Balanced Orbital Refueling Challenge: juggle speed, cost, and payload

[davidparks21]

Resupplying an orbital refueling dock on a regular basis is a laborious but necessary part of a space program of even modest ambition. Knowing how much it costs you to do so is a line item on many challenges to boot. So lets make it not only cost effective, but realistically efficient to do so on a regular basis, then let's throw the bean counters at it.

Goal: Deliver maximal refueling to an orbiting dock, at the lowest cost, and most timely round trip. Balance your payload capacity, delivery/docking time, and expenditures.

Categories:

- Stock parts/aerodynamics,

- Near/Far (stock parts),

- Custom Utility (all non stock stuff regardless of aerodynamics model)

Setup:

- Have a refueling dock in LKO, any orbit you choose (with sufficient space to hold the fuel being delivered, no points awarded here)

Scoring:

- Add points for units of fuel delivered (each resource times the multipliers below)

- Liquid Fuel.........[1.466]

- Oxidizer.............[0.846]

- Monopropellant...[1.733]

- Xeon Gas...........[4.013]

- Subtract 1 point per second of mission time (launch to landing, you may subtract time spent docked to choose a return window)

- Subtract 1 point per \F spent: Launch cost - Recovery value

- Add 10,000 points for completing the challenge

Notes:

- Example: 1 full orange tank delivered in 30 minutes round trip for a cost of 2500:

+ 7200 points (2880 liquid fuel * 1.466 + 3520 oxidizer * 0.846)

- 1800 points (30 min = 1800 sec)

- 2500 points (for 2500 \F spent on trip)

+ 10000 points for challenge completion

= 12900 total score!

- When subtracting points for \F spent, *do* include the cost of fuel that was delivered (the original cost is already accounted for in your points awarded, which is why Xeon Gas has a high multiplier)

- Fuel cost and weights have been normalized so that there is no points benefit of taking one vs. another.

- Negative scores are perfectly valid, and come with no more shame than a late night unintended ignition at the lab.

Rules:

- No part clipping, debug menu, or the like

- No auto pilot assistance (you may use mechjeb for limited planning such as landing at KSC, but piloting and docking must be done manually due to the time constraint)

- Informational mods (mechjeb for planning only, flight engineer, etc) are fair

- Mods Deb Refund and Flight Manager For Reusable Stages may be used, but must be noted if they are (their use will be noted on the leader board and won't go under stock category)

- If your delivery vehicle is recovered (saving cost), time counts until you land for recovery (anywhere on kerbin), if your delivery vehicle is dumped (no cost recovery) time counts until you dock it at your orbital station only.

- Provide screen shots of your endeavor

- I reserve the right to make minor alterations to the scoring system if it's in the long-term interest of the challenge. If such occurs and you lose position, your original accomplishment will be prominently noted.

Challenge Finishers:

Stock:

(1) davidparks21 - 5321 points - Jet based SSTO (trip report)

(2)

(3)

Near/Far:

(1)

(2)

(3)

Custom Utility:

(1)

(2)

(3)

Edited November 15, 2014 by davidparks21
Corrected points multiplier for liquid & Xeon fuels and allowed a deduction for time docked

[Laie]

nonsensical post removed -- I got it all wrong.

Edited September 20, 2014 by Laie

[chengong]

reserved for entry

[Sirine]

Xeon gas will exploit this challenge.

[cakepie]

Scoring:

- Add points for units of fuel delivered (each resource times the multipliers below)

- Liquid Fuel.........[1.448]

- Oxidizer.............[0.846]

- Monopropellant...[1.733]

- Xeon Gas...........[4.133]

- Subtract 1 point per second of mission time (launch to landing)

- Subtract 1 point per \F spent: Launch cost - Recovery value

Xeon gas will exploit this challenge.

Indeed, the multipliers, look pretty arbitrary, I don't see any obvious correlation with cost/mass. Maybe if you could please explain how you chose this points system / arrived at these numbers?

[davidparks21]

Indeed' date=' the multipliers, look pretty arbitrary, I don't see any obvious correlation with cost/mass. Maybe if you could please explain how you chose this points system / arrived at these numbers?[/quote']

The high value for Xeon Gas is almost entirely due to it's very high purchase cost (fully 4 \F per unit), you'll subtract off all those points in you launch-recovery (1 point lost per \F spent). I've normalized all fuels for both cost and weight so there is no benefit to taking one vs. another. Check the math, I hope I didn't get it wrong, but I'm pretty sure they're right. Normalized for weight the points for all fuel delivery is multiplied by 4 (giving what I decided was a fair trade off between cost and time) and then I added in the multiplier for the actual cost of the fuel which cancels out your final launch-recovery cost.

