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Radius of a Racing Circle


Date: 09/20/1999 at 09:34:45
From: Mick Hocking
Subject: Finding the radius of a circle from two known circles

I have two circles, one smaller than the other, both with the same 
center point. A 90 degree arc of these two circles forms a racetrack 
corner (the smaller circle is the inside of the corner, the larger the 
outside). I know the two radii of these corners Ro and Ri. There is a 
racing line through this corner defined by the circle that touches the 
inner circle at its geometric center and the outer circle at 
equidistant points at the entry and exit to the corner. How do I find 
the radius of the circle describing the racing line circle?

I think it's something to do with chords, and I have a solution from 
an article that states that the racing line circle radius Rr is equal 
to  Rr = 3.414 * (Ro - (0.707 * Ri))

I know that 0.707 is sqrt(2)/2 and this sounds like some Pythagorean 
geometry stuff, but I can't figure out how the author of the article 
got this formula.

Any help would be most gratefully received.

Thanks,
Mick


Date: 09/20/1999 at 14:12:22
From: Doctor Peterson
Subject: Re: Finding the radius of a circle from two known circles

Hi, Mick.

I can't see any obvious meaning for your equation. Let's see if we can 
clarify the question so I'll be sure I'm reading it right.

I have several problems with your description. First, what is a 
"racing line" (circle)? Is it a starting line, or the path to be 
followed by a racer, or what? That might help me picture it better.

Second, you say it "touches the inner circle at its geometric center"; 
but the center of a circle is not on the circle, so I'm not sure 
whether you mean the "racing circle" is tangent to the inner circle at 
the center of the arc, or passes through the center of the two 
circles, or something else.

Third, you say it also touches "the outer circle at equidistant points 
at the entry and exit to the corner." Do you mean the end points of 
the outer quarter circle, or some other pair of points?

Here's the best I can do at interpreting your description, assuming 
the racing circle passes through the center of the small arc and the 
endpoints of the large arc:

                         *****o
                   ******o    |\
                ***   o  ****** \
             ***   o ****     |  \
            *    o***         |   \
          **   o* \           |Ro  \
         *   o**    \         |     \
        *  o *        \Ri     |      \Rr
       *    *           \     |       \
      * o   *             \   |        \
      *    *                \ |         \   x
      o----*------------------*----------\----------+
      |   \      Ro             \         \         |
      |       \                   \        \        |
      |           \                 \       \       |
      |               \          Rr-Ri\      \      |
     x|                   \             \     \     |x
      |                    Rr  \          \    \    |
      |                            \        \   \   |
      |                                \      \  \  |
      |                                    \    \ \ |
      |                                        \  \\|
      +---------------------------------------------+
                        Ro+x

The equation for Rr, which I got by considering two right triangles in 
this picture, is:

          Ri^2 + Ro^2 - sqrt(2)*Ro*Ri
     Rr = ---------------------------
               2Ri - sqrt(2)Ro

The equation you gave will give a negative result in this case (with 
the track narrow enough for the "line" to curve inward as shown). When 
my denominator is zero, the "line" is in fact a line, a chord of the 
outer circle, and if the track is still wider it will curve outward.

Let me know how I have to modify my picture to make it right, and I 
can show you more detail once I know we're on the right track.

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


Date: 09/21/1999 at 05:43:22
From: Mick Hocking
Subject: Re: Finding the radius of a circle from two known circles

Dr. Peterson,

Thank you very much for your reply. Considering the rather poor 
description that I gave, you have done extremely well in your 
interpretation. You were very close with your initial solution. In 
answer to your questions relating to the problem:

1) The racing line is the line that a racecar would take through the 
corner to minimize the time through the corner and maximize the speed 
(especially on the exit).

2) Sorry about the phrase 'its geometric center'. That was an error on 
my part; I meant to say that the racing line touches the inner circle 
at the center of the arc and is tangent to the circle as you rightly 
suggested.

3) Here is where your diagram differs from the problem. The racing 
line touches the outer circle at equidistant points on either side of 
the corner, but these points do not lie perpendicular to the center of 
the two circles that form the corner. The entry to the corner touches 
the outer circle at a point before the one on your diagram and the 
exit point touches the outer circle at a point after.

