So one Atomic Rocket Motor is better than four?

[Vanamonde]

I believe the counter argument is that there is more to consider than total dV of a craft.

Alright, follow up question: what argument is that countering? The OP's question was: "So if I'm understanding everything correctly then less is more and if I want a long range craft I'd go further with just one nuke?" When I agreed with this, other people started bringing up the issue that a higher t/w means less game time spent under thrust, as if this somehow refutes my point about delta-V, when in fact those are two unrelated issues. Pack all the engines on there you want, reduce in-game time spent thrusting all you want, if *that* is your goal, but it isn't the OP's goal, because the delta-V of the ship will decline with every engine you add.

Oberth.

The Oberth effect isn't some magical superboost. To get the benefit of the effect, you'd need to add engines to get more thrust. That additional mass is going to reduce fuel efficiency and offset at least part of the benefit gained by the shorter burn. How much is the *net* benefit actually going to be?

[Kulebron]

Which is perfectly doable with the design shown above. If you are using/need more thrust than that, you are probably doing it wrong (ie, less efficiently).

You need 1100 m/s acceleration to circularize around Tylo. With 0.1g it's 18 minutes. Yes, it's within flyby time, you'll still be within SOI, and with MechJeb you can take your dog for a walk meanwhile.

[DrMonte]

I was pointing out that if you are not satisfied with low TWR in space LV-N is not for you. It's good for me, just grab a coffee or a lunch in half hour burn. I don't use MechJeb, 'T' works fine in Kerbol orbit.

[Superfluous J]

Alright, follow up question: what argument is that countering? The OP's question was: "So if I'm understanding everything correctly then less is more and if I want a long range craft I'd go further with just one nuke?" When I agreed with this, other people started bringing up the issue that a higher t/w means less game time spent under thrust, as if this somehow refutes my point about delta-V, when in fact those are two unrelated issues. Pack all the engines on there you want, reduce in-game time spent thrusting all you want, if *that* is your goal, but it isn't the OP's goal, because the delta-V of the ship will decline with every engine you add.

Ah, I see the problem. I used your term, "counter argument" when I really shouldn't have.

I can't speak for the rest of the people advocating a consideration of TWR even when in space but for me, I was not saying you were wrong. You are completely right. If you want the absolute best delta V with an LV-M then yes, you just want 1. Having said that, the question was answered an no further comments were really necessary.

I was answering the question that the OP's question begged: Is Delta-V (ie: "going farther") the only valid consideration? For me, the answer to that question is "no" and I explained why in my comments.

I was not trying to refute your statement. I was trying to add to the conversation, and bring up points that I believe are salient but were not asked by the original poster.

[Fenris]

As I like to point out, go look at the delta-V equation again. You'll find that engine thrust is not a term in the equation.

And I think that's the part that was confusing me .... I'm getting an understanding of it now ... still hard to wrap my brain around thrust not being in the equation.

Just started practising delta-v calculations.

By my estimate you initially used a Rockomax Jumbo-64 fuel tank, and possible a Mk1 cockpit.

You won a space cookie That's exactly what I was experimenting with to try to get a better understanding of delta v ... not something I planned on flying, just the first two components I grabbed for my experiment.

Alright, follow up question: what argument is that countering? The OP's question was: "So if I'm understanding everything correctly then less is more and if I want a long range craft I'd go further with just one nuke?" When I agreed with this, other people started bringing up the issue that a higher t/w means less game time spent under thrust, as if this somehow refutes my point about delta-V, when in fact those are two unrelated issues. Pack all the engines on there you want, reduce in-game time spent thrusting all you want, if *that* is your goal, but it isn't the OP's goal, because the delta-V of the ship will decline with every engine you add.

