5
$\begingroup$

Note!!!!: I am not looking for help with how make a map using Google Maps API (that part is done). I need help with finding data.

I am building a shortwave radio/HF listening log book online, and it uses Google Maps for the plotting of receiving and transmitting stations when logs are made. I wanted to add a layer for real-time critical frequencies to the map, like the one provided by the Australian Space Weather Agency http://www.ips.gov.au/HF_Systems/6/5 (which is an image file). Is there data anywhere that provides plot points for this map?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ I think it might be worth taking the Google Maps mention out of the title. As it is it sounds like you're asking about programming Google Maps, rather than a data source, or what over on Stack Overflow they'd call a "give me the code" question. $\endgroup$
    – Kevin Reid AG6YO ♦
    Dec 5, 2013 at 2:49
  • 1
    $\begingroup$ Also, if it were the only source you found, it wouldn't be hard to take an image like that and use it as your data: just crop it and replace all the black or white pixels with neighboring values. (Check the source's terms of use before actually doing this.) $\endgroup$
    – Kevin Reid AG6YO ♦
    Dec 5, 2013 at 2:51
  • $\begingroup$ @KevinReid It's HARD. I gave it a shot, but unfortunately the projection of the image is way off from the projection if Google Maps. I can't for the life of me get things to match up. $\endgroup$
    – cj5
    Dec 5, 2013 at 6:31

1 Answer 1

Reset to default
4
$\begingroup$

The map projection used in the mentioned webpage is a cylindrical map projection, with the degrees latitude and longitude marked off on the left and bottom edges. Each 10 degrees latitude is about 24 pixels, and each degree longitude is about 27 pixels. It shouldn't be too hard to take each pixel and convert it to a latitude and longitude from this data. Then you'd use the google API to change the latitude and longitude back to the google's projection, which is a variant of the mercator projection. The projections are different, so you can't simply overlay this image on a google map and expect it to match. You'll have to collect the data points in terms of lat/lon pairs, then place them on the google projection using google's tools.

It's reasonably obvious that they are extrapolating this map from limited data, though - the curves and contours suggest that there are only a handful of data points which they then process to generate the map. They themselves suggest this is the case in the text on that page explaining where they get the data from. It appears they don't release their aggregated data in any other form than the map, though. You have two options - contact them and see if they will publish the hourly data in a way you can easily gather it, or go to the sites they list as their sources of data and collect the data yourself.

Neither option is particularly easy, but one or the other should work.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ I agree with @Adam here. I looked at some of the source links and it's clear they are going with like 30 data points in one case and extrapolating from there. I am sure they are using a cylindrical projection because it's a lot easier than a mercator projection math-wise. Though, I expect there must be libraries for conversion in almost any language you'd use. $\endgroup$
    – WPrecht
    Dec 6, 2013 at 15:23
  • $\begingroup$ Okay, now that I know the projection I'll try to see what I can do with converting the pixels to points on the map. Thanks for pointing that out. I have dug through a lot of the FTP sites that they link to, but cannot make heads or tails of the data, and a lot of it seems to show that there is missing data (-1, -1.0, etc...). If I know what the original projection is, I can do the conversion to Mercator. $\endgroup$
    – cj5
    Dec 6, 2013 at 19:52
  • $\begingroup$ This didn't work out. There was no way to determine the projection of the image, if it even is applied, and no matter how I pushed and pulled it, it would not correctly align with the proper coordinate bounds. $\endgroup$
    – cj5
    Jan 10, 2014 at 16:58
  • $\begingroup$ @ChrisWalsh Are you overthinking this a little bit? The map provided is very, very low resolution. Not just in terms of display pixels, but the actual data used for it doesn't provide accuracy to even 10 degrees. Even if you can figure out the projection problem and align it to with a few degrees, does that really buy you anything? Why not just accept that some of the data is going to be off a little bit, but given the source data it probably isn't a big problem, and maybe revisit the projection/data source problem later? $\endgroup$
    – Adam Davis
    Jan 10, 2014 at 17:14
  • $\begingroup$ I might be, but either way the overlay of the image is too messy by my standards. I contacted ASWA via email, and requested if they could provide me with raw data. I'll see where that leads. $\endgroup$
    – cj5
    Jan 11, 2014 at 18:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .