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/
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