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Post by Apotheon on Jul 28, 2018 11:37:08 GMT
Question: I have three almost identical ships, but only one can go to the moon: why? The first has 3.94 km/s dV and 0.4 G acceleration, the second 4.24 km/s and 0.4 G, and the third 4.24 km/s and 0.2 G... witchery! How can I calculate the amount of acceleration necessary to go for instance from Earth low orbit to Moon or Mars low orbit?
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Post by The Astronomer on Jul 28, 2018 12:11:07 GMT
An ion probe can go from Earth to the Moon. You just need to make sure your ship is above the 'balance' between acceleration and delta-v. To find the balance is the hard part, but not so much make a ship that's above it.
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Post by AtomHeartDragon on Jul 28, 2018 13:37:32 GMT
Ideally orbital manoeuvres are impulsive - you apply whatever delta-v you need in an instant, meaning infinite acceleration.
In reality this is obviously not the case - your change of velocity is smeared across part of your trajectory and applied overtime.
This creates inefficiency as parts of your burn happen away from the point where and when it would be ideal and eats more delta-v than you'd theoretically need.
The more acceleration you're capable of, the more your manoeuvres resemble instantaneous changes of velocity - the most stark example would be MPDs and other sorts of ion drive that make it simply impractical to try any sort of impulsive burns, and are generally burned continuously, spiralling in or out to the destination orbit and wasting copious amounts of delta-v in the process (delta-v they have plenty of, mind you).
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Post by Apotheon on Jul 28, 2018 15:55:28 GMT
Whaaa...? No exact equations? Here's my own guesstimate:
One equals 1/5 of the orbital period divided by the escape velocity minus the orbital velocity divided by our acceleration... i.e., we're checking whether we burn the difference between orbital velocity and escape velocity in 1/5 of the orbital period.
1=((1/5*(2*pi*sqrt(r/(G*M))))/60)/((((sqrt((2*G*M)/r)-sqrt((G*M)/r)))/A)/60)
Where G is the gravitational constant, M is the mass of the body, r is distance to the centre of the body, the third r assumes a non-eccentric orbit, A is the acceleration of the spacecraft, and 1/5 is magic*.
The /60 divisors are only there to make the answer on each side of the centre in minutes for convenience.
Insert into Wolfram Alpha with A as variable:
1=((1/5*(2*pi*sqrt(6621000^3/((6.67408*10^-11)*(5.972*10^24)))))/60)/((((sqrt((2*(6.67408*10^-11)*(5.972*10^24))/6621000)-sqrt(((6.67408*10^-11)*(5.972*10^24))/6621000)))/A)/60)
Which is about 300 m/s. Care to validate?
*=1/5 means we must make our burn in 1/5 of the orbital period, which probably depends on M and r.
Edit: this is impossible. It's not scaling with mass, radius, or anything really and Jupiter is an outlier. Here are the accelerations necessary to achieve 80% efficiency compared with infinite acceleration burns: Mercury: 0.125 G Mars: 0.125 G Venus: 0.300 G Earth: 0.350 G Uranus: 0.300 G Neptune: 0.350 G Saturn: 0.400 G Jupiter: 0.750 G
In other words, with 0.4 G acceleration you can go in and out of low orbit around any planet save for Jupiter without worrying about dV waste as long as you have minimum 20% more than the dV map states. 0.35 G also gets you around most places... of course, Martians only need 0.125 G... except this will only work at Mars and Mercury, so if they want to go to Earth or anywhere else, they'll still want 4 G.
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Post by apophys on Jul 30, 2018 10:54:17 GMT
Very low accelerations are generally fine; you just want to avoid low circular orbits. Use highly elliptical orbits (or at least high circular orbits) where possible; you will find that 0.01 G will be more than enough to get you wherever you need to be. With high dV, you get to your destination very quickly using a pseudo-brachistochrone trajectory once you're far enough from significant gravity wells.
If you happen to be in a low circular orbit, you can still use a mere 0.01 G drive to leave without wasted dV, if you have enough time to make multiple passes around the planet, burning prograde at the low point each time (resulting in a flexible, highly elliptical orbit you can easily work with). Yes, this is an extremely slow process, but you didn't mention any time limits.
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Post by phoenixzix on Nov 26, 2018 4:20:16 GMT
Very low accelerations are generally fine; you just want to avoid low circular orbits. Use highly elliptical orbits (or at least high circular orbits) where possible; you will find that 0.01 G will be more than enough to get you wherever you need to be. With high dV, you get to your destination very quickly using a pseudo-brachistochrone trajectory once you're far enough from significant gravity wells. If you happen to be in a low circular orbit, you can still use a mere 0.01 G drive to leave without wasted dV, if you have enough time to make multiple passes around the planet, burning prograde at the low point each time (resulting in a flexible, highly elliptical orbit you can easily work with). Yes, this is an extremely slow process, but you didn't mention any time limits. I think this is called Oberth effect coupled with Bi-ecliptic transfer. And I don’t see even 0.01G as necessary. My fleet supertankers have 1-5mg or something and can escape Earth easily. And they are based at high orbital so no reason to make them gravity capable. I even designed a class of fast tankers (read:blockade runners) but found no use for them in actual combat. More often than not the extra engine wrecked my mass ratio(dv). |
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Post by AtomHeartDragon on Nov 26, 2018 19:49:34 GMT
Fast tankers are absolutely essential for supporting operations in deep gravity wells (think gas giants). Realistically you'd probably still want an MPD on every tanker, especially when staging invasion - being able to slowboat home on fumes rather than being discarded in cold, uncaring void might boost your crews' morale somewhat.
Of course, drone tankers would likely be an even more practical option most of the time (caveat being electronic warfare) - no problem throwing those away.
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Post by doctorsquared on Nov 27, 2018 4:22:45 GMT
Fast tankers are absolutely essential for supporting operations in deep gravity wells (think gas giants). Realistically you'd probably still want an MPD on every tanker, especially when staging invasion - being able to slowboat home on fumes rather than being discarded in cold, uncaring void might boost your crews' morale somewhat. Of course, drone tankers would likely be an even more practical option most of the time (caveat being electronic warfare) - no problem throwing those away. Just a thought, other than the monetary and resource expense of single-use drop tanks, wouldn’t it be a more efficient to build a ‘tug’ ship with an MPD, high-thrust NTR and refueling unit with massive drop tanks to maintain the fleet that can be discarded so they don’t have to haul all of that dead weight back with you? |
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Post by phoenixzix on Dec 4, 2018 10:17:57 GMT
Fast tankers are absolutely essential for supporting operations in deep gravity wells (think gas giants). Realistically you'd probably still want an MPD on every tanker, especially when staging invasion - being able to slowboat home on fumes rather than being discarded in cold, uncaring void might boost your crews' morale somewhat. Of course, drone tankers would likely be an even more practical option most of the time (caveat being electronic warfare) - no problem throwing those away. Wouldn't you equip one of your ship as a tug boat having the best MPD in the fleet (since to my understanding MPD's Isp goes up with power) and tow every other ship? Oh and i see my mistake. I was thinking about defending Pallas in particular when writing that post and assuming decisive battles. If that war drags on then yeah, tankers are good. esp. axial-gunned ships with RCS burning away and drone hordes. I quite liked that drone tanker......... do you know any way of deploying them without a even larger carrier? that drones must deploy off carrier idea is kinda stupid, and gets in the way of testing. Otherwise i think i will rehaul my tankers to drones. Easier on the pocket book. |
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