Help Me Gain an Intuitive Understanding of Lorentz Contraction I'm having a hard time getting an intuitive understanding of Lorentz Contraction. I understand what it is by definition but I don't 'get it.' I'm not a physicist, just an amateur, so sorry if this question comes across as too naïve.
Okay, I was able to understand Time Dilation with the help of the 'light bouncing off two mirrors' experiment (the one that uses Pythagoras' theorem to derive the equation of time dilation) and I've noticed that this same scenario is also used on a lot of websites to derive Lorentz Contraction. So, I'll use it to present my question:
First, to define the setup:
Let's assume person A is "at rest" and person B goes by in a spaceship traveling at velocity $v$, close to the speed of light $c$. Let's call the time measured by A $t$ and that measured by B $t'$. So, we have:
$$t = t'/\sqrt{1-v^2/c^2}$$
I understand that we can exchange A and B (consider B to be "at rest" instead of A) and come up with the same relation. So, for each observer, the other person's watch seems to move slower.
So, if B were traveling at $v = 0.8c$, $\sqrt{1-v^2/c^2} = 0.6$. This means that in the time interval that A counts off 5 minutes on his watch, B counts off only $5*.6 = 3$ minutes.
Okay, this must all be old hat to you guys but I wrote all this just so that you know how much of this I understand.

Now to get to my question, I can't follow most of the explanations of Lorentz Contraction that use this light-bouncing experiment. Here's why:


*

*Lorentz Contraction works in the same direction as the motion of the object. So, immediately, the light-bouncing experiment starts making less sense for me because for time dilation it was the transverse (to the movement of the object) motion of light that created the right-angled triangle and allowed us to use Pythagoras in the first place. If the light is bouncing (off the mirrors) in the same direction as B's spaceship is moving, A would see exactly what B does: a single beam of light bouncing off each mirror alternately, retracing its own path over and over again. Moreover, since light always travels at a constant speed of $c$ in vacuum, if they both happened to time the bouncing of the light beam, they'd both measure the same interval between bounces (I'm not saying the bounces would be synchronized on both their watches, just that the interval would be the same).

*I know that Lorentz Contraction means that A will measure the length of B's spaceship (in its direction of motion) as being smaller than what it actually is (or what it is in B's frame of reference). Since the only constant in all of this is the speed of light, the only way acceptable to all observers is to measure a distance using $c$ as a yardstick. So... imagine that A sees a beam of light start at the 'rear' of B's spaceship and make its way forward to the 'front' of the spaceship. Let's say A times the journey and finds that it takes t seconds on his watch to for the light to cover the distance between the rear and the front. So, for A, B's spaceship is $c*t$ units long. However, since A knows that B's watch is going slower than his own, he can infer that if B sitting in his spaceship had also been timing the beam of light, the time that B measured (say $t'$ seconds) would be $< t$. So, A can deduce that B's spaceship is actually $c*t'$ units long where $$(c*t) > (c*t')$$ or $$\text{A's measure of spaceship length} > \text{B's measure of spaceship length}$$
Now this is precisely the opposite of Lorentz Contraction.
Where did I go wrong?
 A: Please review this answer first: Einstein's postulates $\leftrightarrow$ Minkowski space for a Layman , and understand space-time pictures. Then you get "time dilation" and "Lorentz contraction" from two simple pictures.
Time Dilation
Time dilation refers a line which is in the direction of the time axis, but tilted a little to the right, and which represents the trajectory of the moving observer. This line has equally spaced dots along it's length, which are the ticks of the moving observer's clock. What is the vertical spacing of these equally spaced dots?
In geometry, they would occur with vertical spacing reduced by a factor of $1\over\sqrt{1+v^2}$. In relativity, they occur with vertical spacing increased by a factor of $1\over\sqrt{1-v^2}$ (in units where c=1). It's the same argument, but for the minus sign in the pythagorean theorem.
Length contraction
Length contraction refers to two parallel lines which are both exactly vertical. These are the two ends of a stationary ruler, and their length is measured perpendicularly between them to be L.
Now if you are a moving observer, your t-axis is tilted by a slope of v relative to the original t axis, and your x-axis is also tilted by a slope of v relative to the original x axis. So the actual length of the segment of your x-axis between the two ends of the rulers is the hypetenuse of a right triangle with sides L and Lv. In geometry, it would be longer by a factor of $\sqrt{1+v^2}$, but in relativity, it's length is $L\sqrt{1-v^2}$.
Time contraction
If you turn the Length contraction picture on its side, so that the x-axis becomes the t axis and the t-axis becomes the (negative of the) x axis, then you get a strange picture of horizontal lines. These represent a line of simultaneous flag-wavers. They are everywhere in space, and they lift a flag up once a second.
If you move through these flag wavers in a rocket, and you look out the window to see how often the flag-wavers seem to wave to you, you see a different flag-waver wave a flag each time you see a flag waved. How often do the waves come?
The answer is that they come more frequently by a factor of $\sqrt{1-v^2}$. This is "time contraction", it is the length contraction picture tipped over in time.
Length dilation
If you tip the time dilation picture by 90 degrees into space, it gives you a simultaneity line for a moving observer. This line is marked up by flag-wavers at 1m who wave their flag exactly once.
If you have a moving system of flag wavers, and they all measure the distance between them to be 1m, and they wave their flags once at exactly the same time as measured by them, how far apart are the flag-waving events as measured by you?
Because it is just a tipped over time-dilation picture, the answer is it is longer by $1\over\sqrt{1-v^2}$, the same time-dilation factor for a spatial interval. You can't get any of this straight without a picture, and it is just as obvious as Euclid's geometry, except you have to get used to minus signs in the pythagorean theorem.
A: 
If the light is bouncing (off the mirrors) in the same direction as B's spaceship is moving, A would see exactly what B does: a single beam of light bouncing off each mirror alternately, retracing its own path over and over again.

