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What don't I understand about dV?


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Like the title (and tags) ask, I'm missing something... I do quite well up until I get into the over 300/320 ton area ships, with payloads of generally of at least ~60 tons (the payload is far less heavy once trying to return from my destination, like maybe 15 tons or something).

Obviously, I must not fully understand how dV is calculated or works, because when I get to this point (I always avoid this, I try and make my crafts as minimal in all aspects as Kerbally possible) it comes to a head, and no matter what I do, I can't increase my dV any further...generally not past 3k for the lower stages. My upper stages are fine, just very heavy. My lower stages start becoming inefficient and massive amounts of fuel will only add like 100dV, an obvious indicator I'm Doing It Wrong™.

This specific monstrosity is a result of combining two difficult contracts into one mission, as well as trying to bring 7 kerbals all at once to Minmus for the first time. I have a very similar craft meant for 1 kerbal to return that works just fine.

I can add some pictures in a few hours when I get home if that helps, but I mostly want to understand why I'm doing it wrong, not just how to improve this specific rocket. In the past in similar situations, if I post a picture I tend to only get help with that specific rocket, not how to prevent issues in the future as well.

Thanks KSP :)

Edited by KocLobster
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8 minutes ago, KocLobster said:

My lower stages start becoming inefficient and massive amounts of fuel will only add like 100dV, an obvious indicator I'm Doing It Wrong™.

Welcome to the Tyrrany of the Rocket EquationTM

Others will probably be able to explain it more in depth, but you're basically coming up against the limitations of chemical rockets, and there's not much you can do about it. 

My suggestion would be to take your 70 ton payload, break it up into pieces, launch each part separately, and dock it together in orbit. 

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6 minutes ago, KocLobster said:

My upper stages are fine, just very heavy. My lower stages start becoming inefficient and massive amounts of fuel will only add like 100dV, an obvious indicator I'm Doing It Wrong™.

Not necessarily.  At some point, that's just how it works.  More fuel means your engines can burn longer, but it also means they have more mass to move and thus each second of thrust is less effective.  So you need to start adding more and more fuel to get ever smaller gains in total dV.  And at some point, the added mass becomes so much that you will either start LOSING dV by adding more fuel or your TWR will be below 1 and you'll never get off the launchpad in the first place.

 

You can compensate for this by adding more/better engines, but again these will add more mass.  So at some point, there is still a practical limit to how much you can lift at one time.  Instead of adding more fuel, you're better off trying to do one of three things.

  1. Minimize the mass of each stage, especially the upper stages.  Even a small amount of mass on the last stage can make a big difference later on since that last stage needs to carry more fuel to compensate, but then that adds to the mass the next stage needs, so IT needs to be bigger which adds to the next stage and so on...
  2. Split the payload into two or more smaller modules which can be docked together in orbit or at the destination and launch them separately.
  3. Refuel the craft somewhere along the way(again, effectively splitting the mass you need to carry at any one time up).
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23 minutes ago, FullMetalMachinist said:

Welcome to the Tyrrany of the Rocket EquationTM

Others will probably be able to explain it more in depth, but you're basically coming up against the limitations of chemical rockets, and there's not much you can do about it. 

My suggestion would be to take your 70 ton payload, break it up into pieces, launch each part separately, and dock it together in orbit. 

This had me giggling, haha. I figured it was user error; that's unfortunate, I hadnt realized there was a ceiling to this sort of thing. I guess I'll go with my original plan and lower the weight, and make separate trips. I guess I got too ambitious. I'd love to hear the equation, math and reasoning behind this.

Another thing I don't understand is if I add weight to my top most part of the rocket and the payload, does it exponentially increase the dV requirements of my lower stages, or is it more linear; meaning the where doesn't matter, just the amount of weight I add?

Edited by KocLobster
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1 hour ago, Hodari said:

Not necessarily.  At some point, that's just how it works.  More fuel means your engines can burn longer, but it also means they have more mass to move and thus each second of thrust is less effective.  So you need to start adding more and more fuel to get ever smaller gains in total dV.  And at some point, the added mass becomes so much that you will either start LOSING dV by adding more fuel or your TWR will be below 1 and you'll never get off the launchpad in the first place.

