# Can I Have Some Help From All The Budding Einsteins? (math equation!

Discussion in 'Off Topic [BG]' started by slick519, Aug 31, 2002.

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1. ok, here is the story. I am building a spudgun for my friend, and to build a good one, the volume of the chamber needs to be 1.5 times the volume of the barrel. the part where i am stuck is that after i get the volume of the chamber, how do i get dimensions from only the volume of the chamber? (I need to know what the radius and the length of the chamer need to be)

ok, ill will put my problem in a story problem to clarify

if susie has a 45" long barrel with a radius of 1 inch, and a volume of 141.37167 cubic in. and a combustion chamber with a volume of 212.0575 cubic in, how long is the chamber, and what is its radius?

well, thanks for the help, and please tell me if this wouldnt even work!! I am not even sure that this is possible, but i am no means a math guru!

thanks! slicks

2. ### Selta

Feb 6, 2002
Pacific Northwet
Total fanboi of: Fractal Audio, AudiKinesis Cabs, Dingwall basses
This problem has an indifinite amount of answers, 'cos you can substitue radius with volume, like (just as an example, this sint an asnwer) for having something with 100 cubic feet, you can have 10x10x10 or 20x5x1 etc.etc. We need more details here! Like, what are the restrictions etc.!

3. ### Nick GannTalkbass' Tubist in Residence

Mar 24, 2002
Silver Spring, MD
first off...

the area of a cylander of (pi R ^2) x height (or length)

so a 45 inch pipe with a radius of 1 inch would have a volume of 444.132198 in^2

not 141.37167

________

so that times 1.5= 666.1982971 in^2

so the chamber would have a volume of that number up there...

to find the length, we need to know what the radius will be. If you have the same radius, it will be...

666.1982971 = pi R(which is one)^2 * X

so if you do the algebra, then X will equal 67.5.

if you do a one inch raduis pipe for the chamber, you will need a 67.5 inch pipe.

a 2 inch raduis pipe will be 53.01437603 inches
round it to 53 inches

a 3 inch raduis pipe (6 inch wide) will need to be 23.05619449 inches
round it to 23.5 inches
___________________

ok?

4. ### Nick GannTalkbass' Tubist in Residence

Mar 24, 2002
Silver Spring, MD
if you have any other questions about geometry, just ask. I was good at that. 5. Actually, Einstein never was a mathmatician...  6. that may be so, but i bet that einstein could solve my problems slicks

7. okay you had me up to the "susie" part. is your name susie?

is there something you would like to tell us?

8. Are you sure about that?

(3.14 * 1^2) * (45) =
(3.14 * 1) * (45) =
(3.14) * (45) = 141.3 inches cubed

Munjibunga likes this.
9. You don't need math to solve your problem.

Cut the pipe for the barrel and temporarily cap one end. Fill the pipe with water and measure the volume of water with a measuring pitcher.

Lets say that the volume is one quart.

Start with the pipe for the chamber longer than you'll need. Cap one end of the chamber pipe and pour in one and one half quarts of water. Mark the water level in the chamber and cut the pipe.
That will give you a one and a half to one ratio.

Are you building a pneumatic or hair spray cannon?

Pkr2

PS-who's Susie?

Will_White and blastoff99 like this.
10. I've had good luck sticking to a 3:1 ratio when it comes to Combustion Chamber : Barrel diameters.

And I wouldn't use 2" Diameter Pipe for the barrel....your projectile size gets larger..and gas velocity goes down. 1.25" Sch40 for a barrel, 4" for the combustion chamber.

11. Ok time to whip out the old graphing calculator and complicate the hell out of this problem a bit.

first for the initial math....

(3.14 * 1^2)45 = 141.37in^3

(1.5)(141.37in^3) = 212.06in^3

we then use the formula for the volume of a cylinder
to set up an equation....

212.06in^3 = (3114 * X^2)Y

where x is the radius of the pipe and y is the length.We then solve for y....

y = 212.06in^3 / (3.14 * X^2)

And now we graph it on our trusty Ti-83 and use the trace function to find the lengths needed for the most common pipe diameters.

4" pipe = 16.88 in
3.5" pipe = 22.04 in
3" pipe = 30.00 in
2.5" pipe = 43.20 in
2" pipe = 67.50 in
1" pipe = 270.00 in

there now the problem is solved and we get to pat ourselves on the back for using an overly high tech method for solving a simple problem Will_White likes this.
12. sweet!! thanks alot for all the help!

thanks alot,
slicks

13. 42. Just 42.

Rock on
Eric

15. Hmmm, that makes me think - why not use a taper bore type barrel, should get some nice velocities out of your projectile!!!! But then you'd need to use the equation for a cone............... 16. isn't that just 1/3(h)(pi)(r^2) ?

if it's not, well... ...then i sure look like a jack-ass!

17. Looks good to me! 18. two dollars and fifteen cents

19. Is anyone else starting to worry about the OP making a deadly mistake with the calculations of this potato gun back in 2002?

Munjibunga likes this.
20. ### Gravedigger DavSupporting Member

Mar 13, 2014
Fort Worth, Texas