Radius of a Racing CircleDate: 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 |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/