 

You can compensate for this by adding more/better engines, but again these will add more mass.  So at some point, there is still a practical limit to how much you can lift at one time.  Instead of adding more fuel, you're better off trying to do one of three things.

  1. Minimize the mass of each stage, especially the upper stages.  Even a small amount of mass on the last stage can make a big difference later on since that last stage needs to carry more fuel to compensate, but then that adds to the mass the next stage needs, so IT needs to be bigger which adds to the next stage and so on...
  2. Split the payload into two or more smaller modules which can be docked together in orbit or at the destination and launch them separately.
  3. Refuel the craft somewhere along the way(again, effectively splitting the mass you need to carry at any one time up).

I think you effectively just answered my question I posted probably minutes after your post. So weight up top is exponentially 'more heavy' than lower. That probably explains a lot, my upper stages are 80-140ish tons depending on where you measure from.

Edited by KocLobster
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13 minutes ago, KocLobster said:

Another thing I don't understand is if I add weight to my top most part of the rocket and the payload, does it exponentially increase the dV requirements of my lower stages, or is it more linear; meaning the where doesn't matter, just the amount of weight I add?

Exponentially. If you're interested, you calculate DV by multiplying the Isp of the rocket engine with one standard g and the natural logarithm of the mass with fuel divided by the mass without fuel:
DV = Isp×g×ln mwet/mdry

For instance, say you have a Mk I pod, an FLT-200 tank and a LV909 Terrier. Total mass without fuel: 1465 kg (if I got it right); total mass with fuel: 2465 kg. The LV909 has an Isp in vacuum of 345 (and 85 at 1 atm pressure which is why you don't want to use it in the atmosphere). Plugging in gives you: DV = 345×9.81×ln1.68 = 1755 m/s.

Now, let's put that on top of two FLT-800 tanks and a LVT-45 swivel (and let's not forget the stack decoupler). Empty mass of the first stage: 2250 kg—but wait, we need to add the top stage (full mass) to that as well. That puts the total empty mass at 5015 kg (but it does have a DV of 1755 m/s out of the box). Each FLT-800 holds 4000kg of propellant, so your full mass is 13015 kg. At ground level the Swivel has an Isp of 270, and 320 in vacuum. Let's for argument sake assume that it averages out at 300 for our launch. DV of the bottom stack is then: 300×9.81×ln(13015/5015) = 1428 m/s. Note that you get less DV from this second stage despite having 8× as much fuel! That's because the dry/wet ratio improved only a little bit, and the Isp of the bottom stage is actually less.

So your total DV for the entire stack is 1428 + 1755 = 3183 m/s which will, sadly, not get you into orbit. This is where you can add two SRB's, but now things rapidly get complicated. You'll have to treat the your second stage "with boosters" as one stage (where only a little bit of fuel of the FLT-800's get uses) and then, once the SRB's are gone,use the amount of fuel in your second stage at that point for the calculation of your DV. But I'll leave that as an exercise to the student. This is why most of us resort to Mechjeb or KER for these things. It's not particularly hard, just a lot of work

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1 hour ago, KocLobster said:

Another thing I don't understand is if I add weight to my top most part of the rocket and the payload, does it exponentially increase the dV requirements of my lower stages

Ding ding ding!  :)

The exponential nature of rockets means that doubling the mass of your topmost-stage payload is going to double the mass of your entire rocket.

@FullMetalMachinist said it best, this is a tyranny-of-the-rocket-equation sort of thing.  It requires exponentially bigger rockets to add increments of dV.  It's a fundamental limitation of rockets.

