwoofer.

4*18=74*3.14=226.08" cone area

60*3.14=188.4" cone area

[andym85]

thats not how figure con area...the fomula is pi*r^2

for one 18": 3.14 * 9^2 = 254.34sq" of cone area

so four 18s would be 254.34*4=1017.36sq" of cone area

a 60" sub would be: 3.14*30^2= 2826sq" of cone area

[dwright27]

pwned.

[andym85]

^^lol...wasnt trying to pwn anybody..im just speaking the truth...

[EricM9104]

Sh!t, damn, I guess I didn't remember the formula for figuring out cone area properly. Oh well, we all make mistakes don't we?

[andym85]

yep...what u figured out was actually the length of the circumferene (<-sp?)..which is: 2*pi*r or d*3.14 (d=diameter)

[IH8PunkRok]

well im in precal right now

i wanted to take phsics this year but i didnt since im already takin AP Chem and Honors Bio

i couldve taken calc this year but i didnt wanna do precal over the summer (lazy)

so no one knows if im doin that right?

Edited September 21, 2007 by IH8PunkRok

[IH8PunkRok]

haha just found out sumthin

the hertz level they base 10ˆ-12 (0 db) off of is 100 hz

and meade's box is .57 cubic meters so my calculations are way off

in a one cubic meter box @ 100 hz 1000 watts will create 150 db

i cannot figure out wut i want to figure out

ok for those who think they are up to the challenge check out this website http://www.analogrules.com/dbwatts.html

and the formula to find watts per cubic foot is as follows [db number= 10log ( x/ 10ˆ-12)]

i want to know how many watts it takes to reach 150 db @ 100 hz in steves box (.57 cubic meters)

and if u can figure out how db readings change with increase or decrease in hz then please feel free to explain

i came up with 570 w in a .57 mˆ3 @ 100 hz to hit 150db

honestly that doesnt sound right to me

cause i could hit 150 db in my box with only 137 watts that dont sound right