Basic Relativistic Question - length measurement A while ago we did an easy, introductory exercise on length measurement. Back then it seemed pretty straightforward but now when I look at it I have trouble understanding the assumption which led to the answer in the blink of an eye.
Jack (say his frame is U') flew over a house at the speed of V=0.8c and it took him T=100 ns. What's the length of a house in Earth's reference frame?
We solved this by plugging $\Delta x'$ = 0 and the rest came easily. I don't get it... If we do a measurement in whatever reference frame we should make it simultaneously at two points. My best guess : Jack uses a clock which doesn't move in his frame. 
I'd be really grateful if someone could explain this situation to me or come up with some solution where this assumption is not made or comes up naturally.
 A: Flying at $0.8c$, Jack travels in $100 ns$ the distance $L_J=24m$ $(x=vt)$, where I made the approximation $c=3 \cdot 10^8 m/s$. However, the length he sees is contracted by the factor $\gamma=\frac{1}{\sqrt{1-\beta^2}}$, where $\beta=\frac{v}{c}$.
So taking this length contraction into account, the length $L$ of the house in the reference frame of the earth is $L=L_J \gamma = 40m$.
If you are familiar with Minkowski diagrams, then you just draw the situation and the result is obvious. (As a personal note: almost all special relativity problems can easily solved by drawing the corresponding Minkowski diagram. Unfortunately I don't know how to draw here in SE...)
A: You said 

If we do a measurement in whatever reference frame we should make it simultaneously at two points.

That's not quite right. If you change ''a measurement'' to ''a length measurement'' then you are correct. What you must remember is that $\Delta x$, $\Delta t$, $\Delta x'$, $\Delta t'$, etc are actually differences in spacetime event measurements. The data which we gather is from an event, and we get (position, time). If you measure the length of an object, you actually measure two events which happen simultaneously.
The length not the information given. The information is two events with a stationary clock: when the clock (or some fixed point in U') is at the beginning of the building $\left(\matrix{c T'\\X'}\right)$ and when the clock is at the end of the building (as the clock at rest in U' flies past) $\left(\matrix{c(T'+\Delta t')\\X'}\right)$. [I'm using 4-vectors, but ignoring the y' and z' components].
Now we calculate the difference 4-vector in the U' frame,$\left(\matrix{c\Delta t'\\0}\right)$ . This can be transformed quickly:
$$\left(\matrix{c\Delta t\\ \Delta x}\right)=\left(\matrix{\gamma & \beta\gamma\\ \beta\gamma&\gamma}\right)\left(\matrix{c\Delta t'\\ 0}\right)$$
You know $\Delta t'$=100 ns.
The key here is "What events are being reported to me and what is the reference frame of each event?"