Here's what you can do to make it better:

  1. Isp is king.  TWR matters for launching off the surface, but once you're in orbit, you really don't need it, and what matters above all is fuel efficiency.  In practical KSP terms, what that means is:  nukes.  Make your orbital stages nuke-powered.  You'll get more than double the dV for the same amount of fuel, when compared with the most efficient non-nuclear engine (Poodle).  This will help you far, far more than trying to throw more fuel at the problem, since your dV goes upt linearly with Isp, whereas it goes up only logarithmically with fuel mass.
  2. Refuel in orbit.  Empty fuel tanks are dead weight.  If you can use a fuel tank twice before you throw it away, you've effectively just cut the dead weight in half.  So when you're designing a big interplanetary mission, do it like this:  Make a humongous ship that can arrive in LKO intact but with nearly empty tanks.  Then refuel it in orbit, either with tankers launched from the surface, or with mined fuel brought in from Mun/Minmus.  That way, you can start your interplanetary burn with a full, big ship, and your fuel tanks are doing double-duty (filled up once to get off the surface, then filled up a second time to go interplanetary).  This helps a lot.
  3. For big "mothership" type interplanetary missions:  use ISRU!  Far and away the best way to defeat the rocket equation (even better than raising your Isp) is to get extra fuel along the way.  Design your Eve mission, for example, so that your interplanetary stage arrives at Eve nearly empty of fuel, and mines Gilly to refuel & come home.  (And if you have an Eve lander that needs 50+ tons of fuel to get off Eve?  You can ship it empty, and fill it up at Gilly.)

In summary:  Use nukes; refuel in orbit; refuel via ISRU for the return trip.

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11 minutes ago, Hodari said:

And at some point, the added mass becomes so much that you will either start LOSING dV

This is not true.  Adding fuel will only ever lose dV if you make the wet:dry mass ratio worse.

The main thing to remember is that the amount of dV generated by a (simple) stage is very simple to calculate:

dV = Isp * g * ln(Mw / Md)

...where Mw is the wet mass (i.e. the total mass at the start of the burn), Md is the "dry" mass (i.e. the total mass after the burn), ln() is natural logarithm, Isp is the Isp of the engine (in seconds) and g is the standard value of Kerbin surface gravity (9.80655 m/s^2).  The Mw/Md is the "wet:dry mass ratio" and adding more fuel (in a tank) will add to both Mw and Md.  Once most of the mass of your vessel consists of fuel tanks, adding more fuel tanks will not change the mass ratio by very much at all so there will be very little change in the dV.  Fuel tanks in KSP have terrible mass ratios (mostly around 9:1) which gives an absolute max dV per stage of ln(9) * Isp * g = ~21.5 * Isp but you won't generally get anywhere near that as it assumes that your payload and engines don't weigh anything.

If we start with a 5t payload and 5t of engines and our fuel tanks contain 8t of fuel for every 1t of empty tank mass we could have:

