Efficiency: Most payload per liftoff thrust

[foamyesque]

Rules are simple:

1. Build whatever stock rocket you want.

2. Put a payload into orbit.

3. Orbit is at least 70km x 70km.

4. Person with the best score wins.

5. Score is the orbital energy of the payload divided by liftoff thrust.

6. Payload cannot have any impact on rocket performance beyond mass and drag: RCS, CPs, decouplers, parachutes, tanks etc are all dead weight. You can use whatever you want for payload as long as it satisfies those requirements and you provide the number for its mass.

Have at.

EDIT: Re: Required proof:

1. Screenshot of rocket on the pad

2. Screenshot of the apoapsis and periapsis from the map view

3. Screenshot of the payload in orbit

EDIT: Re: Calculating orbital energy:

dE = u * (2a - R) / (2aR), where R is the radius of Kerbin, a is the semi-major axis of the orbit (i.e, the mean of the periapsis and apoapsis, plus the radius of Kerbin since the map displays apo and peri as altitude ASL and not w.r.t. the centre of the planet), and u is Kerbin\'s standard graviational parameter. It is helpfully provided by the KSP wiki as 3530.461 km^3/s^2.

Divide dEo by liftoff thrust, multiply by payload mass, and that\'s your score.

[Zephram Kerman]

Define orbital energy?

[closette]

I believe that orbital energy would be, in notation that would work in a MS Excel table, m *(0.5*v^2 - 3.5284E9/(altitude + 600000) ) measured at any instant in Kerbin\'s sphere of influence, where m is the total mass of payload (sum of the parts), v is the orbital speed reading in m/s and height is the altitude above Kerbin in meters.

So you can have more payload mass, or a faster orbit, or a larger orbit, or any combination thereof to get a higher orbital energy.

The first term gives the kinetic energy 1/2 mv², while the second gives the gravitational potential energy -GMm/r².

Not sure whether 'efficiency' should be defined as (orbital energy/initial thrust) or a more conventional (orbital energy/fuel mass expended) since the 'take-off energy' is effectively the potential energy contained inside the fuel.

Also, what proof is required?

An interesting challenge, and those of us who loft Mun landers (which are dead weight for the first part of the flight) should be up to it. I\'ll have a go later on.

[Iskierka]

Just a logistic thought, rather than total orbital energy, it should be the delta from the energy it would have at the surface - since orbital energy is negative, and higher orbits get less negative, meaning your number would reduce from getting higher.

IE. Rather than just m *(0.5*v^2 - 3.5284E9/(altitude + 600000) ), use that minus: m *(0.5* 174.55 ^2 - 3.5284E9/(600000) ).

[closette]

Well that would just subtract a constant x mass from the total orbital energy (relative to being at rest at infinity) but that\'s OK, it makes the equation:

delta-E = m *(0.5*v^2 - 3.5284E9/(altitude + 600000) - 9353.18)

delta-E = m *(0.5*v^2 - 3.5284E12/(altitude + 600000) + 5865441.54) where m is the payload mass you end up with in orbit.

(now edited to correct my units mistake pointed out by jebbe below)

The fuel mass you have used (whether solid or liquid or RCS) would be delta-m = m - m0, where m0 is the launch mass, so the efficiency in Joule/kg becomes

delta-E/delta-m = delta-E / (m-m0) - ugly but not too bad to calculate! (We could agree that the command pod mass = 1kg, not 1 tonne, for this challenge).

For those who do not know, you can access the masses of stock parts, including empty/full tank masses, online at the wiki: http://kerbalspaceprogram.com/~kerbalsp/wiki/index.php?title=Parts

I googled around for 'payload efficiency' to see if there was an existing definition one would use but didn\'t find anything suitable.

[jebbe]

Interesting challenge - I\'d like to give it a go when the scoring is figured out!

About that: I think iskierka has a point here: If we were just looking at the orbital energy, the additive constant wouldn\'t matter, but since we want to divide this energy by the initial thrust, we would favor small rockets by not subtracting the surface energy. This is because if we don\'t subtract this surface energy from our final orbital energy, the surface energy is basically score we get for free. But since we have to divide this \'free score\' by our initial thrust (which is larger, the larger the rocket), we get less points for large rockets, so my guess is that this way it would boil down to a 'smallest rocket to orbit' challenge.