Remember that the mirrors are moving. So when the light beam travels from the rear mirror to the forward mirror, observer A would actually see a light beam having to catch up to a receding mirror. Similarly, when the light beam travels from the forward mirror to the rear mirror, observer A would see the mirror catching up to the light. This means that according to observer A, the light beam travels further each time it goes forward than when it goes backward, as this image shows:

Even the two halves of the light beam's trip are not the same length according to A, so clearly A and B have to measure different intervals for at least one of those halves (and in fact both).
Quantitatively, suppose the relative speed of A and B is $v$ and the distance between the mirrors (as seen by A) is $\Delta x_A$. On the forward trip of the light beam, as observed by A, the position of the light beam is described by $x_\text{light} = ct$ and the position of the forward mirror is described by $x_\text{mirror} = \Delta x_A + vt$. The time it takes for the light to reach the mirror is obtained by setting these equal to each other:
$$\Delta t_\text{forward} = \frac{\Delta x_A}{c - v}$$
On the backward trip of the light beam, observer A sees $x_\text{light} = -ct$ and $x_\text{mirror} = -\Delta x_A + vt$, so
$$\Delta t_\text{backward} = \frac{\Delta x_A}{c + v}$$
Adding it up, you get a total round-trip time of
$$\Delta t_{A,\text{total}} = \frac{2c\Delta x_A}{c^2 - v^2}$$
Now, suppose you want to find the relationship between $\Delta x_A$ and $\Delta x_B$, the proper distance (i.e. as seen by B) between the mirrors. Hopefully it should be clear that if you look at this from B's perspective, you get
$$\Delta t_{B,\text{total}} = \frac{2\Delta x_B}{c}$$
If you believe the time dilation formula (and it sounds like you do), you can write
$$\Delta t_A = \frac{\Delta t_B}{\sqrt{1 - \frac{v^2}{c^2}}}$$
and now combining the last three equations leads you to
$$\Delta x_A = \Delta x_B\sqrt{1-\frac{v^2}{c^2}}$$
Basically, time dilation is able to account for part of the factor of $\bigl(1 - \frac{v^2}{c^2}\bigr)$ difference between $\Delta t_{A,\text{total}}$ and $\Delta t_{B,\text{total}}$, but not all of it. We have to attribute the rest to length contraction.

Since the only constant in all of this is the speed of light, the only way acceptable to all observers is to measure a distance using c as a yardstick. So... imagine that A sees a beam of light start at the 'rear' of B's spaceship and make its way forward to the 'front' of the spaceship.

