Posted: 8/19/2016 2:58:04 AM EDT
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I am dumb at the maths. I have no idea how to solve this, but am interested from a literary point of view.
A man has developed an engine that can boost 1.5 gee all day. He wants to go to the asteroid belt, in the middle where it is more dense, about 150 million miles from home. He had built a large craft that will accelerate at 1.5G until 24 hours from mid-point, coast for 48 hours while rotating in free fall, then decel at the same 1.5G. What would be the max velocity at midpoint, and the times to midpoint and destination? |
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Quoted:
It's waaaaay too low. 32.2 fps². Every second it's another 32.2 fps faster. 1900 fps after the first minute, unless I am completely addled. you hit 87k in less than an hour. Quoted:
Quoted:
87,000 fps eta... that was GD response, of course, but I bet it's not far off. It's waaaaay too low. 32.2 fps². Every second it's another 32.2 fps faster. 1900 fps after the first minute, unless I am completely addled. you hit 87k in less than an hour. You're correct. I can barely get Kerbins into a stable orbit. 
And at 1.5 g boost over 48 hours.. |
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Quoted:
Does deceleration occur at "coast for 48 hours while rotating in free fall?" No. Reach "midpoint" then spend 48 hours in freefall while rotating, then start decel. Keeps a standard "down" while under acceleration. Simplifies construction. ETA: And allows 48 hours of zero grav sex. It's a fun book. |
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Quoted:
It's waaaaay too low. 32.2 fps². Every second it's another 32.2 fps faster. 1900 fps after the first minute, unless I am completely addled. you hit 87k in less than an hour. Quoted:
Quoted:
87,000 fps eta... that was GD response, of course, but I bet it's not far off. It's waaaaay too low. 32.2 fps². Every second it's another 32.2 fps faster. 1900 fps after the first minute, unless I am completely addled. you hit 87k in less than an hour. Launching from earth? Gotta figure the well and escape velocity, plus know launch latitude to get specifics normalizing to the ecliptic. 1.5G constant would get you there right quick, though. Out in the beyond, neglecting orbital elements, the equation becomes simpler. With constant thrust, you'll have traveled halfway when you reach 75,000,000km. At constant acceleration from dead stop, d = 1/2*a*t^2. d = 75,000,000M km = 75,000,000,000m, a = 14.7 m/s^2 (1.5G). t^2 = 2d/a = 101015s, or about 28 hours, with peak velocity of a*t = 1484.9km/s (about .005C). Total trip time in flat space about 56 hours, with peak velocity just under 1500km/s. |
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do you understand the kinds of speed you are dealing with? just launching where you are fighting Earths gravity well modern rockets achieve escape velocity in a short time, under 10 minutes to hit 25,000 mph i may be wrong but I think if you did 1.5 Gee it would only take 4 days, 1 day to accelerate, two coasting at 3.1 mil mph and 1 day to slow down |
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Quoted: No. Reach "midpoint" then spend 48 hours in freefall while rotating, then start decel. Keeps a standard "down" while under acceleration. Simplifies construction. ETA: And allows 48 hours of zero grav sex. It's a fun book. Quoted: Quoted: Does deceleration occur at "coast for 48 hours while rotating in free fall?" No. Reach "midpoint" then spend 48 hours in freefall while rotating, then start decel. Keeps a standard "down" while under acceleration. Simplifies construction. ETA: And allows 48 hours of zero grav sex. It's a fun book. |
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Quoted:
you wouldn't reach midpoint and then quit accelerating, midway is some 24hrs after you quit, then another 24 until you light the burner to slow down This - if you want to spend 48 hrs coasting, then you've got somewhere between 18-21hrs at 1.5G, coast, then the same amount of time as before to decel. With a 1.5G drive, I'd guess less than an hour added for takeoff from the surface and orienting to the ecliptic. |
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I get this: Accelerate 18.92 hours, to about 1,002,000 meters/sec Coast 48 hours Decelerate 18.92 hours Total time = 85.84 hours Time to midpoint = 42.92 hours I did it like this: Imagine a graph of velocity vs time: it looks like a trapezoid. You know the length of the top (48 hours), you know the area (150 million miles), and you know the slope of the sides (1.5 g). Make the units agree, and solve for the height (max velocity) and the length of the bottom (total time). |