Google Maps zoom level --> ratio…?

Question

Can anyone provide me with a link (or some details) on the actual ratio to "zoom level" figures for Google Maps?

e.g. Google Maps Level 13 = 1:20000

Answer 1

Not that easy. Given the projection, the size of the tile pixels depends on the latitude of the area you're interested in. Then in terms of transforming tile pixel size in screen pixel size, it depends on the screen and the resolution the data is displayed, the dpi your screen is using.

Answer 2

If you are designing a map you plan on overlaying over google maps or virtual earth and creating a tiling scheme then i think what you are looking for are the scales for each zoom level, use these:

20 : 1128.497220
19 : 2256.994440
18 : 4513.988880
17 : 9027.977761
16 : 18055.955520
15 : 36111.911040
14 : 72223.822090
13 : 144447.644200
12 : 288895.288400
11 : 577790.576700
10 : 1155581.153000
9  : 2311162.307000
8  : 4622324.614000
7  : 9244649.227000
6  : 18489298.450000
5  : 36978596.910000
4  : 73957193.820000
3  : 147914387.600000
2  : 295828775.300000
1  : 591657550.500000

Source: http://webhelp.esri.com/arcgisserver/9.3/java/index.htm#designing_overlay_gm_mve.htm

Answer 3

There is such a table in the documentation of the Virtal Earth Tile System from Microsoft. But as said by GuillaumeC the values depends on the latitude and on the screen resolution. The table gives values as measured at the Equator and at a screen resolution of 96 dpi.

PS: Not sure of that, but the zoom levels by Microsoft might be shifted by 1 in comparison to the zoom levels by Google. But they definitivly use the same projection so that the values remain correct for Google.

Answer 4

To help you understand the maths (not a precise calculation, it's just for illustration):

• Google's web map tile has 256 pixels of width

• let's say your computer monitor has 100 pixels per inch (PPI). That means 256 pixels are roughly 6.5 cm of length. And that's 0.065 m.

• on zoom level 0, the whole 360 degrees of longitude are visible in a single tile. You cannot observe this in Google Maps since it automatically moves to the zoom level 1, but you can see it on OpenStreetMap's map (it uses the same tiling scheme).

• 360 degress on the Equator are equal to Earth's circumference, 40,075.16 km, which is 40075160 m

• divide 40075160 m with 0.065 m and you'll get 616313361, which is a scale of zoom level 0 on the Equator for a computer monitor with 100 DPI

• so the point is that the scale depends on your monitor's PPI and on the latitude (because of the Mercator projection)
• for zoom level 1, the scale is one half of that of zoom level 0
• ...
• for zoom level N, the scale is one half of that of zoom level N-1

Also check out: http://wiki.openstreetmap.org/wiki/FAQ#What_is_the_map_scale_for_a_particular_zoom_level_of_the_map.3F

Answer 5

Just did some calculations and got the following results:

Google Maps shows a 1km ruler (bottom left of the map) which is 90 pixels in length, at zoom level 13. Which means the following:

Assuming the screen resolution is 96 dpi or 36 dpcm, at zoom level 13 we have 0.4km (from 36/90) in 1cm, which yields map scale of 1:40,000 for a 96dpi screen.

For various operations on the screen the best is to take 90px as a basis, as all numbers will be round at all zoom levels, i.e.

• Zoom level 12: 2km in 90px
• Zoom level 11: 4km in 90px
• Zoom level 10: 8km in 90px

and so on.

Note that this is an approximation that should work more or less fine on smaller scales rather than big ones.

(And Google likes round numbers in the end...)

Answer 6

Radius @ Equator 6,378,137 meters exact (WGS-84)

Circumference at Equator = 40,075,017 meters (2πr)

Zoom level 24 uses 2 to the 32 power (4,294,967,296) pixels at circumference.

Equatorial Circumference / 2 32 = .009330692 meters per pixel

Unit at Latitude = (Cosine of Latitude) X (Unit at Equator)

Zoom level doubles each increment.

1 foot (International) = 0.3048 meters

Edit

Well its not really a legitimate question to start with. Scale ratios are relative to printed documents not computer screens. What you need for these images to be used with any accuracy is to know the dimension of each pixel then scale the image according to whatever your overlaying it with.

So back 15-20 year ago someone took WGS-84 as base data. (note in a previous post someone used a value of 40,075,160 I've seen this in Wikipedia a few places and it's incorrect. The correct value is 40,075,017

They then took that and divided it by a full 32 bit integer. This is a logical choice as it yields global accuracy to about one centimeter which is plenty for aerial imagery. 32 bit integers are also efficient to store and process.

Why this is level 24 was chosen I don't know however as someone else here worked out 0 gets you down to one 256 pixel tile for the earth.

Now for an example of how to use the above data. Lets say I have an image at zoom level 20 (as zoomed as they currently let you get) Take 0.009330692 (Zoom 24 at equator) double it for zoom 23, again for zoom 22, again for zoom 21 and one last time for zoom 20. You should now have 0.149231071.

Now lets say our image is at latitude 45. Take the Cosine of that (0.707106781) and multiply it by our 0.149231071 and it will give you 0.105564729 meters. That is the length and height of one pixel from an image at latitude 45 at zoom level 20. If you screen capture a 1000 x 1000 pixel image of that area the dimension are 105.56 meters square. If you want feet divide that 0.3048

As for sources I essencial reversed engineer about 5 years ago from various bit of info and documentation I found on the web including Google and MS mapping support sites.

I have used this hundred of time and overlaying it with actually field survey data and its always been correct. Check it against any to the tables posted here and the numbers will match.