Closest 3dis to 2dis

[US71]

What about 3di's that cross or run concurrent, like I-540 in Arkansas which has a 5 Mile duplex with I-40 ?

[Grzrd]

What about 3di's that cross or run concurrent, like I-540 in Arkansas which has a 5 Mile duplex with I-40 ?

Yet ANOTHER reason to re-designate I-540 north of I-40 as I-49! 

[hbelkins]

What about 3di's that cross or run concurrent, like I-540 in Arkansas which has a 5 Mile duplex with I-40 ?

I think someone mentioned I-87 and I-287 in NY as an example of that.

I can't remember. Are 40 and 540 both signed on the concurrency? Or is it only 40 that's signed?

[US71]

What about 3di's that cross or run concurrent, like I-540 in Arkansas which has a 5 Mile duplex with I-40 ?

I think someone mentioned I-87 and I-287 in NY as an example of that.

I can't remember. Are 40 and 540 both signed on the concurrency? Or is it only 40 that's signed?

WB, both are signed. EB, only 40 is signed, except at Exit 7 (540 South)

[shadyjay]

In NJ, I-95 (NJ Tpke) and I-295 are very close for a few miles.  Further south on the outskirts of Camden, they aren't more than 1/4 mile apart, running parallel with each other.  Of course by that point, I-95 has exited/vanished/etc.  

Then of course in Trenton, the parent (I-95) turns into the child (I-295) NB, and vice versa for the other direction.

But we all knew that!  

[Kacie Jane]

In NJ, I-95 (NJ Tpke) and I-295 are very close for a few miles.  Further south on the outskirts of Camden, they aren't more than 1/4 mile apart, running parallel with each other.  Of course by that point, I-95 has exited/vanished/etc. 

Then of course in Trenton, the parent (I-95) turns into the child (I-295) NB, and vice versa for the other direction.

But we all knew that! 

But apparently we didn't all know the New Jersey Turnpike is not I-95 south of Exit 6, which is where it and I-295 are at their closest.

EDIT: Apologies for the tone of my post yesterday.  In my stupor, I clearly missed the last sentence of your first paragraph.  Although I stand by my point that I-95 and I-295 are only spectacularly close for the two miles from Exit 6 to Exit 7 on the Turnpike, at which point I-295 veers more northwesterly towards Trenton.  The distance between the two is less than a mile for this segment though.

[Perfxion]

Does PRI-2 and PRi-22 count?

[NE2]

No.

[mgk920]

The first one that came to my mind was I-90 and I-790 in New York.  Also I-87 and I-587 in New York.

I-94 and I-794 in Wisconsin don't stray all that far apart, either.

Mike

[empirestate]

I think we can go ahead and just throw in all of the shortest 3dis automatically. :-) But I think the spirit of the question would be, of those 3dis that aren't necessarily the shortest, and in particular among loops and bypasses, which ones never get particularly far from their parent?

To put it another way: what is the shortest maximum distance between a 3di and its associated 2di that isn't achieved primarily as a result of the 3di's overall shortness? (The original example is I-405 in Oregon, to which I might add I-670 in Kansas City.)

In that case, I suppose I-790 might count, since its maximum distance from I-90 is less than its total length. The same can't be said, however, for I-587.

I think that's the question here; otherwise the subject line would have been "Shortest 3di?" and we all know that's pretty easy to look up!

[agentsteel53]

I think this can be expressed as:

what is the interstate with the smallest ratio a/b of:

a) shortest crow-flies distance back to its parent as averaged over every point on its length
b) its overall length

I think that mathematically rigorous definition agrees well with intuition.  It gives a spur like I-375 in Florida a result equal to approximately 1/2, while something like I-405 in Oregon comes out to be well less than 1/2.

this might be interesting if extended over many miles.  what is this ratio if we treat I-94 as a loop of I-90, for example?  

also, what is the "most greatly deviating" 3di, for which this ratio is the largest?  I think in the case of a spur route extending directly perpendicularly away from its parent, the ratio of 1/2 is an absolute maximum - and some spur routes like I-175 and I-375 in St. Petersburg nearly perfectly meet this criterion, but what about for a loop route?  I-435 in Kansas City comes to mind, as does I-495 in Boston.

[KEVIN_224]

Connecticut: I don't think I-384 from East Hartford to Bolton gets too far south of I-84, once it leaves at Exit 59 in East Hartford.

Massachusetts: I-391 isn't all that long in the Chicopee area (just north of Springfield).

Maine: I-395 only goes a few miles east from I-95 in Bangor, heading over the Penobscot River into Brewer.

[vtk]

also, what is the "most greatly deviating" 3di, for which this ratio is the largest?  I think in the case of a spur route extending directly perpendicularly away from its parent, the ratio of 1/2 is an absolute maximum - and some spur routes like I-175 and I-375 in St. Petersburg nearly perfectly meet this criterion, but what about for a loop route?  I-435 in Kansas City comes to mind, as does I-495 in Boston.

I-275 in IN—KY—OH probably scores rather high for a full circumferential...

By the way, a perfect circular beltway bisected by a perfectly straight parent would score 1/π², or about .1; a semicircular loop, twice that; a quadrant of a circle (spur) twice that.

[agentsteel53]

interesting.  I had a post a minute ago that said that your answer does not seem intuitively correct... but upon looking back at it, you are right.  this is because the beltway counts "twice" for the same length of mainline cut off, compared to the loop road, so indeed the beltway's average is half that of the loop.

this was not something which I had considered originally, as to me it leads to the pathology of being able to thread multiple beltways through two points which intersect the mainline (imagine concentric ellipses of increasing obliqueness with one dimension remaining fixed), which brings down the average as the number of ellipses tends to infinity and the number of through roads remains one.

so, then, I offer the caveat that the mainline cuts into segments the beltway under consideration.  there are two segments of I-435 to be considered, then, in my view.

