any of you good in math?

[skife]

this is pretty tricky.


this is made of .050 sheet
has to be 1/64 tolerence.

took awhile for some of us to figure it out.

[paintballfiend]

No, numbers confuse me.

[MoeMag]

Dont want to mess with "tolerence" right now but I think X should be about 9.40864 inches.

("=inch)

LMK


arrrg messed up the # of ci in a quart.

[skife]

Dont want to mess with "tolerence" right now but I think X should be about 9.40864 inches.

("=inch)

LMK

nope.

the top is a fulcrum(sp?) of a cone.

[Pacifist_Farmer]

by my figurings x=1.2056" for volume.

My calculations relied on the supplied dimensions being the internal volume of the container, not the outer dimensions.


If those were the outer dimensions, my number will be off.

"frustrum of a cone"

[skife]

by my figurings x=1.2056" for volume.

My calculations relied on the supplied dimensions being the internal volume of the container, not the outer dimensions.


If those were the outer dimensions, my number will be off.

"frustrum of a cone"

frustrum of a cone, thats right :P

its the outside dimesnsions.

For inside dimensions we got like 4 13/16

[Pacifist_Farmer]

4 13/16? That seems high to me. Is my math wrong?

2.5 qt (fluid) = 144.375 in^3
Vol.FC=0.2618(h)(D^2+Dd+d^2)
h=sqrt(s^2-a^2)
s=4"
a=(D-d)/2 ::Thats 1/2 the difference of the diameters
D=5
d=3
Vol.FC=0.2618(3.8729)(5^2+5*3+3^2)
Vol.FC=49.6823 in^3

Crap, my bad, I used the diameter of the cylinder in the next equation not the radius.

Vol.CYL.=PI*r^2*x
Vol.CYL.=3.14159*(2.5)^2*x

Vol.FC + Vol.CYL. = 144.375

x=4.82

You're right, but that doesn't take into consideration that fact that if the dimensions given are actually O.D. dimensions, the 5" diameter is actually 4.9"+-Tolerance.

[MoeMag]

Round 2.

2.5 qt (fluid) = 144.375 in^3

[jenarelJAM]

^^^ Cheater :P

[MoeMag]

^^^ Cheater :P

Still had to do the math by hand to find x.

[Lenny]

Wait! Find "x"?

I win!

[Swampy]

I can convert liters to gallons but that confuses me please explain.

[Miscue]

Hurm... what else do I have going on at 3am...

Got the same answer... x=4.823"

<a href="http://photobucket.com" target="_blank"><img src="http://i140.photobucket.com/albums/r20/aomiscue/diagram.jpg" border="0" alt="Photobucket - Video and Image Hosting"></a>

I couldn't remember the frustum formula... so I did it the longer way. I also didn't take into account OD... and I'm going to sleep now.

[Miscue]

No it's not within tolerance. You changed the diameter from 5" to 4.9".

[Miscue]

Speaking of right cone frustums... this is how the formula is derived if interested.

<a href="http://jwilson.coe.uga.edu/EMT725/Frustum/Formulas.html">http://jwilson.coe.uga.edu/EMT725/Frustum/Formulas.html</a>