Non-Dimensional Model for Optimal Horizontal Launch Efficiency

[arkie87]

[tl:dr]To obtain efficiency greater than or equal to 90%, a thrust-to-weight ratio (TWR) of at least 1.4 at launch on an atmosphereless, Kerbin-sized body (like Tylo) is required with standard ISP engines; but for planets with smaller orbital velocity or more efficient engines (nuclear or Ion), TWR at launch to obtain efficiency greater than 90% increases. [/tl:dr]

I have made a computer model in Matlab (also available in Excel now! Click here to get it) to simulate optimal TWR (thrust-to-weight ratio), TVR (thrust-velocity-ratio), and DVR (delta-V-ratio). I provide non-dimensional contour plots to show results which are globally applicable for the same non-dimensional parameters.

Link to The Paper: https://www.dropbox.com/s/qud778opmyp4fis/OptimalHorizontalLaunch.pdf?dl=0

Some sample results:

This graph shows the independence of efficiency on DVR. The slope of the curve is zero, until efficiency drops to zero instantly when delta-V-ratio no longer provides enough deltaV to get into orbit due to efficiency.

Since efficiency is not a function of DVR (i.e. how much extra fuel you carry into orbit), we only need to vary TWR and TVR. The above plot contains all the information a player needs when designing craft.

First, increasing TWR increases efficiency while increasing TVR decreases efficiency. Increasing TWR increases efficiency by allowing the craft to aim more horizontal, thereby, using more of its fuel to accelerate into orbit instead of fighting gravity. Increasing TVR decreases efficiency since, for a given DVR, less fuel is burned. This, in turn, results in a more constant TWR during the flight, and therefore, a longer flight as well as a steeper angle above vertical, causing more fuel to be wasted fighting gravity.

Second, for low TVR, lower TWR are needed for a given efficiency; similarly, for a higher TVR, a higher TWR is needed to obtain the same efficiency. This is the direct result of the trends described above, and is the most important result of the simulation: thus, for two crafts with the same ISP engine, the craft on Minmus will require a higher TWR for the same efficiency as a craft on Mun, and so on.

Finally, this model should be easy to test. Since efficiency is not a function of DVR (i.e. size or scale), two craft with the same TWR can be compared on the same planet with different ISP engines or on different planets with the same engine, and the craft with the larger ISP engine or equivalently on the planet with a smaller orbital velocity should have reduced efficiency.

And in case you dont believe that efficiency is independent of DVR for all TWR and TVR:

Edited December 29, 2014 by arkie87

[numerobis]

What are FMR and TVR?

When I coded up basically equation (2) and numerically integrated it, I ended up building up vertical speed.

[arkie87]

What are FMR and TVR?

When I coded up basically equation (2) and numerically integrated it, I ended up building up vertical speed.

FMR is in Eq. 7. TVR = V_e/V_0 = ISP*g0/V_orbit

Hmmm. Let me check that...

[arkie87]

What are FMR and TVR?

When I coded up basically equation (2) and numerically integrated it, I ended up building up vertical speed.

Eq. (1) has a typo. Supposed to be m*(g - V^2/R)...

The typo is only in the paper. The model is correct as is.

Edited December 19, 2014 by arkie87

[GoSlash27]

I've done some testing and arrived at a simple equation that you should find interesting:

For any airless body,

efficiency ≈ 1- [1/2R_tw⁴]

where Rtw= thrust to weight ratio

This curve tracks perfectly with my empirical test results.

[TABLE=width: 100]

[TR]

[TD]T/W at launch[/TD]

[TD]predicted ∆V[/TD]

[TD]actual ∆V[/TD]

[/TR]

[TR]

[TD]0.91[/TD]

[TD]1,058[/TD]

[TD]1,103[/TD]

[/TR]

[TR]

[TD]1.84[/TD]

[TD]639[/TD]

[TD]657[/TD]

[/TR]

[TR]

[TD]3.68[/TD]

[TD]614[/TD]

[TD]630[/TD]

[/TR]

[TR]

[TD]7.37[/TD]

[TD]612[/TD]

[TD]625[/TD]

[/TR]

[/TABLE]

Startlingly, this is completely independent of local gravity, the ∆V of the orbit, or I_sp of the engine.

