Help with calculating Delta-V

[CodaThePortalmaster]

No matter how hard I try I just can't figure out how to calculate Delta-V. Can someone explain it then give an example{and please don't just link me to thewjkrh(or whatever) rocket equation on wikipedia.}

 

Thanks to Zhetaan for helping me understand.

Edited January 26, 2017 by CodaThePortalmaster

[Green Baron]

 

But that is easy. If i may not link to the wiki i must type it myself. dv is a measurement for the ability of a rocket to change its speed. It is nothing else than the amount of fuel thrown out at a certain speed. So the mass difference comes into play and a figure that says something about the speed at which the fuel is thrown out, the specific impulse. The constant doesn't matter for the explanation (propellant weight on earth is used for the ISP).

Here you go: Natural logarithm of wetmass divided by drymass * specific impulse in seconds * earths gravity

dV[m/s] = ln(startmass/drymass) * ISP * 9.81[m/s²]

Example:

Rocket mass before maneuver 100t, afterwards 50t. ISP of engine 300s.

ln(100/50) * 300 * 9.81 = ln(2) * 2943 = ~0.69 *2943 = ~2030m/s.

 

 

Edit: The formatting was done by the crappy forum software on save. Everything is meant as written, just ignore the strikes ...

Edited January 25, 2017 by Green Baron

[tomf]

What are you having difficulty with exactly? The delta v of a rocket is its ability to change its speed or the delta-v of a maneuver is the change in speed required to perform it.

It depends on the amount of fuel carried and the speed with which the engine can propel it out the back. It is made more complicated by the fact that using a certain amount of fuel will accelerate a light rocket, nearly empty of fuel more than a heavy rocket, full of fuel. The maths is a little complicated but the result is that the ln(Mwet/mdry) that you will have seen.

so Delta-v = g* ISP * ln (Mwet/Mdry) where g * ISP = the exhaust speed of the propellant.

The reason ISP is quoted in seconds and you need that g in the equation is an accident of history.

[RCgothic]

DV = Isp * g0 * ln(m0/mf)

I'll unpack that a bit.

DV is change in velocity. It can be calculated for a specific manoeuvre or more generally as the total ability of a vehicle or rocket to change its velocity by.

Isp is specific impulse. It is a measure of engine efficiency - basically how long in seconds an engine takes to exhaust a weight of fuel equal to its thrust. Longer is better because it means you get to burn longer for the same quantity of fuel. It's directly proportional to exhaust velocity, and you'll often find the rocket equation written in terms of exhaust velocity instead of Isp.

g0 is the acceleration due to Earth's gravity, and is required for the conversion between exhaust velocity and Isp. It may seem weird that we use Earth's gravity no matter where the rocket is, but that's because we're implicitly converting from kilograms to Newtons and Newtons are defined in terms of Earth's gravity.

m0 is the mass before the burn.

mf is the mass of the rocket after the burn.

m0/mf is therefore a measure of the mass ratio, or how much of the rocket you've used as fuel. The more fuel you've used, the greater this number, the faster you go.

ln(m0/mf) is the natural logarithm of the mass ratio. It's a non-linear function, but any calculator or spreadsheet can work it out. The reason it's necessary is because the fuel you burn first also has to accelerate the fuel you burn later before that can be used, reducing its efficiency. Burning the 1st kg of a 1000kg spacecraft changes the velocity by less than 1/600th of the amount the 999th kg will get you.

Where m0 is the total mass of a spacecraft topped full of fuel and mf is the dry mass after all fuel has been exhausted, the rocket equation gives you the total ability of the rocket to change its velocity.

If you have the DV required for a manoeuvre, and the Isp of the engines, then the equation can be rearranged to tell you either how much fuel you need to bring along on top of dry mass in order to perform it, or if you have current mass, what your end mass will be after the manoeuvre (And therefore fuel required - but better hope end mass is higher than dry mass or the manoeuvre is impossible).

The rearranged form is:

m0/mf = e ^ (DV / [Isp *g0])

Where e is the exponential function (opposite of natural logarithm) and ^ means "to the power of"

And that's basically all you need to know. Any questions?

Edited January 25, 2017 by RCgothic

[RCgothic]

Or, if you want to calculate the DV required for a particular manoeuvre between orbits, that's an orbital mechanics question, rather than a rocket equation question.

