Math Question

[W P Dixon]

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!
;)

--
Patrick Dixon
student SPL
aircraft structural mech

[Dave S]

sounds like you will need to just fill the tank and measure the output.

Dave

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! ;)
>

[Matt Whiting]

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

[Dan, U.S. Air Force, retired]

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.

Dan, U.S. Air Force, retired

[W P Dixon]

Hee Hee,
No simple answer huh? ;) Thanks guys, I looked more on the web late last
night and I do think I will have to make the shapes into smaller measureable
shapes and add the totals. I do think figuring up something before you
actually build it is alot cheaper,...you don't have to build it but once.
Well we all hope anyway! :)
Also planning to build a set of floats and that's where the volume
formulas really get funky. I would sure hate to spend a grand just to fill
it with water and say, well not right can't use it. Heck my old lady would
kill me if I wasted 200 bucks on a ruined gas tank! HAHA
It won't be to bad figuring it all up "cutting it into basic shapes" ,
just will take some time. For the gas tank, it will be in a VP-1. I am
welding aluminum instead of using the fiberglass. An old high school buddy,
certified nuclear welder is going to weld it up for me. So I need to send
him a drawing of it, thus the need for getting it right. That math stuff is
pretty cool when you can remember the formulas ain't it? :) So for the gas
tank, I just wanted to see how much fuel a aluminum tank would hold with
alittle mod. But the floats , I definitely have to know the volumes of each
compartment before I even think of starting the build there.

Patrick
student SPL
aircraft structural mech

[Luke Scharf]

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... :-)

[Luke Scharf]

W P Dixon wrote:
> Also planning to build a set of floats and that's where the volume
> formulas really get funky.

I just got my seaplane rating. I spent some time looking at the floats
-- it looks like the only way I could calculate the volume of the floats
I was looking at would be a double-integral. Elegant, but could still
be tricky.

One thing to keep in mind is the place where the float will contact the
water at various attitudes. If the contact-point is too far forward
(either because of the attitude of the aircraft or because of the design
of the float), you're flying a taildragger in a soft-sticky-massive
substance that is many times more dense than air... Scary!

Also, floats a have many effects on the aerodynamics of the aircraft.
The side area of the craft is different when it has floats -- a lot more
aerodynamic stuff happening in front of the CG, which can require a
bigger rudder. Also, the instructor told me that in a plow turn, the
change in the amount of the float exposed to the wind was one of the
things that makes the aircraft turn downwing. Lastly, the mass of the
floats would probably change the CG around a bit too.

-Luke

[LCT Paintball]

"Richard Riley" > wrote in message
...
> On Sat, 30 Apr 2005 00:26:32 -0400, "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!
> :;)
>
> I'm not engineer, and I'm not currently playing one on TV, but I think
> it's straightforward. Just so we're clear, I'm assuming -
>
> Each end is a pentagon, each one of *it's* sides is equal length. The
> two ends are parallel to each other. One pentagon is larger than the
> other. They are perpendicular to a line drawn from the center of one
> to the center of the other.
>
> First, find the area of the large pentagon. The formula is
>
> (the length of one side) squared * 1.7
>
> Then multiply by the length of the tank to get the volume if both the
> ends were the size of the large one.
>
> Then do the same thing with the small end.
>
> Now you have 2 volumes. Add them together, divide by 2.
>
> So, if one end is a pentagon with sides that are 8 inches long, and
> the other has sides that are 6 inches, and it's 24" long
>
> Area of end one - 108.8
> Area of end two - 61.2
>
> Volume 1 2611.2
> Volume 2 1468.8
>
> Average 2040
>
> Total 8.83 gallons.

This looks like the winning formula to me. ;)

[Matt Whiting]

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

[Dan, U.S. Air Force, retired]

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

[Dan, U.S. Air Force, retired]

W P Dixon wrote:

> Hee Hee,
> No simple answer huh? ;) Thanks guys, I looked more on the web late
> last night and I do think I will have to make the shapes into smaller
> measureable shapes and add the totals. I do think figuring up something
> before you actually build it is alot cheaper,...you don't have to build
> it but once. Well we all hope anyway! :)
> Also planning to build a set of floats and that's where the volume
> formulas really get funky. I would sure hate to spend a grand just to
> fill it with water and say, well not right can't use it. Heck my old
> lady would kill me if I wasted 200 bucks on a ruined gas tank! HAHA
> It won't be to bad figuring it all up "cutting it into basic shapes"
> , just will take some time. For the gas tank, it will be in a VP-1. I am
> welding aluminum instead of using the fiberglass. An old high school
> buddy, certified nuclear welder is going to weld it up for me. So I need
> to send him a drawing of it, thus the need for getting it right. That
> math stuff is pretty cool when you can remember the formulas ain't it?
> :) So for the gas tank, I just wanted to see how much fuel a aluminum
> tank would hold with alittle mod. But the floats , I definitely have to
> know the volumes of each compartment before I even think of starting the
> build there.
>
> Patrick
> student SPL
> aircraft structural mech

Nothing says you have to build a full size model. Make a fiberglass
model at 1/8 to 1/4 the size you SWAG, fill it with water and measure it
out. You can now scale as needed mathematically.

Don't forget to take into consideration material thickness, baffle
thickness etc as you plan.

