4 ms·

How about deform the physical map by a deformation field defined by ETA (i.e. stretch and compress)? e.g. https://www.slicer.org/wiki/Documentation/4.2/Modules/

How about deform the physical map by a deformation field defined by ETA (i.e. stretch and compress)? e.g. https://www.slicer.org/wiki/Documentation/4.2/Modules/DeformationFieldVisualizer https://www.slicer.org/wiki/Documentation/4.2/Modules/Deform... What would be cool about this is that it is still a map of the map, but the unit becomes time adjusted mile.

msutherl 9y ago
And given a suitable vector representation of the street layout, the stretch-and-compress could be given constraints so as not to warp (i.e. curve streets on) the map
- SubiculumCode 9y ago
  My thought exactly.
darkmighty 9y ago
Perhaps you can go even further and try to make it so that any two points have pairwise correct distances, like in the Azimuthal equidistant projection? https://en.wikipedia.org/wiki/Azimuthal_equidistant_projection https://en.wikipedia.org/wiki/Azimuthal_equidistant_projecti... Perhaps there are metric functions where this is not possible (a quick proof eludes me...[1]), but approximations might be possible anyway. [1] Edit: actually it's quite simple. Any non-isotropic metric cannot have an Euclidean equidistant projection, simply take a pair of points lying in rays (1,0) and (0,1) in an L^p metric, p not 2, as a counterexample.
- darkmighty 9y ago
  Edit2: Forget everything I said :) My geometry is quite rusty I guess. Not only any non-isotropic metric cannot have an Euclidean equidistant projection, any non-Euclidean metric clearly can't have an Euclidean equidistant projection. Metric spaces are defined by pairwise distances! What you can do however, is make the distance to 1 or 2 arbitrarily chosen points correct.
brncsk 9y ago
I actually did this for my bachelor's thesis[1] (maps on p52-53). It's not a pure time-distance representation (constraints on angular distortion to preserve similarity to conventional maps), but anyway. Data is from railway/tram time tables. [1] http://lazarus.elte.hu/hun/digkonyv/szakdolg/2013-bsc/barancsuk-adam.pdf http://lazarus.elte.hu/hun/digkonyv/szakdolg/2013-bsc/baranc... (in Hungarian)

msutherl 9y ago

And given a suitable vector representation of the street layout, the stretch-and-compress could be given constraints so as not to warp (i.e. curve streets on) the map

SubiculumCode 9y ago
My thought exactly.

darkmighty 9y ago

Perhaps you can go even further and try to make it so that any two points have pairwise correct distances, like in the Azimuthal equidistant projection? https://en.wikipedia.org/wiki/Azimuthal_equidistant_projection https://en.wikipedia.org/wiki/Azimuthal_equidistant_projecti... Perhaps there are metric functions where this is not possible (a quick proof eludes me...[1]), but approximations might be possible anyway. [1] Edit: actually it's quite simple. Any non-isotropic metric cannot have an Euclidean equidistant projection, simply take a pair of points lying in rays (1,0) and (0,1) in an L^p metric, p not 2, as a counterexample.

darkmighty 9y ago
Edit2: Forget everything I said :) My geometry is quite rusty I guess. Not only any non-isotropic metric cannot have an Euclidean equidistant projection, any non-Euclidean metric clearly can't have an Euclidean equidistant projection. Metric spaces are defined by pairwise distances! What you can do however, is make the distance to 1 or 2 arbitrarily chosen points correct.

brncsk 9y ago

I actually did this for my bachelor's thesis[1] (maps on p52-53). It's not a pure time-distance representation (constraints on angular distortion to preserve similarity to conventional maps), but anyway. Data is from railway/tram time tables. [1] http://lazarus.elte.hu/hun/digkonyv/szakdolg/2013-bsc/barancsuk-adam.pdf http://lazarus.elte.hu/hun/digkonyv/szakdolg/2013-bsc/baranc... (in Hungarian)