Generally manoeuvres are either Hohmann manoeuvres, which go circular to elliptical to circular in two manoeuvres, plane change, which is negating velocity in one direction and adding it in another by pythagoras, or launch/landing.

Don't try and calculate launch/landing unless you are going advanced. Just look it up.

Plane change DV is easy. How much do you want to change your angular inclination by, and how fast are you going? By the rearranging the cosine rule:

DV = SQRT(2 * v^2 [1-cos(angle)])

Thus for zero inclination change cos(0) is 1 and the DV required is 0.

For a 180, the DV required is 2*v, because you have to negate all of v and add it again in the other direction.

For a 90deg change you need 1.414*v.

And of course any other angle can be calculated in the same way.

Hohmann transfers are a little involved and I'm out of time. Maybe someone else can explain, or I'll be back later.

Edited January 25, 2017 by RCgothic

[Red Iron Crown]

1 hour ago, CodaThePortalmaster said:

I need help understanding the In(Natural Logorithm) part...

A natural logarithm of a number is the power to which e must be raised to equal that number. In equations:

ln(x) = y  means that e^y = x

e is Euler's number, a mathematical constant useful in evaluating things as broad reaching as compound interest, probablilities, and rocket delta-V. It is approximately equal to 2.71828, though it is an irrational number like pi and has an infinite number of digits after the decimal.

It is used in the rocket equation to account for propellant mass having to accelerate remaining propellant mass, a mass that is decreasing as the rocket burns through its fuel. Rather than doing a step wise approximation, we can use the natural logarithm to get an exact measure of the amount of velocity change the rocket can perform.

This can be seen in the derivation of the rocket equation on the wiki page, but I suspect if you haven't encountered natural logarithms yet you haven't got the calculus for that. Fortunately, you can still use the rocket equation by hand using a spreadsheet or calculator's ln function without needing anything more than basic algebra and arithmetic. 

[mikegarrison]

Just don't use "log" (which usually means log₁₀) instead of ln (which means log_e).

[RCgothic]

Log10(y) =x for the same values as y = 10^x

Loge(y) = ln(y) = x for the same values as y = e^x

Similarly there are log#(y) functions for arbitrary values of #^x, but they're rare. I think the most common non-10 non-e values are 2 (binary) and 16 (hexadecimal).

[Green Baron]

3 hours ago, RCgothic said:

Hohmann transfers are a little involved and I'm out of time. Maybe someone else can explain, or I'll be back later.

You explain very well, sir !

Yeah, in the beginning i calculated, now i often just fly. From circular to circular orbit Hohmann is not too frightening.

This link explains Hohmann step by step, better than Wikipedia, i find. So nobody has to recite it again ...

 

Edited January 25, 2017 by Green Baron

[Zhetaan]

@CodaThePortalmaster:

I'll focus on the natural logarithm part because that's what you asked for.  If I'm completely missing the mark and you just don't know how to enter it on a calculator, you should know that a lot of calculators require you to enter the number and then press the 'ln' key.  If you press the key before you enter the number, you'll get an error.

If you want to understand how to calculate needed delta-v for specific orbital manoeuvres, that is covered elsewhere.

If you want to understand how to design and build a rocket to get where you want to go (and back again, if you care about your Kerbals) efficiently (in other words, how to make enough rocket for a given target delta-v), that is also covered elsewhere.

Otherwise, I'll assume that you want to understand something of what the rocket equation itself is about, how it works, why it works, and what the pieces of it mean in relation to one another and the rest of reality.  Here's a tutorial:

 

 

A Non-Calculus Guide to Logarithms and the Rocket Equation

 

Introduction

This assumes that you've never worked with logarithms before; you don't know how they work or what they can do, and, while you can press that 'log' or 'ln' button on a scientific calculator, you have no idea why it's spitting out those strange numbers at you half the time and 'Error' the other half.  It does assume that you can add, subtract, multiply, and divide, that you know what a numerator is, and you know how to find that 'ln' button on your calculator--even if you don't know how it works.

This also assumes that you know what delta-v is, have heard of the Tsiolkovsky Rocket Equation, but are otherwise utterly new to the world of rocketry and KSP--that may be unfair, but the basic assumptions here are that you know to keep the engine bell pointed away from your face at all times, and that a higher delta-v value means your rocket can fly farther.

