Oxidizer/Liquid Fuel Math?

[Ncog Nito]

Got a math question for those more intelligent than me.  I am trying to transfer fuel from my refueling station to another vessel.  I don't want to "fill er' up", I just want to transfer enough to get to where I am going.  I also don't want to over or under estimate the amount of oxidizer I need compared to the liquid fuel.  Can someone give me the math, or direct me in the right direction, to figure up the liquid fuel to oxidizer mix ratio?

[Starman4308]

Look at any of the stock tanks; the ratio is 11 oxidizer to 9 liquid fuel.

The math gets more complex if you have a nuclear rocket mixed in or are using Real Fuels.

[bewing]

When you are transferring fuel, it's hard to get the ratio perfect -- unless you have a small tank available. You have to play games about -- OK, if I fill this tank full, and then transfer as much as possible over to this other tank over here, that'll leave 110 units in the first tank, and then if I add 55 units from that little tank over there, that'll be just about perfect!

 

[FancyMouse]

That's why I have different sizes of tanks on my station so that I can get what I want (each operation would be either empty a tank or fill a tank to full, so that the ratio is always correct).

[Blaarkies]

2 hours ago, Ncog Nito said:

1. "I just want to transfer enough to get to where I am going.  "
2. "I also don't want to over or under estimate the amount of oxidizer I need compared to the liquid fuel.  "

1. To know how much fuel you need to go somewhere, you need to know DV. In cars we measure range in miles/kilometers, in space we measure it in DV(delta-v). a.k.a The amount of extra speed you can add to the spaceship if you were to burn all of your fuel. That dictates if you can go to the Mun, or farther. To find that number, you need to know how much you ship masses when empty, and how much fuel you have onboard. Every 90 LF+110 OX masses at 1ton, but for ballpark numbers i just figure my LF/100 (and always have balanced OX for them) when finding DV for LFO engines.
When doing it for Nuke engines (that use no OX), just do LF/200

2. If you filled up some LF into the small ship, find the balanced OX_needed = LF_onboard * 1.22
    If you filled up some OX into the small ship, find the balanced LF_needed = OX_onboard * 0.82

[Aegolius13]

Assuming you only have regular rocket fuel tanks, and regular rockets, you can also go visually based on the green fill bars.  Since the tanks are designed to have the correct ratio of fuels when full, 75% full of LF will match 75% of oxidizer.  There's a bit of manual error based on how accurately you can click the stop button, but it's usually fairly minor.  

Of course, that's only if you're filling the same size tank with both.  And as mentioned, the math gets more complicated if you're mixing in nuclear engines or something.  

Re: knowing how much you'll need, the delta-v readout on the Kerbal Engineer Redux mod is extremely useful.  It's surprising how much delta-v you can manage with just the last dregs of fuel, but that's the rocket equation in action (a ship with empty fuel tanks is very light, so you can get much more acceleration with every unit of fuel burned).  

[Foxster]

The ratio O to LF is 1.2:1. 

[ForScience6686]

Are you familiar with the rocket equation?  If so all you need is your dry mass and the necessary dv.  Then your wet mass becomes x, solve for x.  Then depending on your propulsion source, you calculate the fuel units to reach your wet mass.

[swjr-swis]

Bill: <reads suggestions, does some numbers>

Bill: <notices more suggestions, with other numbers, recalculates>

Bill: <notices even more suggestions, with yet other numbers, recalculates again>

Jeb, Val and Bob in unison: "Oh heck, just fill 'er up and let's go!"

[GoSlash27]

What @Starman4308 said.

 Take your fuel, multiply by 11 and divide by 9. That gives you the required amount of oxidizer to match.

Best,
-Slashy

[Spricigo]

5 hours ago, swjr-swis said:

Jeb, Val and Bob in unison: "Oh heck, just fill 'er up and let's go!"

And it makes a lot of sense. Empity fuel tanks are deadweight anyway.

[GoSlash27]

28 minutes ago, Spricigo said:

And it makes a lot of sense. Empity fuel tanks are deadweight anyway.

Spirigo,

 I disagree. Those empty tanks are dead weight whether they are filled or not. Fuel that gets brought along but never gets used is a whole lot more dead weight. Plus, it takes time, money and effort to get that fuel up there. No sense in wasting it. The OP is on the right track; take what you need, but use what you take.

