Surface Area/Volume Calc Help Needed

MKSheppard · Post by **MKSheppard** » 2007-06-07 01:16am

Using this as an example:

Basically, it's for tanksharp; I'd like to figure out the following:

Total Volume:
The surface area for each facet.

And I'm pretty much beating my head over it

Fingolfin_Noldor · Post by **Fingolfin_Noldor** » 2007-06-07 01:22am

Is that trapezium perpendicular to the surface?

Kuroneko · Post by **Kuroneko** » 2007-06-07 01:50am

Assuming that the opposing faces, except the top and bottom, are congruent and the top is parallel and centered relative to the bottom (and are both rectangular), let h be the height, L,W be the length and width at the bottom and l,w at the top, respectively. Then the areas are:
bottom: LW; top: lw; front/back: s(w+W)/2; left/right: S(l+L)/2,
where the slant heights are: s = sqrt[(L-l)²/4+h²], S = sqrt[(W-w)²/4+h²].

The volume is V = h(LW+lw)/2.
[Edited to correct s/S reversal.]

MKSheppard · Post by **MKSheppard** » 2007-06-07 02:27am

Kuroneko, is that assuming the slope angle on all the angles are the same? What if the front slope is 25 degrees, while the side slopes are 35, and the rear is 15?

Kuroneko · Post by **Kuroneko** » 2007-06-07 02:41am

MKSheppard wrote:Kuroneko, is that assuming the slope angle on all the angles are the same? What if the front slope is 25 degrees, while the side slopes are 35, and the rear is 15?

The volume formula is actually correct no matter how the angles are skewed, by Cavalieri's principle. However, the area formulas require the opposite sides to be congruent (except the top and bottom), so if the front slopes at 25°, the back has to slope at 25° as well, although the sides could slope at an angle different from that of the front and back.

Hawkwings · Post by **Hawkwings** » 2007-06-07 03:09am

Average of top and bottom times height. That'll work for both surface area and volume. For surface area, use lengths of top and bottom, while for area, use surface area of top and bottom.

Hawkwings · Post by **Hawkwings** » 2007-06-07 03:10am

Hawkwings wrote:Average of top and bottom times height. That'll work for both surface area and volume. For surface area, use lengths of top and bottom multiplied by slant height, while for area, use surface area of top and bottom multiplied by perpendicular height.

Lord Zentei · Post by **Lord Zentei** » 2007-06-07 06:20am

Hawkwings wrote:Average of top and bottom times height. That'll work for both surface area and volume.

If that were true, the volume of a pyramid would be:

(BaseArea + TopArea (=0) ) / 2 * Height

= ½ * BaseArea * Height.

Only it isn't

Post by **Darth Wong** » 2007-06-07 12:37pm

Hmm, it's been so many years since I did geometric stuff like volume and area formulas. Let me see if I can crunch this one up:

EDIT: Hey, Kuroneko's formula looks a helluva lot simpler than mine. Was that integration a big waste of time?

EDIT 2: Oh, I see what happened. He assumed that only two of the faces are sloped.

EDIT 3: Somebody please double-check my work. As much as I like to think I'm infallible, I have this sneaking suspicion that I'm not

Post by **LadyTevar** » 2007-06-07 02:12pm

*fills large container with water, sets inside a tub. Dumps odd-shaped trapezoid into container of water, measures the displaced water.*

TADA!!! VOLUME!

j/k... I have no clue.

brianeyci · Post by **brianeyci** » 2007-06-07 03:36pm

Before Calculus I would've spilt that into nine different shapes. Then using trigonometry, I would've found the angles for all the slants and calculated the volume for the pyramids on the corners, the volume of the "skirts" then the volume of the center base using formulas I looked up for a wedge, pyramid and box. Then added.

Well I would've cheated since the pyramids are on a slant and I would assume Cavalieri's Principle. I didn't hear of it until multivariable Calculus (year two)... all the three-dimensional volume we calculated in year one was solids of revolution. But Cavalieri's Principle seems intuitive to me.

Take a look at this Shep, for some intuition.

Post by **Beowulf** » 2007-06-07 05:34pm

I feel compelled to point out that the trapezoid is specified using the base width and length, height, and angles the armor is sloped at front, rear, and sides.

Eris · Post by **Eris** » 2007-06-07 05:53pm

LadyTevar wrote:*fills large container with water, sets inside a tub. Dumps odd-shaped trapezoid into container of water, measures the displaced water.*

TADA!!! VOLUME!

j/k... I have no clue.

Why in the world are you just kidding? If it's a real world solid you're talking about, your method will actually be more accurate than a mathematical derivation, assuming good instruments. If it's a container, you do the same thing inverted by filling it with water and measuring that. Really, this is a standard MO in a chem lab for testing the actual volume of your glassware.