If you have not heard of the rocket equation, here it is:

delta-v = (engine exhaust velocity) * ln ( wet mass / dry mass )

Definitions will follow.

One thing this tutorial will not do is teach you orbital mechanics and flight.  Everything assumes that you are in free fall, vacuum, and far away from anything that you could crash into.  It won't tell you how to change orbits; it will only tell you about what the elements of the rocket equation mean and how to use them to calculate delta-v.

 

Logarithms

A logarithm is a kind of function, which is maths language for saying essentially that it takes an input, does something to it, and spits out an output.  Examples of functions include trigonometric functions (sine, cosine, tangent, &c.) and, for a simpler one, absolute value.  Many of these functions spit out an output that describes some value or attribute of the input, which is one thing that makes them so useful.  To continue the example, trigonometric functions describe periodic cycles (among other things), and absolute value describes a number's distance from zero on the number line.

Logarithms describe exponential decay, also among other things, but exponential decay is what we're interested in understanding.  What this means is that we're talking about a kind of decay that changes rate, or gets either slower or faster depending on how decayed it already is--it can go either way.  Think of a tree that dies and starts rotting apart; as more rots, it exposes more tree to be rotted, so while it starts breaking down very slowly, the tree's structure completely collapses seemingly all at once at the end.  This is accelerating decay.  An example of decelerating decay is cooling; a piece of red-hot iron can cool by hundreds of degrees in minutes, but once it gets close to room temperature, it can take a very long time for the temperature to lose the last few degrees.

The natural logarithm (given 'ln' in both written and calculator notation) is a special sort that describes continuous decay.  Continuous growth and decay can either accelerate or decelerate, and it appears in a lot of natural processes, such as radioactive decay and population growth.  These two processes are, in turn, described by half-life and doubling rate respectively.  The importance of these terms is that they are constant values that have nothing to do with the actual amount of sample--in other words, to use population growth as an example, if you start with one bacterium, after a single doubling period, you'll have twice as many (in this case, two).  If I start with 100,000 bacteria, then after one doubling period (assuming room and nutrients for them), I'll have twice as many, as well (in this case, 200,000).  It's also important to know that for a given doubling period, you cannot predict exactly when a specific bacterium will double; you only know that it will happen, but for the colony, there are always doubling bacteria, which means that for any point in time, the population is always growing, even though you cannot narrow the process down to specific cells.  In other words, this process describes the whole of the system, but without paying a lot of attention to the individual parts of that system or what they are doing at any specific point in time while the system operates.

This idea of growing or decaying whatever you have without paying attention to how much is actually there or what each piece is doing at any given time is the core of the logarithm's importance in the rocket equation.  In this case, the individual pieces are the drops of fuel; you don't know whether any given one is going to burn next.  You also don't care how many drops there are (well, you do, but not for this) in the sense that if you were to add triple the mass of rocket (copy your rocket and send three docked together) you end up going exactly as far with three identical rockets docked together (or three hundred) as you could go with one rocket alone.  It's easy to say that in this case, it's because you're using three engines instead of one, but it doesn't work that way:  if you were to shut off the two outside engines and only run the middle engine, it would still work the same way.  One engine, one hundred engines--it doesn't matter, because the number of engines is irrelevant.  To call back to an earlier example, if I start with two identical red-hot pieces of iron, they both end up at the same temperature after the same amount of time--the rate of cooling doesn't depend of how many pieces I have, because if they start at the same temperature and end at the same temperature, then they both have to cool by the same number of degrees.  I can't add their temperatures together and say it's the same as one piece with double the extra heat; it doesn't work that way.  Yes, there is more total energy involved if you have more hot parts, but they all have to cool by the same amount, and that's the important part here.

 

Mass Fraction

We can abstract this idea of a system to rockets:  at the end of the day, the important thing is that there is a minimum amount of mass that has to be pushed around.  This is the pod, engine, empty tankage, experiments, Kerbals, and basically everything that isn't thrown out the engine bell.  We call that dry mass.  The other important thing is that there is a maximum amount of mass that needs to be pushed around.  This is everything just mentioned, plus the stuff thrown out the engine bell, i.e., fuel and oxidiser.  We call that wet mass--note that wet mass is dry mass plus fuel rather than fuel alone.  This is important because it gives us both the start point (the full rocket) and an end point (the empty rocket); mass of fuel alone doesn't help us because it doesn't relate to the rest of the rocket until it is actually added to the rest of the rocket.

