Math Question

Dan, U.S. Air Force, retired wrote:
Matt Whiting wrote:

W P Dixon wrote:

Ok Guys and Gals,
I do not remember the formula for this to save my life, so I will
see if yall can come up with it. Yes I did check on the web, but did
not see the formula I need.
I want to figure the volume of a gas tank that will not be round
or square, It will have five sides and then the two ends of the tank.
With one end being larger than the other. I would give exact
measurements , but being as I don't know what they will be yet I
can't:} I need to find the right volume in order to get the right
measurement . Oh the dilemma !
Be gentle math wizards it's been 25 years since I have had to do
this!


Depending on how irregular the tank shape is, you may have to solve
this using numerical integration. However, if the tank shape is the
same in at least one axis (say z or vertical), then figure the area of
the shape in the x-y plane and then simply multiply times the height,
z, and equate that to the volume you desire. Then solve for z.


Matt


That would work if it had the same cross sectional area along Z. He says
otherwise. This leaves to 3 solutions: 1) build it, fill it and measure
the volume coming out, 2) calculus which would be quickest and easiest
or 3) draw a diagram, cut it into solids you can calculate, then add up
the volume of the solids.

If the small end isn't very much smaller than the big end go ahead and
do it Matt's way and add a fudge factor.

I'd find an ME student at a local university and have them create a
solid model of the tank using SolidWorks, ProE or similar. You can then
get an accurate volume with the press of a key. And you can play "what
if" with the design until the cows come home.


Matt

Matt Whiting wrote:
Dan, U.S. Air Force, retired wrote:

Matt Whiting wrote:

W P Dixon wrote:

Ok Guys and Gals,
I do not remember the formula for this to save my life, so I will
see if yall can come up with it. Yes I did check on the web, but did
not see the formula I need.
I want to figure the volume of a gas tank that will not be round
or square, It will have five sides and then the two ends of the
tank. With one end being larger than the other. I would give exact
measurements , but being as I don't know what they will be yet I
can't:} I need to find the right volume in order to get the right
measurement . Oh the dilemma !
Be gentle math wizards it's been 25 years since I have had to do
this!


Depending on how irregular the tank shape is, you may have to solve
this using numerical integration. However, if the tank shape is the
same in at least one axis (say z or vertical), then figure the area
of the shape in the x-y plane and then simply multiply times the
height, z, and equate that to the volume you desire. Then solve for z.


Matt



That would work if it had the same cross sectional area along Z. He
says otherwise. This leaves to 3 solutions: 1) build it, fill it and
measure the volume coming out, 2) calculus which would be quickest
and easiest or 3) draw a diagram, cut it into solids you can
calculate, then add up the volume of the solids.

If the small end isn't very much smaller than the big end go ahead and
do it Matt's way and add a fudge factor.


I'd find an ME student at a local university and have them create a
solid model of the tank using SolidWorks, ProE or similar. You can then
get an accurate volume with the press of a key. And you can play "what
if" with the design until the cows come home.


Matt

Or a cad student if all you want is volume. I use Micro Station 95 left
over from my old collitch days and what he descibes is simple.
Engineering types can do all kinds of neat analysis so they can make
constructive hints. Just bear in mind most engineering students have no
background in auto repair or similar so their grasp of reality may be
limited. When I was going for my EE (I dropped out in 3rd year) in the
1990s I ran into a bunch of kids going for the "big fix."

Dan, U.S. Air Force, retired

A cad program would be really sweet, I have the Volo viewer and have not
meesed with it enough to do much but just open up a file in it! HAHA Any
good programs , cheap mind you,...I am so broke I can't pay attention!!!!
Would like to have cad prints of the finished deal. Heck I may get real
uppity and copyright my design! Volume is important in the floats because
each compartment has to be as equal as possible. And for those interested
they will be aluminum floats. And thinking of a 750 design (standard) for
small planes ultralights? And a 1000-1100 (retract gear).
Yeah yeah I could buy a kit from somewhere, or just buy them, but where
is the fun in that? Just to simple a structure to buy a set of someone
else's design . So I will make the D-750 and D-1000 model! HAHAHA

Patrick Dixon
student SPL
aircraft structural mech

solid model of the tank using SolidWorks, ProE or similar. You can then

I've got solid works. Give me a yell if you want me to draw it up.

If you can find the area of the ends and average them, then multiply it by
the average distance between them, it will give you the volume. If you use
inches, dividing the total cubic inches by 231 will yield the capacity in US
gallons.




