need box help for 2 18s

[Bumpin' Goalie 10+ year member]

for example (r^2)pi = port disand circumference(pi) = port area

right?

Keep trying //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

 

[walled01alero 10+ year member]

but i thought port area was surface area not volume? ^^that formula is volume of the tubes not area???

facepalm from the other side of hell right here.

pi(r^2) IS surface area.. when you multiply the result of pi(r^2) with the length of the port you will get volume displaced..

 

[thericanballer 10+ year member]

That's for surface area' date=' considering it is a flat plane.
Volume has to take length into account.[/quote']

ok, so with the numbers i gave you, can you write them out with the answers so that i can see which formulas give me what?

 

[Bumpin' Goalie 10+ year member]

ok, so with the numbers i gave you, can you write them out with the answers so that i can see which formulas give me what?

Surface area = pi(r^2) = 3.14(5^2) = 3.14(25) = 78.5

Volume = pi(r^2)length = 78.5(23) = 1805.5

Now to make that in cubic feet displaced divide by 1728 = 1.04485 cubic feet displaced.

Someone double check this if you want. Just did it mostly mentally. Should be good though //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

P.s. I'll start charging you for a box design pretty soon here //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif jaykay lol (sorta)

 

[walled01alero 10+ year member]

and btw it isnt that hard to find areas and volumes folks, go look up BASIC MATH on google. the only reason you should ask how to find a volume in an internet post (and still not a reason) is if the object looks like this or something...

cus then it's time for your *** to go back to calculus class..

 

[thericanballer 10+ year member]

facepalm from the other side of hell right here.
pi(r^2) IS surface area.. when you multiply the result of pi(r^2) with the length of the port you will get volume displaced..

ohhhhhhhhhh i get it. my bad. lol. i was thinking too much math

Surface area = pi(r^2) = 3.14(5^2) = 3.14(25) = 78.5
Volume = pi(r^2)length = 78.5(23) = 1805.5

Now to make that in cubic feet displaced divide by 1728 = 1.04485 cubic feet displaced.

Someone double check this if you want. Just did it mostly mentally. Should be good though //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

oh ok i see everything now. thanks. buth then dont i have to multiply the cubic feet of the port by two? im going to have two..

P.s. I'll start charging you for a box design pretty soon here //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif jaykay lol (sorta)

 

[thericanballer 10+ year member]

and btw it isnt that hard to find areas and volumes folks, go look up BASIC MATH on google. the only reason you should ask how to find a volume in an internet post (and still not a reason) is if the object looks like this or something...

cus then it's time for your *** to go back to calculus class..

i no how to fine cubic feet, i just didnt understand the flared port stuff...

 

[thericanballer 10+ year member]

btw, i havent taken calculus...i take it this year

 

[Bumpin' Goalie 10+ year member]

and btw it isnt that hard to find areas and volumes folks, go look up BASIC MATH on google. the only reason you should ask how to find a volume in an internet post (and still not a reason) is if the object looks like this or something...

cus then it's time for your *** to go back to calculus class..

My brain hurts just looking at it. Time and patience and some formulas //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

 

[RAM_Designs 5,000+ posts]

I assumed the port tube wall is 1/4" thick when doing the calculations...makes them take up a little more volume that what Bumpin' did.

 

[Bumpin' Goalie 10+ year member]

I assumed the port tube wall is 1/4" thick when doing the calculations...makes them take up a little more volume that what Bumpin' did.

Depends on if 10" is internal or external. I calculated as if it was external diameter.

Thanks for clarifying, Ryan. //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

 

[walled01alero 10+ year member]

i no how to fine cubic feet, i just didnt understand the flared port stuff...

you didn't say anything about a flared port. and btw the "flare" at the end of the port doesn't actually belong the the port. the volume of air inside the flared sections dont count towards displacement. and when finding port area you have to look at the narrowest part of the port, because it's the narrowest part the determines the restriction in port air velocity. if you're measuring port area at the widest part of the flare just forget that and use the diameter of the tube you have.

8" inner diameter port = 3.14(4in.^2) = 50.24 INCHES SQUARED. look at the units and see if they make sense.. (p.s. 3.14 (or pi) is a dimentionless ratio)

 

[Bumpin' Goalie 10+ year member]

you didn't say anything about a flared port. and btw the "flare" at the end of the port doesn't actually belong the the port. the volume of air inside the flared sections dont count towards displacement. and when finding port area you have to look at the narrowest part of the port, because it's the narrowest part the determines the restriction in port air velocity. if you're measuring port area at the widest part of the flare just forget that and use the diameter of the tube you have.
8" inner diameter port = 3.14(4in.^2) = 50.24 INCHES SQUARED. look at the units and see if they make sense.. (p.s. 3.14 (or pi) is a dimentionless ratio)

x2

 

[thericanballer 10+ year member]

Thanks to everyone for the help.