can someone crunch the numbers for me?

[super spuder]

If i have a 30 inch piece of 3 inch pipe, how long of propane metering tube will i need in 3/4... how long for 1/2? and about 50 PSI... if thats a good pressure. thanks

[psycix]

4.2% of chamber volume divided by pressure in the metering tube (in bars) + 1 for the atmospheric pressure is the meter volume one should have.

Calculate volumes and lengths by:
pi*pi*radius of pipe*length of pipe = volume
volume/(pi*pi*radius of pipe)= length of pipe

Nobody is likely to crunch the numbers for you, but use the above formulas and you can do it yourself.
Check the spudwiki for more information. There is a great page about fuel meters.

[fogus]

4.2% of chamber volume divided by pressure in the metering tube (in bars) + 1 for the atmospheric pressure is the meter volume one should have.

Calculate volumes and lengths by:
pi*pi*radius of pipe*length of pipe = volume
volume/(pi*pi*radius of pipe)= length of pipe

Nobody is likely to crunch the numbers for you, but use the above formulas and you can do it yourself.
Check the spudwiki for more information. There is a great page about fuel meters.

I'm pretty sure the formula should be:
pi*radius*radius and NOT pi*pi*radius.

You see, the formula for the area of a circle can be derived by calculus:

If we were to inscribe a circle with an equation of (y^2+x^2)^(1/2) = r^2 (a circle with radius r, centred on the origin), the the height of the circumference y (from the x axis) (in quadrant 1 and 2) is given by:

y = (r^2-x^2)^(1/2) (by Pythagorean theorem)

thus the area is:

A/2 = def_int(-r to r){(r^2-x^2)^(1/2)}

using the general trigonometric substitution (a^2+x^2)^(1/2) = (a^2+a*sin(t)^2)^(1/2) (from the identity 1-sin^2(t)=cos^2(t) and restricting t from -pi/2 to pi/2:

let x = r*sin(t)
then dx = r*cos(t)dt
and (r^2-x^2)^(1/2) = (r^2-r^2sin^2(t))^(1/2) = (r^2*cos^2(t))^(1/2) = r*cos(t)

then:

A = 4*def_int(0 to pi/2){r*cos(t)*r*cos(t) dt}
= 4r^2 def_int(0 to pi/2){r*cos^2(t) dt}
= 4r^2 def_int(0 to pi/2){(1/2)(1+cos(2t)) dt}

integrating, we get:

= 2r^2[t + (1/2)sin(2t)]

taking the definite integral t, from = 0 to pi/2

= 2r^2(pi/2 + 0 - 0)
= pi*r^2 (or "pie are squared")

I could be wrong though.

[cowkiller]

you know there are programs on this site and some others that can tell you the exact size of meter based on the specs you have and what pressure you want to use. will post a link if i can find it or if some else has them on hand then please post it.


*EDIT* I found one of the sites it has a lot of helpful programs. http://www.burntlatke.com/calc.html

[Hydra]

You know Im seriously weirded out that fogus could be BOTHERED to write all that stuff lol.
But yeah there are a couple of calculators around on the internet.

[starman]

You see, the formula for the area of a circle can be derived by calculus:

Yes it can but totally unnecessary for our purposes....

[fogus]

I was reviewing for a calculus midterm that day.

[psycix]

You typed ALL THAT just because I typo'ed the formula?
Or are you hungry for those 5 spudbucks?

[fogus]

No, it was because I had a calculus exam the next day and all I could think of was, practise practise practise.

What do I do with the spudbux? Can I order food? I'm not hungry now, but I will be later. I know this from previous experience.

[rp181]

A question about spudbux is sure to lead to a locked thread...

They are just for fun, but you can buy a signature,