Now, given that this is not a real world object we're dealing with, you have to do the maths. Even so, you suggested what I would have if you hadn't beaten me to it.

I'd comment on the maths too, but it's been two years since I've had to touch calculus in much detail, so I'd probably bungle it more badly than everyone who has come before me.

Post by **LadyTevar** » 2007-06-07 06:15pm

I was just kidding for the reason you mentioned... It's not a real-world object

Post by **Beowulf** » 2007-06-07 06:39pm

Also, why are people using calculus? It's a mostly simple trig problem!

Rekkon · Post by **Rekkon** » 2007-06-07 07:13pm

Calculus is not simple?

Kuroneko · Post by **Kuroneko** » 2007-06-07 09:02pm

Darth Wong wrote:Hey, Kuroneko's formula looks a helluva lot simpler than mine. Was that integration a big waste of time?Oh, I see what happened. He assumed that only two of the faces are sloped.

Ack, I did. I certainly didn't intend to assume it; I wasn't thinking clearly.

Darth Wong wrote:Somebody please double-check my work. As much as I like to think I'm infallible, I have this sneaking suspicion that I'm not.

It's completely correct. As an additional special case, l0w1 = w0l1 iff the shape is a frustum, which has V = [h/3][A0 + A + A1], where A1, A0 are the areas of the top and bottom faces, respectively, and A = sqrt[A0A1] is their geometric mean.

Post by **Surlethe** » 2007-06-08 09:17am

Beowulf wrote:Also, why are people using calculus? It's a mostly simple trig problem!

Because the solution is simpler with calculus. With trig, as far as I can see, you'd have to break up the shape into several separate sections and grind out the volume, as opposed to a single integration as in Darth Wong's proof.

Fingolfin_Noldor · Post by **Fingolfin_Noldor** » 2007-06-08 11:32am

Surlethe wrote:
Beowulf wrote:Also, why are people using calculus? It's a mostly simple trig problem!
Because the solution is simpler with calculus. With trig, as far as I can see, you'd have to break up the shape into several separate sections and grind out the volume, as opposed to a single integration as in Darth Wong's proof.

Erm.. no need. Assuming the length is uniform, and the trapezium sides are perpendicular to the surface, the old volume = surface area *length still works.

Fingolfin_Noldor · Post by **Fingolfin_Noldor** » 2007-06-08 11:35am

Addendum: I do agree that if it isn't then one has to use Integration however, especially if it isn't a symmetry shape.

Post by **Darth Wong** » 2007-06-08 11:44am

Fingolfin_Noldor wrote:Erm.. no need. Assuming the length is uniform, and the trapezium sides are perpendicular to the surface, the old volume = surface area *length still works.

Did you not bother reading the thread? That assumption is false.

PS. If Beowulf is so sure that a trigonometry solution would be simpler, perhaps he could demonstrate by showing us one. The reason for the invention of calculus in the first place was because it made so many kinds of math problems so much easier.

Fingolfin_Noldor · Post by **Fingolfin_Noldor** » 2007-06-08 11:49am

Darth Wong wrote:
Fingolfin_Noldor wrote:Erm.. no need. Assuming the length is uniform, and the trapezium sides are perpendicular to the surface, the old volume = surface area *length still works.
Did you not bother reading the thread? That assumption is false.

PS. If Beowulf is so sure that a trigonometry solution would be simpler, perhaps he could demonstrate by showing us one. The reason for the invention of calculus in the first place was because it made so many kinds of math problems so much easier.

Yes. My bad. Apologies.

Post by **Surlethe** » 2007-06-08 12:58pm

Fingolfin_Noldor wrote:Addendum: I do agree that if it isn't then one has to use Integration however, especially if it isn't a symmetry shape.

Right: the reason calculus is so powerful is it frees you from those constricting assumptions. In this example, a tank shaped like a cylinder or trapezoidal prism wouldn't be much use, and for more detailed predictions about performance, weight, armor strength, etc., you need more accurate values for the surface area of the armor and the volume of the tank.

MKSheppard · Post by **MKSheppard** » 2007-06-08 03:05pm

So using this

And the following Numbers:

Height 50 cm
Base Length (lo) 100 cm
Base Width (wo) 100 cm
Top Length (l1) 100 cm
Top Width (w1) 100 cm

gives me:

Volume 500,000 cm3

Did I do it right, or did I cock it up horribly?

Gil Hamilton · Post by **Gil Hamilton** » 2007-06-08 04:28pm

Doesn't the top length and width of your shape have to be shorter than the base length to achieve that particular object? If the base length and width and the top length and width are 100cm, then it's just a rectangular prism and you arrive at the same number you got by multiplying length, width, and height.