(A)------------------------------------------------------------------

Assuming a magical stardrive which allows you to accelerate continuously away from Earth at 10 m/s^2, the time taken to reach various distances is:

Earth dist :   Earth time       speed         ship time   ship distance
__________     __________     ____________    _________   _____________

   .05 ly :       0.31 yr     0.312      c      0.31 yr     0.048 ly
   .10 ly :       0.45 yr     0.426      c      0.43 yr     0.091 ly
  0.25 ly :       0.73 yr     0.611      c      0.67 yr     0.198 ly
  0.50 ly :       1.10 yr     0.755      c      0.94 yr     0.328 ly
    1 ly :        1.70 yr     0.873      c      1.28 yr     0.487 ly
    2 ly :        2.79 yr     0.9467     c      1.71 yr     0.644 ly
    4 ly :        4.86 yr     0.9814     c      2.22 yr     0.768 ly
   10 ly :       10.91 yr     0.99623      c    2.98 yr     0.868 ly
   25 ly :       25.93 yr     0.99933      c    3.80 yr     0.915 ly
   50 ly :       50.94 yr     0.999826     c    4.44 yr     0.932 ly
  100 ly :      100.95 yr     0.9999557    c    5.09 yr     0.941 ly
 1000 ly :     1000.95 yr     0.99999955   c    7.27 yr     0.949 ly
10000 ly :    10000.95 yr     0.9999999955 c    9.46 yr     0.950 ly
                                                     

(B)------------------------------------------------------------------

More generally, if a ship can accelerate continuously away from Earth, at constant acceleration a, as measured onboard the ship,

Values as measured by an observer in Earth's frame of reference:

 T: Earth time                   (since the ship's launch)
 D: Earth distance               (from Earth to the ship)
 A: Earth acceleration           (of the ship)

Values as measured by an observer on the ship:

 a: ship acceleration          (of the ship)
 τ: ship time (proper time)    (since the ship's launch)
 d: ship distance              (from Earth to the ship)
 j: distance travelled by ship

 M: mass of ship
 M0: initial mass
 ρ: mass ratio  =  M0/M
 βex: effective exhaust speed(/c)

Relativistic quantities:

${\displaystyle \theta (\tau )={\frac {a}{c}}\tau }$  : velocity parameter

                  a
   theta(tau)  =  - tau        : velocity parameter
                  c
                  

${\displaystyle \beta ={\frac {v}{c}}=\tanh \theta =\tanh {\frac {a}{c}}\tau }$

            v                             a
   beta  =  -  =  Tanh{ theta }  =  Tanh{ - tau }
            c                             c

${\displaystyle \gamma ={\frac {1}{\sqrt {1-\beta ^{2}}}}=\cosh \theta =\cosh {\frac {a}{c}}\tau }$

                    1                                       a
   gamma  =  __________________  =  Cosh{ theta }  =  Cosh{ - tau }
                            2                               c
              Sqrt{ 1 - beta  }

Recurring constants:

${\displaystyle c=2.99792458\times 10^{8}{\mbox{ m/sec}}}$

                        8
   c  =  2.99792458 × 10  m/sec

${\displaystyle {\mbox{year }}=3.155693\times 10^{7}{\mbox{ sec }}(\sim \pi \times 10^{7}{\mbox{ sec, or }}\sim {\sqrt {10^{15}}}{\mbox{ sec}})}$

                         7                 7              15
   year  =  3.155693 × 10  sec  ( ~ pi × 10  sec, ~ Sqrt{10 } sec )

${\displaystyle {\mbox{lightyear }}=9.460530\times 10^{15}{\mbox{ metres}}}$

                             15
   lightyear  =  9.460530 × 10  metres

${\displaystyle {\mbox{for }}a=10{\mbox{ m/sec}}^{2}=1.052626{\mbox{ ly/yr}}^{2},}$

                   2                    2
   for a = 10 m/sec  =  1.052626 ly / yr ,

${\displaystyle {\frac {a}{c}}=3.33564095\times 10^{-8}{\mbox{ sec}}^{-1}=1.052626{\mbox{ year}}^{-1}}$

       a                   -8   -1                -1
       -  =  3.33564095 × 10  sec  =  1.052626 year
       c