Take particular note of this mention:

- When subtracting points for \F spent, *do* include the cost of fuel that was delivered (the original cost is already accounted for in your points awarded)

Here is the breakdown for the multipliers:

- Liquid Fuel........(0.1666)*4 + 0.80 = [1.466]

- Oxidizer............(0.1666)*4 + 0.18 = [0.846]

- Monopropellant..(0.1333)*4 + 1.20 = [1.733]

- Xeon Gas..........(0.0333)*4 + 4.00 = [4.133]

First number is the multiplier that scales the point gain to be equivalent per unit weight.

Second, multiply by 4 which is arbitrary and yields a point value that I deemed a reasonable trade off against both cost and time deductions.

Third number is the cost multiplier, this value is entirely negated by the \F spent when launching the vehicle (only cost of fuel was considered, not cost of the varying tanks)

Edited September 20, 2014 by davidparks21

[cakepie]

First number is the multiplier that scales the point gain to be equivalent per unit weight.

This is the part I don't understand. For example, LF has density 0.005 and unity cost 0.8, so I'm not seeing how you arrived at 0.1620. And if this number is proportional or inversely proportional to density of the fuel, shouldn't LF and Ox be the same? (Same density.)

Cancelling out the initial cost of purchase of the delivered fuel makes sense. But to make it so that there is no advantage to be gained from transporting any kind of fuel over another, it'd probably be most straightforward to use the mass of fuel delivered i.e. amount * density (* your scaling factor of 4 or whatever). Then it really just boils down to the time and monetary cost to provide the necessary delta-V to get said amount of fuel to the refueling station.

[davidparks21]

This is the part I don't understand. For example' date=' LF has density 0.005 and unity cost 0.8, so I'm not seeing how you arrived at 0.1620. And if this number is proportional or inversely proportional to density of the fuel, shouldn't LF and Ox be the same? (Same density.)

Cancelling out the initial cost of purchase of the delivered fuel makes sense. But to make it so that there is no advantage to be gained from transporting any kind of fuel over another, it'd probably be most straightforward to use the mass of fuel delivered i.e. amount * density (* your scaling factor of 4 or whatever). Then it really just boils down to the time and monetary cost to provide the necessary delta-V to get said amount of fuel to the refueling station.[/quote']

Oh bugger, I dropped a 0.4 in my calculations, I see that in my notes, you're right, the density of Liquid and Oxidizer is the same, eagle eyes you have there for numbers!

I've updated the two posts, liquid fuel should have a weight-multiplier of 1.466 (like oxidizer) rather than 1.448 that I originally had.

The rest I've verified with a little example that should flush out the method to the madness:

Looking at an FL-T400 fuel tank, it contains 180 units of liquid fuel with a liquid-mass of 0.90.

Points awarded for delivering that fuel is: 263.88 minus the fuel-only cost of 144.00\F yields 119.88 points for its delivery.

Now let's send up an equivalent mass of mono propellant where 150 units have a liquid-mass of 0.60 in the Stratus-V Cylinder. Or 225 units for a liquid-mass of 0.90. The cost of 150 units of mono propellant is 180\F, or 270\F for our 225 units at a liquid-mass of 0.90.

225 units of mono propellant yields 389.925 points, minus the fuel-only cost of 270\F yields 119.925 points for delivery

The difference only being rounding error.

In both cases you derive the same points given the same liquid mass delivered. The cost of the containers varies, though that cost can be recovered of course.

The oddness of the numbers is because I didn't specifically call out a time-multiplier that I added in (I have it in handwritten notes but didn't call it out in that detailed post). I took the original weight-only multipliers (which are 5, 5, 4, 0.1) and multiplied by 1/25 to account reasonably for time. Then tacked on a multiplier of 4 to that which gave a good balance to cost. Essentially piecemealing the formula together to account for each factor in a way that seemed to strike a nice balance in my thought experiments. I probably should have graphed it all to visually confirm good choices, but hopefully my simple pondering was sufficient for a reasonable value.

My goal being to not benefit extreme solutions, or benefit a single dominate approach, but to benefit a all-around balanced approach. Being that it's hard to see in advance what approach might yield the highest score, I hope keeps it open to creative and varied solutions. We shouldn't end up with monstrosities, and we shouldn't end up with uber efficient but painfully slow beasts, if I did a good job.

Edited September 20, 2014 by davidparks21

[davidparks21]

This is the part I don't understand. For example' date=' LF has density 0.005 and unity cost 0.8, so I'm not seeing how you arrived at 0.1620. And if this number is proportional or inversely proportional to density of the fuel, shouldn't LF and Ox be the same? (Same density.)

Cancelling out the initial cost of purchase of the delivered fuel makes sense. But to make it so that there is no advantage to be gained from transporting any kind of fuel over another, it'd probably be most straightforward to use the mass of fuel delivered i.e. amount * density (* your scaling factor of 4 or whatever). Then it really just boils down to the time and monetary cost to provide the necessary delta-V to get said amount of fuel to the refueling station.[/quote']

Oh bugger, I dropped a 0.4 in my calculations, I see that in my notes, you're right, the density of Liquid and Oxidizer is the same, eagle eyes you have there for numbers!