Dan, U.S. Air Force, retired

[W P Dixon]

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

[W P Dixon]

Now that may not be a bad idea, and it would be pretty cool to build a RC
plane to go on top too! Dang Dan now I have to build another toy!!!!! My
wife is gonna shoot me for sure! ;)
Seriously , that is not a bad idea, may just do that. Will check out the
math stuff first. If it befuddles me to bad I may have to resort to a plan
B, or Plan Dan! ;)

Patrick
student SPL
aircraft structural mech

"Dan, U.S. Air Force, retired" > wrote in message
news:LHOce.1573$aB.391@lakeread03...
>W P Dixon wrote:
>
>> Hee Hee,
>> No simple answer huh? ;) Thanks guys, I looked more on the web late last
>> night and I do think I will have to make the shapes into smaller
>> measureable shapes and add the totals. I do think figuring up something
>> before you actually build it is alot cheaper,...you don't have to build
>> it but once. Well we all hope anyway! :)
>> Also planning to build a set of floats and that's where the volume
>> formulas really get funky. I would sure hate to spend a grand just to
>> fill it with water and say, well not right can't use it. Heck my old lady
>> would kill me if I wasted 200 bucks on a ruined gas tank! HAHA
>> It won't be to bad figuring it all up "cutting it into basic shapes" ,
>> just will take some time. For the gas tank, it will be in a VP-1. I am
>> welding aluminum instead of using the fiberglass. An old high school
>> buddy, certified nuclear welder is going to weld it up for me. So I need
>> to send him a drawing of it, thus the need for getting it right. That
>> math stuff is pretty cool when you can remember the formulas ain't it? :)
>> So for the gas tank, I just wanted to see how much fuel a aluminum tank
>> would hold with alittle mod. But the floats , I definitely have to know
>> the volumes of each compartment before I even think of starting the build
>> there.
>>
>> Patrick
>> student SPL
>> aircraft structural mech
>
> Nothing says you have to build a full size model. Make a fiberglass model
> at 1/8 to 1/4 the size you SWAG, fill it with water and measure it out.
> You can now scale as needed mathematically.
>
> Don't forget to take into consideration material thickness, baffle
> thickness etc as you plan.
>
> Dan, U.S. Air Force, retired

[W P Dixon]

Check out this site it has a bunch of volume calculators on it. This may do
the trick, especially since my math skills are in the lacking
department....from lack of
use.http://grapevine.abe.msstate.edu/~fto/tools/vol/index.html

Patrick
student SPL
aircraft structural mech

[alexy]

Richard: Good post, up until the end, where you used the wrong volume
formula.

Richard Riley > wrote:

>I'm not engineer, and I'm not currently playing one on TV, but I think
>it's straightforward. Just so we're clear, I'm assuming -
>
>Each end is a pentagon, each one of *it's* sides is equal length. The
>two ends are parallel to each other. One pentagon is larger than the
>other. They are perpendicular to a line drawn from the center of one
>to the center of the other.

Good summary of the relevant assumptions!

>First, find the area of the large pentagon. The formula is
>
>(the length of one side) squared * 1.7
I used 1.7205 in my calcs, just to get additional significant digits
in the result.

>Then multiply by the length of the tank to get the volume if both the
>ends were the size of the large one.
So far, so good.

>Then do the same thing with the small end.
>
>Now you have 2 volumes. Add them together, divide by 2.
But this is where your formula falls down. You are basically saying
1/2*h*(B1+B2) where B1 and B2 are the areas of the bases. The correct
formula is 1/3*h*(B1+B2+sqrt(B1*B2)). There is no difference if the
two ends are the same size, and the difference grows with the size
differential between the ends to the limiting case of the tank coming
to a point (zero area "base") where your formula overstates the
correct volume by 50%.

As a "gut check", note that the volume of a cone is 1/3*h*pi*r^2.

For the pentagon examples, this breaks down to .5735*h*(s1^2 +s2^2
+s1*s2)

>Average 2040

And after that, we got almost identical results. Your use of 1.7 in
the pentagon area fromula rather than more significant digits almost
exactly offset the relatively small error in the formula, with ends
this close to the same size. I calculated: .5735*24*(64+36+48)=2037.
>Total 8.83 gallons.
Same
--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

[LCT Paintball]

> 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.

[Cy Galley]

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

[COLIN LAMB]

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

[Morgans]

"W P Dixon" > wrote

.. For the gas tank, it will be in a VP-1. I am
> welding aluminum instead of using the fiberglass. An old high school
buddy,
> certified nuclear welder is going to weld it up for me.

Why not make it round on the ends, and just wrap it, instead of having flat
sides? Less welds to leak. That is, if you are welding each flat to the
other. If you are going continuous, nevermind! <g>
--
Jim in NC

[W P Dixon]

Jim,
If I were welding it, it wouldn't be welded!!! HAHA It would be riveted
and sloshed. However the fellow that will be doing the welding welds at
nuclear facilities and has an xray certification. I know his work very well,
and it won't leak.

Patrick
student SPL
aircraft structural mech
"Morgans" > wrote in message
...
>
> "W P Dixon" > wrote
>
> . For the gas tank, it will be in a VP-1. I am
>> welding aluminum instead of using the fiberglass. An old high school
> buddy,
>> certified nuclear welder is going to weld it up for me.
>
> Why not make it round on the ends, and just wrap it, instead of having
> flat
> sides? Less welds to leak. That is, if you are welding each flat to the
> other. If you are going continuous, nevermind! <g>
> --
> Jim in NC
>

[Peter Dohm]

"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
> :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.
>
> 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. ;->

[alexy]

"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.

[Robert Bonomi]

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 are:
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>

[[email protected]]

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.

[Robert Bonomi]

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*!

[alexy]

(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.

[Dan, U.S. Air Force, retired]

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

[Robert Bonomi]

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.

[Robert Bonomi]

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?