You answered the exact question I asked (or meant to ask) I may not have been entirely clear, but I was interested in distance and not how long it would take me to get there. I took more notice of this when I went to Duna in an unmanned probe and had a huge amount of fuel left over from a Rockomax X 200-32 fuel tank and 1 nuke, later I tried the same flight with double the fuel and four nukes .... It didnt go well fuel wise. Granted I paid no attention to launch windows then and I'm still new enough that no two flights are ever exactly the same, so there are variables that are not accounted for in the second flight.

Thanks for the answers everyone I did learn a lot and the posts were interesting as always.

[SSSPutnik]

One Engine (-5.43 tonnes deadmass, -2LV-N, -2 BZ-52, no girder assembly needed): dV = ln((22.49+18)/(22.49+2)) * 800 * 9.81 = 3945.897 m/s, TWR = 60 / (22.49+18) * 7.207 = 0.206

Nice work! At 11PM local my eyes are glazing over.

[Fenris]

I was answering the question that the OP's question begged: Is Delta-V (ie: "going farther") the only valid consideration? For me, the answer to that question is "no" and I explained why in my comments.

I was not trying to refute your statement. I was trying to add to the conversation, and bring up points that I believe are salient but were not asked by the original poster.

Actually I kind of did bring that up when I threw the word "better" in the topic line even if I didn't mean to. But I'm glad people answered it because I believe now when I design my "do everything ship" I will either have two variants (one set up that will go far and not be worried about the time it takes, and one set up that will get to short and medium ranges quickly). But I think my final design (final for this ship ... I think of 15 different designs on my drive home every day lol) will incorporate interchangeable engines that are parked/docked at a space station.

[Taki117]

...now when I design my "do everything ship"... will incorporate interchangeable engines that are parked/docked at a space station.

Sorry, but this bugs me. it is nigh impossible to have an efficient "do everything ship" your ship should be built around a specific purpose and then optimized for that purpose. I can hear the more experienced players now. "but i use the same launcher for everything!" thats all fine and good, but I can almost guarantee the payload at the top is different based on what you plan to do. Start with the goal in mind and find the most efficent way to achieve that goal.

Now on to the second part, if you are thinking of doing what I think you are thinking of doing it is highly unlikely that it will work. I can just see the headache of trying to swap out engines. The way I see the setup is the new engine is attached directly to a docking port that is attached to a station. Now, i am by no means an expert, but i dont believe fuel will flow freely from a fuel tank through a docking port to an engine. and even then, thats not the strongest of connections, one wrong maneuver and that engine is gone and your ship goes nowhere fast.

this goes back to point 1. Build with a goal. You will avoid lots of headache and inefficiency by not trying to do everything. Want to go to Jool? build a ship with enough dV to get there, but dont use that same ship to go to Eve or Duna, you're just wasting fuel at that point. (I say wasting because the trip to Jool requires more dV than a trip to Duna and more dV means either a more efficient engine, more fuel, or a lighter payload. I also say wasting because if you use your Jool ship to get to Duna you're carrying more fuel than you need.)

My point is worry less about what you want your ship to be able to do (everything) and worry more about what it can do well. (Carry a payload to Laythe and return to Kerbin)

[Seret]

What is the counter-argument here? That loading a ship with a pile of redundant engines is a good idea?

No, I think the counter argument is that if those extra engines increase your enjoyment of the game, then they're not redundant. Not everyone's primary objective is to minmax everything to the highest degree, although I'm quite happy that some people like to play it that way. The OP's question did implicitly ask whether there were situations where adding more LV-Ns conferred some advantage, and some players believe there are.

It's a single player game, play it in the way that gives you most enjoyment, even if that means designing less efficient ships. For a required ÃŽâ€v there are myriad designs that will work. Personally I don't see a problem with adding some extra fuel and another LV-N to a design in order to cut burn times. That's just further optimisation IMO, to try and optimise a parameter other than ÃŽâ€v per part.

Edited November 19, 2013 by Seret

[capi3101]

capi, I only understood 3.14% of your post, but it was beautiful.

Well, thanks.