I've updated the two posts, liquid fuel should have a weight-multiplier of 1.466 (like oxidizer) rather than 1.448 that I originally had.

The rest I've verified with a little example that should flush out the method to the madness:

Looking at an FL-T400 fuel tank, it contains 180 units of liquid fuel with a mass of 0.90.

Points awarded for delivering that fuel is: 263.88 minus the fuel-only cost of 144.00\F yields 119.88 points for its delivery.

Now let's send up an equivalent mass of mono propellant where 150 units have a mass of 0.60 in the Stratus-V Cylinder. Or 225 units for a mass of 0.90. The cost of 150 units of mono propellant is 180\F, or 270\F for our 225 units at a mass of 0.90.

225 units of mono propellant yields 389.925 points, minus the fuel-only cost of 270\F yields 119.925 points for delivery

The difference only being rounding error.

In both cases you derive the same points given the same liquid mass delivered. The cost of the containers varies, though that cost can be recovered of course.

The oddness of the numbers is because I didn't specifically call out a time-multiplier that I added in (I have it in handwritten notes but didn't call it out in that detailed post). I took the original weight-only multipliers (which are 5, 5, 4, 0.1) and multiplied by 1/25 (roughly adjusting for a presumed 25-minute mission). Then tacked on a multiplier of 4 to that which gave a good balance to cost. Essentially piecemealing the formula together to account for each factor in a way that seemed to strike a nice balance in my thought experiments.

My goal being to not benefit extreme solutions, or benefit a single dominate approach, but to benefit a all-around balanced approach. Being that it's hard to see in advance what approach might yield the highest score, I hope keeps it open to creative and varied solutions. We shouldn't end up with monstrosities, and we shouldn't end up with uber efficient but painfully slow beasts, if I did a good job.

[Laie]

That leaves only one grievance: If you eventually want to land at KSC your mission will last one orbit, no matter what. My preferred tanker is a prime example of a painfully slow beast, yet I don't doubt that it can make rendezvous and return within that timeframe. Other vessels may complete the mission in ten minutes, then fast-forward to the re-entry burn. The latter would deserve a better score, IMO.

Maybe only count the time until one undocks from the station? Though even then, I suspect it would be a "reckless docking challenge" more than anything else.

[cakepie]

The oddness of the numbers is because I didn't specifically call out a time-multiplier that I added in (I have it in handwritten notes but didn't call it out in that detailed post). I took the original weight-only multipliers (which are 5, 5, 4, 0.1) and multiplied by 1/25 (roughly adjusting for a presumed 25-minute mission). Then tacked on a multiplier of 4 to that which gave a good balance to cost. Essentially piecemealing the formula together to account for each factor in a way that seemed to strike a nice balance in my thought experiments.

So basically you're saying, if you're hauling more stuff, you can take a bit longer and it'll kind of balance out (in terms of just the fuel points, ignoring the time and financial cost subtraction).

Okay, now I can follow your derivation process and see how you got those numbers.

I think you meant 30 minutes rather than 25 minutes. 5*(1/30)=0.1666...

Also,

Here is the breakdown for the multipliers:

- Liquid Fuel........(0.1666)*4 + 0.80 = [1.466]

- Oxidizer............(0.1666)*4 + 0.18 = [0.846]

- Monopropellant..(0.1333)*4 + 1.20 = [1.733]

- Xeon Gas..........(0.0333)*4 + 4.00 = [4.133]

Ratio of masses between LF:Ox:MP:Xe is 5 : 5 : 4 : 0.1 as you have stated.

Same weight for LF and Ox, and MP = 4/5*0.1666... = 0.1333... <-- looks fine

Xenon should be 0.1/5*0.1666... = 0.00333... <-- you're missing a zero in your figure above. Exploitable!

Once you fix this last discrepancy I think it's safe to say sanity has been restored to your scoring system.

Edit:

That leaves only one grievance: If you eventually want to land at KSC your mission will last one orbit, no matter what. My preferred tanker is a prime example of a painfully slow beast, yet I don't doubt that it can make rendezvous and return within that timeframe. Other vessels may complete the mission in ten minutes, then fast-forward to the re-entry burn. The latter would deserve a better score, IMO.

Maybe only count the time until one undocks from the station? Though even then, I suspect it would be a "reckless docking challenge" more than anything else.

Good point raised -- since the scoring basically demands landing at KSC to recover as much cost as possible, fast-to-rendezvous designs inevitably have to wait for the right re-entry window.

