How far is 1 lap in lane 8 on an outdoor 400m track

Let me re-think that JimFiore. Math doesn't give a damn what IAAF measurement procedures are. You are correct. For staggering purposes lanes 2-8 will be shorter than lane one.

Thanks for that one Malmo.

To those still wondering, I have worked out a couple of examples of weird tracks below. The first one is the "square" track like the one I mentioned. Consider a normal and a square track, each with 100 meter straights (the value is chosen simply because it's nice, other values will also work). For a normal track, that means that you have 200 m of curve, or basically a circle with a 200 m diameter. Again, for ease of calc, assume the outer lane is 10 m beyond. As we have seen, this means that you'll travel 2pi*10m = 62.8 meters further. For the square track, the lane one "turn" is just another 100 meter straight (and let me tell 'ya, if you've never run on one, it's an ugly, ugly thing). Now the outside line is also 100 meters long, but includes two extra 10 meter segments per corner, for a total of 80 extra meters instead of 62.8. This is only true because THESE CORNERS ARE NOT RADIUSED. If they were properly radiused, we'd also get 62.8 meters. (I'm tellin' ya, this was a track from hell).

Example 2. The elliptical track. This is borderline pedantic as I can't imagine why anyone would do this. I mean, the square track is stupid for runners, but I can understand the poor custodian laying out 4 110 yard straightaways. Anyway, the thing to remember is that with a true ellipse, as you add lanes the ellipse becomes less and less eccentric (i.e., more like a circle). This means that the rate of circumferential increase decreases (the lanes don't increase in length in a nice linear fashion). Recall that a circle has the highest area to circumference ratio possible of any 2-D object, and thus by this measure, the more eccentric the ellipse, the less "efficient" it is (similarly a sphere has the least surface area for any given volume contained. This is why you don't see single soap bubbles in the form of cubes, tetrahedrons, and so on, but I digress). I will dispense with the 400 meter requirement and just do a calculation with "nice" numbers to prove the point.

A decent approximation for the circumference of an ellipse is 2pi*sqrt( (a^2+b^2)/2 ) where a and b are the short and long axii lengths. Choose something with moderate eccentricity, say a=1 and b=2 (it's more dramatic with a bigger delta, but you should be able to calc these values in your head). So, the circumference is 2pi*sqrt(2.5). This means that the equivalent circle has a radius of sqrt(2.5) or about 1.581 (yes, I used a calculator to get that to 4 significant digits ;-). Now, consider an outside lane that is (nice number) 1 unit further out. For a circle this yields a circumference of 2pi*2.581, or about 16.2. For the ellipse, this means that a=2 and b=3, so circ=2pi*sqrt( (2^2+3^2)/2 ) or about 16.0. Thus, we see that the outer lane of the ellipse is shorter than the outer lane of the corresponding circle. As they don't agree, it's obvious that the equation derived for a circle is not applicable to an ellipse. Why is it shorter? As I mentioned, the outer lane creates an ellipse that is less eccentric than the inner lane (i.e., it looks more like a circle) and thus will generate smaller length increases. In order to keep the length increase consistant to that of the circle, you'd have to keep the eccentricity of the ellipse constant (basically, keep the ratio of a to b constant). This would necessitate making the lanes wider when they cut the longer "b" axis" than when they cut the shorter "a" axis. Now THAT would truly be a "track from hell"!!

Well, that was fun, but considering what the original poster asked, I'm reminded of a saying about certain professors: "If you ask him for the time, he'll give you the history of clocks."

malmo wrote:

You being the resident Letsrun Acoustic Engineer, I'm suprised you hadn't weighed in on the "mp3 vs CD" sub-thread of the "desert island" thread. Any thoughts?

I haven't noticed the thread, but my thoughts are: MP3's can exhibit very fidelity, however, for highest fidelity you tend to lose their only advantage, namely high file compression ratio (or bandwidth reduction if it's a streaming application). If fidelity is not a concern (and from what I have noticed, it isn't a concern for many people), then the MP3 becomes worthwhile. Personally, I can't deal with HF reduction, dynamic range compression artifacts and weird phasing that occurs from medium to high compression, so I tend not to monkey with MP3's. But then, I can't stand talking on cell phones either for the same reasons. To my ears, the artifacts are too annoying.

For the record, I'm an electrical engineer/computer science guy who also works with acoustics (don't have a degree in acoustical engineering although there is some commonality between these degrees).

Yack! Typo: "For a normal track, that means that you have 200 m of curve, or basically a circle with a 200 m diameter." should read "...200 m circumference". Sorry for any confusion.

malmo wrote:

"it's Pete who hasn't thought this thing all the way through. You are correct (which of course you already knew) that the track could even make right turns and loop-to-loops like a slot-car track. As long as you make a 360 degree journey and none of the other variables (lane width, track length, Pi) has changed the answer will always be:

2*Pi*W*(L-1)"

In the case of squared ends, rather than semi-circles or whatever, the difference in length between the outer track and the inner track is 8 times the distance between the two tracks. Since 8 is not evenly divisible by pi in our version of space-time, one must conclude that:

malmo is wrong

So give is the scoop on Renato. What's he like?