${\displaystyle {\frac {c^{2}}{a}}=8.98755179\times 10^{15}{\mbox{ metres }}=0.9500051{\mbox{ light-years}}}$

        2
       c                   15
       -  =  8.98755179 × 10  metres  =  0.9500051 light-years
       a

(C)------------------------------------------------------------------

In the ship's frame of reference,

${\displaystyle v(\tau )=c\tanh {{\frac {a}{c}}\tau };\quad v(\tau )\rightarrow c\left(1-e^{-2{\frac {a}{c}}\tau }\right)}$

                   a                              -2(a/c)tau
v(tau)  =  c Tanh{ - tau }  ;   v(tau) -> c ( 1 - e         )
                   c

${\displaystyle D(\tau )={\frac {c^{2}}{a}}\left(\cosh {{\frac {a}{c}}\tau }-1\right);\quad D(\tau )\rightarrow {\frac {c^{2}}{2a}}e^{{\frac {a}{c}}\tau }}$

            2                                       2
           c         a                             c   (a/c)tau
D(tau)  =  - ( Cosh{ - tau } - 1 )  ;   D(tau) ->  __  e
           a         c                             2a

${\displaystyle \tau (D)=\cosh ^{-1}\left({\frac {a}{c^{2}}}D+1\right);\quad \tau (D)\rightarrow {\frac {c}{a}}\ln \left({\frac {a}{c^{2}}}D\right)}$

           c           a                          c      a
tau(D)  =  - ArcCosh{ ___ D + 1 }   ;   tau(D) -> - Ln{ ___ D }
           a            2                         a       2
                       c                                 c

${\displaystyle d(\tau )={\frac {c^{2}}{a}}\left(1-{\mbox{sech}}{\frac {a}{c}}\tau \right);\quad d(\tau )\rightarrow {\frac {c^{2}}{a}}\left(1-2e^{-{\frac {a}{c}}\tau }\right)}$

            2                                      2
           c             a                        c         -(a/c)tau
d(tau)  =  - ( 1 - Sech{ - tau } )  ;   d(tau) -> - ( 1 - 2 e        )
           a             c                        a

${\displaystyle j(\tau )={\frac {c^{2}}{a}}\ln \cosh \left({\frac {a}{c}}\tau \right);\quad j(\tau )\rightarrow c\times \tau -{\frac {c^{2}}{a}}\ln {2}}$

            2                                                2
           c           a                                    c
j(tau)  =  - Ln{ Cosh[ - tau ] }    ;   j(tau) -> c × tau - - Ln{2} 
           a           c                                    a
           

${\displaystyle T(\tau )={\frac {c}{a}}\sinh {{\frac {a}{c}}\tau };\quad T(\tau )\rightarrow {\frac {c}{2a}}e^{{\frac {a}{c}}\tau }}$

           c       a                               c   (a/c)tau
T(tau)  =  - Sinh{ - tau }          ;   T(tau) ->  __  e
           a       c                               2a

(D)------------------------------------------------------------------

Alternately, in the Earth's frame of reference, your acceleration is measured as:

${\displaystyle A={\frac {a}{\gamma ^{3}}}}$

         a
A  =  ________
            3
       gamma

${\displaystyle A(v)=a\left(1-\beta ^{2}\right)^{\frac {3}{2}}}$

                      2  3/2
A(v)  =  a ( 1 - (v/c)  )
                              

${\displaystyle D(T)={\frac {c^{2}}{a}}\left({\sqrt {1+\left({\frac {a}{c}}T\right)^{2}}}-1\right)}$

          2
         c               a    2
D(T)  =  - ( Sqrt{ 1 + ( - T )  } - 1 )  
         a               c

${\displaystyle T(D)={\frac {c}{a}}{\sqrt {\left({\frac {a}{c^{2}}}D+1\right)^{2}-1}}}$

         c           a         2
T(D)  =  - Sqrt{ (  ___ D + 1 )  - 1 }   
         a            2
                     c

${\displaystyle v(T)={\frac {aT}{\sqrt {1+\left({\frac {a}{c}}T\right)^{2}}}}={\frac {c}{\sqrt {1+\left({\frac {a}{c}}T\right)^{-2}}}}}$

                 aT                          c
v(T)  =  ______________________  =  ________________________
                      a    2                     a    -2
          Sqrt{ 1 + ( - T )  }       Sqrt{ 1 + ( - T )    }
                      c                          c