This function trends to 100% within 3 digits of resolution by the time t/w has achieved 5.0:1, so no noticeable gains in efficiency will occur above that point.

[TABLE=width: 50]

[TR]

[TD]t/w ratio[/TD]

[TD]loss %[/TD]

[/TR]

[TR]

[TD]1.0[/TD]

[TD]50.0[/TD]

[/TR]

[TR]

[TD]1.2[/TD]

[TD]24.1[/TD]

[/TR]

[TR]

[TD]1.4[/TD]

[TD]13.0[/TD]

[/TR]

[TR]

[TD]1.6[/TD]

[TD]7.6[/TD]

[/TR]

[TR]

[TD]1.8[/TD]

[TD]4.8[/TD]

[/TR]

[TR]

[TD]2.0[/TD]

[TD]3.1[/TD]

[/TR]

[TR]

[TD]2.5[/TD]

[TD]1.3[/TD]

[/TR]

[TR]

[TD]3.0[/TD]

[TD]0.6[/TD]

[/TR]

[TR]

[TD]3.5[/TD]

[TD]0.3[/TD]

[/TR]

[TR]

[TD]4.0[/TD]

[TD]0.2[/TD]

[/TR]

[TR]

[TD]4.5[/TD]

[TD]0.1[/TD]

[/TR]

[TR]

[TD]5.0[/TD]

[TD]0.1[/TD]

[/TR]

[/TABLE]

This suggests that each engine has a single unique optimal t/w point for minimum vehicle mass, minimum vehicle cost, and maximum payload fraction.

I'm working on the equations to establish these points now.

Scratchin' mah head,

-Slashy

Edited December 20, 2014 by GoSlash27

[GoSlash27]

arkie,

I've scoured the paper thoroughly (both this one and the one you linked me on Imgur), and I'm having a hard time understanding what you mean by "TVR" I'm not following the concept of a "specific impulse for a constant planet" */me scratches head*. You define it as V_e/V_o and I was able to deduce that V_o was orbital velocity, but what is V_e? Escape velocity?

I'm having a hard time grasping how 70%+ efficiency would be possible at t/w= 1.0. Would it be because the initial low-efficiency burn time is relatively short compared to the overall burn time?

Also,

When you referred to ignoring the 3rd graph, perhaps that's a bit hasty. While we *do* want more fuel to do things after we achieve orbit, that fuel would technically be payload during the launch. I could easily see how a massive payload being lifted into orbit at 1G with ions would suffer huge penalties in efficiency, since the reduction in mass is negligible and t/w doesn't improve, whereas a SRB would have a much easier time of it because it sheds so much mass.

I neglect this in my estimate.

Thanks and good morning!,

-Slashy

[arkie87]

I've done some testing and arrived at a simple equation that you should find interesting:

For any airless body,

efficiency ≈ 1- [1/2R_tw⁴]

where Rtw= thrust to weight ratio

This curve tracks perfectly with my empirical test results.

[TABLE=width: 100]

[TR]

[TD]T/W at launch[/TD]

[TD]predicted ∆V[/TD]

[TD]actual ∆V[/TD]

[/TR]

[TR]

[TD]0.91[/TD]

[TD]1,058[/TD]

[TD]1,103[/TD]

[/TR]

[TR]

[TD]1.84[/TD]

[TD]639[/TD]

[TD]657[/TD]

[/TR]

[TR]

[TD]3.68[/TD]

[TD]614[/TD]

[TD]630[/TD]

[/TR]

[TR]

[TD]7.37[/TD]

[TD]612[/TD]

[TD]625[/TD]

[/TR]

[/TABLE]

Startlingly, this is completely independent of local gravity, the ∆V of the orbit, or I_sp of the engine.

This function trends to 100% within 3 digits of resolution by the time t/w has achieved 5.0:1, so no noticeable gains in efficiency will occur above that point.