Pete wrote:

Since 8 is not evenly divisible by pi in our version of space-time, one must conclude that:

It would be irrational to assume that an integer divided by pi would result in another integer.

Sorry, but I just couldn't resist a bad math pun. ;-)

(Of course, I'm assuming scalars. If they were vectors, it could work...)

Pete wrote:In the case of squared ends, rather than semi-circles or whatever, the difference in length between the outer track and the inner track is 8 times the distance between the two tracks. Since 8 is not evenly divisible by pi in our version of space-time, one must conclude that:

malmo is wrong

In the case with squared ends the lanes do not have a constant width. At the corners the lanes are wider than they are along the sides. As I have said, and kartelite has said, and Malmo has said, and JimFiore has said, as long as the lane width remains constant, the formula applies for any shape track.

Malmo is right, as are the other posters mentioned above, and none of them has to go back to grade school.

Almost, dukerdog. The shape won't matter as long as the curves are made from simple circular sections. If you use some other shape function, such as an elliptic, you won't get the same results, as I outlined above. BTW, the reason it appeared to work in the example you gave on the prior page is because you chose a pair of ellipses with similar eccentricities. If you had carried the digits out, you'd have seen that the difference in your example was not precisely 2pi (but close). If you chose two ellipses that were a bit different, it would've been obvious. Instead of 10x8 and 8x6, try 6x4 and 4x2. The answer is closer to 6 than 2pi. The greater the eccentricity, the farther off you'll be.

If anyone has run on an elliptical track, please post on the "funky tracks" thread ;-)

All curves are made up of circular sections at the differential scale.

The length of a curve is given by

Integral {r(a)da}

where a is the angle that you are turning through in radians, r(a) is the radius of curvature at angle a, and da is the differential angle. If r(a) defines lane one of your track and lane 2 is always w meters outside lane 1, then the radius of curvature of lane 2 will always be r(a)+w. The difference in the lengths of the two curves will be

Integral {[r(a)+w-r(a)]da} = Integral {[w]da}

Since w is a constant, evaluating this integral from 0 to 2*PI gives

2*PI*w

The reason your example of two ellipses, one with a=1,b=2, the other with a=2, b=3, doesn't follow this formula exactly is because there is not a constant distance between them all the way around. In other words the lane width doesn't remain constant. As you said, they have different eccentricities so it is not surprising they don't stay the same distance apart all the way around. If you drew one ellipse and then drew a second curve that was always w meters outside the first, the second curve you drew would not be an ellipse, but the 2*PI*w would describe the difference between their perimeters.

What happened to kartelite?

JimFiore, Pete: dukerdog is correct. I was waiting for some calculus to enter in the discussion, but since it's been 25 years since I've seen dy/dx I stayed away. But listen up to what dukerdog and your calculus 101 instructor was saying. If you have a curve, either a circular arc, an ellipse or a "random" arc, all three are made up of infinite radii - "of circular sections at the differential scale", said dukerdog. It just happens that the circle has the same infinite radii. Since no matter which of these radii you choose the formula 2*Pi*W holds. Since result of each of the individual radii works the summation of all of the different radii must worjk as well. I know I didn't explain it very well, mathematic proofs are not my gig, but dukerdog is correct.

If you extend your hand straingt out, and rotate in a circle so you come back 360 degrees the difference in the distance your elbow travels compared to the hand is going to be 2*Pi*(delta)r. where (delta)r is the distance from your elbow to your hand. Now, if you stand on the back of a flatbed truck and rotate again, the difference is the same (with respect to the bed of the truck). Relative to the ground you would have two sine curves tracing over one another. The difference in the distance of these to sine curves wouyld be - you guessed it - 2*Pi*(delta)r or 2*Pi*W converted back to the original equation 2*Pi*W(L-1).

It doesn't matter how fast the truck travels or how fast you rotate, or even if both the truck and you move at varying speeds. The answer is still the same.

JimFiore, back to the four-sided track the difference in each lane is (Pi*W)/2. With a four-sided track, radius at the turns in lane eight would be ((Pi*1.067)/2)*(8-1) = 11.7325 meters. 4 x 11.73 = 46.93.

PETE, didn't get to meet Renato today, I'm on day three with a nasty flu and left McVeighs gig early.

Incubation time a few days maybe, makes McVeigh and Fullem and Eddy down for the count by the end of next week.

malmo,

sorry pal, but you're WRONG.

You said it doesn't matter how one makes the path, but it does matter. In the case of the squared ends, the difference is 8 times the space between the inner and outer paths, as outlined above. That gives a very different result than what you've suggested.