${\displaystyle \beta (T)={\frac {v(T)}{c}}={\frac {1}{\sqrt {1+\left({\frac {c}{aT}}\right)^{2}}}}={\frac {1}{\sqrt {1+\left({\frac {a}{c}}T\right)^{-2}}}}}$

            v(T)             1                         1
beta(T)  =  ____  =  _____________________  =  ______________________
             c                    c   2                a    -2   1/2
                     Sqrt{ 1 + ( ___ )   }     { 1 + ( - T )    }
                                  aT                   c
                                  

${\displaystyle \gamma (T)={\sqrt {1+\left({\frac {a}{c}}T\right)^{2}}}}$

                         a    2
gamma(T)  =  Sqrt{ 1 + ( - T )  }
                         c

${\displaystyle \tau (T)={\frac {c}{a}}\ln \left({\frac {a}{c}}T+{\sqrt {1+\left({\frac {a}{c}}T\right)^{2}}}\right)}$

           c     a                 a    2
tau(T)  =  - Ln{ - T + Sqrt[ 1 + ( - T )  ] }
           a     c                 c

${\displaystyle A(T)={\frac {a}{{\sqrt {1+\left({\frac {a}{c}}T\right)^{2}}}^{3}}}={\frac {a}{\left(1+\left({\frac {a}{c}}T\right)^{2}\right)^{\frac {3}{2}}}}}$

                  a                           a
A(T)  =  _______________________  =  _____________________
                      a    2  3               a    2   3/2
          Sqrt{ 1 + ( - T )  }        { 1 + ( - T )   }
                      c                       c

(E)------------------------------------------------------------------

For a trip which goes from standing start to standing finish, calculate the T, τ, etc. to reach the midpoint, then double.

Voyage length ${\displaystyle =2T(D/2)={\frac {2c}{a}}{\sqrt {\left({\frac {a}{2c^{2}}}D+1\right)^{2}-1}}={\sqrt {{\frac {D^{2}}{c^{2}}}+{\frac {4D}{a}}}}}$

                                 2c          a          2
Voyage length  =  2 T( D/2 )  =  __ Sqrt{ ( ____ D + 1 )  - 1 }
 (Earth time)                     a            2
                                             2c
                                            
                              =  Sqrt{ D^2 / c^2 + 4 D / a }
                             
                             
                                         2
                                        D      4 D  
                              =  Sqrt{ ____ + _____ }
                                          2      a
                                         c      

Voyage length ${\displaystyle =2\tau (D/2)={\frac {2c}{a}}\cosh ^{-1}\left({\frac {a}{2c^{2}}}D+1\right)=?{\frac {2c}{a}}\sinh ^{-1}\left(2(T(D/2){\frac {a}{2c}}\right)}$

                                   2c           a
Voyage length  =  2 tau( D/2 )  =  __ ArcCosh{ ____ D + 1 }
 (ship time)                        a             2
                                                2c
                                               
                     =? ( 2 c / a ) ArcSinh{ 2 T(D/2) a / 2 / c }
                    
                             

${\displaystyle V_{\mbox{max}}=V(T(D/2))=c{\sqrt {1-\left({\frac {a}{2c^{2}}}D+1\right)^{-2}}}}$

                                         a          -2
Vmax  =  V( T(D/2) )  =  c Sqrt{  1 - ( ____ D + 1 )    }
                                           2
or                                       2c

${\displaystyle V_{\mbox{max}}=V(\tau (D/2))=c\tanh {\cosh ^{-1}\left({\frac {a}{2c^{2}}}D+1\right)}}$

                                             a
Vmax  =  V( tau(D/2) )  =  c Tanh{ ArcCosh( ____ D + 1 ) }
                                               2
                                             2c

Voyage length ${\displaystyle =2j(\tau (D/2))={\frac {2c^{2}}{a}}\ln \left({\frac {a}{2c^{2}}}D+1\right)}$

                                        2
                                      2c       a 
Voyage length  =  2 j( tau(D/2) )  =  __  Ln{ ____ D + 1 }
(ship's odometer)                      a         2
                                               2c
                            