[TABLE=width: 50]

[TR]

[TD]t/w ratio[/TD]

[TD]loss %[/TD]

[/TR]

[TR]

[TD]1.0[/TD]

[TD]50.0[/TD]

[/TR]

[TR]

[TD]1.2[/TD]

[TD]24.1[/TD]

[/TR]

[TR]

[TD]1.4[/TD]

[TD]13.0[/TD]

[/TR]

[TR]

[TD]1.6[/TD]

[TD]7.6[/TD]

[/TR]

[TR]

[TD]1.8[/TD]

[TD]4.8[/TD]

[/TR]

[TR]

[TD]2.0[/TD]

[TD]3.1[/TD]

[/TR]

[TR]

[TD]2.5[/TD]

[TD]1.3[/TD]

[/TR]

[TR]

[TD]3.0[/TD]

[TD]0.6[/TD]

[/TR]

[TR]

[TD]3.5[/TD]

[TD]0.3[/TD]

[/TR]

[TR]

[TD]4.0[/TD]

[TD]0.2[/TD]

[/TR]

[TR]

[TD]4.5[/TD]

[TD]0.1[/TD]

[/TR]

[TR]

[TD]5.0[/TD]

[TD]0.1[/TD]

[/TR]

[/TABLE]

This suggests that each engine has a single unique optimal t/w point for minimum vehicle mass, minimum vehicle cost, and maximum payload fraction.

I'm working on the equations to establish these points now.

Scratchin' mah head,

-Slashy

Looks like a reasonable quick approximation, though my equations predict efficiency varies with TVR and FMR as well (slightly).

arkie,

I've scoured the paper thoroughly (both this one and the one you linked me on Imgur), and I'm having a hard time understanding what you mean by "TVR" I'm not following the concept of a "specific impulse for a constant planet" */me scratches head*. You define it as V_e/V_o and I was able to deduce that V_o was orbital velocity, but what is V_e? Escape velocity?

I'm having a hard time grasping how 70%+ efficiency would be possible at t/w= 1.0. Would it be because the initial low-efficiency burn time is relatively short compared to the overall burn time?

Also,

When you referred to ignoring the 3rd graph, perhaps that's a bit hasty. While we *do* want more fuel to do things after we achieve orbit, that fuel would technically be payload during the launch. I could easily see how a massive payload being lifted into orbit at 1G with ions would suffer huge penalties in efficiency, since the reduction in mass is negligible and t/w doesn't improve, whereas a SRB would have a much easier time of it because it sheds so much mass.

I neglect this in my estimate.

Thanks and good morning!,

-Slashy

TVR = V_e/V_0

V_0 = V_orb = sqrt(g*R)

V_e = ISP*9.81 i.e. exhaust velocity

TVR can vary by increase in V_0 i.e. bigger planet or by changing exhaust velocity i.e. ISP. For a constant planet i.e. constant v0, ISP can be used to change TVR.

Yes, for high FMR, burn time is very long since most of the mass is fuel. Thus, even though TWR is 1, fuel mass accounts for most of this, thus, TWR changes HUGELY during burn (TWR in my equations is a constant during flight as it is TWR at start)

Edited December 20, 2014 by arkie87

[GoSlash27]

Looks like a reasonable quick approximation, though my equations predict efficiency varies with TVR and FMR as well (slightly).

http://i.imgur.com/vBXohtc.jpg

TVR = V_e/V_0

V_0 = V_orb = sqrt(g*R)

V_e = ISP*9.81 i.e. exhaust velocity

TVR can vary by increase in V_0 i.e. bigger planet or by changing exhaust velocity i.e. ISP. For a constant planet i.e. constant v0, ISP can be used to change TVR.

Yes, for high FMR, burn time is very long since most of the mass is fuel. Thus, even though TWR is 1, fuel mass accounts for most of this, thus, TWR changes HUGELY during burn (TWR in my equations is a constant during flight as it is TWR at start)

Ah... Okay, I'm following you now.

But then how does having a higher Isp or leaving a more massive planet help you achieve improved efficiency at 1G initial thrust? I would expect that to be the opposite... I mean, if you have a high Isp, then it would also mean that you have a low FMR, which I would expect to hinder efficiency at 1G rather than help it since your mass isn't reducing as much to help you.

Still confuzzled

-Slashy

[GoSlash27]

Oh, wait...

