Ends of track exactly semi-circular?

I thought about this when someone stated that tracks were oval (not a track athlete) and I thought, "Tracks are two semi-circles joined by straight-aways..."

and it's been law ever since

boston one day wrote:

To prepare for a spring of speedwork, I measured the lane widths and calculated the lane offsets for my local track (Cobb Track at Stanford) ... and only then realized that my calculations assume that the two ends of the track are exactly semi-circular.

Is that true in general? Is it a requirement for a standard 400m track that each end be in the shape of a half-circle?

No. Google "broken back track." At all levels, (HS, NCAA, IAAF) a track with curves that aren't matching semicircles can satisfy regulatory requirements. This is an issue for schools (and European stadia) that wish to put a track around a wide soccer field rather than the narrow American football field (the latter being better suited to the "true oval").

As followup, if you go here, the compiler generally mentions if a track is a broken-back (not a true oval):

http://www.trackinfo.org/tracks.html

I suspect Cobb probably is an oval (having run on it quite a bit myself).

There is a PDF somewhere on IAAF.org - don't have time to search for it now, that gives all the dimensions required for true oval and the modified dimensions required to include sports fields of various football codes inside the oval. I suspect that a large percentage might not be a true oval.

boston one day wrote:

To prepare for a spring of speedwork, I measured the lane widths and calculated the lane offsets for my local track (Cobb Track at Stanford) ... and only then realized that my calculations assume that the two ends of the track are exactly semi-circular.

Is that true in general? Is it a requirement for a standard 400m track that each end be in the shape of a half-circle?

It doesn't matter what the shape of the turns- it could be semi circle or and oval - the calculations will still be the same. Just as long as the lane widths are constant.

Use a wheel to measure the distance from the start/finish of the 400/800/1600/3200 to the start of the 1500/300 hurdles. If you're at a track that does have semicircular turns, then the turns together form a circle with a diameter of 200 meters, meaning that the diameter of this circle--that is, the distance you'll wheel as directed above--is 200/Pi or 63.66 meters.

malmo wrote:

boston one day wrote:

To prepare for a spring of speedwork, I measured the lane widths and calculated the lane offsets for my local track (Cobb Track at Stanford) ... and only then realized that my calculations assume that the two ends of the track are exactly semi-circular.

Is that true in general? Is it a requirement for a standard 400m track that each end be in the shape of a half-circle?

It doesn't matter what the shape of the turns- it could be semi circle or and oval - the calculations will still be the same. Just as long as the lane widths are constant.

For that matter, the lanes could snake back and forth the other way for a while as well...not sure what being a perfect circle has to do with it, or why someone will always vehemently deny this is the case.

I ran once at a school with a "square" track: 4 straights, and 4 rounded corners.

kartelite wrote:

For that matter, the lanes could snake back and forth the other way for a while as well...not sure what being a perfect circle has to do with it, or why someone will always vehemently deny this is the case.

I don't know why people don't understand this, but I think this might be a way to explain how it works.

I think that people take C=2*Pi*r as a literal end all. If the radius changes, they say, how can I calculate the circumference? What they don't see is that any curve, whether smooth (consistent radius) or not, is just the summation of many smaller arc segments, in fact, it is the summation if many infinitesimally small arc segments. It's just that with a smooth circle the radius of all of those infinitesimal segments is the same, therefore, adding all of the smaller segments is unnecessary. Simply use 2*pi*r.

Real life example at Stockholm. That track has sharp turns on the corners then a wide radius mid turn. It looks similar to this drawing.

Stockholm

The IAAF has a rule that the maximum acceptable radius for records is 50m in the lane that the record was set. That means that the maximum (inside lane) radius on an IAAF certified track (all 8 lanes) would be 50 - (8*1.22m) = 40.24m (inside lane) or 126.41m curves/73.59 straights. Realistically you are looking at tracks with 35m radius turns. 110m curve/90m straights.

An additional rule is that a track radius in lane 8 can exceed 50m if it does so for not more than 1/3 of the turn. I'm guessing that lane 8 at Stockholm might fall under this rule.

If you have a measured track then the circumference has already been measured, so you don't need make any messy calculations via 2*pi*r. Since the distance of the inside lane has already been determined, it doesn't matter what the actual curvature looks like. You already know that the two curves plus the two straights adds up to 400m (at least on a nominal count).

1) If the radius of the curves is consistent, then the difference between and lane 1 and lane x is simply 400m + 2*pi*(Lx). In simple language, it is the lane width more than the track radius, or the lane width more than the radius of all of the incremental arc segments.

2) If the radius of the curves varies wildly -- lets say 35m for part, 45m for part, 60m for part, and 40m for part -- the difference between lane 1 and lane x is still 400m + 2*pi*(Lx).

malmo wrote:

The IAAF has a rule that the maximum acceptable radius for records is 50m in the lane that the record was set. That means that the maximum (inside lane) radius on an IAAF certified track (all 8 lanes) would be 50 - (8*1.22m) = 40.24m (inside lane) or 126.41m curves/73.59 straights. Realistically you are looking at tracks with 35m radius turns. 110m curve/90m straights.

Correction

50 - (7*1.22m) = 41.46m or 130.25m curves/69.75m straights

I don't believe your explanation is correct. If the ends are not circular, we need more information to calculate lane lengths.