Thanks again, the last thing I would ask, although I am very grateful 
for the time you must have already spent on helping me, is: could you 
show me your work so that I can see how your formula was derived from 
the diagram. Sorry if this seems obvious to you but I think I really 
need to refresh my basic geometry rather soon.

Once again, any help will be most gratefully appreciated.

Mick


Date: 09/21/1999 at 09:05:49
From: Doctor Peterson
Subject: RE: Finding the radius of a circle from two known circles

Hi, Mick.

I'll see if I can modify my picture to fit your description.

I just realized that you might mean that the racing circle is tangent 
to the straight part of the outside edge of the track, as well as to 
the center of the inside arc. That may give me the missing 
information, and it fits with the physical requirements.

Here's my picture, using my realization that your racing circle is 
tangent to the sides (for which the knowledge that this was supposed 
to be the fastest path, and that it goes beyond the corner, helped a 
lot):

                       *****************o  o o o o o o
                 ******     |    o                   |
              ***           |o  racing circle        |
           ***            o |                        |
          *            o******************************
        **          o**     |                        |
       *         o**\       |                        |
      *        o *    \Ri   |                        |
      *         *       \   |                        |
     *      o  *          \ |center of corner        |
     *---------*------------*------------------------+
     *   o     *            |  \                     |
     *         *            |    \                   |
     * o       *            |      \                 |
     *         *            |        \x*sqrt(2)      |
     *o        *            |          \             |
     *         *            |            \           |
     o         *            |              \         |
     o         *            |                \       |
     o         *            |                  \     |
     o         *            |                    \   |
     o         *            |                      \ |
     o---------*------------+------------------------*
               Ro                       x      center of racing circle

The radius Rr is equal to Ro + x, from the bottom of my picture, and 
also to Ri + x*sqrt(2), from the diagonal. (The o's are a quarter 
circle, which is tangent to the edges of the track at both ends and 
the middle; x is the distance from the point at which it leaves the 
edge of the track to the start of the corner.)

Setting these two expressions equal, I can solve for x:

     Ri + x*sqrt(2) = Ro + x

      x*sqrt(2) - x = Ro - Ri

     x(sqrt(2) - 1) = Ro - Ri

                        Ro - Ri
                  x = -----------
                      sqrt(2) - 1

                    = (sqrt(2) + 1)(Ro - Ri)

(The last simplification was done by multiplying numerator and 
denominator by sqrt(2) + 1.)

Now I can calculate Rr:

     Rr = Ro + x

        = Ro + (sqrt(2) + 1)(Ro - Ri)

        = Ro + Ro*sqrt(2) + Ro - Ri*sqrt(2) - Ri

        = Ro*(2 + sqrt(2)) - Ri(sqrt(2) + 1)

                                 1 + sqrt(2)
        = (2 + sqrt(2)) (Ro - Ri*-----------)
                                 2 + sqrt(2)

                                (1 + sqrt(2))(2 - sqrt(2))
        = (2 + sqrt(2)) * (Ro - --------------------------)
                                (2 + sqrt(2))(2 - sqrt(2))

        = (2 + sqrt(2)) * (Ro - Ri*sqrt(2)/2)

Well, 2 + sqrt(2) = 3.414 and sqrt(2)/2 = 0.707, so this is your 
formula. Let me know if I've left anything out.

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


Date: 09/21/1999 at 10:02:57
From: Mick Hocking
Subject: Re: Finding the radius of a circle from two known circles

Doctor Peterson,

I'm amazed! I was just replying to your last mail when you go and send 
me the answer! Doctor Math... solutions faster than you can type a 
reply! I'm so impressed. Please accept my most sincere thanks for 
solving this puzzle for me. I think that you have the complete answer.

I'm off to write a physics model for the racecar now and then some AI 
to get the car to race around the track calculating its own racing 
line. If it would be of any interest to you or your students I can let 
you know how I get on (and any further puzzles).

Many thanks again,
Mick
    
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Physics/Chemistry

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