Since you were wondering about calculus and such, without getting into the actual mathematics, imagine, if you will, ends of the track constructed as triangles which extend some condiderable distance from the straights. If you extend them out to infinity, the difference in length between the inner and outer path will become smaller, until it approaches 4 times the space between the inner and outer paths.

Dukerdog's blather about consistently equal path widths is a red herring. We opened up the discussion to consider paths with angles, which implicitly excludes consistently equal path widths.

malmo,

you are WRONG, WRONG, WRONG!

I don't know how you can sleep at night, giving financial advice with such pitiful basic mathematical abilities. I suppose it helps when you possess such absolute conviction.

Did anyone actually witness the arrival of Renato? Surely someone must have something to report.

Of course, I'm wrong too.

My infinite triangle example is a good one, but the analysis is wrong. The difference actually increases to infinity, rather than decreasing to 4 times the space between lanes. But, infinity does not equal malmo's handy solution, at least not in this universe.

In any event, while I was briefly wrong, malmo is still WRONG. Stubborn and wrong.

OK, lets hear 'em. Riordan HS in San Francisco had a goofy track, also around baseball field. Best perspective is to descirbe as 400m relay. First leg was OK, then second leg disappeared behind dugout, make a hard left turn at the exchnge. Any meet, you could almost gaurantee some team would not emerge... Third leg makes another very sharp turn and heads for home. Fun part was the big wall about 30m after finish, and only 20m after 110HH finish. what a circus! Amazing thing is a few years back they made the track all-weather, and kept the goofy shape. All to please the damn baseball folk! While Riordan Track has won section titles, Riordan baseball has done no such thing, and are normally about 3-9 in league play each year.

OK Pete, show me the calculations:

1) four 100 meter straights, eight 42 in lanes (1.067m), radius drawn from each corner.

each corner the distance travelleds on the outside lane is

(2*Pi*r(L-1))/4

(2*Pi*1.067(8-1))/4 = 11.73267

four turns equals 4*11.73 or 46.73

Are you saying that 2*Pi*r is not the circumference of a circle?

Show me the calculations to disprove, Pete. Let's see the "actual" mathematics?

Pete wrote:

Dukerdog's blather about consistently equal path widths is a red herring. We opened up the discussion to consider paths with angles, which implicitly excludes consistently equal path widths.

We've been saying from the start that the formula is

2*PI*w

where w is the width of the lane. What did you think we meant by w if it wasn't a constant? It's been pretty obvious that everybody has been talking about the case where the lane width is constant. But if it makes you happy, yes, if the lane width doesn't remain constant then the formula doesn't apply. Lane 1 could go to Los Angeles and back and Lane 2 could go to New York and back and the difference will not be 2*PI*w. Very insightful.

What this all boils down to is the difference between continuously curved tracks (which dukerdog was implying early on) and tracks with hard corners.

Dukerdog was actually right (I was too quick in reading his initial post to see the weaselly qualification about rounded corners). malmo was not right.

dukerdog, what I'm getting at about uniform width tracks is just that the width does not remain constant at a hard corner, but could do do in any sort of curved track. Therefore when you impose the limitation of uniform width tracks, it means no hard corners.

malmo, you said it didn't make any difference whether the tracks had corners, and this is where you demonstrated your ignorance of basic geometry. Sorry, but I won't do the math for you. You do the math for yourself. Start by drawing yourself a picture.

PETE SAID:

"Since you were wondering about calculus and such, without getting into the actual mathematics, imagine, if you will, ends of the track constructed as triangles which extend some condiderable distance from the straights. If you extend them out to infinity, the difference in length between the inner and outer path will become smaller, until it approaches 4 times the space between the inner and outer paths."

"Dukerdog's blather about consistently equal path widths is a red herring. We opened up the discussion to consider paths with angles, which implicitly excludes consistently equal path widths."

Who knows what Pete is talking about dukerdog? He frequently gets lost in discussion here. "triangles extending considerable distance..." "We opened up the discussion to consider paths with angles, which implicitly excludes consistently equal path widths."

If you go back to the beginning of this thread you'll find that no one, including Pete, has said anything resemebling this. Perhaps Pete, who can't be bothered with the math, might be bothered by showing any post that even resembles this nonsense?

You're cracking me up old man.

I guess you don't need to actually grasp fundamental concepts to advise people about money, so long as you can impose your thinking on those around you through sheer willfulness.

Pete, you are are absolutely idiodic. I did the math - show me where this forth grade math problem is wrong - PLEASE?

Perhaps what you are missing, or what I'm not making clear, is that inside lane has corners and the 2nd through 8th lanes an arc is drawn.

You're so obsessed with stalking me you are not even engaged in the discussion.

BTW, if the track had perfectly square corners the difference between lanes would be 8.536 meters, or 59.75m in llane eight. But we are not talking aboutt that.