Rather than stop the clock once undocked, another alternative is to not include "idle time" -- start timing at launch, pause timing once docked and fuel transfer completed*. The refueler must stay docked while the clock is paused. Restart the clock when refueler undocks from the station* to begin re-entry, timer stops when mission ends and refueler is recovered. (Both * timepoints can be easily validated by screenshot)

This will promote both speedy and efficient launch, redezvous, and return and recovery, rather than ignoring the latter part of the mission.

Edited September 20, 2014 by cake>pie

[davidparks21]

So basically you're saying' date=' if you're hauling more stuff, you can take a bit longer and it'll kind of balance out (in terms of just the fuel points, ignoring the time and financial cost subtraction).

Okay, now I can follow your derivation process and see how you got those numbers.

I think you meant 30 minutes rather than 25 minutes. 5*(1/30)=0.1666...

Also,

Ratio of masses between LF:Ox:MP:Xe is 5 : 5 : 4 : 0.1 as you have stated.

Same weight for LF and Ox, and MP = 4/5*0.1666... = 0.1333... <-- looks fine

Xenon should be 0.1/5*0.1666... = 0.00333... <-- you're missing a zero in your figure above. Exploitable!

Once you fix this last discrepancy I think it's safe to say sanity has been restored to your scoring system.

Edit:

Good point raised -- since the scoring basically demands landing at KSC to recover as much cost as possible, fast-to-rendezvous designs inevitably have to wait for the right re-entry window.

Rather than stop the clock once undocked, another alternative is to not include "idle time" -- start timing at launch, pause timing once docked and fuel transfer completed*. The refueler must stay docked while the clock is paused. Restart the clock when refueler undocks from the station* to begin re-entry, timer stops when mission ends and refueler is recovered. (Both * timepoints can be easily validated by screenshot)

This will promote both speedy and efficient launch, rendezvous, and return and recovery, rather than ignoring the latter part of the mission.

1. Thanks for the correction on the Xeon Gas, those pesky zeros, basically worthless until you put them in front of something useful...
   
2. I didn't actually realize that you recover more at KSC than any other landings (I don't use mechjeb myself and only occasionally manage a KSC landing). Given that point (thanks Laie) I think Cake's solution seems most in line with the goal and is simplest. I'll add a rule allowing you to deduct "docked" time. Part of the challenge is indeed rendezvous & docking, as large refueling tankers are beasts to rendezvous and dock, having a vessel that is as realistically manageable as it is efficient is a challenge goal.
   
3. Yes, I meant 1/30th, I had played with a variety of combinations and just noted the wrong one.
   

Edited September 20, 2014 by davidparks21

[cakepie]

Great!

You might also want to set policy on plugins such as DebRefund or FMRS for those who might want to make non-SSTO attempts. These plugins don't add parts or modify aero; would you accept such entries under stock category?

[davidparks21]

Great!

You might also want to set policy on plugins such as DebRefund or FMRS for those who might want to make non-SSTO attempts. These plugins don't add parts or modify aero; would you accept such entries under stock category?

I haven't used either of those, they look fantastic, and do look to be in keeping with the general purpose of the challenge. I'll allow them, but will specifically note they're use on the leader board.

[davidparks21]

At long last I've completed my own challenge, so without adieu, here is the first entry:

A twin jet engine SSTO that ejects on jets alone and uses minimal oxidizer to dock with the station. It's quite competitive on cost, using just 600 units of jet fuel, a bit of oxidizer, and a touch of mono propellant, but takes a good long time to eject on jets. Nor did I do a masterful job of docking, so ding me for a little non Jeb-level piloting.

Details:

Delivered:

1384.01 units Liquid Fuel (arrived with 1455.08, undocked with 71.07 for return)

1708.56 units Oxidizer (arrived with 1755, undocked with 46.44 for return)

56.41 units Mono Propellant (arrived with 67.72, undocked with 11.31 for return)

Timing:

Docked @ 1:19:08

Undocked @ 5:31:09

Landed @ 5:53:56

Total time (excluding docked time): 1:41:55 (6115 seconds)

Cost:

74,365 \F (cost at lift off)

72,229 \F (recovery value)

= 2,136 \F total round trip spend

Score:

+ 2028.95866 (liquid fuel)

+ 1445.44176 (oxidizer)

+ 97.75853 (mono-propellant)

- 6115 (total time)

- 2136 (total cost)

+ 10000 (challenge completion)

----------------------

= 5321

Notable challenges:

- Ejecting on *two* jet engines required quite a learning curve, it's something I've been trying to work out for a while now. Ejecting a lighter plane on one jet engine is a lot more doable, but you won't deliver much of a payload on a single jet. Two engines requires that they be stacked vertically (so as to handle inevitable burnouts and misaligned thrust gracefully), and requires tweaking of the thrust limiter on the more dominant of the two jets as the other teeters on the edge of burnout between 30-40km.

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Edited November 12, 2014 by davidparks21