Has anyone managed to create a miner capable of net-gain-to-orbit on Moho?

[Himynameisjake]

I've really been struggling with this, trying to get more ore to orbit than I use in fuel/monoprop to dock to an orbiting refinery. I've come within about 5% of break even, but I can't even get to 1:1, much less profit. Any suggestions would be very helpful!

I've been using a single poodle engine and experimenting with different tank combinations for ore and LF/O, because nukes overheat right now no matter how I build them.

[wasmic]

Have you tried using a stationary miner on Moho and a transport vessel to ferry the fuel?

[Himynameisjake]

Have you tried using a stationary miner on Moho and a transport vessel to ferry the fuel?

I haven't, I figured that would be less efficient due to the added weight of docking ports and extra engines. I'd like to return all ship parts to orbit.

[Captain Sierra]

I haven't, I figured that would be less efficient due to the added weight of docking ports and extra engines. I'd like to return all ship parts to orbit.

Generally, you want to refuel on the ground and in orbit. That means you only need fuel for 1 leg and thus have less tank mass, less total weight, and less engine power needed (again more weight savings). What you are asking for is a moho lander with ~40% payload fraction (50% fraction on wet mass, plus tankage, engines, legs, docking ports, etc.). I need to verify that is mathematically possible (there's a chance it might not be)

I'd drop down a wellhead which has the drills and the ISRU convertor on it, and use a tanker lander. The wellhead can easily refuel itself, jettison remaining ore, and make it back to orbit (it got down in that configuration did it not? ).

[adinfinitum]

I just ran a few quick calculations, and it looks like it will be very difficult, but not impossible, to build a miner that can get a slim profit from Moho. Going off a deltaV map, it takes 870 m/s dV to land on Moho, so let's say you want 1000m/s just to be safe. It will take the same amount to get back into a low Moho orbit, so you need 2000m/s. Using only poodle engines, this means 44% of your mass has to be fuel. If you want to hit break even for ore, you need the same mass in ore, since it's a 1:1 mass conversion. So we've got 44% fuel, and 44% ore, that leaves you 12% of the mass for other stuff like engines, drills, solar panels, legs, and all that stuff.

If you make a ship that converts the ore into fuel on planet and then just takes fuel back up, you can refill on planet, and only need 25% of your mass to be fuel, however much percent you have leftover after the drills, convertor, etc. is just profit fuel.

In this case I'd say bringing ore up is not the best way to go about it.

[Himynameisjake]

Generally, you want to refuel on the ground and in orbit. That means you only need fuel for 1 leg and thus have less tank mass, less total weight, and less engine power needed (again more weight savings). What you are asking for is a moho lander with ~40% payload fraction (50% fraction on wet mass, plus tankage, engines, legs, docking ports, etc.). I need to verify that is mathematically possible (there's a chance it might not be)

I'd drop down a wellhead which has the drills and the ISRU convertor on it, and use a tanker lander. The wellhead can easily refuel itself, jettison remaining ore, and make it back to orbit (it got down in that configuration did it not? ).

That doesn't sound like returning more to orbit than used to get down and back up.

[KerikBalm]

"Generally, you want to refuel on the ground and in orbit."

Well, not really in orbit... that sort of defeats the purpose of bringing up more fuel from moho than you used.

Any "refuel" in orbit will be subtracted from the fuel brought up, as far as the getting better than 1:1 is concerned.

The ground refuel is valid though.

I note that the long process is the ore extraction, the conversion is pretty fast.

So 1 converter should serve for a large amount of ore. The large your ore ferry, the smaller the fraction of mass the converter is.

Keep in mind that when ascending, you're carrying lots of ore, so the extra mass of the converter doesn't affect it that much.

When descending, because you have the converter, you're not carrying the fuel needed to go back up.

Also when starting your ascent, all your tanks can be full (ore and LF+O).

I would definitely go for refiing on the surface.

Now, you only need 870 dV to ascend to orbit.

Using a poodle, lets find your mass fraction needed:

870= 350*9.81* ln (X)

ln(x) = 0.2534

x = 1.288

Only 22.4% of your payload needs to be fuel for the ascent, if you refine on the surface.

1x poodle produces 250 kN of thrust. Lets assume you want a starting TWR of 2.0...

