Thinking about the Twin Paradox, what would a Twin Traveling at .8c, calculate as the amount his EBT aged at the mid-point in a 4 lightyear trip? Referring to the twin paradox, where we have an Earth Bound Twin “EBT”  and a Traveling Twin “TT” who travels away from Earth at velocity= .8c, lets say to Alpha Centauri, when the TT is half way there, what would the TT calculate as how much his EBT’s aged at that point?
To simplify, round the distance to Alphas Centauri to say 4 lightyears.

I am thinking .8c yields a lambda of .6 and so the TT’s  4 lightyear trip, reduced to 2.4 lightyears due to space contraction. Therefore the midway point is only 1.2 light years. Traveling at .8c, means the TT must himself be 1.2/.8…. 1.5 years older at the midway point.
Since his EBT, relatively, is “traveling” at v=-.8c during this time, his EBT must appear to age relatively slower compared to him.  Using the lambda of .6, the TT would calculate as the amount aged by his EBT to be 1.5*.6=.9 years.
Is this correct?
I find it easier to just write down the coordinates of the events of interest.
Given coordinates $(t, x)$ in the frame of the Earth and choosing units such that $c=1$, the Lorentz transform to the frame of the traveller is $(t', x')$, where
$$\begin{align}
t' &amp;= \gamma(t - \beta x) \\
x' &amp;= \gamma(x - \beta t)
\end{align}$$
In your scenario, $\beta = \frac{v}{c} = \frac{4}{5}$, giving $\gamma = (1 - \beta^2)^{-1/2} = \frac{5}{3}$. The relevant events, in the reference frame of the Earth, are then:

Traveller leaves Earth at $(0, 0)$.
Traveller reaches halfway point at $(\frac{5}{2}, 2)$.
Earth is at the origin for all $t$, giving coordinates $(t, 0)$.

Transforming these into the traveller's frame, these become:

Traveller leaves Earth at $(0, 0)$.
Traveller reaches halfway point at $(\frac{3}{2}, 0)$ — as expected, the space coordinate is zero because the frame is moving with the traveller.
Earth is located at $(\frac{5}{3}t, -\frac{4}{3}t)$ — note this is expressed in terms of $t$, the time coordinate in the reference frame of the Earth. As expected, the Earth falls behind the traveller as time advances.

Equating the time coordinates in the traveller's frame gives $\frac{5}{3}t = \frac{3}{2}$, which gives $t = \frac{9}{10}$ years for the time that has passed on Earth, as viewed in the traveller's frame (who would have experienced a duration of $\frac{3}{2}$ years up to this point).