I did the proof of the backwards Tsiolkovsky equation a couple of weeks ago on a different thread. The idea is that given a "target delta-V", you can use it to figure out exactly how much fuel you need. It assumes that a fuel tank of a given mass when it's dry weighs nine times as much when it's full, which is a true assumption for every liquid fuel tank in KSP except for the Round-8 and Oscar-B. Everything besides the fuel tank is "dead mass", mass that doesn't change over the course of the lifetime of the stage. Now, getting a fuel amount that directly corresponds to a combination of fuel tanks is nearly impossible, so what I do is divide the amount I need by .05625 tonnes, the mass of a full FL-T100 tank (the smallest tank for which the basic 9:1 assumption still holds true), and get "FL-T100 equivalents", from which you can get the other tanks (FL-T200 = 1 FL-T100, FL-T400 = 4 FL-T100, FL-T800/X200-8 = 8 FL-T100, X200-16 = 16 FL-T100, X200-32 = 32 FL-T100, Jumbo64 (Orange Tank) = 64 FL-T100).

TWR was probably the equation I was vague on, but I assumed everybody knew that one: TWR = T / Mg, with g ≡ (GM_planet)/R², T being the amount of thrust available (an LV-N outputs 60 kN of thrust at full throttle, of course), and M being the total mass of the rocket (deadmass + fuel).

Incidentally, GM_planet is what's known as the "gravitational parameter"; it's just the gravitational constant of the universe times a planet's mass. According to the wiki, this value for Kerbin is 3.5316*10¹² m³/s². It's planetary radius is 600,000 m, and if you do the math you get surface g = 9.81 m/s². A 100 km orbit is at a radius of 700,000 m, from which I got that 7.207 m/s² figure.

I'm probably not helping matters here, am I?

Nice work! At 11PM local my eyes are glazing over.

Same here...

And here I was thinking that someone would point out that when you send a twenty tonne lander to Eve, it better damn well not still be twenty tonnes when it comes back up...

Edited November 19, 2013 by capi3101

[antbin]

has anyone tried quantifying the gains from (exploiting Oberth effect with) higher TWR versus the losses from a lower mass ratio?

I'd be interested in this too. My suspicion is that the TWR break-even point would be very low. But having flown an ion-powered SSTO to Laythe and back again, the TWR was so low (0.05?) that I had to perform multiple boosting orbital passes in order to have a hope of making it efficiently.

The TWR was so low on that ion craft that I would be climbing out of Kerbin's gravity well, speed would be *decreasing* at full throttle, and MechJeb's required Delta-V indicator would just keep increasing. Nothing to do but to cut throttle and coast until the next orbital pass. Made intercept planning very difficult.

Anyone who starts with "assuming the Delta-V written on this chart" is missing the point of the Oberth effect. Rocket power per fuel burned increases as speed increases, and speed increases (temporarily) as you fall into a planet's gravity well. Got to burn while the speed is hot!

Edited November 19, 2013 by antbin

[RRoan]

The Oberth effect isn't some magical superboost. To get the benefit of the effect, you'd need to add engines to get more thrust. That additional mass is going to reduce fuel efficiency and offset at least part of the benefit gained by the shorter burn. How much is the *net* benefit actually going to be?

A reduction in vehicle ÃŽâ€v due to added weight is fine if it results in a burn that requires less ÃŽâ€v in the first place. And a one minute orbit insertion burn is going to require a lot less ÃŽâ€v than a seventeen minute one.

[capi3101]

http://en.wikipedia.org/wiki/Oberth_effect

http://www.projectrho.com/public_html/rocket/mission.php -- Lucky us, I found a formula that calculates the Oberth delta-V bonus.

Maybe it's just the way I'm reading it, but it looks like for gameplay purposes you'd want to place your periapsis right over the spot you want to eject from, and you want it to be at 70,000 meters on the nose at that point.

Orbital velocity at 70,000 meters is 2,295.86 m/s; that's roughly as fast as your speed can initially be without dragging the atmosphere (technically atmosphere ends at 69,078, thus 69,079 is as low as you can go and the velocity is 2,297.46 m/s, but I don't see anybody really wanting to get down that close).