Best,
-Slashy

 

[Spricigo]

1 minute ago, GoSlash27 said:

Spricigo,

 I disagree. Those empty tanks are dead weight whether they are filled or not. Fuel that gets brought along but never gets used is a whole lot more dead weight. Plus, it takes time, money and effort to get that fuel up there. No sense in wasting it. The OP is on the right track; take what you need, but use what you take.

Best,
-Slashy

 

IMHO that is why, in a tortuous way, it make sense.  Ideally those tank should never been there to be left empty or to be filled with extra dead weight. So, ideally, one don't have space for more fuel than necessary.

Off course you are correct for what probably its the actual (non-ideal) circumstance of a vessel that need to be moved as it is. By now may be difficult to remedy heavier than necessary fuel tanks.

Also, there is the (slight) possibility to find a good use for this extra fuel at the destination and the whole question if the optimization its actually worth the trouble.

[bewing]

A slight possibility? I always have uses for every last drop once I get to my destination, and wish I had twice as much. This is the main reason to always travel with full tanks. You'll find uses for the fuel if you try.

 

[Spricigo]

1 hour ago, bewing said:

A slight possibility? I always have uses for every last drop once I get to my destination, and wish I had twice as much. This is the main reason to always travel with full tanks. You'll find uses for the fuel if you try.

 

Well, having a use don't means having a good use.

Maybe it only means you usually don't bring enough fuel while I usually bring too much.

[Red Iron Crown]

On 5/23/2017 at 10:40 AM, Foxster said:

The ratio O to LF is 1.2:1. 

You should probably specify that this is rounded to one decimal place, the error can be significant when dealing with large quantities of propellant. 

[Foxster]

16 hours ago, Red Iron Crown said:

You should probably specify that this is rounded to one decimal place, the error can be significant when dealing with large quantities of propellant. 

No need. You've done it for me. 

You'd need a hell of a lot of fuel for the difference to matter and if you are messing about with those kinda quantities then chances are you are using full LF+O tanks anyway. 

Edited May 25, 2017 by Foxster

[weebiboi.ksp]

I don't know the equation either, but I know the LF+OX ratio is 11 units ox for 9 units liquid fuel. I did the equation in calculator and 9 x 1.23 = 11.07. So just use 1.23 as the multiplier, despite the fact that it might leave you with a little extra oxidizer. 

Edited May 9, 2018 by weebiboi.ksp

[Arugela]

There is a ratio of LF to OX. You don't need to use estimates.

Oxidizer = 0.55 (11/20)

LF = 0.45 (9/20)

FYI, 9+11=20=denominator.

If you can use that to determine the total weight of LF+OX combos for calculating delta V.

Example:

550 OX = 550 / 0.55 = 1000 LF+OX * 0.45 = 450 LF

1000 LF+OX = 1000 * 0.005 = 5 tons.

LF and OX = 0.005 tons per unit.

Mono = 0.004 tons per unit

Ore = 0.01 tons per unit.

Reverse Example:

450LF / 0.45 = 1000 LF+OX * 0.55 = 550 OX

etc.

Determine whichever thing you need from this.

Use the total LF+OX as your weight value. Then use the time to burn and either oxidizer or lf and it's total consumption per second to determine burn time etc. If you use Nervs with your rocket engines you can use burn times or calculate Ox seperately and add the Lf time on top or whatever is convenient. Overall weights are for delta V. There are many ways to calculate it.

Electricity has weight based on how much the batteries weight, but I forget the numbers. It's either 0.05 or something. Divide the weight by the capacity. I think it's the same all around. You can use this for xeno base stuff. It can be used to calculate obtainable delta V based on efficiency of the solar panels and their weight to gain or how much you can get from generators etc effectively overall.

 

That ratio is, btw, why the original smallest tanks in the game were 9 LF to 11 OX. It was to demonstrate the ratio partially.

This is very basic algebra. And fractions are more accurate. They force you to use exact values. IT IS SUPEEEERRRIIIIOR!!!! 8D

That and geometry for the ability to represent irrational numbers!! 8d Geometry is ratios... So it is all SUPEERRRRIIOOORRR!!

 

 

Edited May 10, 2018 by Arugela