Light bulb just went on (I think)...

I believe eq.7 should not be defined as " (M_w-M_d)/M_w ", but rather " e^{(ÃŽâ€V/I_spG_o)} ".

In this case, the fuel consumed would be a function of the I_sp, thrust, and time, not the amount of fuel we put in it to start.

Likewise, we would have already defined G_o when we set our t/w, so I suspect this may be an unnecessary variable.

Yes? No?

-Slashy

Edited December 21, 2014 by GoSlash27

[GoSlash27]

Okay, I think I just got it for sure:

FMR by your definition (7) is really just e^{(ÃŽâ€V/I_spG_o)}

Likewise, TVR is really just I_spG_o/V_o

e is a constant

ÃŽâ€V and V_o are inputs we already have.

Therefore, neither of these values need definition. In this model, for each FMR there can only be one TVR, and each of these have already been defined by the values we plugged in to start.

It would vary with Isp and V_o ,but not with FMR or TVR.

Pretty sure that's right,

-Slashy

Edited December 21, 2014 by GoSlash27

[K^2]

Has it been proven that zero vertical velocity landing/takeoff are optimal efficiency? I know all empirical tests point that way, but I'm wondering if anyone managed to confirm it with equations.

[arkie87]

Has it been proven that zero vertical velocity landing/takeoff are optimal efficiency? I know all empirical tests point that way, but I'm wondering if anyone managed to confirm it with equations.

Not sure what you mean by this. As opposed to crashing into the ground at 100 m/s? I would think 0 m/s is more efficient

[GoSlash27]

Not sure what you mean by this. As opposed to crashing into the ground at 100 m/s? I would think 0 m/s is more efficient

Hehe...

IRT K^2's question, I read it as launching at precisely 1.0:1 T/W ratio.

If that's the case, then no. It could arguably be efficient for SSTOs in atmosphere (I actually do that myself), but on an airless planet you are better- off launching at a higher t/w because there's no such thing as terminal velocity. This is actually what arkie is trying to pin down.

Best,

-Slashy

[arkie87]

Oh, wait...

Light bulb just went on (I think)...

I believe eq.7 should not be defined as " (M_w-M_d)/M_w ", but rather " e^{(ÃŽâ€V/I_spG_o)} ".

In this case, the fuel consumed would be a function of the I_sp, thrust, and time, not the amount of fuel we put in it to start.

Likewise, we would have already defined G_o when we set our t/w, so I suspect this may be an unnecessary variable.

Yes? No?

-Slashy

What makes you say this? We cannot change how we define FMR, without rewriting all the equations to compensate (unless you can prove that your definition is equivalent). The only thing we can redefine is how we define efficiency. Anyway, what is deltaV in this case: expended deltaV or craft deltaV?

The fuel consumed is a function of the amount of fuel we put into it at the start, since the more fuel we have at the start, the more we expend to accelerate the extra fuel we carry.

The amount of deltaV we can get from one stage has a maximum since as we add fuel, we also add structural mass to hold it. Even if we add infinite fuel tanks, such that payload and engines are negligible mass, our m_wet/m_dry approaches a constant given how well we can build fuel tanks (approximately 9/1 in KSP). The only way to increase deltaV beyond this point in a single-stage rocket is to increase Isp.

[GoSlash27]

What makes you say this? We cannot change how we define FMR, without rewriting all the equations to compensate (unless you can prove that your definition is equivalent). The only thing we can redefine is how we define efficiency. Anyway, what is deltaV in this case: expended deltaV or craft deltaV?

I can prove that my definition is equivalent.

e^{(ÃŽâ€V/I_spG_o)}= M_w/M_d. <-- Rocket equation. You can freely substitute one for the other. I'll derive this form if you'd like.

ÃŽâ€V would be "expended" in this case, since that's that's the context you're using it in.

*edit* unless it's not. In which case... it's uhh... not (sheepish grin)... It's in whichever sense you intended the other form of it, 'cuz they really both mean the same thing.

The fuel consumed is a function of the amount of fuel we put into it at the start, since the more fuel we have at the start, the more we expend to accelerate the extra fuel we carry.