As the fuel burns, it is continuously consumed.  As you have less and less fuel, the better and better that fuel is able to push the rocket around because each particle of the mass remaining in the rocket receives a larger and larger share of the energy of motion:  in other words, the most efficient drop of fuel is the last drop of fuel--it doesn't matter what your rocket looks like or how big it is; for all rocket systems, this is true.  This situation in reverse is what the decay in the rocket equation describes:  as you add more fuel to the rocket (wet mass), the fuel you add becomes less and less effective at pushing that mass.  It still adds capability (you're producing thrust regardless) but the amount of capability you add per drop of fuel keeps falling because each drop of fuel you add has to push all of the other drops, plus the dry mass of the rocket.

Going back to the equation, the logarithm part is of the ratio of wet to dry mass; that is, wet mass (which will be higher) is the numerator and dry mass is the denominator.  This value of wet mass divided by dry mass is called the mass fraction.  The dry mass is constant for a given rocket, so we can ignore it for now; let's consider the wet mass in terms of the dry mass.  That is to say, let's divide the wet mass by the dry mass to get a single number and see what happens.

 

Fun with Graphs and Calculators

You can follow along with this graph.

Let's start with zero fuel.  Wet mass then equals dry mass, so the mass fraction is equal to one.  The natural logarithm of one is zero, so our delta-v is also zero--which you would expect from having no fuel.  Now let's add fuel equal to the dry mass of the rocket, so the fraction is equal to two.  The natural logarithm of two is approximately .693, which doesn't mean much to you yet, but it's not zero and that's what I want you to look at.  Let's add double the fuel:  the dry mass is still there, so double fuel plus dry mass gives a fraction of three.  The natural logarithm of three is about 1.097--this is definitely more, but it's not quite double the value even though it's double the fuel.  The efficiency of the added fuel is decaying, as we said it would.

 

Exhaust Velocity

To understand the y-axis (and what the .693 and 1.097 values mean), let's look at the other part of the rocket equation.  We've well covered the logarithm; the other part is the exhaust velocity of the engine.  KSP converts this to specific impulse which is a measure of exhaust velocity working against one gravity of acceleration (that is, engine exhaust velocity / 9.81 metres per second per second), but both refer to the engine's efficiency.  I will use the traditional version (exhaust velocity) because it's easier to understand intuitively.  The idea is that if a rocket engine functions by throwing mass out the back, a more efficient engine will throw that mass out with more energy:  rockets function by Newton's Third Law (equal and opposite reactions), so a more energetic exhaust means more energy of motion to push the rocket.  The energy of motion, in turn, depends on two things:  mass and velocity.  If two different engines are running on two test rockets with the same dry mass and with the same amount of propellant, then the mass is constant between them, so the only way for an engine to give more energy to the propellant mass is for it to give that mass more velocity.  Therefore, the more efficient engine has a higher exhaust velocity, which gives its test rocket more final velocity, which, by definition, is more delta-v.

The tie-in to the y-axis of the graph above is that there is some mass fraction that has a natural logarithm of one.  This is not the same as the mass fraction being one--this is the output, not the input.  What it means is that there is some amount of wet versus dry mass that gives the total rocket a delta-v equal to the engine's exhaust velocity.  For a test rocket--or any rocket--this mass fraction is the same, and no matter what the engine is, if it has this mass fraction, the delta-v will equal the engine's own efficiency in terms of exhaust velocity.  The value of this mass fraction is a number called e and that is equal to approximately 2.718.  The applications and theory behind the number e are many and whole university courses can be devoted just to the study of this number, so I will simplify it by saying that e is the base rate of decay for all continuously decaying functions.  Why this is true is not a subject to discuss here, but suffice it to say that it is true, just the same as π is the ratio of circumference to diameter of all circles of all sizes.  Thus, it not only makes sense that e should appear in the rocket equation (said earlier to be a continuous decay function), but it also makes sense that, if it is the base rate of decay, that using it in the rocket equation will output the rocket's base efficiency.

To finally define the y-axis, then, you should understand that it represents the delta-v of the rocket, but expressed as multiples of the engine's exhaust velocity.  This makes it somewhat similar to the half-life of a radioactive isotope or the doubling rate of a population.  In other words, a mass fraction of 2.718, e, will yield a delta-v of one engine exhaust.  What the means in terms of metres per second depends on the engine you're using; a rocket moved by a Nerv will end up moving much faster than one moved by a Skipper, all else being equal (which it isn't; they use different propellants, but it could be figured out).

