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Mar 28, 2014
04:36 PM

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Mar 28, 2014
04:36 PM

Model with simple frustrum - volume calculation wrong

I make a lot of models that involve pyramidal frustrums (transition of a rectangular section to another rectangular section, linearly). Rather than use the measured value calculated by Creo, I thought I'd have a go at calculating the volume directly from the dimensions.

When I check my calculations, the volume calculated directly from the dimensions differs from the Creo analysis result.

I minimized the accuracy value, but still get the exact same result.

Here are my simple values

Base rectangle: 5 X 3 -> Base area: 15

Top rectangle: 4 X 2 -> Top area: 8

Height: 1

Frustrum of a pyramid volume is

Volume = ( Area Base + Area Top + SQRT ( Area Base * Area Top ) ) * Height / 3 = 11.3182

Creo, on the other hand, gives a volume of 11.3333.

Anyone have any insights into this?

This is simple shape, and kind of scares me about what might be happening with more complicated geometry.

Solved! Go to Solution.

1 ACCEPTED SOLUTION

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Mar 28, 2014
07:56 PM

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Mar 28, 2014
07:56 PM

I think Creo is right.

I'm not sure where the formula came from but ...

L = l_base - ((l_base-l_top)/delta_z) * Z

W = w_base - ((w_base-w_top)/delta_z) * Z

A = L * W =

l_base*w_base

- (l_base*((w_base-w_top)/delta_z) + w_base*((l_base-l_top)/delta_z)) * Z

+ ((l_base-l_top)/delta_z)*((w_base-w_top)/delta_z)*Z*Z

Integrating:

V =

l_base*w_base*Z

- (l_base*((w_base-w_top)/delta_z) + w_base*((l_base-l_top)/delta_z)) * Z*Z/2

+ ((l_base-l_top)/delta_z)*((w_base-w_top)/delta_z)*Z*Z*Z/3

Plugging it all in I get 11.33...

6 REPLIES 6

Mar 28, 2014
07:56 PM

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Mar 28, 2014
07:56 PM

I think Creo is right.

I'm not sure where the formula came from but ...

L = l_base - ((l_base-l_top)/delta_z) * Z

W = w_base - ((w_base-w_top)/delta_z) * Z

A = L * W =

l_base*w_base

- (l_base*((w_base-w_top)/delta_z) + w_base*((l_base-l_top)/delta_z)) * Z

+ ((l_base-l_top)/delta_z)*((w_base-w_top)/delta_z)*Z*Z

Integrating:

V =

l_base*w_base*Z

- (l_base*((w_base-w_top)/delta_z) + w_base*((l_base-l_top)/delta_z)) * Z*Z/2

+ ((l_base-l_top)/delta_z)*((w_base-w_top)/delta_z)*Z*Z*Z/3

Plugging it all in I get 11.33...

Mar 28, 2014
09:40 PM

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Mar 28, 2014
09:40 PM

I should be able to simplify the integral to be the same as the original formula, but so far no luck. I looked at the Wikipedia page that showed the derivation of the original formula, but it doesn't yield terms I can yet convert.

I get 11.3182 from the original formula, but don't see the reason for the discrepancy.

Mar 29, 2014
12:48 AM

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Mar 29, 2014
12:48 AM

I got the original formula from the CRC math tables. A handy book from olden times which has all sorts of nice answers to unusual geometry questions. It's an odd conundrum. One would, reasonably, expect any sound mathematical method to yield the same results.

I can see where your integration comes from, and how it gets the number Creo gets. But still, I wonder if I'm missing something crucial about the frustrum...

Mar 29, 2014
12:19 PM

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Mar 29, 2014
12:19 PM

I put the integration solution and original into Excel and tried some experiments.

Case 1: the top and bottom the same; the original formula and mine agree.

Case 2: the top dimensions both zero; the original formula and mine agree.

Case 3: the top length equal to the bottom and the top width to zero the formulas disagree. Since it's simulating a triangle section, the volume should be 1/2 that of Case1. Mine is, but the original formula gives 1/3.

What is interesting is that the original formula doesn't distinguish between the case where the top surface area is zero because it is a point and the case where it is a line.

It may be that the original formula is built on the assumption that the sides all could extend to a single point and is shape independent while the integration handles sides of differing slopes, but only rectangles.

Mar 29, 2014
02:08 PM

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Mar 29, 2014
02:08 PM

I think I figured it out. The frustrum equation, though not described as such, must only apply to a square pyramid, i.e. where L and W are the same for each height, z. When that assumption is applied to the integrated solution, the resultant equation collapses to the one from the CRC tables.

I wish I'd noticed that earlier.

Jul 11, 2017
07:53 AM

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Jul 11, 2017
07:53 AM

Just in the interest of completeness, I needed to document this for some training stuff, so I have a nice illustration of the problem with the mathematical solution employed. Attached is picture of the results.