The Answer Archive
Providing students and professionals with answers to their toughest software, technology, math, and science problems. Browse the Archive!
Search Questions and Answers:

Show only unanswered questions

Real-World Carpentry and Trigonometry

Posted by admin on 07 Nov 2008 at 12:00 AM
$5.00

Math and Statistics / Trionometry

Dear ,

I'm stumped in trying to come up with a formula to calculate the
height of an arc at the midpoint of the chord that defines it knowing
only the length of the arc and the length of the chord. I have
attempted to figure it out with trig formulas with no success. I have
formulas for similar calculations to find the radius of a given an arc
given its width and height but am stumped on this one.

I use these formulas for calculations in my custom millwork and
construction business. If one of the formulas you have in your
"segments of circles" section covers this, I wasn't able to figure out
how to apply the formula. It may be in the "example #1" since (if I
understand it) I know "s" and "c" and need to come up with "h" but I
don't understand the part about for "x" numerically as shown in your
text:

Case 1: You know s and c. Then solve
c/s = sin(x)/x,

for x, which must be done numerically (see below). Then
theta = 2x,
r = s/theta,
d = r cos(x),
h = r - d,
K = r2[theta-sin(theta)]/2.

I apologize if my question as written is confusing, but this problem
is keeping me up nights trying to figure it out!

Thanks,
Matt Jackson

Accepted Answer:

Answer provided by admin on 07 Nov 2008 at 12:00 AM

Hi, Matt.

The page you reference is:

Segments of Circles:
The Arc, Chord, Radius, Height, Angle, Apothem, and Area

Yes, it's case 1 that you want, and it is one of the difficult cases.
There is no closed-form solution; you must use a method of successive
approximations. Did you click on the link "see below"? It gives an
example of how to solve the equation sin(x)/x = k, where (in case 1)
k is c/s. Once you have the solution for x, you can plug this value
of x into the formula below:

h = r - d
= s(1 - cos(x))/(2x)

Thanks, Doctor Rick, for the (very) quick response!

I saw the "see below" section on your site but haven't made the
connection on how to apply it. Would it be possible to show me an
example with a "simple" problem, say, with an arc length of 52 and a
chord length of 51 or something? I could then work through real-world
problems that I encounter by substituting different numbers in place
of the numbers in your example. If it is too much typing, I
understand, so don't work overtime just for this! Life's too short :-)

Either way, thanks for your help... I am much further along than I
was before your reply. And I think I will sleep better tonight knowing
that there isn't some real easy solution that I was just not thinking
of.

Respectfully,
Matt Jackson
The Timber Tailor

P.S.: If you have any woodworking or remodelling questions that YOU
need help with, let me know... it would be good to return a favor.

Hi, Matt.

The "see below" section does go through an example. Let me connect
that example to your problem. It shows how to solve

sin(x)/x = 3/4

As I said, k = 3/4 is c/s in case 1. If the chord length (c) is 30
feet and the arc length (s) is 40 feet, then c/s = 30/40 = 3/4, and
we need to solve the equation above to find x. As the example shows,
the solution is x = 1.27570. Then, as I said, we use this value of x
in the formula

h = s(1-cos(x))/(2x)
= 40(1-cos(1.27570))/(2*1.27570)
= 40(1-0.2908320)/2.5514
= 11.1180996 feet

There is one tricky point here: when you take the cosine of x, you
must treat x as an angle in RADIANS. (If you're using the calculator
in Windows, for example, click the Radians radio button before "cos".)

Does this help?

If you'd like to give us something in return, you might tell us of
other ways in which you have needed math to do your work (regardless
of whether you needed help or you could do it on your own). Students
ask us how math is needed in the "real world," and we can point to
questions like yours, but there may be some quite different uses that
haven't come to our attention.

Dear Dr. Rick,

Thanks again for the quick attention to my dilemma!

I use math of all sorts: algebra, geometry, arithmetic, and trig,
nearly every day in my construction/custom carpentry business. I
wasted a lot of time in my early years as a carpenter figuring angles
and curves by trial and error. This costs much extra time and
wastes materials. As it started to dawn on me that I could use a
little trig to help solve circle and angle problems, it became easier
to get things right the first time.

Roof slopes and stairs are nothing more than right angle problems.
Circular stairs add the elements of circles and angles to the
problems. It turns out to be a HUGE advantage to be able to figure a
diagonal measurement when doing layout work for a foundation, and it
usually really impresses other tradesmen if you have an accurate
answer ready for a problem they can't solve by trial and error.

As for a problem I encounter, this would be typical: In a historic
restoration project, a large arch-top window needs the trim molding
replaced because the original has rotted away. I go to the jobsite
to take measurements so I can make a new piece of trim in my shop.
What I need is the radius for the arch, but trial and error is not
practical with the size and height of the window. However, I can
easily measure the width of the arc and the height of the arc at the
center.

Along with all my other tools, i.e.: hammer, saw, drill, tape measure,
etc., I carry "cheat sheets" with various formulas with me to every
jobsite. As a tool, they are fully as important as any. Using one of
these formulas with y as the arc width and x as the height, I
calculate radius r: r=(y squared/8x)+ (x/2) (which is simple math for
you!)

With the radius length I draw out the appropriate arc on a piece of
wood and cut it on a bandsaw, and presto, I have a replacement piece
of trim for the old building. Simple with a little math, a real "hat
fire" for a trial and error approach!

I recently built a room with an arched ceiling and used similar
formulas to do all the calculations for everything from the roof
framing to the curved crown molding trim. At the risk of sounding vain
I'll say that being the person on a jobsite who knows how to use math
is like being superman and playing football with a team of "regular"
people... it's almost like cheating to have such an unfair advantage
8^).

Thanks for your help on my problem.

Matt Jackson
The Timber Tailor

P.S.: When I have worked through this current problem to the point
where I can understand it, I will make a cheat sheet that shows it as
well, and carry it with the others.

Hi, Matt.

It's my turn to thank you. I'll see if I can get this to our
archives so students can see it.

Rating: * * * * *      Awarded: $5.00
Ask a Question
Most Recent Questions
Most Recent Answers
Home Page

Login
Register

Information Technology
Software, Hardware, Design

Programming Languages
Database
Graphic Design
Operating Platforms
Hardware and Networking

Math and Science
Homework Help

Biology and Geology
Math and Statistics
- Algebra I
- Algebra II
- Basic Arithmetic
- Boolean Algebra
- Calculus I
- Calculus II
- Combinatorics
- Differential Equations
- Geometry
- Linear Algebra
- Logic
- Multi-Variable Calculus
- Pre-algebra
- Probability
- Statistics
- Trionometry
Chemistry
Physics
Engineering

Are you an information
technology or
math/science expert?

You can earn money by
providing answers to questions
at AnswerArchive.com.
Learn How

Browse Archive | Ask a Question | Most Recent Questions | Most Recent Answers | Home Page | Contact Us | Help
(c) 2008 The Answer Archive