Math Question

"Richard Riley" wrote in message
...
On Sun, 01 May 2005 03:00:55 GMT, "COLIN LAMB"
wrote:

:Cut the shape of the tank with a piece of balsa or other solid wood.
Then
lace 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.

Great idea. Cheaper with styrofoam.

He'll need to allow for a little less fuel, due to the volume of the tank
material;
but this is exactly the method I might have suggested if I had read the
question
earlier. A great additional benefit is that you end up with a model of the
tank
so that it will be much less likely to be built inverted--if there are
irregular
shapes.

Of course, _I_ would never make such an obvious mistake. ;-

"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.

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

Submerge the tank in a barrel of water. Measure the rise and calculate.

--
Gene Seibel
Confessions of a Pilot - http://pad39a.com/publishing/
Because I fly, I envy no one.

In article .com,
wrote:
Submerge the tank in a barrel of water. Measure the rise and calculate.


Todays Language Lesson:

Classical Greek: Eureka!
the translation: Ye gods, that bathwater is *HOT*!

(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.

Robert Bonomi wrote:

Todays Language Lesson:

Classical Greek: Eureka!

Proper response: you don't smell so great yourself.

Classical Greek: Euripides
Modern translation: you are too fat to be wearing these


Dan, U.S. Air Force, retired

In article ,
alexy wrote:
(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?

the answer to that is 'yes'. as long as you're measuring the 'height'
as the perpendicular distance between the ends.

And can the
shapes be rotated?

Yuppers, but proving it takes calculus. consider an infinite series
of plane sections, parallel to the base. nothing changes if they are
'aligned' relative to each other, or skewed. thus the integral (volume)
will be the same.

(I think the answer to both of those is "yes", but
am I thinking of it correctly?)

It would appear so. grin

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.

eek! you're absolutely correct.

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.

lessee, with all the intermediate steps:
k* b^3/(b-l)*d - k*l^2*(b/(b-l)-1*d
k*(b^3/(b-l)*d - l^2*(b/(b-l)-1)*d)
k*(b^3/(b-l)*d - l^2*(b/(b-l)-b/b)*d)
k*b*(b^2(/b-l)*d - l^2(1/(b-l)-1/b)*d)
k*b*d*(b^2(/b-l) - l^2(1/(b-l)-1/b))
k*b*d*(b^2(/b-l) - l^2(1/(b-l)-((b-l)/(b-1))/b)))
k*b*d*(b^2(/b-l) - l^2(1/(b-l)-(b-l)/((b-1))*b)))
k*b*d/(b-l)*(b^2 - l^2(1-(b-l)/b))
k*b*d/(b-l)*(b^2 - l^2(1-(1-(l/b)))
k*b*d/(b-l)*(b^2 - l^2(l/b))
k*b*d/(b-l)*(b^2 - l^3/b)
k*d/(b-l)*(b^3 - l^3)

Yup. I got it wrong first time. I lost a parenthesization, at step 4. *sigh*

Definitions: USENET -- open mouth, insert foot, echo internationally.


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!

That's cheating! grin

Appreciate that you caught my errors, nonetheless..



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.

In article mZtde.4739$aB.1570@lakeread03,
Dan, U.S. Air Force, retired wrote:
Robert Bonomi wrote:

Todays Language Lesson:

Classical Greek: Eureka!

Proper response: you don't smell so great yourself.

Classical Greek: Euripides
Modern translation: you are too fat to be wearing these


Man walks into a tailor shop.
Proprietor looks up and says: Euripides?
The customer responds: Yes. Eumenides?