With 8t of fuel, Mw = 19, Md = 11, ratio = 1.73 => dV = 5.4 * Isp

With 16t of fuel, Mw = 28, Md = 12, ratio = 2.33 => dV = 8.3 * Isp

With 24t of fuel, Mw = 37, Md = 13, ratio = 2.85 => dV = 10.3 * Isp

With 32t of fuel, Mw = 46, Md = 14, ratio = 3.28 => dV = 11.6 * Isp

With 40t of fuel, Mw = 55, Md = 15, ratio = 3.67 => dV = 12.8 * Isp

With 48t of fuel, Mw = 64, Md = 16, ratio = 4.00 => dV = 13.6 * Isp

With 56t of fuel, Mw = 73, Md = 17, ratio = 4.29 => dV = 14.3 * Isp

With 64t of fuel, Mw = 82, Md = 18, ratio = 4.56 => dV = 14.9 * Isp

With 72t of fuel, Mw = 91, Md = 19, ratio = 4.79 => dV = 15.4 * Isp

With 80t of fuel, Mw = 100, Md = 20, ratio = 5.00 => dV = 15.8 * Isp

With 120t of fuel, Mw = 145, Md = 25, ratio = 5.80 => dV = 17.2 * Isp

With 160t of fuel, Mw = 190, Md = 30, ratio = 6.33 => dV = 18.1 * Isp

With 1600t of fuel, Mw = 1810, Md = 210, ratio = 8.62 => dV = 21.1 * Isp

With 8000t of fuel, Mw = 9010, Md = 1010, ratio = 8.92 => dV = 21.46 * Isp

Of course, in practice, there comes a limit (usually quite early on in this table) where your engines no longer provide enough thrust to give acceptable acceleration to all that extra fuel (either making it impossible to lift off for a launch stage or making the burn take excessively long for orbital manoeuvres) so you have to start adding more engine mass which makes your mass ratio worse and lowers the dV, e.g. with the 80t of fuel example you may find that you actually need 20t of engines to give it enough TWR to lift off which changes the ratio to 115/35 = 3.29 for a dV of only (11.7 * Isp).

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It helped me to see a graphical representation of this.  The important thing to realize is that the natural logarithm function in the rocket equation is limiting the gains from adding more fuel.  See this simple graph of Y = ln(X):

 graph-ln.gif

As you can see, as we linearly increase X we get diminishing increases in Y.  In the rocket equation 'X' is related to your fuel & payload mass, and 'Y' is related to your delta V & Isp.

Edited by illumiz
graph was wrong
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@Snark

Thank you for posting; as always your posts are incredibly helpful and very easy to understand. ISRU was actually already being utilized on this mission; I think people aren't realizing I simply didn't have the dV to both get into Kerbin orbit and land on Minmus in one shot like I was trying to do. I was under the impression that nukes were essentially wasteful if not used for interplanetary missions. I thought for the most part, you didn't want to use nukes if you weren't going past Minmus. Even if that isn't the case, my main issue was not having enough dV to get into orbit, or just not being able to be efficient enough getting into LKO that I could make it to Minmus and land. I suppose it is possible that I could of if I had used nukes.

@illumiz

Helpful graph, thank you.

@Waxing_Kibbous

I'm realizing now that this will be necessary. I am able to do both contracts in the same mission, but bringing 7 Kerbals just wasn't possible. The first contract was essentially landing on Minmus with a space station consisting of antenna, docking port, power generation, room for at least 5 Kerbals, 1000 mono propellant, and 4000 liquid fuel. It's pretty much impossible to have that enormous amount of fuel after all the dV needed to land on Minmus from Kerbin, so an ISRU and drills were necessary. This ties into the second contract quite well, which requires you to mine 1050 ore from Minmus, and bring it back to Kerbin and land it.

I have already successfully done this mission and completed the two contracts in one go as I originally planned. However, I got to ambitious trying to actually return a bunch of Kerbals (ie. a lot of extra weight via Hitchhiker/Mk2 command pod). I wanted to do this because I was already having to make room for 5 Kerbals, and being my first manned trip to Minmus, I wanted to bring as many Kerbals as possible since they'd all gain experience being the 1 stars that they were. It may have been possible with infinitely more tweaking and maybe if I had an absolute perfect ascent profile, but it was just not worth it in the end, so I did it with just a single engineer.

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2 hours ago, KocLobster said:

This helps a lot, thank you. I don't like having to do too much math, so I always use KER.

Most people don't, especially when it's tedious and time-consuming. I do recommend going through the exercise once or twice on a simple rocket (nothing with radially attached boosters as it complicates things without adding extra insights) just to get a better feeling for "what makes" delta-V.

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8 minutes ago, KocLobster said:

I think people aren't realizing I simply didn't have the dV to both get into Kerbin orbit and land on Minmus in one shot like I was trying to do. I was under the impression that nukes were essentially wasteful if not used for interplanetary missions. I thought for the most part, you didn't want to use nukes if you weren't going past Minmus. Even if that isn't the case, my main issue was not having enough dV to get into orbit, or just not being able to be efficient enough getting into LKO that I could make it to Minmus and land. I suppose it is possible that I could of if I had used nukes.

Landing on Minmus requires the least Δv of any destination in the game.  I one way trip to the surface of Minmus shouldn't take more than 5,000 m/s.  That should be well within the reach of chemical rockets.

