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  1. Standard member sonhouse
    Fast and Curious
    05 Aug '09 09:45
    I made some math errors on my first go round, now correcting them and introducing simplified calculations: The formula for time using distance and acceleration is T=Sqrt (distance/acceleration).
    Now using feet per second squared, for 3 milliG's is .096 (1 G =32 Ft/S^2)
    So for interstellar distances, if you remember two numbers, 6.33E-8 and 3.13E-18, the new formula is: 6.33E-8/(Light years/3.13E-18. This gives the answer directly in years for that one acceleration figure of 3 milli G's.
    so 6.33E-8*(sqrt (LY/3.13E-18))=74 years. A bit simpler, eh!
    The only caveat is this works for nearby stars, say within 20 LY or so because after that relativistic time changes due to getting too close to C. But for say, 8LY, (distance to Sirius), it is 6.33E-8*(8/3.13E-18)^0.5 = 101.2 years.

    For closer distances, miles first: Since the formula is T=(2S/A)^0.5, which represents how far you get for a certain acceleration, you have to change it around to just T= 2*(S/A)^0.5 which represents accelerating half way to the ultimate distance and then decelerating the second half.

    Using miles and feet per second squared you first multiply miles by 5280 to get feet then use .096 feet/second squared as the # for 3 Milli G's, so it goes (2* (miles*5280/.096)) /86400 = answer in days.

    So to simplify that for one accel figure of 3 milli G's with distance in millions of miles, you use two constants, 2.3E-5

    and 1.8E-5 like this:

    2.3E-5*sqrt (miles/1.8E-5) so for 50 million miles it is 2.3E-5*(50E6/1.8E-5)^0.5=38.33 days, accurate to 2 decimal places, good enough for our work, eh!

    For Kilometers, it is 2.3E-5*sqrt(kilometers/2.9E-5) so for 80 million Kilometers it is the same, 38.2 days, also within 2 decimal places.

    So 500 million Km=95.5 days. It can't get much simpler mathematically than that for sure!

    So to reiterate, for light years, answer in years: 6.33E-8*(LY/3.13E-18)^0.5

    For miles, answer in days: 2.3E-5*(miles/1.8E-5)^).5

    For kilometers, answer in days: 2.3E-5*(kilometers/2.9E-5)^0.5

    Try these formula's for different distances, see what you get!
  2. Standard member sonhouse
    Fast and Curious
    05 Aug '09 22:00 / 8 edits
    A deeper look at those equations and I see how it can be a more general solution:

    The original simplified version for LY, Time in years=6.33E-8*(LY/3.13E-18)^0.5 is for 3 Milli G's. The # 3.13E-18 is the stand-in for 3 milliG's, so if I just divide it by three, it now is 1.04E-18, which is the stand-in for ONE milliG. So now that part (LY/1.04E-18 times the # of milli G's) gives a general solution, not just one accel rate. So the new formula is:

    Time in Years=6.33E-8*((LY/(1.04E-18*Mg))^0.5), where LY=Light years, Mg= the # of milliG's of acceleration.
    So for 20 LY and 20 Mg's, 6.33E-8*((LY/(1.04E-18*20))^0.5)=62 years.

    30 LY and 20 Mg's=76 years.

    Same thing for distance in miles and time in days:
    2.3E-5*((miles/(6E-6*Mg))^0.5)=time in days.

    So 2.3E-5*((50E6/(6E-6*10))^0.5)=21 days (50 million mile trip at 10 MilliG's)

    For Km, now it's Time in Days=2.3e-5*((80E6/(9.66E-6*10))^0.5)=21 days (same thing, 80 million kilometers, 10 MilliG's)

    These steps are worked out exactly as shown for an algebraic calculator. On my Casio Fx-300ES, the E is the x10^x button next to the Ans button.

    These formulae conk out somewhere around 30 LY and 30 Milli'G's, becoming less accurate in trip times because beyond that, the max velocity gets close enough to C for relativistic effects to be felt. So for 20 LY and 20 Mg's, you get pretty close to C but not so close for time dilation to change the destination times by much. I think peak velocity for a 20 LY trip at 20 MG's is about 120,000 miles per second, 192,000 Kilometers per second. That is about 2/3C, not much happening there relativisticly speaking.
    I have to go to bed, maybe tomorrow I will show the exact steps for an RPN calculator.