Escape velocity at 70,000 meters is 3,246.86 m/s - that sounds about right.

And the formula is:

dV = √(V_f² + V_esc²) - √(V_h² + V_esc²)

Where V_f is final velocity, V_h is initial velocity, and V_esc is escape velocity. dv in this case refers to delta-V at periapsis.

Spent most of the morning just trying to research this bit; it will take a while before I can work on trying to quantify it. I will need to adjust my case examples to account for the lower starting altitude for TWR. Fun for the afternoon local.

[Death Engineering]

But what do you expect at 11 PM local, anyway?

I have an urge to look for my slide-rule and pocket protector after reading this...

[capi3101]

I have an urge to look for my slide-rule and pocket protector after reading this...

I'm ashamed to admit that I have ready access to both......

[Traches]

For all y'all complaining about long burns-- try using physics time warp (Alt+.). You only get 4x, and it can get squirrely, but it makes a 20 minute burn take 5 minutes, or a 4 minute burn take a minute. You can have efficiency AND not spend half your playtime doing a burn.

[capi3101]

Okay... here's my array from last night, adjusting for the higher gravity for TWR (at 70,000 meters, g = 7.867 m/s²)

1 Engine: 3945.897 m/s, TWR = 0.188

2 Engines: 3585.303 m/s, TWR = 0.350

3 Engines: 3361.88 m/s, TWR = 0.498

4 Engines: 3165.016 m/s, TWR = 0.633

6 Engines: 2833.879 m/s, TWR = 0.867 (had the TWR wrong last night - should've been 0.957; six nukes output 360 kN, not 300)

Let's normalize those delta-V values, using the three engine case as our baseline (remember, we designed the three-engine case to give us almost exactly what we needed for the whole mission):

1: +584.017 m/s

2: +223.423 m/s

3: +0 m/s

4: -196.864 m/s

6: -528.001 m/s

Once again, we're starting with orbital velocity = 2,295.86 m/s and escape velocity at that altitude is 3,246.86 m/s

We want to go to Eve; that takes 1,030 m/s (from the example last night), and we're steely-eyed enough to get it perfect right at the periapsis.

So we want our final velocity to be 3,325.86 m/s (2295.86 + 1030 = 3325.86).

So we have:

dV = √(3325.86^2 + 3,246.86^2) - √(2,295.86^2 + 3,246.86^2)

dV = 671.3851028 m/s

If we apply 671.385 m/s of delta-V at a 70,000 meter periapsis, we get 1,030 actual delta-V. So the Oberth Effect bonus in this case is 358.615 m/s (1030 - 671.385). That's applies regardless of the number of engines involved. If we apply that as a bonus to our normalized values above, we can see that we'll still come out ahead with four engines, but by the time we get up to six engines we've given up too much delta-V via the spacecraft's mass to make it worth our while.

I may have misunderstood what the issue is regarding the usefulness of the Oberth effect; I think I'll go re-read the posts on this thread. Hopefully this has been helpful to some of y'all, at least.

[Nao]

Adding to the TWR discussion.

When exiting Kerbin from LKO the effect of TWR becomes important below ~0,15 TWR. For example comparing a burn at 0,5 twr from 70x70 orbit to Ap of 18000km (close to Kerbin escape) with a burn at 0,16 TWR to the same Ap the difference in dV requirement is only ~5% (900 to 947m/s). And from comparison i did the other day in another discussion for 1561m/s burn with TWR 1,05 and 9,45 respectively (that's a burn to Moho from 70km Kerbin orbit) the difference is only around 0,68%.

@capi3101 As you say yourself, the 671.385 m/s of "bonus" dV from Oberth adds to evey craft no matter the desingn, so its kind of pointless to compare the dV value to changes in total dV for different number of engines.

What would be worth exploring is impact of thrust on altitude at which the burn occurs, as low TWR craft would burn while flying slower at higher altitudes, assuming same starting orbit.

