**EINSTEINIAN LENGTH CONTRACTION.**A 1 km long train (1 km long in the observer's stationary frame) goes past the observer at 50 km/s. Einstein would say that the train appears shorter by 0.014 mm. Here the standard FitzGerald-Lorentz factor of (1-vv/cc)^0.5 for 50 km/s is 0.999 999 986 111 111.

**AETHERIAN LENGTH CONTRACTION.**If the aetherwind is 00 km/s, ie if the observer is in the absolute frame, then Mac, an aetherist, will agree with Einstein's 0.014 mm. Using the same factor.

**AETHERWIND.**But if the aetherwind is blowing at 500 km/s vertically upwards through the observer then Mac will say that the train will appear shorter by 0.140 mm, ie approx 10 times the Einsteinian number. This is because Mac draws a triangle 500 km/s high & 50 km/s long with a hypotenuse of 502.5 km/s, the 502.5 km/s being the aether wind blowing through the train. Mac uses the 502.5 for the v, giving a factor of 0.999 998 597. This gives a 1.40 mm contraction along the almost vertical angle of the 502.5 km/s. But the train is not on that angle, it is horizontal, so Mac calculates the horizontal component of the 1.40 mm by using the factor 50/502.5 which gives 0.140 mm.

**EINSTEINIAN TIME DILATION.**An atomic clock (which Mac knows ticks at 1,000,000 cps in the observer's stationary frame) goes past the observer at 50 km/s. Einstein would say that the clock appears to tick slower by 0.013 888 889 cps. Here the standard Lorentz time dilation factor of (1-vv/cc)^0.5 for 50 km/s is 0.999 999 986 111 111, ie the same as for length contraction.

**AETHERIAN TICKING DILATION.**If the aetherwind is 00 km/s, ie if the observer is in the absolute frame, then Mac will agree with Einstein's 0.013 888 889 cps. Using the same factor.

**AETHERWIND.**But if the aetherwind is blowing at 500 km/s vertically upwards through the observer then Mac will say that the clock will appear to tick slower by 0.013 888 927 cps.

That is almost identical to the Einsteinian 0.013 888 889 cps. The question arises, why is this so close to the Einsteinian number, when the earlier similar scenario for length contraction gave an Aetherian number 10 times the Einsteinian number?

The answer is that in the case of length contraction Mac could ignore the 500 km/s vertical aetherwind because the horizontal component was zero km/s. But Mac cant ignore it for ticking, because ticking is they say a scalar. In other words, ticking depends on the speed of the aetherwind, whereas length contraction depends on the velocity of the aetherwind. Therefore Mac knows that the 1,000,000 cps in the stationary frame (K) will have an actual value of 999,998,611 cps in the observer's frame (k), & an actual value of 999,998,597 cps in the clock's frame (k'). And Mac knows that the observer in frame (k) will believe/feel that his own identical atomic clock has an apparent ticking of 1,000,000 cps, because it too is affected by the 500 km/s wind.

**CRAWLER'S LAW**says that the apparent ticking of an atomic clock is identical in every inertial frame (even though the actual ticking will usually be different)(the actual will depend on the speed of the aetherwind)(but the apparent wont, it never varies).

Where were we? Ok, as it turns out, 0.999,998,611 divided by 0.999,998,597 is 0.999,999,986.111 072, very close to the Einsteinian 0.999 999 986 111 111. The end result is that the difference amounts to only 3.4 nano seconds in a day.

That’s the way it goes. Einstein's silly STR gives goodish numbers very easily & quickly in some scenarios. When i say goodish, in fact i believe that both theories are wrong. I reckon that the factor for length contraction & the factor for ticking dilation are both wrong. But that’s another story.