3 ms·

I don't think visibility would be all that hard to compute for the purposes of this site; I did similar calculations for radio "visibility" once (four thirds ea

I don't think visibility would be all that hard to compute for the purposes of this site; I did similar calculations for radio "visibility" once (four thirds earth radius etc.): - Choose a grid of square cells of appropriate size (can be quite large here) - Put it on a plane tangential to an idealized flat earth ground at the center (easy in ECEF) - Offset by the desired height - Perform a "raycast" from the center cell to the center of each other cell - Calculate elevation from interpolated SRTM data along the way and store the maximum elevation* along the path (idea: something similar to DDA traversal through the SRTM data? Wouldn't be exactly linear, though…) - You can pretty much derive anything via linear algebra from there * "Elevation" from the viewpoint of the chosen plane, obviously.

jvanderbot 4y ago
Visibility is not hard to compute. I said you could use "viewshed", which is the usual method of calculating line of sight to a given point from all points in an area. But it is a computationally intensive task to take a large enough area to accommodate any arbitrary asteroid impact. Orders of magnitude more than a web response, even accounting for down sampling and heuristics to restrict the size of the compute area. Almost any GIS software package can do it for you. https://en.m.wikipedia.org/wiki/Viewshed https://en.m.wikipedia.org/wiki/Viewshed What you're suggesting is that, except you didn't account for curvature of earth.
- btschaegg 4y ago
  > Orders of magnitude more than a web response That's likely where our different assesments come from, then. I was thinking more along the lines of "it's parallalelizable and easily offloaded to the client". An addendum for completeness' sake, though :) > What you're suggesting is that Not entirely. I'd build a LUT of "how high up must you be to be visible here" per "pixel", instead of "is the ground here visible", which I'd argue is more useful for modelling here. > except you didn't account for curvature of earth I didn't spell it out explicitly, but that's not the case if you do your projections right. (And you have to project, as SRTM data is always relative to the EGM96 geoid.)

jvanderbot 4y ago

Visibility is not hard to compute. I said you could use "viewshed", which is the usual method of calculating line of sight to a given point from all points in an area. But it is a computationally intensive task to take a large enough area to accommodate any arbitrary asteroid impact. Orders of magnitude more than a web response, even accounting for down sampling and heuristics to restrict the size of the compute area. Almost any GIS software package can do it for you. https://en.m.wikipedia.org/wiki/Viewshed https://en.m.wikipedia.org/wiki/Viewshed What you're suggesting is that, except you didn't account for curvature of earth.

btschaegg 4y ago
> Orders of magnitude more than a web response That's likely where our different assesments come from, then. I was thinking more along the lines of "it's parallalelizable and easily offloaded to the client". An addendum for completeness' sake, though :) > What you're suggesting is that Not entirely. I'd build a LUT of "how high up must you be to be visible here" per "pixel", instead of "is the ground here visible", which I'd argue is more useful for modelling here. > except you didn't account for curvature of earth I didn't spell it out explicitly, but that's not the case if you do your projections right. (And you have to project, as SRTM data is always relative to the EGM96 geoid.)