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27 minutes ago, OhioBob said:

Landing on Minmus requires the least Δv of any destination in the game.  I one way trip to the surface of Minmus shouldn't take more than 5,000 m/s.  That should be well within the reach of chemical rockets.

Depend on tech level and spread sheeting skill.  Minimus with safety margins is around 6000 dv.  3800 Kerbin orbit, 250 inclination change, 1000 transfer, 150 capture, 300 landing, 150 taking off, 150 return. For a total of 5800.  Yes it can be done for a lot less but margins are always nice.

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(edit: sorry about repeating stuff everybody else has said... this post was interrupted several times by RL and was only actually posted a long, long time after it was started)

 

4 hours ago, KocLobster said:

...I'd love to hear the equation, math and reasoning behind this.

Another thing I don't understand is if I add weight to my top most part of the rocket and the payload, does it exponentially increase the dV requirements of my lower stages, or is it more linear; meaning the where doesn't matter, just the amount of weight I add?

The rocket equation itself is simply:

delta-V = ln(wet mass / dry mass) * Isp * 9.81

where "ln" is the natural logarithm:

300px-Log.svg.png

Wet mass (i.e. total rocket including fuel) divided by dry mass (same, with no fuel) is the mass ratio - which can't ever be less than one.

So basically your delta-v rises dramatically as you start adding fuel. If you have a ship with a Poodle (Isp = 350) and a mass ratio of 2.7, you can get 350 * 9.81 * 1 delta-v = 3433 m/s. But even if you add infinite fuel it'll never exceed 3.50 * 9.81 * e = 9300 m/s.

The reasoning behind this is simply that if you add fuel, you have to lift that same fuel. You get diminishing returns for every additional bit of fuel, and the mathematical definition of "diminishing returns" is the natural logarithm.

 

So for the second part of your question, about adding weight: any weight added will reduce the dv of the stage where it is added and all lower stages.

To maximise dv, you want to minimise what you are carrying, and maximise the amount of fuel in that carried weight. This is why drop tanks and asparagus staging are by far the best way of maximising efficiency: you need tanks and engines to carry the fuel load, but you want to get rid of the weight of the tanks when they are empty, and get rid of the engines when you no longer need so much thrust.

Therefore, you add drop tanks (I often add drop tanks to my SRBs) and route fuel to the engines so that they use that first. You put a number of 2x radial stages around your rocket and route fuel from one to the next and then to the core, so that when each set of tanks (and engine, needed to carry them) is empt, then you abandon them leaving the rest of the rocket fully fueled with the optimum level of thrust.

And of course, you use vertical staging so that a large lower stage (with a mass ratio of 2 or 3 or so) carries a smaller upper stage (again with a mass ratio of 2 or 3 or so) and so on - giving each stage an optimal amount of delta-V before it drops.

So if you add mass to the final stage, it'll severely reduce the final stage's dv, significantly reduce the dv of the stage below it, and slightly reduce the dv of the stage before that...

Edited by Plusck
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Some tips:

1. You don't have to send actual kerbals for station/base contracts. Just enough space to hold that many kerbals and that's enough

2. (cheaty tip) The ore you mine does not have to be the same ore you land on Kerbin. So to "land" that amount of ore on Kerbin, you just launch another ship with enough amount of fuel, use a few sepratrons to lift off for a few centimeters, then contract system will be happy about it.

On the other hand, you should definitely optimize your payload first. 2 hitchhikers+ISRU+drill should be just about 10t. Empty tank doesn't count as payload as you can just bring the upper stage to the surface of Minmus empty.

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2 hours ago, Nich said:

Depend on tech level and spread sheeting skill.  Minimus with safety margins is around 6000 dv.  3800 Kerbin orbit, 250 inclination change, 1000 transfer, 150 capture, 300 landing, 150 taking off, 150 return. For a total of 5800.  Yes it can be done for a lot less but margins are always nice.

