Street grid pattern
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From: alexy on 19 May 2010 15:25 Les Cargill <lcargill(a)cfl.rr.com> wrote: >Is this because the worst-case is SQRT(2) for running the sides >of an equilateral right triangle Times, they are a-changing. They hadn't even invented equilateral right triangles when I was in eighth grade! ;-) -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.
From: Mike Terry on 19 May 2010 15:49 "alexy" <nospam(a)asbry.net> wrote in message news:ht1cte$kof$1(a)news.eternal-september.org... > "Mike Terry" <news.dead.person.stones(a)darjeeling.plus.com> wrote: > > >"alexy" <nospam(a)asbry.net> wrote in message > >news:ht17h0$6lb$1(a)news.eternal-september.org... > >> "Mike Terry" <news.dead.person.stones(a)darjeeling.plus.com> wrote: > > > >> >- max inefficiency for long trip is about 15% > >> > >> I don't get this. Assume the streets are laid out (using compass > >> points) 0-180, 60-240, and 120-300. If you want to go due east, you > >> are forced to take roads heading either 60 or 120 to make eastward > >> progress, and N-S roads to bring you back to the intended due east > >> course. How can you do that with only 15% inefficiency? > > > >You go either 60 or 120. Your intended direction is 90, so you are only > >travelling at 30 degrees "off course". 1/cos(30 deg) ~= 1.155 (i.e. about > >15% overhead) > > > Well, if you want to go 1 mile east, I agree that after traveling on a > 60-degree road for 1.155 miles, the eastward component of your travel > will have been 1 mile. But you will be .58 miles north of the point > you were aiming for. Travel down the N-S road to your target, and you > will have traversed 1.73 miles to go one mile due east. After travelling on a 60-degree road for a while, to get rid of your North component, travel on a 120-degree road for the same distance. Both directions have the same 15% overhead so overall you get an overhead of just 15%. (Remember we're considering *long* journeys at this point, i.e. they're large compared to the grid size...) > >> > >> > > >> >For a square grid, both are about 41% > >> > >> How do you get from 13th street, half-way between 1st and 2nd avenues, > >> to 14th street, half-way between 1st and 2nd avenues while walking > >> only 1.41 blocks? > > > >Hey you're right! Worst case for square grid is 100% overhead. > > > >But long distance worst case is about 41%, no? You are travelling at worst > >at 45 degrees from your intended direction, and 1/cos(45 deg) ~= 1.414. > > > >> > > >> >So the triangles are better for LONG trips, but worse for SHORT ones?? > >> > >> No, with your correction of my shart-trip inefficiency for the > >> triangle, I believe the triangular layout is worse for both. > > > >Square grid: > > Worst case : 100% > > Long distance worst case: 41% > > > >Triangular grid: > > Worst case : 100% > > Long distance worst case: 15% > I still think the 15% should be 73%. > > > >Basically, with the triangular grid you can travel closer to the direction > >you want in the worst case. > But your "correction" is along a road that has no component > contributing to your intended direction of travel, while for the > square, you are making progress even on the correction leg. > > I think tiling the city with hexagons may actually be the most > efficient, but I haven't tried to put any numbers to paper yet on > that. (Enough trouble just getting it right on squares and triangles!) I'll have to think of that - my first stab would be: Worst case : 73% Long distance worst case: 33% I expect 73% is the minimum achievable for absolute worst case, but long distance would seem to be better on a triangular grid. (The long distance case giving 33% is in the direction of one of the lattice lines, which had no overhead in the triangular lattice, but on the hexagonal lattice you're forced to keep diverting off and back again, so you have to travel 4 units to get 3 units further away. But maybe there's a worse direction to have to travel?) Regards, Mike.
From: Rod Speed on 19 May 2010 15:51 alexy wrote > Rod Speed <rod.speed.aaa(a)gmail.com> wrote >> Ron Peterson wrote >>> Some cities have a rectangular grid pattern for their streets and >>> in the worst case, trips can be 40% longer than a straight line trip. >> But thats only with the shortest trips, so isnt as significant as it looks. > If a city is laid out on a N-S / E-W grid, how can you > move in a northeasterly direction without traveling sqrt(2) > times the straight-line distance between the points. You cant, but that is only for just one block usually, so that isnt as significant as the 40% looks. >> And it isnt 40% either. > I assumed he was rounding for sqrt(2) - 1 Sure, but its not that very often. >>> Other cities, have streets that follow the geography >>> because of mountains, streams, and lakes with some >>> trips being several times as long as a straight line trip. >> Yes, mostly with lakes and big rivers etc. >>> Is there a non-Cartesian grid pattern for a flat community that >>> does better for the worst case than the Cartesian grid pattern? >> Yes, a higher density cartesian grid, smaller blocks. > Won't make any difference, Yes it does with the actual extra distance travelled as opposed to the percentage. > assuming we are talking about idealized zero-width streets and point intersections. Stupid assumption. > Obviously, with real-world street widths, you can save a little by > walking diagonally up each block, and this effect would improve with > smaller blocks, but I took his question to be more the idealized case. More fool you. And if he was talking about walking and not the roads he actually mentioned, one obvious way to reduce the distance walked is what happens in the real world, walkways that arent roads. >>> Am I asking the right question? >> Nope. > Well, why don't you help him frame the question you would like to answer? What I would like to answer is completely irrelevant. His question wasnt specific enough, so it isnt the right question.
From: Rod Speed on 19 May 2010 15:53 alexy wrote > Les Cargill <lcargill(a)cfl.rr.com> wrote >> Is this because the worst-case is SQRT(2) for >> running the sides of an equilateral right triangle > Times, they are a-changing. They hadn't even invented > equilateral right triangles when I was in eighth grade! ;-) Yes, you are way past your useby date. Thats the new maths, silly.
From: Les Cargill on 19 May 2010 16:27
alexy wrote: > Les Cargill<lcargill(a)cfl.rr.com> wrote: > >> Is this because the worst-case is SQRT(2) for running the sides >> of an equilateral right triangle > > Times, they are a-changing. They hadn't even invented equilateral > right triangles when I was in eighth grade! ;-) Doh! Isosceles, then - 45/45/90 being the angles in units of degress. -- Les Cargill |