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Arc of a Circle

Date: 08/16/2003 at 21:37:31
From: Frank Anderson
Subject: Arc of a circle

I am planning a model railroad, using Lionel section track. Lionel's 
wide radius curve track forms a 6' diameter circle when the 16 
individual pieces are assembled. I would like a formula to calculate 
the distance between parallel lines if I use one right-hand curve 
section to turn away from a straight line followed by one left-hand 
curve section to return to a straight line, running parallel to the 
original line.  

I understand there are 360 degrees in a circle, and each of the 16 
pieces of curve track forming the 6' diameter circle will represent a 
22.5 degree turn.  My goal is to calculate the distance between 
straight lines after I turn 22.5 degrees to the right of the first 
line followed by a 22.5 degree turn back to the left.


Date: 08/16/2003 at 22:40:37
From: Doctor Peterson
Subject: Re: Arc of a circle

Hi, Frank.

You can solve this and related problems using formulas found in our 
FAQ:

   Segments of circles
   http://mathforum.org/dr.math/faq/faq.circle.segment.html 

But that takes a little ingenuity. I'll show you how to attack it 
directly with a basic knowledge of trigonometry. Here's a picture:

                    ooooooooooooo
               ooooo             ooooo
            ooo                       ooo
          oo                             oo
        oo                                 oo
       o                                     o
      o                                       o
     o                                         o
    o                                           o
    o                                           o
    o                     +---------------------+
    o                     |\                    o
    o                     | \                   o
     o                    |  \                 o
      o                  x|   \               o
       o                  |   \ r            o
        oo                |22.5\           oo
          oo              | deg \        oo   |
            ooo           |      \    ooo     v
               ooooo      +-------+ooo ---------
                    oooooo+oooooo---------------
                                              ^
                                              |

Consider the 22.5 degree arc at the bottom. The vertical rise 
indicated will be r-x; and x=rcos(22.5). When you put two opposite 
curves together, the total vertical rise will be 2(r-x). So the 
distance between the two parallel tracks will be

  d = 2(r-x) = 2r(1 - cos(22.5)) = 0.1522r
    = 0.1522 * 3 ft = 0.4567 ft = 5.48 in

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

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


Date: 08/18/2003 at 23:15:54
From: Frank Anderson
Subject: Arc of a circle

I appreciate your fast reply. I lack the basic knowledge of 
trigonometry this requires, but I am determined to understand this 
material, and would like to dissect your reply for a better 
understanding.

It appears you are using a right triangle as a basis for your 
calculations. Is this accurate?  If so, 'r' looks like the radius of 
the circle, as well as the hypotenuse of the right triangle.  In the 
right triangle, how are you identifying the two non-90-degree angles 
of the right triangle, as well as the length of the undefined side 
and side x?  

Also, in the text of your response, you supplied r-x as the formula 
for the vertical rise of the first curve track section, with x being 
equal to rcos(22.5). In r-x, does 'r' represent the radius of the 
circle? Also, in layman's terms, does 'rcos(22.5)' represent 'r' 
multiplied by the result of 'cos' multiplied by 22.5? 

I greatly appreciate your assistance,

Frank


Date: 08/18/2003 at 23:36:28
From: Doctor Peterson
Subject: Re: Arc of a circle

Hi, Frank.

I'll repeat the picture with more labels:

                    ooooooooooooo
               ooooo             ooooo
            ooo                       ooo
          oo                             oo
        oo                                 oo
       o                                     o
      o                                       o
     o                                         o
    o                                           o
    o                                           o
    o                     O---------------------+
    o                     |\                    o
    o                     | \                   o
     o                    |  \                 o
      o                  x|   \               o
       o                  |   \ r            o
        oo                |22.5\           oo
          oo              | deg \        oo   |
            ooo           |      \    ooo     v
               ooooo      C-------Booo ---------
                    ooooooAoooooo---------------
                                              ^
                                              |

You are interpreting things right. In the big right triangle COB in 
the picture, the hypotenuse OB is r, the radius of the circle; and I 
have called the vertical leg OC x. The angle AOB at the top of the 
right triangle is your 22.5 degree angle, and the arc AB at the 
bottom is your track section, which represents 1/16 of a circle.

The key to trigonometry is the definition of certain "trigonometric 
functions", one of which is the cosine. When we write

  cos(AOB)

we mean the cosine of the angle AOB, which is defined as the ratio of 
the "adjacent side" OC to the hypotenuse OB. So

  cos(AOB) = x/r

Solving for x, we multiply both sides by r:

  r cos(AOB) = x

That's the formula I gave: to find x, you multiply r times the cosine 
of 22.5 degrees. And the value of that cosine (which you can find 
using a scientific calculator, such as the one supplied with Windows) 
is 0.9239, which you can then use to find the value of r-x.

The basics of trig are really not too hard to understand, but it's 
certainly mystifying if you've never seen it. We have a brief 
introduction here:

   Trigonometry in a Nutshell
   http://mathforum.org/library/drmath/view/53942.html 

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


Date: 08/21/2003 at 12:16:08
From: Frank Anderson
Subject: Arc of a circle

Your new explanation and labels were enough to get me going, and I 
have been able to identify the 'Vertical rise' of all of the diameter 
track circles I need.  Thank you very much for breaking this down 
into an understandable format.

Now that I can calculate the distances between parallel lines offset 
by arcs, I would like to expand my capability to include straight 
lines in the formula. In plain words, from a straight line, I will 
use a 22.5 degree right hand turn, followed by a 6", 10", or 14" 
piece of straight track, followed by the offsetting 22.5 degree turn 
back to the left.

I assume you are going to introduce another right triangle to 
calculate the new distance, with the 90-degree angle at the bottom 
right, the 22.5 degree angle at the bottom left, the 67.5 degree 
angle at the top, and the straight track being the hypogenous. This 
will leave the bottom side of the right triangle as my new extension 
to the distance between the track.  Am I on the right track?

Again, I appreciate your assistance with my elementary problem.  I am 
sure you had more interesting issues, unless you are also into trains.

Frank


Date: 08/21/2003 at 12:31:46
From: Doctor Peterson
Subject: Re: Arc of a circle

Hi, Frank.

Yes, you are on the right track, and I'll pardon the pun.

You start with a 22.5 degree curve, and then continue in that 
direction for a distance d before turning again:

  +...-----+
      ...  |a
         ..|
           +
           | \
           |   \  d
          h|     \
           |       \
           |     22.5\
           +-----------+
                       |..
                      b|  ...
                       +-----...+

We've calculated distances a and b (which are equal); now we use even 
simpler trig to find h:

  h/d = sin(22.5)

  h = d * sin(22.5)

So id d=10 inches, h is

  h = 10 * sin(22.5) = 10 * 0.38268 = 3.8268 in

Then the total offset is the sum of a, h, and b.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Trigonometry

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