In order not to scare away people by these formulas, let me suggest a different scoring: The challenge already specifies that we want an orbit with a periapsis of at least 70km. That should be enough to give us equal potential energy, and due to orbital mechanics, everyone would need the same minimum velocity, letting us drop the kinetic energy as well. We could then just divide the total mass we got into orbit by the initial thrust.

Oh by the way: Would an ASAS that gets switched on during ascent still count as payload? Because that would make flying the thing much less annoying, and get at least me much more motivated...

[foamyesque]

Orbital energy is: http://en.m.wikipedia.org/wiki/Specific_orbital_energy#section_3

Multiplied by the mass of the payload, of course.

This is done to even out the various orbits that might be achieved.

Proof: Screenshot showing apoapsis & periapsis, picture of rocket & payload on the pad, picture of payload in space.

ASAS do not count as payload.

[jebbe]

Sorry, but if you really take that definition, you get higher score the lighter your payload. That is because the orbital energy is negative, and if you multiply it with the payload mass, you get a \'more negative\' value for higher masses - lower score. So the best entry would be lifting nothing, giving zero points - which is what you would get by acually lifting something infinitely far away.

[foamyesque]

Sorry, but if you really take that definition, you get higher score the lighter your payload. That is because the orbital energy is negative, and if you multiply it with the payload mass, you get a \'more negative\' value for higher masses - lower score. So the best entry would be lifting nothing, giving zero points - which is what you would get by acually lifting something infinitely far away.

I linked to a very specific equation: The difference between the orbital energy and ground.

[sal_vager]

I feel so stupid, I used to know how to use these formulas but now it\'s all gone

Anyway, I used the ship pictured, the payload is the bottom middle tank which will still be full when in orbit, with a weight of 2.5 kilo, in fact the top middle tank will also have about 1/5th of it\'s capacity remaining.

The pipes transfer fuel to the center tanks, and the RCS is only for getting back down, but this ship can be reduced in weight a lot if you don\'t mind killing Kerbals.

All three engines are the gimballed type for a total launch thrust of 525n

If you can put the orbital energy calculation in a format I can understand I\'ll be happy to use it.

Edit:

I revised my ship to leave 3 Kerbal skeletons with the fuel delivery, now it only needs 5 fuel tanks to put 1 full tank into orbit, and with only 2 gimballed engines.

Edit 2:

I only counted my full tank as payload, everything else was affecting performance including the capsule as it provided the SAS.

(2.5 * ((0.5 * (2 294.8^2)) - (3.5284E12 / (70 849 + 600 000)) + 5 865 441.54)) / 350 = 23 134.937 (Thanks jebbe and Closette)

Lower numbers are better right?

[jebbe]

Ok, so basically it would be orbital energy of payload in orbit minus orbital energy of payload on the ground divided by initial thrust? I would then just take closettes formula, divide it by the initial thrust instead of the consumed fuel and get:

m *(0.5*v^2 - 3.5284E9/(altitude + 600000) - 9353.18) / thrust_initial

edit 4: wrong formula, see later post! Correct would be:

m *(0.5*v^2 - 3.5284E12/(altitude + 600000) + 5865441.54) / thrust_initial

My first shot would then be 19,390 m, with a payload mass of 1+0.9+0.9=2.8kg, an altitude of 489400m, a velocity of 1564.9m/s and an initial thrust of 175N. Clearly that can be improved! Also, funny to have fuel efficiency measured in meters...

@sal_vager:

Basically all you need to do is copy this equation into the google search field and replace m by your payload mass, v by your velocity in m/s, altitude by your altitude in meters (not kilometers) and thrust_initial by the sum of the max-thrust of all engines. Doesn\'t matter where you are, just take a screenshot at some random point in orbit and get velocity and height from there.

From what you posted I don\'t get enough information to do the calculations for you, unfortunately...

edit: For your modified ship I get 26,184m - good job! edit 4: wrong formula, see later post!

edit 2: If you include the mass of the pod, you should get the same result as I do. And, no higher numbers are better.

edit 3: Ok, I think we need a ruling here - can we include the mass of the pod? If not, I only get to 12,464m. Which ever way, I will have to try again tomorrow!