Your equation would be useful if we could count what is the mean speed at which the burn occurs (2,295.86 for infinite TWR) and probably something around 2,150m/s for TWR ~0,15 (guesstimate). We could then compare bonus dV from the two instances.

It would be around 612m/s bonus dV for the second rocket, so if the rocket with TWR 0,15 would have 59+m/s more than the 0,5 TWR one it would be better for Eve transfer.

The 59m/s for 1030m/s burn is close to the one tested in game (47m/s for 900m/s burn at little higher TWR) so my rough math results look reasonable.

Edited November 19, 2013 by Nao

[RadHazard]

A reduction in vehicle ÃŽâ€v due to added weight is fine if it results in a burn that requires less ÃŽâ€v in the first place. And a one minute orbit insertion burn is going to require a lot less ÃŽâ€v than a seventeen minute one.

Not quite. ÃŽâ€v is independent of burn time (assuming you discount factors such as inefficiencies due to burning during less optimal times during your orbit, which are non-trivial, but can be reduced by using multiple 'kicks' rather than one long one). If you want to change your orbit to a specific orbit for a transfer, you need to have a specific amount of velocity. The change in velocity you need to achieve at the burn point is known as required ÃŽâ€v. This is number is independent of the mass of a craft, and is derived only from it's current orbit and the orbit you want to achieve. This is how the ÃŽâ€v maps work. It takes (roughly) 2000 m/s to reach Jool, regardless of whether you are flying a 2 ton probe or a 200 ton colony ship.

You are correct in saying increased mass decreases your available ÃŽâ€v. However, increasing thrust while keeping ISP constant (e.g. by adding more engines) will never increase your available ÃŽâ€v, even if you had zero-mass engines available. ISP is, quite simply, a measure of how much ÃŽâ€v you can get out of a given amount of fuel. To make it clear why this is, imagine you have a spaceship that consists of nothing but fuel tanks and two identical, weightless engines. If you were to run both engines, you would get 2X kN of thrust, where X is one engine's worth of thrust. If you were to shut off one of the engines, you would have only X thrust, meaning it would take twice as long to complete a burn. However, since you are only running one engine instead of two, you consume only half as much fuel in a given period of time, say 1 second, as you would with both engines running. At the end of the burn, both factors cancel out completely, meaning you have burned the exact same amount of fuel.

This is consistent with basic physics, as you have hurled the same amount of fuel backwards at the same speed (albeit over a longer time period), regardless of the number of engines. Thus, the total momentum change of your rocket is equal.

I'd also like to point out that the program in my sig uses the same calculations as what capi3101 is posting. You can play around with it using different number to verify the results posted (although you'd have to do it backwards, starting with a minimum TWR and ÃŽâ€v). The results are quite significant.

For example, for a 5 ton payload, in order to achieve a ÃŽâ€v of 4000 and a (Kerbin-surface-relative) TWR of 1 when full, you'd need 4 LV-Ns and 9.29 tons of fuel, for a total mass of 23.29 tons.

That same 5 ton payload with 4000 ÃŽâ€v but a relaxed Kerbin-surface-relative TWR of 0.5 requires only 1 LV-N and 4.81 tons of fuel, and has a total mass of 12.06 tons.

The reason 1/4 the number of engines provides 1/2 the thrust is because of the tyranny of rocketry: You need more fuel to carry the increased engine mass, which needs more engines to get thrust needed for the same TWR, which needs more fuel... etc.

Also note that Kerbin-surface-relative TWR (KSRTWR?) is simply an alternative way to represent a given level of acceleration a ship is capable of, since Kerbin surface gravity will always remain at 9.81m/s² regardless of where you go in the universe. A more logical expression would be the Thrust-to-Mass ratio, where a TMR of 1 would represent the ability to accelerate at 1 m/s per second (1m/s²), but I figured most players would have an intuitive feel for how "thrusty" a ship with ~1-2 KSRTWR feels in space, since most early ships they would design would have about that much thrust.

