Calculating length of summer solstice day using tilt and latitude
If you are reducing everything to plain geometry, you can compute the fraction of the "sunny" arc of the circle of latitude for a given location. (Hint: draw a diagram showing the tropic of Cancer, the circle of latitude, the terminator and the axis, all as straight lines.) The height of this segment of a circle of latitude (at $\lambda$) is

$$h = \frac{\cos(\lambda) + \sin(\lambda)\times \tan(\epsilon)}{\cos(\lambda)}$$

The angle of this segment is

$$a = 2\times \arccos(1 - h)$$

I get 


New York: 14.9297092807291
London: 16.4195095586004


which is fairly close to the "expected" numbers you have provided. The difference is very likely due to Mathematica using a geoid rather than a sphere and including atmospheric refraction.