where I swim all the lanes are the same length.
where I swim all the lanes are the same length.
Bear of Bad News wrote:
A square with rounded corners is not a circle.
The four rounded corners form a circle. The circumference of this circle is part of the lane length.
this is a lame site, but the calculator might help...
ovaltine wrote:
No, it is not the circumference of any circle in the problem. W is the lane width, which is not a radius. Just because there is a 2*pi*width doesn't make it a circumference. pi is just a mathematical number that appears in many mathematical expressions, too many to be enumerated. The fact that the formula can be easily shown to be true for semi-circles doesn't mean it only works for semi-circles.
The formula you cite is indeed dependent on the circumference of a circle and therefore a radius of which W is a component. Allow me to derive it for you....
The length (L) of a lane on a standard oval track is defined as L = 2S + 2π(R + (n-1)W) where S is the length of the straight and R is the radius of the turn (n and W are the same as yours). Distributing 2π results in L = 2S + 2πR + 2π(n-1)W. Now 2S + 2πr is the length of lane 1, so you can substitute your expression for that and end up with your formula.
Thus the formula is based on the circumference of a circle, and therefore also the raduis/W.
whether this is dependent on track shape/lane width on a per-track basis, the difference (on a precisely marked track) between the 400m (two-turn) starting line in lane 1 and the 400m starting line in lane is the answer on every track (with two turns). "extend" the lane 5 400m start line to intersect the inside lane 1 line and the distance between that point and the point where the lane 1 400m start line touches the lane line is what you would need to add to the track distance times how ever many laps you do.
Math Mann, as has been already stated, just because something is true for a specific case doesn't mean it is not true for a more general case.
The 2*Pi*w solution is "dependent on the circumference of a circle," a circle of radius=w. Early in the original thread on this topic I provided a proof of this 2*Pi*w formula for any shape track, but then later Malmo's professor friend provided a much more intuitive explanation. Basically, from the perspective of the person in lane 1, the person in lane 2 is simply running in a circle around him. That circle has a radius of w.
dukerdog wrote:
Math Mann, as has been already stated, just because something is true for a specific case doesn't mean it is not true for a more general case.
The 2*Pi*w solution is "dependent on the circumference of a circle," a circle of radius=w. Early in the original thread on this topic I provided a proof of this 2*Pi*w formula for any shape track, but then later Malmo's professor friend provided a much more intuitive explanation. Basically, from the perspective of the person in lane 1, the person in lane 2 is simply running in a circle around him. That circle has a radius of w.
Thank you. I was interpreting "wondering" as seeming to think that there was a physical circle (or arcs thereof) that one could see on the track that made the formula true. This intuitive explanation provides an excellent way to get at the true answer without invoking some higher math. So to both your credits (and Malmo's prof. friend), yes there is a circle, but it is a relative circle, not an physical one.
Arc length is found by integrating sqrt( (dx)^2 + (dy)^2)
If the radius is non-constant, we can write r = f(A) where A is the angle.
With some very basic calculus you can establish that the arc length in lane 1 would be found by integrating
sqrt( f(A)^2 + f'(A)^2)dA
and the arc length in lane two would be found by integrating
sqrt( f(A)^2 + 2f(A)w + w^2 + f'(A)^2) where w = lane width
If r = constant, then f'(A)= 0 and the difference in these expressions becomes 2*pi*w after integration. If r is not constant then there are some situations where the difference is 2*pi*w, but those are special cases.
dukerdog wrote:
Math Mann, as has been already stated, just because something is true for a specific case doesn't mean it is not true for a more general case.
The 2*Pi*w solution is "dependent on the circumference of a circle," a circle of radius=w. Early in the original thread on this topic I provided a proof of this 2*Pi*w formula for any shape track, but then later Malmo's professor friend provided a much more intuitive explanation. Basically, from the perspective of the person in lane 1, the person in lane 2 is simply running in a circle around him. That circle has a radius of w.
Thank you for your insight. It appears we are in agreement. Also, thank you for mentioning the other thread. I look forward to reading it.
As previously mentioned, the following website can provide the answer:
As has been stated here a few times, the shape of the track is irrelevant, it could be a circle, it could have long straights, it could have short straights, it could be a figure eight with loop-de-loops, or it could randomly meander, just as long as you make one complete revolution and/or aligned in the same direction as where you started, the difference is 2*pi*d.
If you were to circumnavigate the earth in any direction at any lattitude and meander in any direction, the top of your head will travel exactly 2*pi*h farther than your feet.
(where h is your height)
There is one variable to be thrown in here. A 400 meter track is a nominal measurement, no tracks are actully 400m around, the actual distance is shorter than 400m. IF the track has a curb the inside lane is measured 30cm in, or 2*pi*30cm short of 400m. Lanes with no curb are measured 20cm in, or 2*pi*20cm short of 400m.
Curb to lane one the difference is: 2*pi*(L+10cm)
and: lane one to lane two the difference is: 2*pi*L
A track without a curb, all lanes are 2*pi*L different with respect to each other.
Math Mann wrote:
Thank you for your insight. It appears we are in agreement. Also, thank you for mentioning the other thread. I look forward to reading it.
Be prepared for some high comedy.
http://www.letsrun.com/forum/flat_read.php?thread=250143&page=01.3 miles
malmo wrote:
Be prepared for some high comedy.
http://www.letsrun.com/forum/flat_read.php?thread=250143&page=0
Absolutely hilarious! As I said (albeit more briefly), this issue has been hashed, rehashed, used and abused many times before. Thanks for the LR history lesson. I still stand by my initial response to the OP.
ovaltine wrote:
Absolutely hilarious! As I said (albeit more briefly), this issue has been hashed, rehashed, used and abused many times before. Thanks for the LR history lesson. I still stand by my initial response to the OP.
Your initial response was correct, except in the case of a track with a fixed curb. Then you have to make an adjustment for the difference in measuring for curb vs line. In that case, lanes 1-2 will have a different offset than lanes 2 through 8.
In Excel syntax the formula would be as follows:
S_L =IF(S_1="curb",S_1 + 2*π*W*(L-1.10),S_1 + 2*π*W*(L-1.0))
malmo wrote:
In Excel syntax the formula would be as follows:
S_L =IF(S_1="curb",S_1 + 2*π*W*(L-1.10),S_1 + 2*π*W*(L-1.0))
That should be
S_L =IF(S_1="curb",S_1 + 2*π*W*(L-0.90),S_1 + 2*π*W*(L-1.0))
Where W is the lane width in meters.
malmo wrote:
That should be
S_L =IF(S_1="curb",S_1 + 2*π*W*(L-0.90),S_1 + 2*π*W*(L-1.0))
Where W is the lane width in meters.
Oh heck, might as well get this thing perfect.
S_L =IF(S_1="curb",S_1 + 2*pi()*W*(L-0.90),S_1 + 2*pi()*W*(L-1.0))
Where W is the lane width in meters, and S_1 is the NOMINAL measurement of the inside lane.
malmo wrote:
Oh heck, might as well get this thing perfect.
S_L =IF(S_1="curb",S_1 + 2*pi()*W*(L-0.90),S_1 + 2*pi()*W*(L-1.0))
Where W is the lane width in meters, and S_1 is the NOMINAL measurement of the inside lane.
Thanks for the correction due to curbs.
EOT...please?
malmo wrote:
Be prepared for some high comedy.
http://www.letsrun.com/forum/flat_read.php?thread=250143&page=0
What a moron that Pete was. And not just on this thread.
OK, now EOT.
EOT, not quite yet. The odd thing was that Pete is an engineer by day. EOT.