To see why, let's say that the ends of the track are elliptical. I looked up the formula to calculate the circumference for an ellipse, and it is quite complicated. In fact, there may not be an exact "formula."

Specifically, the formula is not linear in r, so when calculating lane lengths, the different radii do not subtract out, as they do when calculating lane lengths under the assumption of semi-circular lane ends.

For practical values of a and b associated with a track the Ramanujan formula reduces to pi*(a+b) with an error of about an inch so it's not worth worrying about. I.e. it is linear in a and b for small values of 3((a-b)/(a+b))^2.

boston one day wrote:

I don't believe your explanation is correct. If the ends are not circular, we need more information to calculate lane lengths.

To see why, let's say that the ends of the track are elliptical. I looked up the formula to calculate the circumference for an ellipse, and it is quite complicated. In fact, there may not be an exact "formula."

Specifically, the formula is not linear in r, so when calculating lane lengths, the different radii do not subtract out, as they do when calculating lane lengths under the assumption of semi-circular lane ends.

This is a layup for me.

If the track shape is elliptical it doesn't matter. An ellipse is a specific kind of curve defined by two focal points. The fact that to solve the circumference of an ellipsis requires a more complex integration, rather than a simple 2*Pi*r is irrelevant, because you are physically measuring the circumference of this hypothetical elliptical track. It is 400m.

You are going around that ellipse once. That is 360°. While each point is defined by those two focus points it can also be represented by an imaginary radius, but is doesn't matter what that imaginary radius is, the line in Lane X is 2*Pi*r longer.

Secondly, if you were to add lanes on this ellipical track you wouldn't do so by merely increasing the ratio between those focal points (which means that the lanes would not have a constant width) you would simply measure the lanes and equal distance from the inside lane (line) of the ellipse.

In other words, only lane 1 would be an ellipse, by definition. Lanes 2, 3, 4, ... would not be. You don't need calculus to measure lane 2, 3, 4 etc. you need 2*pi*r. You don't need calculus to measure lane 1 because you are going to physically measure it. (see note at the bottom of the page) Lane 2 would be 2*pi*r more than the distance on the inside lane.

Another way of looking at an ellipse is that it can not only be described as a ratio between to focus points it can also be described as a constantly changing radius. Each infinitesimally small point along that path has a radius. As long as you make one revolution, no matter what your path, the distance traveled L meters from that inside path is always 2*pi*L more.

If you walked around the world your head would travel exactly 2*pi*r more than your feet.

NOTE. Now if you were going to measure out lanes 2, 3, 4...etc as an ellipse then that is another story. But what sense does it make to have 1.5m wide lanes at two points on the track and .35m wide lanes at the tightest curves?

RWIW, this question has been answered before by two physicists.

Phil. wrote:

For practical values of a and b associated with a track the Ramanujan formula reduces to pi*(a+b) with an error of about an inch so it's not worth worrying about. I.e. it is linear in a and b for small values of 3((a-b)/(a+b))^2.

Good point, but here's my followup.

Please note, that is for calculating the circumference on paper. Since lane two is NOT an ellipse (by defintion) then difference between lane 1 and lane x is simply 2*pi*r carried out to however many significant figures you want.

Again, my original "so what?" point is that the formula for calculating circumference on paper doesn't matter because you will be actually physically measuring your elliptical track in lane 1 and that lane 2-8 would not be ellipses anyway. Your mathematical "so what?" point that the difference (referring to lane 1 calculations of an ellipse) would only be off by 1" in 400m!

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This is a layup for me.

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malmo I didn't know you hooped

This 400m track is legit and has four 90 degree turns and four straights..

http://maps.google.com/maps?f=q&source=s_q&hl=en&q=middle+schools&aq=&sll=40.101033,-83.117774&sspn=0.001793,0.003449&gl=us&ie=UTF8&rq=1&ev=p&split=1&t=h&radius=0.11&hq=middle+schools&hnear=&ll=40.101595,-83.117597&spn=0.001793,0.003449&z=18

I thought there must be an online calculator for this task (adjusting outside lanes to desired distance) and wondered whether it would include any qualificaton for curve shape (i.e., semi-circles or not). North American Racewalk Foundation has one, widely linked, which states as follows:

"Have you ever been frustrated by having to use the outer lanes of the local high school or college track. You know what you want to do in Lane 1 but are not sure how to adapt your plan to the outer lanes. You could give up on using the track, you could bully your way onto Lane 1, OR you could use our Track Calculator to develop confidence in using any lane of the track for almost any workout. . . If your workouts are on a track that has an odd shape (e.g., curves are not circular arcs), please see NOTE 2 before proceeding. . . .NOTE 2 - TRACK SHAPE: This calculator is only accurate on tracks that are circular, or which consist of 2 or more straightaways connected by curves that are arcs of a circle. For tracks of other shapes, the calculations will be only reasonable approximations -- though very useable for most people."

The technical advisor for the website is mentioned as Dr. Wayne Armbrust, Ph.D. physics, who operates a consulting company that does track marking,

www.computomarx.com

. Since everyone reads letsrun, or knows someone who does, someone should tell Dr. Armbrust to weigh in.

link to calculator

Dr. Armbrust should weigh in that he was wrong, or more likely, that the person who wrote that didn't consult him.

This was covered in a thread several years ago. This is the link to it, but it doesn't seem to be working anymore.

http://www.letsrun.com/forum/flat_read.php?thread=250143