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!

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!