Moho's Gravity is 2.7 m/s^2.

Thus with on poodle, your craft mass on ascent should be 46 tons.... A refinery at <4.5 tons represents less than 10% of this mass.

poodle: 1.75 tons. Refinery: 4.2 tons (?)

Fuel for craft ascent+ tanks: 11.6 tons

Excess mass budget: 28.45 tons for ore/mining equipment/batteries/solar panels/reaction wheels/command pods/whatever

That should get you over a 50% payload fraction

That doesn't sound like returning more to orbit than used to get down and back up.

Actually, it does. He's just separating the mining equipment from the tanker to get a higher payload fraction

-update-

Lets calculate a payload fraction assuming no self-refuel on the surface.

dV cost is doubled: 870*2 ->

1740= 350*9.81* ln (X)

ln(x) = 0.5068

x = 1.66

Now 40% of you payload needs to be fuel for the deorbit and ascent.

which means 9/8* 40 = 45% percent of your lander's mass needs to be fuel+ tanks for the descent+ascent.

Of those 46 tons a poodle can lift, 20.7 tons need to be fuel+ tanks just to get down and up.

You've added 10 tons of fuel to avoid carrying a 4 ton ISRU convertor

*Something seems wrong with my math, it shouldn't take less than double the fuel to get double the dV... what have I done wrong :/

*ah, I see, in the 2nd case, its assuming a much lower dry mass. Use more fuel to push less mass.

Just like the fuel won't double each time, especially when you get to really really bad mass ratios, like 40:1 or 80:1

Edited May 17, 2015 by KerikBalm

[KerikBalm]

Ok, so I read adinfinitum's post, and I see an obvious criteria that I left out (although he left soemthing out too: fuel tank mass).

I'm not being conservative, and not planning for 2,000 dV, just 1740.

So fuel needs to be 40% of your mass, not 44% as he calculated.

He pointed out that obviously, to break even, if 44% of your mass is fuel, then 44% of your mass needs to be ore to break even.

So, cutting the dV budget tighter, we'll use 40% fuel, and 40% ore to payback that fuel.

The best mass ratio on the LFO tanks is 9:1. That means 9/8 * 40% -> 45% of you mass needs to be full LFO tanks.

The best ore tank ratio is 17:2 (8.5:1 -> so its actually a better LF tank)

If 40% of your craft mass is Ore, then 8.5/7.5 * 40 = 45.3% of your mass needs to be ore+ore tanks.

That means tanks+ Fuel to go down and back up + ore to break even will be 90.3% of your mass at first glance...

When you add reaction wheels, solar, drills, etc... It gets to be quite a load- may not be possible to break even

But we're forgetting something, we go down with empty Ore tanks!

So the fuel for the first part of the trip can be 40% less, because Ore, being 40% of the craft mass, is missing at that time).

So, lets say that each part of the trip would take the same amount of fuel if it weren't for adding ore after landing.

Then we only need 80% of the fuel we previously calculated (60% for the first part, 100% for the 2nd)

So only 36% of the mass needs to be fuel + tanks, and thus only 36.25% needs to be ore + tanks.

--- gah, actually we need to consider the ascent and descent seperately ---

Descent-Start: Empty ore tanks, full fuel tanks

Descent-Finish: Empty ore tanks, partially full fuel tanks

*Mine*

Ascent-Start: Partially full fuel tanks, full ore tanks

Ascent-Finish: Empty fuel tanks, full ore tanks

We assume for break even that Empty fuel tank+ full ore tank mass = Full fuel tank + Empty ore tank mass

Thus Descent-Start mass = Ascent Finish mass

So we see the craft is heaviest at the start of Ascent -> by the amount = to the fuel remaining in the LFO tanks for ascent

We also assume that the the fuel used for each leg of the journey is 22.4% of the mass at the start of that leg.

Assume the craft is 46 tons at its heaviest point- the start of the ascent- (46 tons being the point at which a poodle achieves a 2:1 TWR on Moho).

10.304 tons of that is fuel for the ascent.

The craft would then be 35.7 tons at the start of the descent.

Thus 8 tons of fuel would be consumed on the descent.

A total of 18.3 tons of fuel would be consumed, so your ore tanks would need to hold 18.3 tons.

LFO tanks would then weigh: 18.3 *1/8 = 2.29 tons

Ore tanks would weigh: 18.3* 1/7.5= 2.44 tons

On ascent, minimum mass would be:

1.75 tons (poodle) + 10.3 tons (LFO for ascent) + 18.3 tons (Ore to break even)+ 4.73 tons (tank weight)

= 35.1 tons

46-34.5= 10.9 tons

With one poodle, you should be able to break even with 10.9 tons left in the budget for things like: drills + solar arrays + command pod/probe core + landing leg+ reaction wheels/RCS + docking port, etc.

Any mass left over can be used to haul up Ore in excess of break even-> which also means the craft mass is even lighter during the descent, meaning you'll save part of that 8 tons of fuel (but less than 3 tons).

Lets say of those 10.9 tons, you need only 2 tons for that other stuff (No mk1-2 command pod here! that sucker is 4 tons on its own!)

Thats 8.9 tons excess (keep in mind, thats not all going to be ore... the tanks themselves have noticable weight).... Your craft starts at about 36 tons, and brings up ~ 7.7 tons of ore in excess of what is needed to replace losses.

7.7/36 = 21.4% payload fraction

(only 16.7% if you count the mass after mining, and then subtract what is needed to break even).

It seems to me that you should have an easy enough time breaking even.

I'd still recommend bringing the converter with you. Then you only need to replace the <8 tons of fuel for the descent, not 18 tons of fuel.

Having to haul up an extra 10.3 tons of ore to replace the 10.3 tons of fuel that you burned during the ascent does not make sense when the converter masses less than 4.5 tons. It also reduces the "waste" due to empty tank mass, as you ascend with full LFO and ore tanks.

Hauling a piece of equipment <4.5 tons down to the surface is clearly worth it when it means not having to haul 10.3 tons of fuel to the surface for ascent. Thus you'll need even less than 8 tons for the descent.

You can use mostly LFO tanks which have better mass ratios, and you won't be running dry when you get to orbit. (but for maximum efficiency, you will be running dry when you set down on Moho, but you should have enough of a margin that you can take a little extra fuel for the landing)

Bring the refinery with you!!!

Edited May 19, 2015 by KerikBalm

[Geschosskopf]

While Ore-to-fuel results in the same mass of resources, the empty mass of the ore tank is rather more than the empty mass of a fuel tank. Thus, it's always cheaper on fuel to move product than to move Ore. (This is since 1.0.2--in 1.0, Ore tanks were too light so it was better to move Ore than product). This is especially true because you can't burn Ore directly, so anything moving Ore also needs fuel tanks, so having both on the same ship addes even more mass.

So, the most effient way to turn Ore on the surface into fuel in orbit is to refine on the surface and have a dedicated, minimalist tanker to carry the desired amount of product plus whatever fuel it needs for the round trip.

[moronwrocket]

Can this information get added to an ISRU page on the wiki? Be great to see math for the major bodies.

[Himynameisjake]

This has been an awesome discussion in theory, I'll try to put it to use tonight and see what I can come up with for a real design!

[Himynameisjake]

Ok, so I read adinfinitum's post, and I see an obvious criteria that I left out (although he left soemthing out too: fuel tank mass).

I'm not being conservative, and not planning for 2,000 dV, just 1740.

So fuel needs to be 40% of your mass, not 44% as he calculated.

He pointed out that obviously, to break even, if 44% of your mass is fuel, then 44% of your mass needs to be ore to break even.

So, cutting the dV budget tighter, we'll use 40% fuel, and 40% ore to payback that fuel.

The best mass ratio on the LFO tanks is 9:1. That means 9/8 * 40% -> 45% of you mass needs to be full LFO tanks.

The best ore tank ratio is 17:2 (8.5:1 -> so its actually a better LF tank)

If 40% of your craft mass is Ore tank, then 8.5/7.5 * 40 = 45.3% of your mass needs to be ore.

That means tanks+ Fuel to go down and back up + ore to break even will be 90.3% of your mass at first glance...

When you add reaction wheels, solar, drills, etc... It gets to be quite a load- may not be possible to break even

But we're forgetting something, we go down with empty Ore tanks!