Edited November 19, 2013 by RadHazard
clarification

[Nao]

@RedHazard I agree with RRoan here, because while getting to Mun or Minmus you can use the "multiple kicks" it falls off hard as we go to other planets. Jool needs 2000m/s dV and in practice you can only multiple kick the fist 800-900m/s.

[capi3101]

@capi3101 As you say yourself, the 671.385 m/s of "bonus" dV from Oberth adds to evey craft no matter the desingn, so its kind of pointless to compare the dV value to changes in total dV for different number of engines. What would be worth exploring is impact of thrust on altitude at which the burn occurs, as low TWR craft would burn while flying slower at higher altitudes, assuming same starting orbit.

Your equation would be useful if we could count what is the mean speed at which the burn occurs (2,295.86 for infinite TWR) and probably something around 2,150m/s for TWR ~0,15 (guesstimate). We could then compare bonus dV from the two instances.

Well, first off, the bonus dV from Oberth I calculated was only 358.615 m/s, and that was at 70,000 meters. Second, the formula for Oberth effect does not take into account the rate of acceleration, just the initial velocity, final velocity and escape velocity. All three of these values (and thus the Oberth effect bonus) should be the same for all craft that burn the same amount of delta-V at the same altitude, regardless of their TWR. What would affect the bonus is a slower initial speed, i.e. starting the burn with a higher periapsis.

The formulas I used assume an instantaneous burn occurring right at periapsis; same as the maneuver node system KSP uses. Granted that's an incorrect assumption and may have an effect on what we're talking about. I'd wager the correction for that assumption would involve a pretty nasty looking differential equation; I could be wrong about that, of course. I'll do some homework and get back to y'all about that.

[RadHazard]

@RedHazard I agree with RRoan here, because while getting to Mun or Minmus you can use the "multiple kicks" it falls off hard as we go to other planets. Jool needs 2000m/s dV and in practice you can only multiple kick the fist 800-900m/s

I partially agree, although I have managed a 2-kick burn to Jool using a ship that had a combined burn time of 11 minutes. However, unless I misunderstood, he wasn't talking about the difficulty of hitting a transfer window with a low-thrust craft. He was saying that higher-thrust craft needs to expend significantly less ÃŽâ€v to make the same transfer, which is untrue.

Edited November 19, 2013 by RadHazard

[Nao]

@capi3101

Ah yes, i've forgot to subtract the values from 1030 (either way the difference result of 59m/s remains the same). Somehow the idea of "bonus" dV seems hard to grasp for me. I'm used to think of it in terms of energy.

The basic equation does not count for rate of acceleration, but we can integrate it (or, since i'm bad at it) just iterate several impulsive burns with changing starting conditions. I've just used only one and picked Pe that felt would be close to integrated median from experiences in 0,16TWR flight i've done earlier today. So yes, I've started from higher periapsis, but to correct the increased Pe of such orbit i've just lowered the required speed so that in both cases final orbits had the same energy (same speed when exiting SOI).

(sorry if i used the wrong terminology there, never learned math in English really:P)

Edit:

@RadHazard

Welp he mentions 17minutes, so i did think about a far trip. You don't really make such burns in KSP in anything other than either interplanetary stage on 1 LV-N or a probe on ION's, and Ion engine doesn't have much design space to choose from either way so arguing twr feels pointless (just slap one of them and burn away with whatever you have).

You are 100% right with how it works in theory and the multiple kick thing is indeed very useful, just not always.

Edited November 19, 2013 by Nao

[RRoan]

Not quite. ÃŽâ€v is independent of burn time (assuming you discount factors such as inefficiencies due to burning during less optimal times during your orbit, which are non-trivial, but can be reduced by using multiple 'kicks' rather than one long one). If you want to change your orbit to a specific orbit for a transfer, you need to have a specific amount of velocity. The change in velocity you need to achieve at the burn point is known as required ÃŽâ€v. This is number is independent of the mass of a craft, and is derived only from it's current orbit and the orbit you want to achieve. This is how the ÃŽâ€v maps work. It takes (roughly) 2000 m/s to reach Jool, regardless of whether you are flying a 2 ton probe or a 200 ton colony ship.