[sal_vager]

There seems to be a sweet spot with thrust to weight ratio, I always seem to get the best overall performance with 1 gimballed engine per 3 fuel tanks or equivalent weight.

I haven\'t really used the big engine much after I found I could reach a marginally higher altitude with the more efficient engine.

Edit:

I\'ll just put this here so as not to clog up the thread, I think foamyesque needs to be clear about what we can count as payload, I thought as you have to have a capsule no matter what, it shouldn\'t be counted, nor should empty tanks as they held the fuel you used to climb, even something like the ASAS is affecting the rocket as it helps keep you on course, wobbling loses efficiency.

He said 'dead weight', so I think that means anything that didn\'t and couldn\'t contribute to your flight, except negatively.

[thorfinn]

My try at this (pictures and calculation of exact score will follow later):

3 gimballing engines lit on ground <- 515 kN

82x190 km orbit

2x RCS tanks ( 1.8 ) + 8x RCS blocks (0.4) + 1 decoupler ( 0.8 ) + 1 capsule (1.0) + 1 chute (0.3) + 1 ASAS ( 0.8 )= 5.1 tons payload

I see that the ASAS isn\'t counted: not interested, then... I don\'t think I\'m able to fly ascents manually (and that ASAS got to orbit and could be used there, so why isn\'t it 'payload' anyway? )

[jebbe]

I\'m really sorry, but I think there\'s something wrong with the formula: If I\'m not mistaken, the gravitational parameter is 3.5284E12 and not 3.5284E9. That also changes the specific surface energy to 5865441.54, and we have to add it instead of subtracting it, because we subtract a negative quantity. The formula should then read:

m *(0.5*v^2 - 3.5284E12/(altitude + 600000) + 5865441.54) / thrust_initial

That would then give 39 610.8m for my ship without pod and 23 134.9m for yours, sal_vager. Still, larger is better.

I really hope I didn\'t screw things up again, but I\'m really tired by now, so I can\'t guarantee anything anymore...

[closette]

Thanks for the correction jebbe and my apologies to all for putting out the wrong equation.

[foamyesque]

Okay, to answer a few questions now that I\'m working with a computer and not a phone:

Payload is stuff that cannot be used to help you get into orbit-- that is to say, if you removed it, the rocket would still fly. So, no, full tank dry weights don\'t work, because you can\'t delete that fuel tank; command pods don\'t count, because they provide control; likewise with ASAS and RCS and so on.

You\'re perfectly entitled to add them to your rocket to improve your control abilities, of course; they just don\'t count as payload. Think of it like trying to install a satellite in orbit-- you\'re being paid to put the satellite there. The guys hiring you aren\'t going to pay you for an extra ten tons of random strutwork.

As for the calculation, it\'s simple and there\'s a reason I wanted the apo/peri-apsis screenshot

(here\'s the non-mobile-Wiki link: http://en.wikipedia.org/wiki/Specific_orbital_energy#Additional_energy):

dE = u * (2a - R) / (2aR), where R is the radius of Kerbin, a is the semi-major axis of the orbit (i.e, the mean of the periapsis and apoapsis, plus the radius of Kerbin since the map displays apo and peri as altitude ASL and not w.r.t. the centre of the planet), and u is Kerbin\'s standard graviational parameter. It is helpfully provided by the KSP wiki as 3530.461 km^3/s^2.

Divide dEo by liftoff thrust, multiply by payload mass, and that\'s your score.

[Entroper]

This was just my first attempt, but my basic strategy was to modify one of my lift vehicles and remove the boosting SRBs, to lower the takeoff thrust. Then I added fuel until the thing would barely make it off the ground, then added payload until it barely made orbit. I could probably add another fuel tank or two and still make orbit, but it\'d be really close. I\'m fairly satisfied with this run until I see what other people can do. For my next attempt, I\'d be curious to try a much smaller design and see if I can get close to the same efficiency.

Alright, screenshots attached of the rocket on the pad and the payload in orbit. I didn\'t get a screenshot of the ap/pe altitudes, but I\'m rendering a video to YouTube which should have them.