Well, I did say a "one way" trip to Minmus, so I wasn't counting the return.  The numbers I used were:  3500 Kerbin orbit + 100 plane change + 930 transfer + 160 capture + 300 landing = 4990 m/s.  I usually try to launch as close as possible into the correct orbital plane, so my 100 m/s plane change is just a contingency to fine tune it.  For the return trip, I generally budget:  230 launch + 170 transfer = 400 m/s.

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2 hours ago, Nich said:

Depend on tech level and spread sheeting skill.  Minimus with safety margins is around 6000 dv.  3800 Kerbin orbit, 250 inclination change, 1000 transfer, 150 capture, 300 landing, 150 taking off, 150 return. For a total of 5800.  Yes it can be done for a lot less but margins are always nice.

Woah, wait.

If you just go equatorial for LKO, then fix the inclination when you're 3/4 of the way to Minmus (and re-fix the inclination on the capture burn when you get there) it'll cost you a few dozen m/s at most. Nothing like 250 m/s.

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1 hour ago, Plusck said:

Woah, wait.

If you just go equatorial for LKO, then fix the inclination when you're 3/4 of the way to Minmus (and re-fix the inclination on the capture burn when you get there) it'll cost you a few dozen m/s at most. Nothing like 250 m/s.

I haven't done an inclination change burn to Minmus in years. I just target to hit it at the AN or DN. That means my inclination change is actually done AT Minmus where I'm going the slowest, am coupling it with my slowdown burn, and am getting a little help from Oberth because I'm also close to Minmus' surface. I have no idea how much it saves but wouldn't be surprised that it wasn't even worth accounting for.

I figure this because I don't account for it. I just ignore that number on the charts.

 

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Same here I just figured the posted was a beginner.  Like I said I had a lot of margin.  I normally Launch into the plane or wait for a transfer to AN/DN.  I guess if you want a bare minimum I would say 3225 to LKO (from cheap and chearfull challenge), 946 direct transfer (my transfer stages are pretty low TWR so a loss of efficiency there), I thought I normally capture for about 154 but dv map says it is 160 min.  Perhaps I am burning more on transfer and saving some on capture.  200 For landing as my landers normally have 20-50 TWR for minimus because I like the Terrior which is way over powered for minimus. Totaling 4525.

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1 hour ago, Nich said:

Same here I just figured the posted was a beginner.  Like I said I had a lot of margin.  I normally Launch into the plane or wait for a transfer to AN/DN.  I guess if you want a bare minimum I would say 3225 to LKO (from cheap and chearfull challenge), 946 direct transfer (my transfer stages are pretty low TWR so a loss of efficiency there), I thought I normally capture for about 154 but dv map says it is 160 min.  Perhaps I am burning more on transfer and saving some on capture.  200 For landing as my landers normally have 20-50 TWR for minimus because I like the Terrior which is way over powered for minimus. Totaling 4525.

Yeah, my numbers have a little bit of margin in them as well.  The way I build and fly my missions, a total of 4800 m/s is a reasonable expectation (that's landing only).  I am surprised, however, by your return trip numbers.  For takeoff and return you have only 300 m/s.  I usually require about 400 m/s.  What are you doing that allows you to return to Kerbin using only 300 m/s?

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35 minutes ago, OhioBob said:

Yeah, my numbers have a little bit of margin in them as well.  The way I build and fly my missions, a total of 4800 m/s is a reasonable expectation (that's landing only).  I am surprised, however, by your return trip numbers.  For takeoff and return you have only 300 m/s.  I usually require about 400 m/s.  What are you doing that allows you to return to Kerbin using only 300 m/s?

The accepted numbers for Minmus are 180 to low orbit from the flats and 160 from low orbit to to a trajectory that will intercept Kerbin's atmosphere. That mans it should take about 340m/s. In my experience this is pretty right on. I'd be surprised to see someone do it for 300, surprised enough that I'd want proof :)

It could (and probably is) a simple case that Nich rounded down while you round up. I mean, in the whole vast scheme of things 300 m/s isn't that much.

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Ha ha to be honest it was just a guess I have not done a take off and return in forever.  I normally meet up with MInimus station and use an escape pod for Kerbin return.

Edited by Nich
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