Let's look again at what this graph means.  Here's the link again so you don't have to scroll up.  A mass fraction of two (the rocket's dry mass plus an equal mass of fuel) gives a delta-v of .693 times the engine exhaust velocity.  If you want to go farther on that fuel, use an engine with a higher exhaust velocity.  If you use the mass fraction of three (dry mass plus double the mass in fuel) you get a delta-v of 1.097 times the engine exhaust velocity.  If you use a mass fraction of e (dry mass plus 1.718 times that mass in fuel), you get a delta-v of one engine exhaust.

Let's look at it the other way:  how much mass fraction do we need to get multiple engine exhausts of delta-v?  It's hard to see on the graph, but to get a delta-v of two engine exhausts, we need a mass fraction of about 7.389.  That is the dry mass plus 6.389 times that mass in fuel.  To get a delta-v of three engine exhausts, we need a mass fraction of just over 20.  To get four engine exhausts, we need a mass fraction of nearly 55.  To get five, we need a mass fraction of just over 148--in other words, a rocket that is one part payload and 147 parts fuel.  This is the decay of fuel efficiency spoken of earlier:  each time we add one more engine exhaust's worth to our total delta-v, it requires that we add nearly three times the mass fraction that is already there.  More specifically, it requires that we add 2.718, or e, times the mass fraction that is already there--see how that number keeps popping up?

 

Practical Problems and Applications

We should understand that we cannot keep adding mass fraction to our rockets without limit.  We can add propellant mass, but there is a definite limit to the mass fraction because for all the mass of fuel that you add, you must also add the mass of the tankage to contain it.  In KSP, the absolute theoretical limit is 9, that being the mass fraction of the fuel tanks themselves (8 parts fuel plus 1 part tank).  Of course, this is only hypothetical because a tank of fuel cannot fly without at least an engine and some kind of control, both of which add mass, but one could, for example, set up a giant assemblage of fuel tanks all pushed by a tiny Ant engine and controlled by an OKTO2 that approaches, but does not reach, that limit of nine.  In the case of a theoretical maximum mass fraction, the available delta-v is approximately 2.197 times the engine exhaust.  Actual delta-v obtained depends on the engine you choose.

To calculate it, it's relatively easy to drain all the fuel out of a stage and figure the mass while still in the VAB.  Once you have the masses both wet and dry, you can divide the wet by the dry to get mass fraction.  Then you just need to take the natural logarithm of that number (break out that calculator!) and multiply it by the engine exhaust velocity to get delta-v.  Remember, KSP reports Isp, not exhaust velocity, but that means your only change is to multiply the Isp by 9.81 metres per second per second.

For the Ant plus giant tank stack I pointed out earlier, the engine has an exhaust velocity of 3090.15 m/s (315 Isp * 9.81 m/s/s).  Assuming that it's got a mass fraction of nine, the natural logarithm we want is 2.197.  2.197 times 3090.15 gives a total delta-v of 6790 m/s.  It will also take a very long time to burn all that propellant, but there it is.

The ant, however, is not a terribly efficient engine; it's really just a small engine.  Efficient, low-cost, low-part-count KSP rockets can have a mass fraction that approaches 5--meaning a rocket that is 20% rocket and 80% fuel, and real-life rockets usually have much higher mass fractions.  The Space Shuttle clocked in at 15.4, the Saturn V (of Apollo fame) had 23.1.  Soyuz comes in at 38 and a bit:  that means just over 2.63% of the rocket is capsule; the rest is fuel.  Earth-based rocketry is hard.

In KSP, the most efficient LFO engine is the Poodle, followed closely by the Terrier, and then the Rhino and Dart.  The Terrier is the lightest of these, so it adds the least to the dry mass and consequently hobbles the mass fraction the least (the Rhino shines as a high-thrust vacuum engine; its mass--9 tonnes!--is pretty crippling for operations that don't need the thrust), but let's assume that we can make a Poodle with zero mass.  The Poodle's exhaust velocity is 3433.5 m/s (350 Isp * 9.81 m/s/s), so theoretical maximum delta-v is then about 7540 m/s.  Actual, achievable delta-v on a useful rocket at the far end of tank-heavy (thus at the far end of useful; a mass fraction of five is still a lot of fuel per unit rocket, especially if you're already in orbit) is more near 5500 m/s.

