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I'm looking for some information on what the fuel/propellant mass per kg of payload is for ion drives. Hall X3 or VASIMR or whichever is currently the 'best' that we have available.

This isn't for any specific 'mission', though I know distance is an important factor. Kg of fuel per Kg of payload per AU would be my ideal response, if there's a relatively straightforward way of calculating this?

All the data I can find is either obscure or specific. As in I can find technical specs (which I don't know what to do with), or I can find fuel data for very specific missions but with no way to extrapolate to something useful for myself.

Thanks for any help :)

Actually in case it is necessary I'll give an example of something I'd like to calculate with this - 500,000kg payload, 1au travel. Actually I guess it might involve multiple drives... which might make things more complicated. So I'll leave this as-is for now and see what kind of response might turn up.

Edit: Correcting this due to being very tired and asking the question badly. As has been pointed out, I should be using deltaV not distance.

As such, what I'm looking for is the derived equation from the rocket equation, where I can enter the target deltaV and the specific impulse of the engine, and the equation will spit out the fuel mass required per kg of payload. This may actually be within my ability to solve... but I've never been confident in rearranging formulae.

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    $\begingroup$ AU or any other space distance measure would not be good I think. As an example, once you reach Sun escape velocity you would be able to travel many AU for free without expending any further mass. It would be more meaningful Kg of fuel per Kg of payload per dV (delta velocity). dV is the actual measure of "distance" between orbits so to speak, that accounts for how much far a rocket can go consider its propellant mass and efficiency. $\endgroup$ – BlueCoder Nov 2 '19 at 11:01
  • $\begingroup$ See this map which tells how much "far" are different planets and orbits in dV reddit.com/r/space/comments/29cxi6/… $\endgroup$ – BlueCoder Nov 2 '19 at 11:01
  • $\begingroup$ Indeed often rockets are compared for how much payload they can bring to a certain orbit e.g. payload to LEO or payload to Moon for example. Which is in a way what you are asking, just with a fixed propellant mass (a specific rocket) and a fixed dV (dv of LEO or Mars). It must also be added that since thrust/weight ratio is very low in Ion engines, they cannot lift off planets (i.e. 0 payload from Earth to LEO), so the answer would be restricted to conditions where those engines can operate (i.e being already lifted in space by something else). $\endgroup$ – BlueCoder Nov 2 '19 at 11:08
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    $\begingroup$ In a comment to the other answer space.stackexchange.com/questions/39687/… Russel Borogrove has rightfully pointed out that due to the rocket equation "Fuel requirements increase exponentially with velocity requirements." So even an answer like payload for propellant for dV does not exist (since it changes if dV is increased). The engine efficiency measure that you can find is specific impulse (I_sp). $\endgroup$ – BlueCoder Nov 2 '19 at 11:18
  • $\begingroup$ Turned the comments into an answer :) By the way, Welcome to Space Exploration SE! $\endgroup$ – BlueCoder Nov 2 '19 at 11:39
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As such, what I'm looking for is the derived equation from the rocket equation, where I can enter the target deltaV and the specific impulse of the engine, and the equation will spit out the fuel mass required per kg of payload. This may actually be within my ability to solve... but I've never been confident in rearranging formulae.

Here’s the rocket equation, rearranged to “mass fraction form”:

$$M_f = 1 - \frac{m_1}{m_0} = 1-e^{-\Delta V / v_\text{e}}$$

This is "what fraction of the entire spacecraft is fuel". From there it's trivial to get to "what fraction isn't fuel" -- that is, dry spacecraft plus payload:

$$e^{-\Delta V / v_\text{e}}$$

So the total fueled ship mass per dry spacecraft mass is:

$$ \frac{1}{e^{-\Delta V / v_\text{e}}}$$

or just:

$$ {e^{\Delta V / v_\text{e}}}$$

Getting from there to payload mass requires you to make some engineering assumptions about tankage, structure, and engine mass, which vary widely. For chemical rocket stages we often see between 5% and 10% of fuel mass in dry structure, but we can see both these extremes in two otherwise rather similar designs: the Zenit 3SL and the Falcon 9.

Structural mass requirements put a practical upper limit on delta-v per stage, even though the rocket equation itself doesn't impose a theoretical limit.

Assuming 210000 m/s exhaust velocity and 6600 m/s required delta-V (LEO to low Mars orbit, e.g.), the total fueled mass is 1.032 units for every unit of dry mass. That is, only about 3% of the ship is fuel: with such a high exhaust velocity, Earth-to-Mars shuttles would not be dominated by fuel mass. You can plug those values back into the conventional rocket equation:

$$v_\text{e} \ln \frac{m_0}{m_f}$$

to check; $m_0$ (initial, fueled mass) is 1.032 and $m_f$ (final, dry mass) is 1.0 here.