Actually, that's not the best way to go about measuring distances, for exactly the reason I described above. As you saw, if you time a light beam traveling from the back of the spaceship to the front (or from a rear mirror to a forward mirror), the time you will actually measure is $\Delta t = \frac{\Delta x}{c - v}$, not $\Delta t = \frac{\Delta x}{c}$ as you thought. (Of course, in a reference frame where the distance being measured is at rest, this works fine since $v = 0$.)
The easiest and recommended way to measure the distance of a moving object is by sitting still at a point and recording the times when the front and back of the object pass you. Once you have the time difference, you can determine the length of the object in your reference frame by $\Delta x = v\Delta t$, where $v$ is the object's speed relative to you. Since boosts between different reference frames "mix" time and space, it's easiest to keep your spatial coordinate fixed when you're measuring time, and vice-versa.
A: Two very good answers both keeping rigidly to mainstream thinking on special relativity.
However, there is a flaw in this thinking.
In 1971, Hafele & Keating did the experiment of comparing the time measured by previously synchronised atomic clocks with one moving and the other "stationary". Since this experiment, the mainstream has ignored the proven fact that time really does pass more slowly for the entity that accelerates and moves with constant velocity. Time dilation is not simply an artefact of relative motion, it is a reality, the only reality. 
The two clocks really did show the difference in elapsed time for the two frames of reference. This is at varience with the accepted mainstream idea of there being no preferred reference frame, ie, either clock must appear slow to the other, whichever frame you observe from. Since one clock really does show less elapsed time than the other, this cannot be the case.
The only difference between the frames is the fact that only one of them accelerated and incresed its relative speed. The other did not and its clock therefore keeps normal undilated time.
Time in the "accelerated" frame is proven to pass more slowly than in the stationary frame and this is in accordance with special relativity. But, we cannot simply say that because we can consider either frame to be stationary, then it does not matter which one moved. It definitely does matter which one moved as the clock goes slower in that frame. This logical assumption is wrong because it ignores the history of how the relative motion was produced. It was produce by the one frame being accelerated and so the relative time dilation is NOT symmetrical.
To me, this is not surprising when you consider the four dimensions involved. The three physical spacial dimensions are symmetrical in that they extend in two directions from any point in space to +infinity to -infinity.
The dimension of time however, is anti-symmetrical or bi-direction, or assymetrical. It can only extend in one direction, into the future.
Without writing a book on this ("TIME DILATION The Reality"), it can be shown that length contraction is merely a relative illusion but that time dilation is indeed a reality and Lorentz mathematics proves just that if you hold this to be the case.
A: The Lorentz time dilation explains how to transform the time when it was observed in different reference systems. Let us now as space becomes. Try to understand how they relate as reference systems $S (x, y, z, t)$ and $S´ (x ', y', z ', t')$  To better understand the system assume that S is the set of Ivete, blue and S 'of Jan, red.
Using the length of a rigid rod for measurements of space. We believe that this bar is along the x axis at rest in frame S (IVETA), blue in the initial image. At rest in S, position measurements are made at the same time we can measure the length of the bar by their extreme position $x1$ and $x2$. We call this length L0 is given by:
$$L_0 = X_2 -X_1   $$
!
This measure of the length of the bar resting length will call or proper length of the bar. Whenever we measure the length in the system where it is at rest length will call their own.
Similarly we can have another bar identical to the previous rigid located in the S '(Jan) along the axis x' and at rest in this system. The rod length is measured by Jan:
$$L_0 = X_2´ -X_1´   $$
and is the resting length and proper length of the bar to Jan (S ').
But what happens when Jan wants to measure the bar Ivet Ivet want to measure or bar Jan? That is, measuring the lengths of these bars seen from a moving frame. Not forget that for Ivet, Jan moves to Jan Ivet velocity v moves at speed-v.
Now we have to include the time the measurement is carried. The length of the bar seen from Jan (S ') is measured at time t' is important to note that the time is the same as that x'2 x'1. It is easy to understand, Jan takes a ruler and measure the bar by looking at the same time in the initial and final position, the difference is the length in S '. Ivet will do the same for the bar at S.
Well, how do you measure the bar Ivet Jan?. The measurement result is provided by the Lorentz transformation.
hope i help you. 
