How long is the outside lane of a 400m track?

Assuming it's lane 8, and each lane is about one metre in width, I think a full lap in lane 8 should be right on 488m.

The easiest way to figure it out would be to stand on the line the 4x100 uses. then just measure the distance back to the starting line. so it would be 400 + x = distance of 1 lap in outside lane.

trackhead wrote:

Assuming it's lane 8, and each lane is about one metre in width, I think a full lap in lane 8 should be right on 488m.

Yes, lane 8 (looks like 1 m width).

I calculated 450 meters....not sure if this is correct.

I too am not very strong in math, but I know that 3.5 laps in lane 8 is equivalent to a mile. Actually, I'm not totally sure if it's 1600m or 1609m (1M), but either way if that's the case I would think taking 1600m/3.5 (laps) would give you the answer which is just about 450+ meters. Hope that helps, but then again I don't even know if my math is right, but at the very worst 3.5 laps in lane 8 is a mile.

I think there was a huge thread on this a while back, but all you have to do is measure the start of the 400m dash in lane 8 back to the start of the curve and Bob's your uncle.

Of course it depends on the track (Occidental College for ex. has unusally long turns and short straighaways; the 1500m starts way down on the turn), but if we assume 100m straights and 100m turns:

We can make the 100m turn into a semi-circle, and using 2πr, we can estimate the radius of a circle with a 200m circ. at 31.830995m. So we're going to add 7m to the radius as we are 7 lanes out from the original radiusm, giving us a new radius of 38.830995m. Put this back into the formula of 2πr and we get the total distance of both lane 8 turns as 243.98228m, add this with the 2 100m straights and we get 443.98228m per lap in lane 8.

Of course I stopped taking math a long time ago so it's likely I screwed up somewhere.

use the search function. We ended up having a very in depth discussion a couple of months ago.

The formula you want to use to get the closest approximation to the distance is:

D = 400 + (400-2S)*pi*w*(L-1)

D is the Distance in lane L

S is the length of the straight

w is the width of the lane (1.22m on a standard track, as per IAAF regulations, I think 1.21m and 1.23m are also acceptable).

Remember to use a measurement for the length of the straight that has at least as many significant figures as the lane width eg 96.7m if using 1.22m width etc

the straight away is 100m x2 = 200m

the curves are taken together as one circle, whose circumference is calculated as such:

200m = circumference for inside lane

therefore the diameter is 200/pi = 63.66m

for the diameter of lane 8 we add 16m which is = 79.66 diameter * pi = 250.26m

therefore the lane 8 total distance around the track is 200m + 250.26m = 450.26 m

You just listed a specific case of the general formula in my post, but messed it up. You don't add 16m to the diameter (assuming non-existant 1m wide lanes). From lane one to eight you only move out SEVEN lanes (hence the L-1 in the formula). So trackhead is correct but anymouse has overestimated the distance.

i am sorry, i was just trying to help

i have failed letsrun math 101

Ok, if I use 1.22m for the width I get 40.370995m for the new radius, and 253.65838m for the distance of both curves for a final distance of 453.65838m for a full lap in lane 8, or, when rounding to signficant figures, 453.66m using sig figs.

measuring the distance from the 400m start line in lane eight back to the finish line may not always work. The UNCC track has longer curves and shorter straights.

It doesn't matter how long the straights are or even if they're equal or straight, nor does it matter if the curves are smooth, all that matters is that you make one 360 degree revolution and the lane width is equidistant. The radius of the arch doesn't mater either.

"The answer is 2*PI*d, the difference in length between the first lane and any other lane is 2*PI*d.

The track could be crooked, and meandering, or be a figure 8. It could have loop-the-loops in it, as long as the net result is that the track turns 360 degrees, the difference will be 2*PI*d.

IF you walked around the world in any direction, east-west or north-south, across mountains or on flatland, your head will travel exactly 2*PI*d further than your feet."

http://www.letsrun.com/forum/flat_read.php?thread=524853&page=1

Simple Math is correct. So if the lanes are truly 1m wide, the additional distance is 2*PI*8m which is about 50m.

446.2 meters, lane 8 standard outdoor track.

Of course it depends on the track. Try thios calculator:

http://www.philsport.com/narf/atrack.htm

No it doesn't grandpa. The only thing that matters is the width of the lanes. The difference between the inside lane and the outside lanes, if it's a 200 meter or a 400 meter track, or one lap around your house, or one lap around your city, will always be 2*pi*d

engineering professor wrote:

if the lanes are truly 1m wide, the additional distance is 2*PI*8m which is about 50m.

Slight (and incomplete) correction. Your formula should be 2*pi*(7 + x), not 8m, unless you plan to run in the very outside of lane eight. Here x represents the distance between the inside of lane eight and the imaginary line along which the runner travels, which I believe should be one foot. So 2*pi*7.31m would be very close. I reckon Mark has it right.

usually about 8 metres per lane if it's a standard track.... so, obviously not counting lane 1 of course.....7 x 8=56 so it would be 456 metres.