Bicycle Tag of bike

Took me 5 minutes to find on google maps based on what I could see in the picture.

Canary Wharf, the Shard to the West of it, so it must have been taken from the North. Shard is taller than One Canada Square but Shard smaller in the picture, so we must be closer to CW than the Shard, so somewhere North and East of CW. Next to a bend in a big road and then started looking on Google Maps. Any suitable bend in the A12 would have had the Olympic Park in the foreground so tried further round to the East. A13. Look at terrain view as it's obviously something higher than the surrounding area. Beckton Alps. Bingo.

Tempted to try SCIENCE to see how close I can get by going all trigonometrical on the two obvious structures in the photo (using their known heights and distance apart, etc). Will post my attempt once someone has tagged it properly.

Didn't get that far, but what I found seems to tally with reality.

No surprise that the further away from an object you get the smaller it will be in a picture. With the types of distances we're talking about here we can assume this is linear and ignore the curvature of the earth.

Looking at the image a quick mess around with MS Paint gave One Canary Wharf as 38 pixels high, and the Shard as 27 pixels high. (These are very rough approximations.) The distance between the two is ~108 pixels (which I don't end up using).

OCW is 235m high. The Shard is ~285m high.

38/235 = 0.1617 (4dp)
27/285 = 0.0947 4dp)

Ratio between the two is:-

(38/235)/(27/285) = 1.71 (2dp)

So this says that we're 1.71 times further away from the Shard than we are One Canary Wharf, but no idea how far.

The distance between OCW and The Shard is near enough 4.6km (as measured using an online tool).

So the first point we could consider is a point 4.6/(2.71) = 1.7km on a direct line from OCW to the Shard (i.e. it is 2.9km from the Shard directly towards OCW).

We know that with it being 1.71 times further to the Shard than OCW we can't be further than a certain point otherwise we can't create a triangle, e.g. if we consider a point 171km from the Shard then we must be 100km from OCW, but we can't complete a triangle as it's only 4.6km from Shard to OCW (the third side of the triangle) and the two shortest sides (100km+4.6km) aren't longer than the longest side (171km). If the sides of the triangle are x, 1.71*x and 4.6km then the maximum value for x is when 1.71*x-x = 4.6km. 4.6km/0.71 = 6.5km (roughly). And so we're somewhere within 6.5km of OCW and 11.1km of the Shard.

Luckily the Shard and One Canary Wharf are pretty much at the same latitude so we can consider the line between the two as an x-axis. And so I'll use the Shard as (0,0) and OCW is (4.6,0).

Knowing the lengths of all 3 sides of a triangle (n, 1.71*n, 4.6) we can use the cosine rule to calculate all of the angles involved and therefore the coordinates of any of the points as n varies.

Doing this parametrically we can do the following in gnuplot:-

set parametric
set trange [0:6.5]
fx(t)=t*1.71*((((1.71*1.71*t*t)+(4.6*4.6)-(t*t)))/(2*1.71*t*4.6))
fy(t)=t*1.71*sin(acos(((((1.71*1.71*t*t)+(4.6*4.6)-(t*t)))/(2*1.71*t*4.6))))
plot [0:12][0:12][0:12] fx(t),fy(t)

And we get the following graph of possible points that are 1.7 times further away from (0,0) than from (4.6,0):-

(x and y axes are in km. the origin is the Shard)

Bonus points to anyone who can overlay this (to scale and fix the proportions) on a google maps image). I've run out of time.

The image position should be somewhere on/near that line (given the roundings and approximations). A fatter line would indicate a greater margin for error.

To see how close my guess was I used http://www.freemaptools.com/measure-distance.htm to measure the distance between the Shard/OCW and the real photo point:-

Shard to Beckton Alps = 10.313km
OCW to Beckton Alps = 5.782km

10.313/5.782 = 1.78 (2dp) which isn't far off my 1.71 approximation.