Spaceflight in straight lines

[raxo2222]

I assume such flight needs fission/fusion reactors or beam network to power engines of such ships.

Would be constant acceleration and deceleration at 0.1 g good enough for lunar/solar system colonization/tourism?

How much DV would use such flight to Moon/Mars/ other planets?

Higher acceleration would mean higher top speed at midway. But you spend less time in flight.

Edited June 16, 2017 by raxo2222

[cubinator]

.1 g for an hour gets you going over 3 km/s, so you could probably go to the moon in less than a day. In short, yes, that high of a constant acceleration will get you places fast. Jupiter would be available for colonization if the ship is large enough.

[Bill Phil]

If we go to Mars at its closest approach, 54.6 million km, then, at  0.1 g, it would take about five and a half days to get there in a brachistochrone trajectory. Maximum velocity (at the mid point) is about 231 km/s. Total journey Delta v is twice that, at about 462 km/s. Assuming a mass ratio of e, we need an exhaust velocity equal to our Delta v. Now, at 0.1 g, we need to generate .98 newtons per kilogram of ship. Using the kinetic energy equation, we can solve for power by dividing both sides by time. Then, by multiplying the exhaust velocity by the force and dividing by two, we get the power output for the engine. For our engine here, we need about 226.7 kilowatts per kilogram of ship. For a 100 Tonne ship, that's about 22.7 Gigawatts of energy. Considering how energy works, you'd get anywhere from a few percent of that (using beamed power?) to twice that in waste heat. The major issue here will be radiators... unless we use nukes. Not nerva engines, but actual bombs. As in project Orion...

All this is back of the napkin. Feel free to criticize.

[magnemoe]

1 hour ago, Bill Phil said:

If we go to Mars at its closest approach, 54.6 million km, then, at  0.1 g, it would take about five and a half days to get there in a brachistochrone trajectory. Maximum velocity (at the mid point) is about 231 km/s. Total journey Delta v is twice that, at about 462 km/s. Assuming a mass ratio of e, we need an exhaust velocity equal to our Delta v. Now, at 0.1 g, we need to generate .98 newtons per kilogram of ship. Using the kinetic energy equation, we can solve for power by dividing both sides by time. Then, by multiplying the exhaust velocity by the force and dividing by two, we get the power output for the engine. For our engine here, we need about 226.7 kilowatts per kilogram of ship. For a 100 Tonne ship, that's about 22.7 Gigawatts of energy. Considering how energy works, you'd get anywhere from a few percent of that (using beamed power?) to twice that in waste heat. The major issue here will be radiators... unless we use nukes. Not nerva engines, but actual bombs. As in project Orion...

All this is back of the napkin. Feel free to criticize.

Note that burning to turnover is very inefficient, you will not do it unless you have an reaction less drive as an solar sail. 
Problem is that the middle of the burn has little impact on travel time as you start braking shortly after. 
If you increase acceleration and just burned the 50% of the time and cruise the rest you will save a lot on either reaction mass or travel time. 

Interesting in that an nuclear pulse drive will have multi g acceleration so you would cruise most of the time. 
its also one of the most kerbal thing ever. 
Use them a lot in my game. 

[ChrisSpace]

http://www.projectrho.com/public_html/rocket/appmissiontable.php

[kerbiloid]

Brachistochrone trajectory according to the book of Iosif Shklovsky, quoting Carl Sagan, referring to Peshek(?) and Senger(?).

Copy this into a html file and open it in browser.
Acceleration and distance are to be edited inside.

Spoiler

<script>

var
    AU = 1.49598e11,
    c = 2.99792458e8,
    SecondsInYear = 86400 * 365.2422,
    LY = c * SecondsInYear;
    
var
    acceleration = 9.81,
    distance = 5.203 * AU
//    distance = 30000 * LY
    ;

/*
t = (2c/a) * arch(1+aL/c^2) -> (2c/a) * ln(aL/c^2)

v = c (1 - (1+aL/c^2)^-2) ^ (1/2)
*/    
    
var
    k = acceleration * distance / (c * c);
    duration = (2 * c / acceleration) * Math.acosh(1 + k),
    duration_rough = (2 * c / acceleration) * Math.log(k),
    max_speed = c * Math.sqrt(1.0 - Math.pow(1.0 + k, -2));

    
var
    s = "Given:"
        + "\nAcceleration, m/s2: " + acceleration
        + "\n\nDistance, m: " + distance
        + "\nDistance, AU: " + distance / AU
        + "\nDistance, LY: " + distance / LY
        + "\n\nResult:"
        + "\nDuration, s: " + duration
        + "\nDuration, d: " + (duration / 86400)
        + "\nDuration, y: " + (duration / (86400 * 365.25))

        + "\n\nDuration (relativistic, rough), s: " + duration_rough
        + "\nDuration (relativistic, rough), d: " + (duration_rough / 86400)
        + "\nDuration (relativistic, rough), y: " + (duration_rough / (86400 * 365.25))

        + "\n\nMax speed, km/s: " + max_speed / 1000.0
        + "\nMax speed / c: " + max_speed / c
        + "\n1 - max speed / c: " + (1 - max_speed / c)
        + "\n";
    
alert(s);    

</script>

 

This is a common formula for relativistic and non-relativistic movement under constant acceleration to a given distance.

Duration - duraion according to the onboard clock. (Important for LY distances, of course. For AU distances - just a duration).
Duration (relativistic, rough) - rough approximation without hyperbolic functions, only for relativistic distances.
Max speed - it is.

Edited June 19, 2017 by kerbiloid

[wumpus]

On 6/16/2017 at 5:03 PM, magnemoe said:

Note that burning to turnover is very inefficient, you will not do it unless you have an reaction less drive as an solar sail. 
Problem is that the middle of the burn has little impact on travel time as you start braking shortly after. 
If you increase acceleration and just burned the 50% of the time and cruise the rest you will save a lot on either reaction mass or travel time. 

Interesting in that an nuclear pulse drive will have multi g acceleration so you would cruise most of the time. 
its also one of the most kerbal thing ever. 
Use them a lot in my game. 

Burning to turnover generally assumes that energy is free and crew comfort is the primary goal.  Another possibility is time is most critical (and energy isn't really on the table): you burn whatever the crew can take during the initial burn, but there is never any reason to drop below 1g.  L.E.Modest Jr. wrote a book that had a lot of this, but unfortunately it came out after KSP.  I'm pretty sure he botched all the orbital mechanics (I *might* be wrong as he was mostly concerned with direct burns, but I really suspect my KSP-tuned orbital senses got it right).

One related idea that might possibly be done would be "turnover at escape" (presumably for unmanned probes).  This would involve using gravity assists to fling a probe to Mars, and then using ion motors for capture.  In cases like this it is entirely possible that the ions are never used to get to Mars, only to "stop" once there.  They start the slow change from Hohmann transfer to spiral orbit once slingshotted past the Moon, and don't stop until aerobraking.