"Matt Whiting" wrote in message
...
W P Dixon wrote:

Ok Guys and Gals,
I do not remember the formula for this to save my life, so I will see
if yall can come up with it. Yes I did check on the web, but did not see
the formula I need.
I want to figure the volume of a gas tank that will not be round or
square, It will have five sides and then the two ends of the tank. With
one end being larger than the other. I would give exact measurements ,
but being as I don't know what they will be yet I can't:} I need to find
the right volume in order to get the right measurement . Oh the dilemma !
Be gentle math wizards it's been 25 years since I have had to do this!



Depending on how irregular the tank shape is, you may have to solve this
using numerical integration. However, if the tank shape is the same in at
least one axis (say z or vertical), then figure the area of the shape in
the x-y plane and then simply multiply times the height, z, and equate
that to the volume you desire. Then solve for z.


Matt

"Cy Galley" wrote:

If you can find the area of the ends and average them, then multiply it by
the average distance between them, it will give you the volume.

That's an excellent approximation, and is probably what I would use
for a calculation in my head or on the back of an envelope. But the
actual formula is just about as easy with a calculator or spreadsheet.

actual formula (assuming ends in parallel planes, I believe) is 1/3 *
h *(b1+b2+sqrt(b1*b2)), where b1 and b2 are the areas of the bases.

Your formula is exact if b1=b2. If one end is 4 times the area of the
other (twice the linear dimensions), your formula overstates the
volume by about 7%. If one end is twod=ce the area of the other, it
overstates the correct answer by less than 2%. So the bases don't have
to be very close in size for your approximation to give pretty good
results.

As a worst case, your approximation approaches a 50% overstatement as
the shape gets close to a "pyramid" in which one end has zero area.
--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

W P Dixon wrote:
Ok Guys and Gals,
I do not remember the formula for this to save my life, so I will see
if yall can come up with it. Yes I did check on the web, but did not see
the formula I need.
I want to figure the volume of a gas tank that will not be round or
square, It will have five sides and then the two ends of the tank. With
one end being larger than the other. I would give exact measurements ,
but being as I don't know what they will be yet I can't:} I need to find
the right volume in order to get the right measurement . Oh the dilemma !
Be gentle math wizards it's been 25 years since I have had to do
this!


Depending on the shape of the tank, you can divide it into prisms (the
area of a triangle, multiplied by the height) and add them up.

This method should work for any tank that can be composed of
non-overlapping easy-to-calculate volumes. Or, if they are overlapping,
you can subtract the intersection betweeen the two overlapping volumes
-- but that starts getting a little hairy, since negative volumes make
my head hurt when I'm working with something real. :-)

If that doesn't work, I'm with the guy who suggested integration. I
would try regular calculus first -- I would use the
rotate-around-the-axis trick (if you've done it, the previous phrase
should make sense?), and then subtract off any volumes that aren't
there. After that, I'd try for for a numerical solution. A program
like Mathematica (http://www.wolfram.com) can make the plug&chug parts
easier -- especially for a numerical solution.

The complexity really depends on how irregular your tankis. If your
tank just has an irregular footprint but is the same height all-around,
(a puzzle piece that is 10" thick), all you need is the area of the
footprint and the height -- multiply them together and you're done.

I hope this helps,
-Luke

P.S. I am not a math wizard! But, I do drink beer with math wizards... :-)

Cut the shape of the tank with a piece of balsa or other solid wood. Then
place it in a tub of water until submerged and measure the volume of water
displaced. If you have something like a laundry sink, it is pretty easy
math.

Even easier is to build it into the shape that will fit into the VP and then
measure the liquid it holds and rewrite the manual.

As they proved with the Hubble telescope, you cannot really trust the math
guys, anyway.

Colin

In article ,
W P Dixon wrote:
Ok Guys and Gals,
I do not remember the formula for this to save my life, so I will see if
yall can come up with it. Yes I did check on the web, but did not see the
formula I need.
I want to figure the volume of a gas tank that will not be round or
square, It will have five sides and then the two ends of the tank. With one
end being larger than the other. I would give exact measurements , but being
as I don't know what they will be yet I can't:} I need to find the right
volume in order to get the right measurement . Oh the dilemma !
Be gentle math wizards it's been 25 years since I have had to do this!



It's *not* a single formula.

That said, it's not a difficult problem to solve.

There are at least two approaches that work for your situation:

1) for the 'general case' --
You 'decompose' the object into some simpler ones, calculate the volume
of each of those simpler ones, and 'add em up'.

the 'simpler' forms you want to get to a
A) a cylinder -- top/bottom are parallel to each other, and the same size,
shape doesn't matter, as long both are identical.
B) a wedge -- bottom a parallelogram, tapers to a _line_ at the top
C) a cone -- bottom any shape, tapers to a point at the top.