[vtk]

interesting.  I had a post a minute ago that said that your answer does not seem intuitively correct... but upon looking back at it, you are right.  this is because the beltway counts "twice" for the same length of mainline cut off, compared to the loop road, so indeed the beltway's average is half that of the loop.

I'm not sure I get your little explanation there, but I explain it like this: the full beltway has the same average distance a from its parent as the half-circle loop, but its total length b is greater, so the score a/b is less.  Either way, I agree that full circumferentials should be divided at their parent to be compared fairly under this metric.

[agentsteel53]

yep, we're approaching the same intuitive result from slightly different angles - that to have the metric be fair, the paths should be split by the mainline.

btw your math (2/pi^2 for the loop road) is right.  it's a bit shocking to see how rusty I am at integral calculus: took me about 15 minutes to get the answer!  

the real question is, can we prove that the score approaches 1/2 as the spur road gets shorter and shorter of a semicircular segment and approaches zero length? 

[agentsteel53]

the real question is, can we prove that the score approaches 1/2 as the spur road gets shorter and shorter of a semicircular segment and approaches zero length?  

it does.  dang, this took me even longer to do.  if we let x be the portion of the semicircle "built" (x==1 is beltway, x==1/2 is loop road), our integral to calculate the sum of all the infinitesimal distances is:

from theta=0 to theta=2*pi*x, integral of r*abs(sin(theta))*dtheta.  

knock off the absolute value because 0 < x < 1/2 for this exercise, and set r == 1 for simplicity; evaluate to get -cos(2*pi*x) - -cos(0), as the sum.  write that more simply as 1 - cos(2*pi*x)

to get the average, divide by the total arc length 2*pi*x, and then to get our score, divide again by total arc length.

so we need to figure out lim(x->0) (1 - cos(2*pi*x))/((2*pi*x)^2)

let z = 2*pi*x for simplicity; as x->0, z->0 as well, so we can swap out the limit to z->0.

this is of the form 0/0, so derivative of numerator and denominator yields sin(z)/(2z)

this is more usefully expressed as 1/2 * sin(z)/z as this is a well-known limit.  (proving why lim(z->0) sin(z)/z == 1 is a bit beyond my recollection at this time!)

1/2 * 1 == 1/2. QED.

[bugo]

All of this math is literally giving me a headache.

[Alps]

The Inner Loop in Rochester is a 3/4 loop off of I-490 (1/4 is multiplexed, technically). NOT COUNTING the multiplexed part, which really isn't signed as such anyway, the average distance would be roughly 2/9 if it were considered a square. This is a strange but fun exercise...

[vtk]

to get the average, divide by the total arc length 2*pi*x, and then to get our score, divide again by total arc length.

That's why I think this metric is a little weird, because we're actually dividing by the length squared.  Then again, didn't some of the previously-discussed metrics divide by the square of the length, too?

this is of the form 0/0, so derivative of numerator and denominator yields sin(z)/(2z)

Yes, L'Hopital's Rule is quite useful.

proving why lim(z->0) sin(z)/z == 1 is a bit beyond my recollection at this time!

I forget the actual proof, but I know it involves the Sandwich Theorem.

[NE2]

proving why lim(z->0) sin(z)/z == 1 is a bit beyond my recollection at this time!

I forget the actual proof, but I know it involves the Sandwich Theorem.

This is also L'Hopital's Rule, innit?

[vtk]

proving why lim(z->0) sin(z)/z == 1 is a bit beyond my recollection at this time!

I forget the actual proof, but I know it involves the Sandwich Theorem.

This is also L'Hopital's Rule, innit?

Yes, apparently that also applies.  I think I must have learned about the Sandwich Theorem before L'Hopital's Rule, or maybe the former is used to prove the latter?  This is stuff I learned a decade ago, and typing that really makes me feel old.

[NE2]

Hmmm, I don't think I ever learned the 'sandwich theorem' per se - it was just "duh, of course that's true". I mean Archimedes used the bloody thing.

[agentsteel53]

That's why I think this metric is a little weird, because we're actually dividing by the length squared.  Then again, didn't some of the previously-discussed metrics divide by the square of the length, too?

I think it makes sense because the score ends up as a dimensionless figure, meaning that we can scale the world and the score remains the same.  thus, we can compare something like I-90/I-94 from Wisconsin to Montana, to something like I-10/I-610 in New Orleans, and have the relative results be meaningful.

proving why lim(z->0) sin(z)/z == 1 is a bit beyond my recollection at this time!

I forget the actual proof, but I know it involves the Sandwich Theorem.

Wikipedia calls it the squeeze theorem, which now I seem to vaguely recall my calculus teacher calling it that as well:

http://en.wikipedia.org/wiki/Squeeze_theorem

from there, we get that we can prove the sin(x)/x limit by using the inequality cos(x) < sin(x)/x < 1 for the two bounds.  

I don't remember - how do we prove that cos(x) < sin(x)/x as x approaches 0?

[NE2]

I don't remember - how do we prove that cos(x) < sin(x)/x as x approaches 0?

Start with tan(x)>x (I don't remember how best to prove this, but I think you can use the fact that the derivative of tan(x) (which is sec^2(x)) is greater than or equal to 1, to the curve is above a line with derivative equal to 1).