Negatory, Kind Sir.

Putting more fuel in the stage than it will burn doesn't mean that it will burn it, it just means that the leftover fuel and structure is dead weight (i.e. "payload"). Unburned fuel has already been accounted for as weight in the t/w ratio.

Consider a tanker lifting a payload of Kethane from Tylo to orbit. It doesn't matter whether you're lifting a ton of kethane and tankage or a ton of lead. All that matters is that it's a ton.

The amount of deltaV we can get from one stage has a maximum since as we add fuel, we also add structural mass to hold it. Even if we add infinite fuel tanks, such that payload and engines are negligible mass, our m_wet/m_dry approaches a constant given how well we can build fuel tanks (approximately 9/1 in KSP). The only way to increase deltaV beyond this point in a single-stage rocket is to increase Isp.

Agreed.

-Slashy

Edited December 21, 2014 by GoSlash27
Me an' mah stoopid typos...

[GoSlash27]

Oh, wait!

I think I just spotted a disconnect.

You said "(Mw-Md)/Mw" rather than "Mw/Md". Is there a particular reason you phrased it that way, or was that a typo?

-Slashy

Edited December 21, 2014 by GoSlash27

[K^2]

Not sure what you mean by this. As opposed to crashing into the ground at 100 m/s? I would think 0 m/s is more efficient

That's just matching velocity at one point on trajectory. Instead, you maintain constant zero radial until you are on Hohmann. That's the same strategy as has been used for landings on many of the maximizing efficiency challenges. It beats suicide burn by a significant margin, even if the later is perfectly executed. And, needless to say, it's much easier to execute and much safer than a suicide burn.

Hohmann or bi-elliptical are clearly optimal under infinite TWR limit. But under finite TWR, getting to that initial Hohmann is a non-trivial optimization problem. I'm just wondering if anyone managed to actually solve it.

[arkie87]

I can prove that my definition is equivalent.

e^{(ÃŽâ€V/I_spG_o)}= M_w/M_d. <-- Rocket equation. You can freely substitute one for the other. I'll derive this form if you'd like.

ÃŽâ€V would be "expended" in this case, since that's that's the context you're using it in.

*edit* unless it's not. In which case... it's uhh... not (sheepish grin)... It's in whichever sense you intended the other form of it, 'cuz they really both mean the same thing.

Oh, yeah, its just raising both sides to e.... yeah. Well, for consistency, if dV is total, then m_dry is fully dry. if dV is expended at time t, then m_dry is actually mass at time t.

Negatory, Kind Sir.

Putting more fuel in the stage than it will burn doesn't mean that it will burn it, it just means that the leftover fuel and structure is dead weight (i.e. "payload"). Unburned fuel has already been accounted for as weight in the t/w ratio.

Consider a tanker lifting a payload of Kethane from Tylo to orbit. It doesn't matter whether you're lifting a ton of kethane and tankage or a ton of lead. All that matters is that it's a ton.

Agreed.

-Slashy

I'm not sure where the disconnect is. I agree.

Oh, wait!

I think I just spotted a disconnect.

You said "(Mw-Md)/Mw" rather than "Mw/Md". Is there a particular reason you phrased it that way, or was that a typo?

-Slashy

I used that form because it simplifies the equation for dimensionless mass: m = 1 - FMR t

That's just matching velocity at one point on trajectory. Instead, you maintain constant zero radial until you are on Hohmann. That's the same strategy as has been used for landings on many of the maximizing efficiency challenges. It beats suicide burn by a significant margin, even if the later is perfectly executed. And, needless to say, it's much easier to execute and much safer than a suicide burn.

Hohmann or bi-elliptical are clearly optimal under infinite TWR limit. But under finite TWR, getting to that initial Hohmann is a non-trivial optimization problem. I'm just wondering if anyone managed to actually solve it.

hmmm. I think i misunderstood you. You are saying people land by doing the reverse of how my model says to take off? hmmm... and it's supposed to be better than a suicide burn? I guess that makes sense since doing the reverse of suicide burn isnt efficient unless you have high TWR....