Then, you simply have to remember one formula for each form:
A) volume of cylinder == area of base * height
B) volume of 'wedge' == (area of base * height)/2
C) volume of 'cone' == (area of base * height)/3


2) for the 'special case' where the two ends are the same shape, differing
only in size, and are parallel to each other --

You can 'extend' the edges that join the ends, until those edges meet.
(they are guaranteed to all come together at a single point.)

You now have a *cone*. You can calculate the volume between the two
"ends" by calculating the volume of the entire cone above the larger
end, and then subtracting the volume of the cone above the smaller end.

the 'height' of the cone, from the 'big end', can be computed by
using _any_ *linear* dimension of the big end, and the corresponding
dimension from the little end, as follows:

To keep things manageable, lets invent some names:
b is a linear dimension of the big end
l is a linear dimension of the little end
d is the distance between the ends, measured perpendicular to the
planes of the ends.
c is the distance that the 'convergence point' is above the big end
A(b) is the _area_ of the big end
A(l) is the _area_ of the little end

c = b/(b-l) * d

so, the volume of the full cone above the big end is:
A(b)*c
and the volume of the full cone above the little end is:
A(l)*(c-d)
thus, obviously, the volume of the 'truncated cone', between the
two ends is:
A(b)*c - A(l)*(c-d)
eliminating the intermediate calculation, by back-substituting for c, gives:
A(b)*(b/(b-l)*d) - A(l)*((b/(b-l)-1)*d)
The area for any given shape is proportional to the square of the
linear dimension of that shape, differences in shape are accounted for
by a difference in the proportionality constant, *only*.

so we get:
k*b^3/(b-l)*d - k*l^2*(b/(b-l)-1)*d
or
k*b*d/(b-l)*(b^2 - (l^2+l-b))

doing it "step-wise" (computing height of cone, and end areas, separately)
is easier grin

(Robert Bonomi) wrote:



2) for the 'special case' where the two ends are the same shape, differing
only in size, and are parallel to each other --
This seems to be the case the OP was asking about. At least that is
the prevailing assumption in this thread (which of course does not
make it valid). This specificity is good, since I had been falling
into the trap of assuming this formula for any two end shapes, not
just identical ones leading to a "cone". BTW, is it accurate to say
that this does not have to be a frustrum of a right cone? And can the
shapes be rotated? (I think the answer to both of those is "yes", but
am I thinking of it correctly?)


You can 'extend' the edges that join the ends, until those edges meet.
(they are guaranteed to all come together at a single point.)

You now have a *cone*. You can calculate the volume between the two
"ends" by calculating the volume of the entire cone above the larger
end, and then subtracting the volume of the cone above the smaller end.

the 'height' of the cone, from the 'big end', can be computed by
using _any_ *linear* dimension of the big end, and the corresponding
dimension from the little end, as follows:

To keep things manageable, lets invent some names:
b is a linear dimension of the big end
l is a linear dimension of the little end
d is the distance between the ends, measured perpendicular to the
planes of the ends.
c is the distance that the 'convergence point' is above the big end
A(b) is the _area_ of the big end
A(l) is the _area_ of the little end
To keep it even more manageable (for me) I used "s" for "small"
instead of "l" for "little", to keep me from assuming that (k-1)*(k+l)
is k^2-1. g

c = b/(b-l) * d

so, the volume of the full cone above the big end is:
A(b)*c
You forgot the 1/3 factor. In a sense, it gets "picked up" in your k
later, but probably better not to mix the two, IMHO.

and the volume of the full cone above the little end is:
A(l)*(c-d)
thus, obviously, the volume of the 'truncated cone', between the
two ends is:
A(b)*c - A(l)*(c-d)
eliminating the intermediate calculation, by back-substituting for c, gives:
A(b)*(b/(b-l)*d) - A(l)*((b/(b-l)-1)*d)
The area for any given shape is proportional to the square of the
linear dimension of that shape, differences in shape are accounted for
by a difference in the proportionality constant, *only*.

so we get:
k*b^3/(b-l)*d - k*l^2*(b/(b-l)-1)*d
or
k*b*d/(b-l)*(b^2 - (l^2+l-b))
I don't follow this step. And the term on the right does not add up
dimensionally.

I think this step should be k/3*d/(b-s)*(b^3-s^3) (I also reinjected
the 1/3 factor here, so my k is 3 times as large as yours.)

Dividing by (b-s) gives me 1/3 *d *k *(b^2+bs+s^2)
And substituting back to the easily calculated or measured areas using
A(b) = kb^2 and A(s) = ks^2, we get
1/3 *d *(A(b)+sqrt(A(b)*A(s))+A(s))
Which conveniently agrees with the formulas in my CRC Standard
Mathematical Tables book. Glad those formulas haven't changed in the
last 35 years!

doing it "step-wise" (computing height of cone, and end areas, separately)
is easier grin



--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.