Time and Acceleration question

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phavoc
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Time and Acceleration question

Postby phavoc » Wed Mar 02, 2011 5:11 pm

Ok, for those of you who are good with acceleration and time/distance equations, I haz a question.

Assume you have an object being boosted at 1G for 8 hrs, and then released on a ballistic trajectory. At the opposite end something else will 'catch' it and then decelerate for 8 hrs to bring it to a relative rest. The object needs to travel about 1.2 billion Km (or approximately the average distance from Earth to Saturn).

Can someone a) post the formula to calculate this, and b) roughly speaking tell me how long the object would be in flight from point A to point B?
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Postby Jeraa » Wed Mar 02, 2011 5:32 pm

All of this is going from memory, so may be totally wrong.

The Interplanetary Transit Times Table (page 145) assumes constant acceleration to the midpoint, then constant decelleration to the end. Since we know we are acceleration and deceleration for a total of 16 hours, thats close enough to the 10,000,000 km number. So we cover a total of 10 million kilometers during our acceleration/deceleration phases combined.

No is where it gets a little fuzzy. I believe that a 1-g drive accelerates at 10 meters per second. After 8 hours of thrust (28,800 seconds), it should be moving at 288,000 meters per second, or 288 kilometers per second. Dividing the remaining distance not covered by our acceleration/decelleration phases, and we get 4,131,944 seconds, or 1147 hours, or 47.8 days.

So, 47.8 days, plus 16 hours. Call it 48 or 49 days, give or take a few hours. Now if the ship were to accelerate/decellerate all the way there as normal, it would only take 7 to 8 days.
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lastbesthope
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Postby lastbesthope » Wed Mar 02, 2011 7:24 pm

The basic equations of motion in their various forms are all you need.

s=ut+(0.5*a*t^2)
v=u^2+2as
s=v*t

s=distance travelled
u=initial velocity
v=fiinal velocity
t=time
a=acceleration

All expressed in a matching set of units

so in your example, the object is accelerated at 1g for 8 hours
I'll assume g=10 metres per second to keep the sums easy, though this will reduce the time required to travel the distance

during intial acceleration:
u=0
t=28800 seconds
a=10 m/s^2

s=ut+(0.5*a*t^2)
u=0 so ut=0
0.5*a=0.5*10=5
so it simplifies to s=5t^2=5*28800*28800=4147200000m

so it travels that distance

The final velocity is v=u+at=0+10*28800=288000 metres per second

during the deceleration phase the numbers are a little harder

s=ut+(0.5*a*t^2)
This time u=288000 (the final speed after acceleration phase at the beginning of the journey
t=28800 seconds again (8 hours)
a=10 again
s=(288000*28800)*(0.5*10*28800*28800)=4147200000

Which you'll notice is the same as the accel phase, which it should be, it's just the reverse of the same flight

Now you add those 2 distances, subtract that from the total distance to be travelled and calculate how long it takes to travel that dfistance by dividing the "cruising distance" (approx 1.2*10^12 m) by the "cruise velocity" we calculated earlier of 288000 metres per second.

Anyway, the answer assuming g at 10 m/s^2 and total distance as you specified is just over 48.5 days.

At g=9.8065, a more accurate value it takes just over 49.5 days, the difference being due to the lower cruise speed attained due to the lower value of g used.

LBH
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hdan
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Postby hdan » Wed Mar 02, 2011 8:20 pm

lastbesthope wrote:At g=9.8065, a more accurate value it takes just over 49.5 days, the difference being due to the lower cruise speed attained due to the lower value of g used.
I'm not sure if MgT expressly states this, but in other Travellers, 1G = 10.0m/s^2 instead of Earth's 9.8.
/hdan
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far-trader
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Postby far-trader » Wed Mar 02, 2011 8:40 pm

hdan wrote: I'm not sure if MgT expressly states this, but in other Travellers, 1G = 10.0m/s^2 instead of Earth's 9.8.
Only in GURPS Travller iirc :) A rounding simplification for easier memorization and to speed up such calculations I guess.
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phavoc
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Postby phavoc » Wed Mar 02, 2011 9:16 pm

Thanks Gents! This definitely helps out.
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Postby lastbesthope » Thu Mar 03, 2011 7:16 pm

Just remember the equations of motion I gave in my post, you can figure it all out from there

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