That gives me an idea: someone should make a plugin that puts that marker on your navball... to show you where you have to burn to counter gravity at current thrust level and velocity

Edited December 21, 2014 by arkie87

[GoSlash27]

Has it been proven that zero vertical velocity landing/takeoff are optimal efficiency? I know all empirical tests point that way, but I'm wondering if anyone managed to confirm it with equations.

you maintain constant zero radial until you are on Hohmann. That's the same strategy as has been used for landings on many of the maximizing efficiency challenges. It beats suicide burn by a significant margin, even if the later is perfectly executed. And, needless to say, it's much easier to execute and much safer than a suicide burn.

Hohmann or bi-elliptical are clearly optimal under infinite TWR limit. But under finite TWR, getting to that initial Hohmann is a non-trivial optimization problem. I'm just wondering if anyone managed to actually solve it.

Aha! Now I'm pickin' up what you're puttin' down

This model assumes that to be true even at extremely low t/w ratios, but does not go out of it's way to prove it.

I have no doubt that the zero vertical velocity landing is the most efficient (since it's the most efficient launch), but it's also the least accurate option for landing at a specific location and most prone to smack you into a crater if you're not careful.

Best,

-Slashy

[K^2]

If you keep TWR constant during burn, rather than total thrust, optimal takeoff has to be time-mirrored copy of optimal landing. So even in more realistic case, you do expect strategies to be similar, if not the same. But it's kind of hard for me to insist on that without having formal proof of optimal solution here.

Constructing such proof looks to be a mess, however. For starters, you are trying to optimize fuel consumption while consuming it at a constant rate. And your destination isn't a point in space, but rather a state. That suggests that what we want to do is find the shortest path in the E x L space, since energy and angular momentum fully determine the desired orbit. However, dE/dt will depend on current velocity and dL/dt will determine on current altitude. That's in addition to thrust direction, which is what we want to optimize. And velocity and altitude on given orbit will depend on mean anomaly. Additionally, we have a restriction that altitude can't drop bellow surface elevation.

Zero vertical velocity ascent implies T = t, which means mean anomaly always stays zero. (It jumps to 2 pi past circularization.) That has the right feel to it, but I can't think of a simple way to prove that it's the answer, and I'm afraid to even sit down and write out all of the formulae to do this properly.

Edit: Do you have a proof that it's optimal strategy for ascent, Slashy? Again, it seems like it should be, but I'm interested in formal proof. I would be happy with first order optimal condition verified.

Edit2: Actually, with above in mind, I definitely can do a check for first order optimal condition on this rigorously. That would prove that it's at least locally optimal, and that'd be a great start. (Except that altitude constraint is going to be a pain. Gah.)

Edited December 21, 2014 by K^2

[GoSlash27]

Oh, yeah, its just raising both sides to e.... yeah. Well, for consistency, if dV is total, then m_dry is fully dry. if dV is expended at time t, then m_dry is actually mass at time t.

*quote snipped*

I'm not sure where the disconnect is. I agree.

*quote snipped*

I used that form because it simplifies the equation for dimensionless mass: m = 1 - FMR t

My point being (I guess... I've confused myself so badly by this point...) is that we wouldn't bother plugging in FMR as a global constant or display results as varying with it. We have already been given the FMR when we input the t/w, Isp, and Go. We just assume that we have *enough* fuel at the outset and whatever mass we lost was fuel.

Accordingly, there's no need to display results in the y axis as FMR. So long as you have enough fuel, the results don't vary with it.

In the case of Ve/Vo, This is definitely a parameter that impacts things, but I'm not understanding how it concludes that launching from Tylo, (for example) would dramatically improve the efficiency of a grossly underpowered launcher as compared to (say) Gilly.

I'll see if I can confirm this through an empirical test; I'll launch 2 vehicles from Tylo to 60km orbit. One will launch at 1G (or at least as close as I can get it), while the other will launch at 4G. Isp will be 290.

I'll post the test results here.

Then later, I'll try the same sort of test from Gilly using PB-ion and see what that shows.

Best,

-Slashy

[GoSlash27]

First test results are in, and it definitely supports what the model predicts, so proof-of-concept dance!

