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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

trigonometry this requires, but I am determined to understand this
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?

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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