Velocity Time Dilation Confusion So I'm trying to understand velocity time dilation, even still after brushing it multiple times throughout university. I am doing so with the example of a spaceship travelling to Alpha Centauri [AC].
Lets approximate both the Earth and AC to be stationary, the distance between Earth and AC to be be 4ly, the speed of our relativistic spaceship to be 0.95c, the reference frame of the stationary Earth to be S, and the reference frame of the spaceship to be $S^{'}$.
As I understand, the equation $T^{'} = \gamma T$ (with $T^{'}$ = observed time in S, T = rest time in $S^{'}$, $\gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$, v = speed of moving body ) will tell us the dilated time that an observer on Earth will measure for our spaceship.
At a speed of 0.95c, it will take the spaceship $\approx$ 4.2yrs to get to AC. Thus, setting T=4.2yrs, the equation above tells us the an observer on Earth will instead say that it took the spaceship 13.5yrs. This seems to fit with the famous phrase "moving clocks run slow".
I know special relativity and time dilation are supposed to be counter-intuitive, but I understand the above example as saying that going faster (approaching the speed of light) also makes other observers perceive you to be going slower than you really are.
Thoughts? Explanations? Is my use of  $T^{'} = \gamma T$ incorrect? Did I assign variables incorrectly or miss anything integral? Is SR and time dilation really that weird?
 A: If T' is the observed time in S, then T' ~ 4.2 yrs as the observer in S measures the time to reach AC as 4.2 ly divided by the 0.95c speed. Then T = T'/ which is <T' so the clock in S' is going slower.
A: I don't understand your interpretation of the time dilation equation. The equation is (in words) the relationship between the elapsed coordinate time and the elapsed proper time of a relatively moving clock.
$$\mathrm{elapsed\,time\,according\,to\,synchronized\,clocks\,at\,rest\,in\,S} = \gamma\cdot \mathrm{elapsed\,time\,according\,to\,clock\,moving\,in\,S}$$
The clock moving in $S$ is the spaceship's clock. Denote the elapsed time for the journey from Earth to AC according to this clock as $\tau$.
There are two synchronized clocks at rest in S, one on Earth (to record the time of the event that the spaceship leaves Earth) and one at AC (to record the time of the event that the spaceship arrives at AC). From the time of these two events recorded by these two clocks, the elapsed coordinate time in $S$ for the spaceship's journey is $4.2\,\mathrm{yr}$.
Thus
$$4.2\,\mathrm{yr} = \gamma\cdot\tau\quad\rightarrow\quad\tau \approx 1.3 \,\mathrm{yr}$$
This is the reason we say moving clocks run slow. Note that this is symmetrical. The clocks at rest in $S$ run slow as observed from the spacecraft and, further, the clock at Earth is observed to not be synchronized with the clock at AC.