Given that it takes somewhere above 8,000 m/s to return from the surface of Eve at sea level, we can therefore conclude that, unless you can do it with Nervas or ion engines (not likely!), it is impossible to return from Eve's sea level surface.

 

Extra Stuff

... For a single stage.  Incidentally, this is the proof that Eve single-stage-to-orbit from sea level is impossible, but that is not the case with multistage rockets.  It turns out that you can cheat at the rocket equation a bit by getting rid of your extra tanks as soon as they are empty.  While they don't count as propellant mass, once they are gone, they don't count as dry mass, either, so what happens is your mass fraction jumps up every time you stage.  Generally, the payload is more than just a fuel tank, so the mass fraction never jumps to quite the level it was before staging.  In other words, as you stage away empty tanks, you end up with a better mass fraction than you had but still a rocket that is more and more payload, so the mass fraction gets smaller and smaller until you get to the last stage, which, if it's a drifting free-fall satellite, may well have a mass fraction of one because it's never supposed to go anywhere again.

The ideal version of staging is to discard your empty fuel tank as soon as it is empty--in essence, we're looking at a tank that burns away as we use the fuel in it--and that would be another continuous decay function.  In KSP, that can be approximated with the Oscar-B, the smallest fuel tank.  Ten Oscar-Bs make one FL-T400 ... but ten stage-able Oscar-Bs require nine more decouplers than that one FL-T400, as well.  There is eventually a point at which the mass needed to handle discarding the fuel tanks overtakes the amount of propellant saved by discarding them more frequently.  What this means in terms of practical application is that it is slightly more efficient to have a single fuel tank with a two-engine cluster at the bottom of the rocket that is staged by one decoupler than it is to have two half-sized fuel tanks, each with a single engine and staged by radial decouplers on either side of the rocket, provided that the masses of the radial and inline decouplers are the same.

[PB666]

18 hours ago, Zhetaan said:

@CodaThePortalmaster:

The ant, however, is not a terribly efficient engine; it's really just a small engine.  Efficient, low-cost, low-part-count KSP rockets can have a mass fraction that approaches 5--meaning a rocket that is 20% rocket and 80% fuel, and real-life rockets usually have much higher mass fractions.  The Space Shuttle clocked in at 15.4, the Saturn V (of Apollo fame) had 23.1.  Soyuz comes in at 38 and a bit:  that means just over 2.63% of the rocket is capsule; the rest is fuel.  Earth-based rocketry is hard.

In KSP, the most efficient LFO engine is the Poodle, followed closely by the Terrier, and then the Rhino and Dart.  The Terrier is the lightest of these, so it adds the least to the dry mass and consequently hobbles the mass fraction the least (the Rhino shines as a high-thrust vacuum engine; its mass--9 tonnes!--is pretty crippling for operations that don't need the thrust), but let's assume that we can make a Poodle with zero mass.  The Poodle's exhaust velocity is 3433.5 m/s (350 Isp * 9.81 m/s/s), so theoretical maximum delta-v is then about 7540 m/s.  Actual, achievable delta-v on a useful rocket at the far end of tank-heavy (thus at the far end of useful; a mass fraction of five is still a lot of fuel per unit rocket, especially if you're already in orbit) is more near 5500 m/s.

Given that it takes somewhere above 8,000 m/s to return from the surface of Eve at sea level, we can therefore conclude that, unless you can do it with Nervas or ion engines (not likely!), it is impossible to return from Eve's sea level surface.

So let me add my two cents on this. Regarding ISP and mass fraction, etc.

If you are taking off from a high g environment with alot of turbulence and drag, ISP is less of a concern and brut thrust is what you want, with the typical booster its only in operation from 5 seconds to 1.5 minutes, you are not carrying it to the mun so this whole consideration does not have much value.

When you are turning to achieve a positive rate of climb and horizontal velocity (in other words to place yourself on a trajectory that lasts more than a couple of minutes in the inertial state) you really want alot of thrust and ISP is important but less of a concern.

