Just thought of the answer to my own question. The answer came to me while I was eating a frozen yoghurt at the shopping mall. How obvious !
Two reasonable assumptions:
1. The centres of the key shanks, black or white, are equally spaced.
2. The borders of the white notes are equally spaced.
For these particular tenths therefore, what matters for stretching purposes is the distance between the upper border of the lower white notes and the centre of the upper black note. Well, strictly speaking , the lower border of the upper black notes, 1/24 of an octave less, but for comparative playing purposes in this example this doesn't matter.
Let the distance of an octave be unity. Then the centres of the key shanks are therefore at 1/24 + 2n/24 and the borders of the whites at m/7.
For an E tenth:
The upper edge of the E is at 3/7 and the centre of the G# is at 41/24
Stretching distance = 41/24-3/7=215/168
For a D tenth:
The upper edge of the D is at 2/7 and the cente of the F# is at 37/24
Stretching distance=37/24-2/7=211/168
For an A tenth:
The upper edge of the A is at 6/7 and the centre of the C# is at 51/24
Stretching distance=51/24-6/7=214/168
For a B tenth:
The upper edge of the B is at 0 and the centre of the D# is at 31/24
Stretching distance=31/24-0=217/168
Which is precisely what I have observed. You know, I never thought of this. How bloody obvious !