Making some logical assumptions about the coordinate systems being used (which should really be spelled out explicitly), the equation
t(B) - t(A) = r(AB) / (c-v)
does not say that the photon moves at a velocity c-v in any frame.
What it says that if a photon is approaching another moving object, the time it takes in the frame where the object is moving is given by solving
d = c*t +A (a light beam starting out at d=A at t=0)
d' = v*t+B (an object starting at d=B at t=0 with velocity v)
for the time when d=d'. This means c*t+A = v*t+B, or (c-v)*t = B-A
When you go to the frame of the moving object, you find (of course) theat the velocity of the photon in the moving frame is equal to c. You finding that in the moving frame, the light beam and the object do not start out at the same time 't'. that the distance is lorentz contracted, and that time is dilated.
[add]
I'm going to explain this more fully, in the hope that it will get through. (I'm guessing that this is the same person I talked to before that was confused about this issue, though of course it's a resaonably common confusion)
.
Anyway...
We have two frames of reference. One is stationary, one is moving at velocity v. The origins of the two frames coincide. We have an object, who is also moving at velocity v (the same velocity that the two frames are moving).
in frame 1, a light beam starts out at t=A, v=c
In frame 1, the equation of the light beam is
x = A + c*t
To convert this to frame 2, we use the Lorentz transform
x' = gamma*(x - v*t) t' = gamma*(t - v*x/c^2)
where gamma = 1/sqrt(1-v^2/c^2)
since x=A and t=0, we find
in frame 2, a light beam starts out at
x' = gamma*A, and t' = -gamma*v*A/c^2.
Since the velocity of light is a constant, 'c', in any frame, the equation for the motion of the light beam in frame 2 is
1) x = gamma*A + c*(t' + gamma*v*A/c^2)
(the only equation for an object moving at 'c' that passes through the given point).
Similarly, in frame 2, we find that an object starts out at
x' = gamma *B and t' = -gamma*v*B/c^2
The relativistic velocity equation will give a velocity for the object in frame 2 as zero, because I am making the assumption that our frame is comoving with this object (but has an origin that's displaced).
Thus the equation of motion for the moving object as a function of time in frame 2 is
2) x' = gamma*B
So, the two objects will meet when eq 1) equals eq 2)
gamma*A + gamma*A*v/c
gamma*A + c*(t' + gamma*v*A/c^2) = gamma*B
which boils down to gamma*A(1+v/c) + c*t' = gamma*B
or t' = gamma*(B-A)/c - gamma*A*v/c^2
which is quite consistent, it says that the distance is contracted by gamma, and you have a time offset of -gamma*A*v/c^2 due to the relativity of simultaneity.