Cubic Footage of a Tapered Log

Date: 08/07/2002 at 18:42:15
From: Christopher Watkins
Subject: Cubic Footage of a Tapered Log

Is there an equation that will calculate the cubic footage of a log 
given the diameters of each end and the length?

Currently, I am using Smalian's Formula:
(Area1 + Area2)/2 x Length (ft)
Area = .005454 x diameter^2

Is there a formula that is more accurate by accounting for the taper 
in the log?

Example:
Diameter1 = 9.2 inches
Diameter2 = 10.3 inches
Length    = 102.84 inches
Smalian's calculates 4.443 cubic feet.

Is this the best answer?

Thank you for your time.

Date: 08/07/2002 at 23:06:05
From: Doctor Peterson
Subject: Re: Cubic Footage of a Tapered Log

Hi, Christopher.

A "tapered cylinder" like this is properly called a "frustum of a 
cone," and the formulas for it can be found in the Dr. Math Formulas 
FAQ:

http://mathforum.org/dr.math/faq/formulas/faq.cone.html#conefrustum 

  V = Pi(R^2+rR+r^2)h/3

For your numbers, this gives

  V = 3.14(4.6^2 + 4.6*5.15 + 5.5^2)*102.84/3 = 7686 in^3

Dividing this by 12^3, we get 4.448 ft^3. So your approximation is 
pretty accurate in this case - particularly considering that a log is 
probably not exactly conical anyway.

Out of curiosity, I looked up "Smalian" to see where your formula 
comes from. It's clear why it would be close, since it is simply the 
average of the volumes of cylinders with the maximum and minimum 
diameters. The following page

Week 2 Mensuration Notes - Hans Zuuring, School of Forestry,
   Univ. of Montana
http://www.forestry.umt.edu/academics/courses/For202/Week2_Notes.htm 

explains that Smalian's formula applies to a frustum of a paraboloid, 
which is slightly convex compared to a cone. It also gives a formula 
equivalent to ours for the frustum of a cone, and one ascribed to 
Newton that applies more generally, by using a measurement at the 
middle to account for the actual curvature of the taper. I haven't 
looked into just when each of these would be exact. This page gives 
some more detail you may find useful:

http://sres.anu.edu.au/associated/mensuration/BrackandWood1998/ 
  Select LOGS.HTM (Measurement of logs), and then volume.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 08/08/2002 at 11:33:12
From: Christopher Watkins
Subject: Thank you (Cubic Footage of a Tapered Log)

Dr. Peterson,
Thank you so much for your time, and for the quick lesson.
For your information, I work for a timber company, and a 
computer scanner at our lathe (plywood) is calculating a 
higher cubic ftg number than I was physically calculating.  
I believe that this is due to an offset in the computer 
system, but I wanted to make sure that I was using the most 
accurate formula before I submitted my report.

I really appreciate your time, and apologize for not 
finding the formula before writing.  I was not aware that 
the situation or example is termed a "frustum".  I was 
looking in other areas.  
Wonderful Job!!

Date: 08/08/2002 at 12:09:59
From: Doctor Peterson
Subject: Re: Thank you (Cubic Footage of a Tapered Log)

Hi, Christopher.

I was intrigued by the formulas I found researching your question, 
because I found information about formulas for the parabolic frustum 
only in forestry sites, which didn't make it clear whether they are 
exact or approximations (since they seem so different). Our site 
doesn't even list the formula for volume of a segment of a paraboloid,

    V = pi r^2 h/2

Looking a little deeper than I did last night, I found that 
"Smalian's formula" is correct for a parabolic frustum

    V = pi (r1^2 + r2^2) h/2
or
    V = h(A1 + A2)/2

(where h is the height, r1 and r2 are the bottom and top radii, and 
A1 and A2 are the bottom and top cross-section areas), while "Huber's 
formula"

    V = pi rm^2 h
or
    V = h Am

(where rm and Am are the radius and area of the section halfway 
between the top and bottom) is also correct, because for a paraboloid

    rm^2 = (r1^2 + r2^2)/2

Further, "Newton's formula"

    V = pi (r1^2 + 4rm^2 + r2^2) h/6
or
    V = h(A1 + 4Am + A2)/6

is valid for both the parabolic frustum, the conical frustum, and in 
fact any shape (called a general prismatoid) whose cross-sectional 
area follows a linear, quadratic, or cubic law, which also includes 
spherical segments and more:

      http://mathworld.wolfram.com/GeneralPrismatoid.html 

All this is beyond what you need, of course; I just wanted to fill in 
some gaps in my list of formulas.

If you need any further help figuring out where the errors are 
arising, feel free to send me additional information.


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/