Real-World Carpentry and Trigonometry

Date: 11/19/2002 at 02:21:20
From: Matt Jackson
Subject: Arc/chord formula

Dear Dr. Math,

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

Date: 11/19/2002 at 10:01:23
From: Doctor Rick
Subject: Re: Arc/chord formula

Hi, Matt.

The page you reference is:

   Segments of Circles: 
   The Arc, Chord, Radius, Height, Angle, Apothem, and Area
   http://mathforum.org/dr.math/faq/faq.circle.segment.html 

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)

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

Date: 11/20/2002 at 01:16:55
From: Matt Jackson
Subject: Arc/chord formula

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.

Date: 11/20/2002 at 10:09:05
From: Doctor Rick
Subject: Re: Arc/chord formula

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.

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

Date: 11/20/2002 at 16:40:01
From: Matt Jackson
Subject: Arc/chord formula

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.

Date: 11/21/2002 at 08:09:34
From: Doctor Rick
Subject: Re: Arc/chord formula

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.

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