Figuring the travel time and acceleration of a sud-lightspeed spacship.

[Blue]

Thanks in great part to Atomic Rockets, I've been doodling and dabbling with the numbers for making a spaceship that could travel up to 0.5c for a story I'm making.

However I can't seem to get the math right for figuring out it's travel time, acceleration time, or grade of acceleration. I want the vessel to accelerate somewhere between 0.5G and 1.5G.

Using my *cough* knowledge of physics, I know that

d = (v

0

t + 0.5*a*t²) to solve for distance if acceleration, time and initial velocity are known

and

d = v*t to solve for distance if velocity is known and is constant, and time is known

So basically I just needed to plug these numbers together, or so I thought.

T = time total = 2t

a

+ t

t

//(where t

a

means time spent accelerating, and t

t

means time spent traveling at a constant velocity.)

v

0

= 0 m/s //(effectively, in comparison to...)

v

t

= 0.5c = 1.49896229*10

8

m/s //(traveling speed)

(0.5 * 9.80665m/s²) ≤ a ≤ (1.5 * 9.80665m/s²) //(possible acceleration values)

d = 1.12365412 Lightyears = 1.06303617*10

16

meters //(distance to be covered in my story)

So I thought the solution for time to accelerate, time spent coasting at half-light speed and acceleration values could be shown in a three-variable equation, which could result in a 3D graph showing a peak.

d = (v

0

t + 0.5*a*t

a

²) + (v

t

* t

t

) + (v

t

t + 0.5*(-a)*t

a

²)

(The reason why the second acceleration value would be because the vessel must decelerate approaching the destination. For the second part of accelerated motion, the velocity initial would inherit from the travel velocity of 0.5c.)

So,

1.06303617*10

16

= (0.5*a*t

a

²) + (1.49896229*10

8

* t

t

) + (1.49896229*10

8

*t

a

+ 0.5*(-a)*t

a

²) , 4.903325 ≤ a ≤ 14.709975

Resulting in a three variable problem, solving for a, t_a and t_t, where a would be answered in meters per second, and the time would be solved in seconds.

However when I tried it out on Wolfram Alpha, a (acceleration) cancelled out and it gave me a linear function of t_a and t_t, which made no sense to me. Does anyone know what I'm doing wrong?

Edited August 14, 2014 by Blue

[K^2]

For starters, all of this will be wrong, because by 0.5c, you are dealing with relativity.

But ignoring relativity, t_a = v_t / a. Then you can use the t_t to be whatever you need to get the total distance traveled.

[KOCOUR]

umm? This or this

And this one is quite fun.

Edited August 14, 2014 by KOCOUR

[Z-Man]

And to pinpoint the specific mistake, a also enters with a plus sign into the deceleration phase. You have it with a minus. (If you want it with a minus, use your first equation, but with a v_0 equal to your max velocity.)

[KerikBalm]

Indeed, you don't subtract that. The time spend and distance covered accelerating is the same as that which you spend decelerating.

d = (v0t + 0.5*a*ta²) + (vt * tt) + (vtt + 0.5*(-a)*ta²)

v0 = 0, so get rid of that

d = 1/2(a*t²) + 1/2(a*t²) +(v*t) ----- sorry that I lost the subscripts

d = (a*t²) +(v*t)

v=1.49896229*108 m/s (just going to call it 0.5c)

for acceleration, t= 0.5c/a

d = (a*(0.5c/a)²) +(0.5c*t) = 0.25c^2/a + 0.5c*t

d is fixed, so t just varies with a. Note that you get negative values for t if a is insufficient to allow for acceleration to 0.5c before reaching the halfway point.

[Blue]

For starters, all of this will be wrong, because by 0.5c, you are dealing with relativity.

I've just remembered that I'm an idiot!

Indeed, you don't subtract that. The time spend and distance covered accelerating is the same as that which you spend decelerating.

d = (v0t + 0.5*a*ta²) + (vt * tt) + (vtt + 0.5*(-a)*ta²)

v0 = 0, so get rid of that

d = 1/2(a*t²) + 1/2(a*t²) +(v*t) ----- sorry that I lost the subscripts

d = (a*t²) +(v*t)

v=1.49896229*108 m/s (just going to call it 0.5c)

for acceleration, t= 0.5c/a

d = (a*(0.5c/a)²) +(0.5c*t) = 0.25c^2/a + 0.5c*t

d is fixed, so t just varies with a. Note that you get negative values for t if a is insufficient to allow for acceleration to 0.5c before reaching the halfway point.

Helpful! I guess I could've replaced (v0t + 0.5*a*ta²) + ... + (v0t + 0.5*(-a)*ta²) with just 2(v0t + 0.5*a*ta²) + ...

umm? This or this

And this one is quite fun.

D'oh