                                                2
for instance, for a   =  1 kgal  ( = 1000 cm/sec   ~ 1 "gee" )

                                                    15
Distance to Alpha Cen =  4.3 ly  =  41 Pm  =  41 × 10 m
 
                                 6
Tau to Alpha Cen      =  111 × 10 sec  =  3.5 yr

                                 6
Time to Alpha Cen     =  187 × 10 sec  =  5.9 yr

V_max                 =  0.95 c

distance travelled    =  2.3 ly

(F)------------------------------------------------------------------

For a perfectly efficient photon rocket,

${\displaystyle \theta =\tanh ^{-1}{\beta }=\ln {\frac {M_{0}}{M}}={\frac {a}{c}}\tau ,}$ so

                                  Mo       a
theta  =  ArcTanh{ beta }  =  Ln{ __ }  =  - tau,  so    
                                   M       c 

${\displaystyle M(\tau )=M_{0}e^{{\frac {a}{c}}\tau }}$

                  -(a/c)tau
    M(tau)  =  Mo e

For a perfectly efficient photon rocket, accelerating from v = 0 to β(×c),

${\displaystyle \rho ={\frac {M_{0}}{M}}=\gamma (1+\beta )={\sqrt {\frac {1+\beta }{1-\beta }}}}$

        Mo                                   1 + beta
rho  =  __  =  gamma ( 1 + beta )  =  Sqrt{ __________ }
         M                                   1 - beta

or alternately,

${\displaystyle \beta (\rho )={\frac {\rho ^{2}-1}{\rho ^{2}+1}}}$

                      2
                   rho  - 1
    beta(rho)  =  __________
                      2
                   rho  + 1

${\displaystyle \gamma (\rho )={\begin{matrix}{\frac {1}{2}}\end{matrix}}\left(\rho +{\frac {1}{\rho }}\right);\quad \gamma (\rho )\rightarrow {\frac {\rho }{2}}}$

               1          1                          rho
gamma(rho)  =  - ( rho + ___ )   ;   gamma(rho)  ->  ___
               2         rho                          2

For an imperfect rocket, with effective exhaust speed(/c) of β_ex,

${\displaystyle \theta =\beta _{ex}\ln \rho }$, so ${\displaystyle M(\tau )=M_{0}e^{-{\frac {a\tau }{c\beta _{ex}}}}}$

                                                 -(a tau)/(c B_ex)
theta  =  B_ex Ln { rho } ,  so    M(tau)  =  Mo e

${\displaystyle \beta (\rho )={\frac {\rho ^{2\beta _{ex}}-1}{\rho ^{2\beta _{ex}}+1}}}$, or ${\displaystyle \rho (\beta )=\left({\frac {1+\beta }{1-\beta }}\right)^{\frac {1}{2\beta _{ex}}}}$

                 2B_ex 
               rho    - 1                          1 + beta  1/2/B_ex
beta(rho)  =  ____________ ,  or   rho(beta)  =  { _________ }
                 2B_ex                             1 - beta
               rho    + 1

${\displaystyle \gamma (\rho )={\begin{matrix}{\frac {1}{2}}\end{matrix}}\left(\rho ^{\beta _{ex}}+\rho ^{-\beta _{ex}}\right);\quad \gamma (\rho )\rightarrow {\frac {\rho ^{\beta _{ex}}}{2}}}$

                                                               B_ex
               1      B_ex    -B_ex                         rho
gamma(rho)  =  - ( rho    + rho     )  ;   gamma(rho)  ->  _____
               2                                             2
                            

(G)------------------------------------------------------------------

FAQ page for The Relativistic Rocket: http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html also http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

John Walker's relativity-exploring ship: http://www.fourmilab.ch/cship/cship.html

The Oh-My-God Particle: http://www.fourmilab.ch/documents/ohmygodpart.html

Erik Max Francis's frontpage: http://www.alcyone.com/max/noframes.html

Wayne Throop's frontpage: http://www.sheol.com/throopw

Chris Hillman's relativity page: http://www.math.washington.edu/~hillman/relativity.html

Jason Hinson's FAQ site: http://www.physics.purdue.edu/~hinson/ftl/index.html