Test #1:

Comparing the effect of orbiting a large planet on a low I_sp launcher's efficiency loss due to low t/w

M_i- 3.77 tonnes Isp= 290 s Launch from Tylo surface to 60kM orbit

[TABLE=width: 500]

[TR]

[TD][/TD]

[TD]Thrust (kN)[/TD]

[TD]T/W initial[/TD]

[TD]M_o (tonnes)[/TD]

[TD]ÃŽâ€V expended (m/sec)[/TD]

[TD]Efficiency (%)[/TD]

[/TR]

[TR]

[TD]Test1[/TD]

[TD]28[/TD]

[TD].946[/TD]

[TD]1.28[/TD]

[TD]3,076[/TD]

[TD]73.8[/TD]

[/TR]

[TR]

[TD]Test2[/TD]

[TD]112[/TD]

[TD]3.78[/TD]

[TD]1.66[/TD]

[TD]2,340[/TD]

[TD]97.0[/TD]

[/TR]

[/TABLE]

As the results show, yeah the initial state of being underpowered hurts you, but not as much as I would've expected. Running the same test on the Mun yielded a 50% drop in efficiency for a 1G launch using the same engine, so Tylo itself is responsible for this improvement.

Lookin' good so far!

-Slashy

[arkie87]

First test results are in, and it definitely supports what the model predicts, so proof-of-concept dance!

Test #1:

Comparing the effect of orbiting a large planet on a low I_sp launcher's efficiency loss due to low t/w

M_i- 3.77 tonnes Isp= 290 s Launch from Tylo surface to 60kM orbit

[TABLE=width: 500]

[TR]

[TD][/TD]

[TD]Thrust (kN)[/TD]

[TD]T/W initial[/TD]

[TD]M_o (tonnes)[/TD]

[TD]ÃŽâ€V expended (m/sec)[/TD]

[TD]Efficiency (%)[/TD]

[/TR]

[TR]

[TD]Test1[/TD]

[TD]28[/TD]

[TD].946[/TD]

[TD]1.28[/TD]

[TD]3,076[/TD]

[TD]73.8[/TD]

[/TR]

[TR]

[TD]Test2[/TD]

[TD]112[/TD]

[TD]3.78[/TD]

[TD]1.66[/TD]

[TD]2,340[/TD]

[TD]97.0[/TD]

[/TR]

[/TABLE]

As the results show, yeah the initial state of being underpowered hurts you, but not as much as I would've expected. Running the same test on the Mun yielded a 50% drop in efficiency for a 1G launch using the same engine, so Tylo itself is responsible for this improvement.

Lookin' good so far!

-Slashy

It's cool when model matches reality! :feelin good:

[GoSlash27]

arkie,

Yeah it is! Great job on the model!

I decided to change up the second test. I'm afraid that an ion powered flight from Gilly would use so little fuel that I won't see a change in mass, so I'm using an LV-1 instead. I'll go low Isp and low gravity and see how that shakes out.

Best,

-Slashy

[GoSlash27]

Okay, test #2 was definitely a bust.

Even at a low Isp, It just didn't show enough mass consumption to get an accurate read on DV.

Nevertheless, I was able to see the reasons why a large planet penalizes underpowered lifters less than small planets:

#1) Inertia. Thrust- to- mass is the name of the game for changing the direction of an object, and the low gravity of a small body means a ridiculously low thrust- to mass.

#2) Lack of centripedal lift. I spent the entire push to orbit from Gilly at a high pitch angle, whereas on Tylo it was a small portion of the entire launch stuck in the "this lifter is an underpowered pig" regime of the launch.

While the smaller gravity well penalized more heavily (presumably) for being underpowered, it was a larger percentage of a miniscule DV budget, so not really worth discussing. If a "perfect" launch is 30 m/sec, who cares if you wind up spending 60 m/sec?

I think these tests confirm that the model is sound in this area. Next up, varying Isp.

If this test pans out, then I think it's safe to confirm this model with empirical testing.

I will build a munar lifter at high Isp and confirm that the model predicts it's behavior. I expect it to be penalized more severely than the low Isp lander for being underpowered..

Best,

-Slashy

Edited December 21, 2014 by GoSlash27