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  • $\begingroup$ Funnily enough, I found this not long ago - medium.com/teamindus/… Which has a very similar version of the formula you wrote first, except that they write it as mass ratio = e ^ (delta-v/exhaust velocity). I'm assuming the two negatives cancel out and result in the same end result. If I'm doing it correctly, then a ship with specific impulse of 210000, going from LEO to low mars orbit, will have a mass ratio of about 1.03. Meaning for every 1kg of payload, you'd need 1.03kg of fuel. Does this seem correct? $\endgroup$ – nirurin Nov 3 '19 at 4:21
  • $\begingroup$ Changing the sign in the exponent cancels reciprocation, not subtraction. I'll address your example in an edit, but note again that dry mass is not the same as payload mass. $\endgroup$ – Russell Borogove Nov 3 '19 at 4:27
  • $\begingroup$ "but note again that dry mass is not the same as payload mass" - Yeh sorry, I mostly say 'payload mass' as a shorthand, for my own uses when I refer to payload mass I mean "the entire rest of the ship that isn't fuel". I guess I should use 'ship dry mass' instead to be less confusing for people. $\endgroup$ – nirurin Nov 3 '19 at 4:30
  • $\begingroup$ Did you intend 210000 m/s exhaust velocity or 210000 second specific impulse? $\endgroup$ – Russell Borogove Nov 3 '19 at 4:31
  • $\begingroup$ that is m/s. The specific impulse would be 21400seconds. $\endgroup$ – nirurin Nov 3 '19 at 4:32
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AU or any other space distance measure would not be good. As an example, once you reach Sun escape velocity you would be able to travel many AU for free outside the solar system without expending any further mass.

It would be more meaningful Kg of fuel per Kg of payload per dV (delta velocity).
dV is the actual measure of "distance" between orbits so to speak, that accounts for how much far a rocket can go consider its propellant mass and efficiency.
See this map which tells how much "far" are different planets and orbits in dV: Solar System subway map

Indeed often rockets are compared for how much payload they can bring to a certain orbit e.g. payload to LEO or payload to Moon for example. This is in a way what you are asking, just with a fixed propellant mass (a specific rocket) and a fixed dV (dv to LEO or Mars). It must also be added that since thrust/weight ratio is very low in Ion engines, they cannot lift off planets (i.e. 0 payload from Earth to LEO), so the answer would be restricted to conditions where those engines can operate (i.e being already lifted in space by something else).

However, due to the Rocket Equation, fuel requirements increase exponentially with velocity requirements. (See also Russell Borogove's answer and comments to your other question).

So even an answer like payload for propellant for dV does not exist (since it changes if dV is increased).

The engine efficiency measure that you can find is specific impulse ($I_{sp}$).
(See also the rocket equation and how this term actually depends on the engine).

So here are the $I_{sp}$ for the engines you mentioned:
Hall X3 - between 1000s and 8000s (Wikipedia)
VASIMR - about 5000s for VX-200 version (Wikipedia)

Consider that chemical engines have efficiency ($I_{sp}$) that reaches around 450s in space (or even less) and you see why Ion engines can be so good (once you have lifted them in orbit).

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  • $\begingroup$ Aha yes, it was very (very) late when I wrote this original question, so what I had in mind was orbital distances (in AU) but what I really -meant- was deltaV in order to reach those distances. My bad. $\endgroup$ – nirurin Nov 2 '19 at 17:14
  • $\begingroup$ "It would be more meaningful Kg of fuel per Kg of payload per dV (delta velocity)." Along with the subway map you provided, this would be ideal. I'm a little lost on the derivations of the rocket equation though (Borogrove's version has an 'energy' requirement but I don't know where he's drawing that figure from, for example). I suspect it's all basic for people who have dealt with this data before, but it's new to me. I have two specific impulse values in mind (12,000, and 22,000), I don't suppose you could give me some tips (or point me to a guide) on how to convert that to KGf/KGp/dV $\endgroup$ – nirurin Nov 2 '19 at 17:22
  • $\begingroup$ Just to correct that, what I actually would need (I think?) is a version of the rocket equation that would allow me to plug in the target deltaV and the specific impulse of my engine, and the result of the equation is the kg of fuel required per 1kg of payload. $\endgroup$ – nirurin Nov 2 '19 at 20:12
  • $\begingroup$ @nirurin the rocket equation has the "initial mass" and "final mass" baked into it. The difference being the propellant that is expended. You're going to have to know more about your vehicle to know how much of the "final mass" is useful payload and how much is engines, structure, etc. So your problem appears to me to be under-specified. $\endgroup$ – Organic Marble Nov 2 '19 at 22:57
  • $\begingroup$ My problem is that I am unable to work out the design of the ship (the specifics, at least) without knowing what the ratio between fuel mass and payload mass is. Without knowing that, I can't know what my ballpark upper limit for payload mass would be. $\endgroup$ – nirurin Nov 2 '19 at 23:35

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