Payload mass = 28 LFTs = 70 kg/tonnes/whatever the mass unit is.

Liftoff thrust = 18 LFEs + 1 vectoring LFE = 3 775 N/kN/whatever the thrust unit is.

Apoapsis = 407 235 m, Periapsis = 76 045 m

Energy = u*(2a - R)/(2aR) = 3 560.461 * (2*841.64 - 600)/(2*841.64*600) = 3.819 km^2/s^2.

Score = 3.819 km^2/s^2* 70 kg / 3 775 N = 70.8 km

Cool. Score units are km. That was unexpected.

For some reason, my ship really wanted to pitch down near the end of my burn. The middle vectoring engine had run out of fuel long ago, and the capsule\'s cheat-steering couldn\'t keep up. Oh well, it didn\'t drift too far from prograde.

Video:

[jebbe]

There seems to be a sweet spot with thrust to weight ratio, I always seem to get the best overall performance with 1 gimballed engine per 3 fuel tanks or equivalent weight.

That\'s a very keen observation: if you want to optimize fuel efficiency, there really is a sweet spot at a thrust weight ratio of two, at least near the surface. There\'s a very detailed discussion going on about this in the Goddard problem challenge: http://kerbalspaceprogram.com/forum/index.php?topic=7161.0

In your case, if you have 175N of thrust, ideally you carry 82.5N divided by g(~=9.81m/s^2), which is a weight of 8.9kg (or whatever unit you want to give it), including the engine\'s weight.

However, we are not really optimizing fuel efficiency in this challenge: dividing by the liftoff thrust instead of the consumed fuel, it\'s more of an \'engine efficiency\' or whatever you want to call it. That also explains the strange unit of meters for the efficiency.

As of the scoring: My new formula should now be equivalent to the one given by foamyesque, giving the same score; only difference is kilometers vs meters, which is a factor of 1000.

[sal_vager]

@ jebbe

I didn\'t do anything more complex than see what kind of boosters gave the best performance, 4 tanks to 1 engine was good but a lot slower than Closettes ideal speed range, 2 tanks was great but burned out too early.

@ Entroper

Can you show me the formula with your numbers plugged in?, I can\'t figure it out.

@ foamyesque

Damn your unofficial intellectual barrier to entry I\'m trying hard to learn though.

@ All

Okay I am resubmitting the K-S-F-D-V as it can fly without the payload, If you removed jebbe\'s RCS tanks you\'d have a large gap (It should still count, I\'m just having fun)

I\'ll use jebbe\'s formula as he kindly told me where to put my numbers, this is what google gives me.

(2.5 * ((0.5 * (2 299.7^2)) - (3.5284E12 / (70 605 + 600 000)) + 5 865 441.54)) / 350 = 23 201.6714

(Running the tanks dry only got me 24 740.1119)

Edit:

(2.5 * ((0.5 * (2 201.2^2)) - (3.5284E12 / (152 705 + 600 000)) + 5 865 441.54)) / 200 = 45 005.6954

(5 tanks and the large engine only got me 53 644.4579, not enough to beat Entroper)

(6 tanks only got me to 54 846.3035, that\'s the limit of what a single large engine can handle)

This is probably the best I can do with my flying, I tried a gimballed engine version and a 5 tank + gimball version of the ship below, both failed to make stable orbit, they were just too heavy for the available thrust.

It turned out I needed the extra thrust of the more powerful engine, and the number is high because I ran the tanks dry this time, I didn\'t think of that before so wasted potential energy.

The payload is of course the top tank.

[jebbe]

I didn\'t do anything more complex than see what kind of boosters gave the best performance, 4 tanks to 1 engine was good but a lot slower than Closettes ideal speed range, 2 tanks was great but burned out too early.

Oh, yes, that makes sense now - simplifying, one could say that closette\'s optimal speed range uses the TWR = 2 condition (plus some more complicated stuff, see other challenge).

(2.5 * ((0.5 * (2 201.2^2)) - (3.5284E12 / (152 705 + 600 000)) + 5 865 441.54)) / 200 = 45 005.6954

Nice - I\'ll have to try again I guess. Also, we\'re both still far from Entropers 70,800m - that must be beatable as well! We\'ll probably need much larger rockets though, since we have to carry the dead weight of the pod.

