jellyfish420

11-10-2004, 11:18 PM

i was reading on another forum that the formula was 1.463x10^8....blah,blah...

i've always used 1.84...this is what jmac told me. who is right???

i've always used 1.84...this is what jmac told me. who is right???

View Full Version : port formula

jellyfish420

11-10-2004, 11:18 PM

i was reading on another forum that the formula was 1.463x10^8....blah,blah...

i've always used 1.84...this is what jmac told me. who is right???

i've always used 1.84...this is what jmac told me. who is right???

DBfan187

11-10-2004, 11:19 PM

Big Mac.

Acidburn

11-10-2004, 11:22 PM

JL Audio agrees with Jmac

jellyfish420

11-10-2004, 11:35 PM

thats guys...gotta go spread the good news. or just rub in someone else was wrong!!!

jellyfish420

11-12-2004, 02:06 AM

heres the response i got. someone help me interpert.

The speed of sound in air @ 18 degrees centigrade is 1.347*10^4 inches per second that value squared is 1.815*10^8 that squared value divided by 4*pi is 1.45*10^7. The particular equation he is referring to assumes that the port diameter is in inches. The equation you may be using could assume the port diameter is in some other type of unit.

The speed of sound in air @ 18 degrees centigrade is 1.347*10^4 inches per second that value squared is 1.815*10^8 that squared value divided by 4*pi is 1.45*10^7. The particular equation he is referring to assumes that the port diameter is in inches. The equation you may be using could assume the port diameter is in some other type of unit.

sumone

11-12-2004, 04:00 AM

there's two equations I've found:

Lv= [(1.463*10^7*R^2) / (Fb^2 x Vb)] - 1.463*R

and

Lv = Av*1.84*10^8/[Vb*1728*(Fb/0.159)^2] - 0.823*sqrt(Av)

so your 1.84 is not in the same context as the 1.463 and shouldn't be considered "right" over a different system of equations.

Lv= [(1.463*10^7*R^2) / (Fb^2 x Vb)] - 1.463*R

and

Lv = Av*1.84*10^8/[Vb*1728*(Fb/0.159)^2] - 0.823*sqrt(Av)

so your 1.84 is not in the same context as the 1.463 and shouldn't be considered "right" over a different system of equations.