Once you establish a good rate of climb while exiting all drag environment then ISP is very important and high fuel mass to engine ratios are important, BUT also important you need a maximum burn time to reach orbit that is only slightly higher that the time it takes to reach apogee from the previous stage separation. If your engine is incredibly efficient (800s) but you cannot circularize your orbit in time, well thats something we will all enjoy watching on U-tube. There is a caveot, which is the seat of your pants scenario in which you put just enough fuel in to get close to orbit and then after entering space run the ion drive engines to assist and then finish off the orbit with ION drives (or faster with NTR). I have done this many time and you have to launch essentially perfectly and maintain a perfect launch trajectory.

Once you are in orbit, you can have ION drives stacked with a new elements lets call it Mg and give the ION drive 150000 ISPvel (instead of ISPsec). So the basic problem here is that in orbit you are at perigee +/- 20' for only a few minutes. And your ION drives are producing at this ISP are producing mN of thrust so basically you are kicking and an oblongated spiral out of orbit. This is not very time intelligent or efficient, its better to have enough thrust that in one kick and a 40' span you take advantage of oberth-like physics to get you from your circle to the desired. In fact with the New Horizons mission, one of the fastest space craft leaving earth, the best dynamic really involves kicking before circularization. IOW your take off pitch rolls you onto a holman trajectory that takes to deep into interstellar space and flies by pluto, its not a verticle shot but also never circularizes. Think of New Horizons as having a trajectory after its final LEO burn as intercepting the surface of the earth, not several 100 kilometers above the earth. The closer your (-time) trajectory is to the point mass that defines the gravitational body, the more impact the oberth effect has.
  So in this circumstance you want an engine and an efficiency most of the time that does not require more than say 3 or 4 kicks (IIRC orbit around kerbin is 30 minutes, earth is 80 minutes). So the engine should burn about 900dv acceleration at least in 15 min or your match and planning is going to be nuts.

  So now lets change the argument, lets suppose you are at on mojo and you want to then return back, we all know how difficult that is because it requires costly plane changes also. But with ion drive you have the benefit of sol, mojo you can change planes around mojo in the manner you lift off and then after circularlization use the ion drives to essentially drive the craft along a polar orbit until it suits a transition to whatever planet. Then of course you have to exit mojo. At the point you leave mojo you can forget about the planet and consider the position at perigee around the star (though depending on Mojo's perigee it may not be, but for a highly efficient low mass ION drive the good thing is mojo's period is 80 days, not 30 minutes which means a slow burning engine has an advantage of oberth-like physics on a long orbit and Ion drive has advantage of being close to the Sol. If you are deep in space and trying to leave the system, you have very long orbital periods and a long time to conduct burns, but the problem is that you have a tiny amount of energy.

So a clever idea for getting way out in space fast is getting a very high ISP engine & fuel very close to sol, doesn't need to be on a circular, estimate how many burn days it would take, start the burn days/2 before perigee and burn all the fuel and wee you will be souring into deep space with no hope of return. Heres the problem in reality, in KSP the ION drives produce way more thrust than Earthmade ION drives and require way less energy. In reality if you had a really efficient ION drive, you would need dozens of heavy football field size solar panels to power it to burn that fuel fast enough to take advantage of oberth physics, even relatively close to the sun. At earths orbit the power requirements and weight and structural issues with the panels are too high, you really need something like nuclear power (which currently fission power offers the same weight problem and not as reliable in space as solar panels). So that leave fusion and fusion drives.

So NTRs are the thing they expend nuclear fuel but they run on hydrogen that is difficult to store. In the game the ISP is around 800, in reality a modern age NTR the ISP would be close to 1000. In the game the tank holds reductant and oxidant side of the tank is wasted weight. In reality you would have for space circular shaped hydrogen tanks to minimize surface area to weight ratios and they probably would be mounted on a lattice of somesort that would release tanks as they expired. In the game tanks can hold the reductant indefinitely, in reality you would need a wrap around each tank and fuel lines that captured the hydrogen that leaked out and repressurized the tanks with it, and this would not last indefinietly. SO NTRs would only give a decent mass ratio but if used relative soon after lift-off, so if you wanted to use oberth affect around say jool, you need another less efficient propulsion system (fusion coupled to ion does not have enough thrust). I actually designed an NTR that could land and take off from the mun more efficiently than any in game rocket, the engine was large and tanks were mounted around the engine instead of above, and essentially I assumed that Kerbals were immune to radiation sickness or high gamma burns of any kind. 

 

 

 

 

Edited January 26, 2017 by PB666