So the fuel for the first part of the trip can be 40% less, because Ore, being 40% of the craft mass, is missing at that time).

So, lets say that each part of the trip would take the same amount of fuel if it weren't for adding ore after landing.

Then we only need 80% of the fuel we previously calculated (60% for the first part, 100% for the 2nd)

So only 36% of the mass needs to be fuel + tanks, and thus only 36.25% needs to be ore + tanks.

--- gah, actually we need to consider the ascent and descent seperately ---

Descent-Start: Empty ore tanks, full fuel tanks

Descent-Finish: Empty ore tanks, partially full fuel tanks

*Mine*

Ascent-Start: Partially full fuel tanks, full ore tanks

Ascent-Finish: Empty fuel tanks, full ore tanks

We assume for break even that Empty fuel tank+ full ore tank mass = Full fuel tank + Empty ore tank mass

Thus Descent-Start mass = Ascent Finish mass

So we see the craft is heaviest at the start of Ascent -> by the amount = to the fuel remaining in the LFO tanks for ascent

We also assume that the the fuel used for each leg of the journey is 22.4% of the mass at the start of that leg.

Assume the craft is 46 tons at its heaviest point- the start of the ascent- (46 tons being the point at which a poodle achieves a 2:1 TWR on Moho).

10.304 tons of that is fuel for the ascent.

The craft would then be 35.7 tons at the start of the descent.

Thus 8 tons of fuel would be consumed on the descent.

A total of 18.3 tons of fuel would be consumed, so your ore tanks would need to hold 18.3 tons.

LFO tanks would then weigh: 18.3 *1/8 = 2.29 tons

Ore tanks would weigh: 18.3* 1/7.5= 2.44 tons

On ascent, minimum mass would be:

1.75 tons (poodle) + 10.3 tons (LFO for ascent) + 18.3 tons (Ore to break even)+ 4.73 tons (tank weight)

= 35.1 tons

46-34.5= 10.9 tons

With one poodle, you should be able to break even with 10.9 tons left in the budget for things like: drills + solar arrays + command pod/probe core + landing leg+ reaction wheels/RCS + docking port, etc.

Any mass left over can be used to haul up Ore in excess of break even-> which also means the craft mass is even lighter during the descent, meaning you'll save part of that 8 tons of fuel (but less than 3 tons).

Lets say of those 10.9 tons, you need only 2 tons for that other stuff (No mk1-2 command pod here! that sucker is 4 tons on its own!)

Thats 8.9 tons excess (keep in mind, thats not all going to be ore... the tanks themselves have noticable weight).... Your craft starts at about 36 tons, and brings up ~ 7.7 tons of ore in excess of what is needed to replace losses.

7.7/36 = 21.4% payload fraction

(only 16.7% if you count the mass after mining, and then subtract what is needed to break even).

It seems to me that you should have an easy enough time breaking even.

I'd still recommend bringing the converter with you. Then you only need to replace the <8 tons of fuel for the descent, not 18 tons of fuel.

Having to haul up an extra 10.3 tons of ore to replace the 10.3 tons of fuel that you burned during the ascent does not make sense when the converter masses less than 4.5 tons. It also reduces the "waste" due to empty tank mass, as you ascend with full LFO and ore tanks.

Hauling a piece of equipment <4.5 tons down to the surface is clearly worth it when it means not having to haul 10.3 tons of fuel to the surface for ascent. Thus you'll need even less than 8 tons for the descent.

You can use mostly LFO tanks which have better mass ratios, and you won't be running dry when you get to orbit. (but for maximum efficiency, you will be running dry when you set down on Moho, but you should have enough of a margin that you can take a little extra fuel for the landing)

Bring the refinery with you!!!

This was awesome, just tried it out and it worked great! The best part is, if you bring enough panels with you and launch in the daylight, you can refine as you go, to make sure you don't run out of gas! My first attempt at building with this design lead to a net profit of 148 ore, not incredibly remarkable considering that it took almost a year to mine at one of the higher density locations, but it is a proof of concept that the idea works!

Edited May 19, 2015 by Himynameisjake

[Himynameisjake]

I was able to bump to 500 spare ore on my second build, it's looking very doable to hit reasonable numbers. Now I need to figure out how to modify this guy for duna.