For my next attempt, I\'d be curious to try a much smaller design and see if I can get close to the same efficiency.

That was my original idea, but sadly I think that we have no chance with a smallish design as long as the pod weighs us down. Anyway, I\'d be happy to be proven wrong!

[Entroper]

That was my original idea, but sadly I think that we have no chance with a smallish design as long as the pod weighs us down. Anyway, I\'d be happy to be proven wrong!

That\'s true. You can use a lighter pod than I did, though. Sadly it\'s a pain to switch out upper parts of the craft, as you currently have to go back and reconnect all your struts. But a simple RCS tank plus 2 blocks would weigh a lot less than an LFE+LFT.

[sal_vager]

Okay, this is my last attempt, these ships are horrible to fly.

(7.5 * ((0.5 * (209.7^2)) - (3.5284E12 / (80 422 902 + 600 000)) + 5 865 441.54)) / 600 = 73 048.5051

It wont make orbit if you take tanks off the boosters, and I can\'t fly any higher than this because I can\'t zoom the map on linux, the orbit lines and \'apsis wont show up on the default zoom once you get too far out.

Someone else can see what the final energy of this ship is, but be warned, although it\'s stable it flies like a double decker bus.

Edit: I didn\'t burn up all the fuel, meaning there\'s potential energy wasted, the score would be higher with this ship if I could see to fly it further out.

[jebbe]

Someone else can see what the final energy of this ship is, but be warned, although it\'s stable it flies like a double decker bus.

You already did calculate the final energy, divided by the thrust, didn\'t you? As I said, using my formula gives you the exact same result as following foamyesque\'s instructions (at least within the numerical precision).

edit: Oh, I get it now - you didn\'t spend all the fuel. You seem to be pretty close to escaping already, not sure if there still is any circular orbit at those energies where all your fuel is spent. What you could do is just keep burning to a parabolic trajectory and then calculate your energy - although I\'m not sure if that would still count as \'putting payload into orbit\'.

Oh, btw - how do you run KSP on Linux? Using wine? I tried that, but it wouldn\'t load, missing some libraries.

That\'s true. You can use a lighter pod than I did, though.

A lighter pod? If you go for stock parts only, isn\'t there just one standard pod?

Sadly it\'s a pain to switch out upper parts of the craft, as you currently have to go back and reconnect all your struts.

That\'s so true - and my staging gets messed up as well, all the time.

Anyway, I\'m still amazed how you got such a high score and actually managed to go back! But, with my Kerbonauts forever stuck with their payload, I was able to creep past both of you to 76,504m. I even used an additional ASAS, which probably cost some points, but made flying a LOT easier...

(4*7*2.5+6*0.05) *(0.5*2453.1^2 - 3.5284E12/(89169+ 600000) + 5865441.54) / (175*6+200*12)

[Entroper]

A lighter pod? If you go for stock parts only, isn\'t there just one standard pod?

I meant a lighter return stage. Though it now appears that returning to Kerbin is optional.

[sal_vager]

The challenge is to lift a payload with the least possible thrust, I should try another ship with a bigger load and not carry it so far out.

The trouble with that is I already tried my biggest rocket, and didn\'t even come close to your score, and I had to endure 1 frame per second as it flew.

To be honest I don\'t really see the point to this challenge from an efficiency standpoint, you\'d want to loft the most weight per unit of fuel, not unit of engine power, as you can always put your engines on boosters with chutes at the top so you can recover them later, fuel burned is fuel gone for good however.

The most a single large engine can lift is 7 fuel tanks plus capsule, and the most a small engine can lift is 6 fuel tanks plus a capsule, so all you really can do is multiply that, you wont get any better.

Something I have found is that for us Munshotters, the lander is often just dead weight, so now I like to strap my boosters directly to my lander and use the landers engine as well, I have a nice range of direct ascent Skif\'s that I\'ll upload as soon as I think of a story to go with them

And decent names, I can\'t think of good names.

Edit:

Here is a sneak preview of my direct ascent Skif range, with 0.14\'s persistence, launch windows now become important unless you want to fly them the old way and go around Kerbin first

I was thinking of naming each after a musical instrument, the small one might be the pan pipe for instance