Street grid pattern
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From: alexy on 19 May 2010 16:31 "Mike Terry" <news.dead.person.stones(a)darjeeling.plus.com> wrote: >"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%. DOH! [dope slap] > >(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. Yeah, thinking of it another [equivalent] way, with the triangular grid, you have choices of roads going toward six different compass points, while for the square, roads only hear toward four compass points. So with the triangular configuration, you can, on average, get closer to a target direction. And the hexagonal tiling gives you no wider choice of road directions, while throwing larger blocks in the way of your progress. Thanks for bearing with me on this. -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.
From: alexy on 19 May 2010 17:14 "Rod Speed" <rod.speed.aaa(a)gmail.com> wrote: >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. Why just one block? Assume a city is a ten-mile by ten-mile square, laid out in a 1-mile grid. How do you get from the southwest corner to the northeast corner while traveling less than 14 miles? > >>> And it isnt 40% either. > >> I assumed he was rounding for sqrt(2) - 1 > >Sure, but its not that very often. No, just worst case, which was what he was asking about. > <snip> > >>>> 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. Back to the example above, assume the same city is laid out in a 1/10 mile grid. How can you get from SW to NE in less than 14 miles? What difference did the smaller grid make? -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.
From: Rod Speed on 19 May 2010 17:38 alexy wrote: > Rod Speed <rod.speed.aaa(a)gmail.com> wrote >> 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. > Why just one block? Because you hardly ever go from one corner to another over multiple blocks and when you do it from say one mid side to the furthest over over multiple blocks, you dont see that 40% extra with those etc. > Assume a city is a ten-mile by ten-mile square, laid out in a 1-mile grid. In practice most blocks arent that big in real citys. > How do you get from the southwest corner to the > northeast corner while traveling less than 14 miles? Its an academic scenario which isnt seen in practice enough to matter. >>>> And it isnt 40% either. >>> I assumed he was rounding for sqrt(2) - 1 >> Sure, but its not that very often. > No, just worst case, which was what he was asking about. Yes, but the worst case isnt really what matters in the real world. >>>>> 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. > Back to the example above, assume the same city is laid > out in a 1/10 mile grid. How can you get from SW to NE in > less than 14 miles? What difference did the smaller grid make? None with that particular example which is a tiny subset of real world trips. With real world trips, with say a trip from the middle of one of the sides to the middle of the opposite side, it helps substantially etc. And if you just want to fix the problem with the worst case, then you should have a couple of diagonals as well as the basic square grid of streets etc. But that really fucks up the intersections and thats the reason no one bothers when that worst case is so rarely seen in practice.
From: Mike Terry on 19 May 2010 17:51 "Les Cargill" <lcargill(a)cfl.rr.com> wrote in message news:4bf449c8$0$15824$9a6e19ea(a)unlimited.newshosting.com... > 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. > You can have an equilateral right triangle on a spherical surface - I expect that's what you meant :-) > -- > Les Cargill
From: Ostap Bender on 19 May 2010 18:32
On May 19, 11:39 am, "Mike Terry" <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > "alexy" <nos...(a)asbry.net> wrote in message > > news:ht17h0$6lb$1(a)news.eternal-september.org... > > > > > "Mike Terry" <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > > >"alexy" <nos...(a)asbry.net> wrote in message > > >news:ht11oe$l3t$1(a)news.eternal-september.org... > > >> Ron Peterson <r...(a)shell.core.com> 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. > > >> Actually, that is true only for a long trip. Worst case is a trip from > > >> x street, midway between y avenue and y+1 avenue to the same place on > > >> x+1 street. As the crow flies, it is 1 block, but by road it is 2 > > >> blocks, a 100% longer trip. > > > >> >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. > > > >> >Is there a non-Cartesian grid pattern for a flat community that does > > >> >better for the worst case than the Cartesian grid pattern? > > > >> With streets laid out in equilateral triangles, the worst case is a > > >> 75% longer (actually sqrt(3)-1) trip, but unlike the square blocks, > > >> the potential inefficiency doesn't decrease with distance. So the > > >> triangles are better for short trips, but worse for long ones. > > > >This seems the wrong way round to me. Also the 75% seems wrong... > > > >For a triangular grid: > > >- max inefficiency for short trip is 100% > > > You are correct. I was thinking from the midpoint of one side to the > > opposite apex. But the greater inefficiency is from points on two > > sides, equidistant from the junction of their two sides (and not too > > close to the other end of the street). > > > >- 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) > > > > > >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% > > Basically, with the triangular grid you can travel closer to the direction > you want in the worst case. But a larger percentage of your land is wasted on roads. |