Does constant acceleration (from starship crew pov) to relativistic speeds look newtonian? The last 25 years I thought that time dilation, mass increase and length contraction is "scaled" that way that on the first glance the crew of an accelerating starship could think everything is newtionain. I thought that when a starship accelerates constantly with 1g for 10 years (from the crews pov) it reaches roughly the speed that gives them a time dilation, mass increase and length contraction of factor 10, so they expierience a 10 light years journey (from stationary pov) in just one year of their personal time. From stationary pov they need 10 years and time ist slowed down to 0.1, from ships pov they need one year because the distance is contracted by 0.1.
So I thought that you have to accelerate for 100 years with 1g (ships pov) to get a time dilation of factor 100.
Yesterday I read that article.
https://en.wikipedia.org/wiki/Space_travel_using_constant_acceleration
which states:

At a constant acceleration of 1g, a rocket could travel the diameter
  of our galaxy in about 12 years; if the last half of the trip involves
  deceleration at 1g, the trip would take about 24 years.

1g constant acceleration for 12 years could only be meant from the ships pov.
But on the other hand it says:

A half-myth: It gets harder to push a ship faster as it gets closer to
  the speed of light This is a half-myth because it depends on the frame
  of reference. It is true for those watching from the planetary
  reference frame. For those experiencing the journey (in the ship's
  reference frame) it is not true. For both the planetary frame and the
  ship's reference frame, the ship will change speed in a Newtonian
  way—push it a little and it speeds up a little, push it a lot and it
  speeds up a lot. However, in the planetary frame the ship will appear
  to be gaining mass due to its high kinetic energy, and the mass-energy
  equivalence principle. Should the engines be giving a constant thrust,
  this will result in progressively smaller acceleration due to the
  higher mass it is required to accelerate. From the ship's frame, the
  acceleration would continue at the same rate.

which seems to backup my thoughts. But apparently I was wrong that the scaling of the effect acts the way I thought.
Am I wrong, or the article?
 A: Ok. I think I found the answer here:
http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html
Which is from "John Baez's Stuff". Afaik he is an expert on this topic. :-)

First of all we need to be clear what we mean by continuous
  acceleration at 1g.  The acceleration of the rocket must be measured
  at any given instant in a non-accelerating frame of reference
  travelling at the same instantaneous speed as the rocket (see
  relativity FAQ on accelerating clocks).  This acceleration will be
  denoted by a.  The proper time as measured by the crew of the rocket
  (i.e. how much they age) will be denoted by T, and the time as
  measured in the non-accelerating frame of reference in which they
  started (e.g. Earth) will be denoted by t.  We assume that the stars
  are essentially at rest in this frame.  The distance covered as
  measured in this frame of reference will be denoted by d and the final
  speed v.  The time dilation or length contraction factor at any
  instant is the gamma factor γ.

...

T          t         d          v                γ
1 year     1.19 yrs  0.56 lyrs  0.77c            1.58  
2          3.75      2.90       0.97             3.99
5          83.7      82.7       0.99993          86.2
8          1,840     1,839      0.9999998        1,895
12         113,243   113,242    0.99999999996    116,641


...

In the rocket, you can make measurements of the world around you. One
  thing you might do is ask how the distance to an interesting star you
  are headed towards changes with T, the time on your clock.  At
  blast-off (t=T=0) the rocket is at rest, so this distance initially
  equals the distance D to the star in the non-accelerating frame.  But
  once you are moving, however you choose to measure this distance, it
  will be reduced by your current distance d travelled in the
  non-accelerating frame, as well as the whole lot contracted by a
  factor of γ, your Lorentz factor at time T.