You are correct in saying increased mass decreases your available ÃŽâ€v. However, increasing thrust while keeping ISP constant (e.g. by adding more engines) will never increase your available ÃŽâ€v, even if you had zero-mass engines available. ISP is, quite simply, a measure of how much ÃŽâ€v you can get out of a given amount of fuel. To make it clear why this is, imagine you have a spaceship that consists of nothing but fuel tanks and two identical, weightless engines. If you were to run both engines, you would get 2X kN of thrust, where X is one engine's worth of thrust. If you were to shut off one of the engines, you would have only X thrust, meaning it would take twice as long to complete a burn. However, since you are only running one engine instead of two, you consume only half as much fuel in a given period of time, say 1 second, as you would with both engines running. At the end of the burn, both factors cancel out completely, meaning you have burned the exact same amount of fuel.

This is consistent with basic physics, as you have hurled the same amount of fuel backwards at the same speed (albeit over a longer time period), regardless of the number of engines. Thus, the total momentum change of your rocket is equal.

I'd also like to point out that the program in my sig uses the same calculations as what capi3101 is posting. You can play around with it using different number to verify the results posted (although you'd have to do it backwards, starting with a minimum TWR and ÃŽâ€v). The results are quite significant.

For example, for a 5 ton payload, in order to achieve a ÃŽâ€v of 4000 and a (Kerbin-surface-relative) TWR of 1 when full, you'd need 4 LV-Ns and 9.29 tons of fuel, for a total mass of 23.29 tons.

That same 5 ton payload with 4000 ÃŽâ€v but a relaxed Kerbin-surface-relative TWR of 0.5 requires only 1 LV-N and 4.81 tons of fuel, and has a total mass of 12.06 tons.

The reason 1/4 the number of engines provides 1/2 the thrust is because of the tyranny of rocketry: You need more fuel to carry the increased engine mass, which needs more engines to get thrust needed for the same TWR, which needs more fuel... etc.

Also note that Kerbin-surface-relative TWR (KSRTWR?) is simply an alternative way to represent a given level of acceleration a ship is capable of, since Kerbin surface gravity will always remain at 9.81m/s² regardless of where you go in the universe. A more logical expression would be the Thrust-to-Mass ratio, where a TMR of 1 would represent the ability to accelerate at 1 m/s per second (1m/s²), but I figured most players would have an intuitive feel for how "thrusty" a ship with ~1-2 KSRTWR feels in space, since most early ships they would design would have about that much thrust.

I did not say that reducing burn time via adding more engines increases ÃŽâ€v, and did in fact say the opposite, so I am unsure why you felt the need to write a multi-paragraph post lecturing me about things I already know. I merely noted that there are cases where an increase in thrust leads to a reduction in the required ÃŽâ€v relative to a design with less thrust, which is completely correct.

TWR and TMR are the same if using proper units.

[RadHazard]

I did not say that reducing burn time via adding more engines increases ÃŽâ€v, and did in fact say the opposite, so I am unsure why you felt the need to write a multi-paragraph post lecturing me about things I already know. I merely noted that there are cases where an increase in thrust leads to a reduction in the required ÃŽâ€v relative to a design with less thrust, which is completely correct.

TWR and TMR are the same if using proper units.

I apologize for coming off as lecturing. I did see where you mentioned that increased mass results in less ÃŽâ€v. It seems I misunderstood when you mentioned that you can save ÃŽâ€v with higher thrust values. I had assumed that you had confused an increase of burn time with an increase in ÃŽâ€v required, but after re-reading it I'm guessing you were talking about the ÃŽâ€v savings you get by reducing the amount of time you spend burning far from the node.