Effect of initial TWR on orbit dV cost

[arkie87]

I already know C++, thank you. The trouble is it's a lot more work to set up drawing pretty pictures with complex equations in C++

Well then you will have to trust my result

[GoSlash27]

I'm glad that everybody's in agreement.

I personally don't care about absolute precision AFA what I'm planning on doing with this info, but I am bothered when my empirical results aren't predicted by the model.

To that end... my empirical results are exceeding *both* of these models on smaller bodies and extremely low t/w ratios.

There's something else at work, and I don't know what it is or how significant it is... but it's there and neither model is accounting for it.

For example, Using o-10 engines (Isp=290s) from the Mun and an initial t/w of 0.91, I'm establishing 10km orbit with .526 efficiency. Neither model would predict that's possible, especially with the necessity of clearing terrain on the way.

And if my results are suspect, arkie actually confirmed it on one of his launches as well; predicted maximum efficiency 69%, actual result 72%.

(arkie, please correct me if I'm mistaken here)

Best,

-Slashy

Edited December 26, 2014 by GoSlash27

[arkie87]

For example, Using o-10 engines (Isp=290s) from the Mun and an initial t/w of 0.91, I'm establishing 10km orbit with .526 efficiency. Neither model would predict that's possible, especially with the necessity of clearing terrain on the way.

So when you start with TWR of 0.91, what exactly happens for the first few minutes when you burn with TWR < 1? You should stay in place and just waste fuel. Is that what happened?

And if my results are suspect, arkie actually confirmed it on one of his launches as well; predicted efficiency 69%, actual result 72%.

http://i.imgur.com/vCZ0OYW.png

(arkie, please correct me if I'm mistaken here)

Best,

-Slashy

I dont see 69% or 72% anywhere in that graph that you quoted. Please explain where you are getting these numbers from....

[Anquietas314]

So when you start with TWR of 0.91, what exactly happens for the first few minutes when you burn with TWR < 1? You should stay in place and just waste fuel. Is that what happened?

Just a thought, but if you're on a particularly steep slope (in terms of what you can safely land on) that might just be enough to lift off (before falling further down the hill/crater/... and possibly going boom )

[GoSlash27]

So when you start with TWR of 0.91, what exactly happens for the first few minutes when you burn with TWR < 1? You should stay in place and just waste fuel. Is that what happened?

Yeah, it's pretty hilarious to watch (and difficult to control). At first it just sits there. Then it slowly picks up every leg but one, and drags/bumps around the surface trying to tip over. Eventually it rises barely off the surface and I can *very gingerly* begin to slowly accelerate east while maintaining altitude.

I dont see 69% or 72% anywhere in that graph that you quoted. Please explain where you are getting these numbers from....

The 72% would be the LFO numbers. v_0/deltaV= 571.8/797 = 71.7% efficiency.

Referring to your contour plot, TVR=4.98 and t/w=1.1 should only yield about 68% efficiency. Please run the numbers and confirm this if you don't mind. Maybe I'm reading the chart incorrectly?

Thanks,

-Slashy

Edited December 26, 2014 by GoSlash27

[GoSlash27]

arkie,

Oh, wait... I think I screwed that up. I shouldn't be using Vo, huh? *facepalm*

You never posted the minimum DV for your final orbit, but it's got to be more than your final Vo, right?

Confusin' myself again,

-Slashy

[Iskierka]

Does the model include effects of sidereal rotation velocity? 9 m/s on its own is only a fraction of the difference, but having initial v_theta may have sufficient knock-on effects to explain the difference.

[GoSlash27]

Does the model include effects of sidereal rotation velocity? 9 m/s on its own is only a fraction of the difference, but having initial v_theta may have sufficient knock-on effects to explain the difference.

I'm thinking the same.

Of course...Tylo has a sidereal rotation of 17.8 m/sec and I wasn't exceeding the prediction there at extremely low t/w. Then again, that would be a much smaller fraction of my orbital velocity as well...

Best,

-Slashy

[arkie87]

Yeah, it's pretty hilarious to watch (and difficult to control). At first it just sits there. Then it slowly picks up every leg but one, and drags/bumps around the surface trying to tip over. Eventually it rises barely off the surface and I can *very gingerly* begin to slowly accelerate east while maintaining altitude.

The 72% would be the LFO numbers. v_0/deltaV= 571.8/797 = 71.7% efficiency.

Referring to your contour plot, TVR=4.98 and t/w=1.1 should only yield about 68% efficiency. Please run the numbers and confirm this if you don't mind. Maybe I'm reading the chart incorrectly?

http://i52.photobucket.com/albums/g13/GoSlash27/CP2B_zps82948542.jpg

Thanks,

-Slashy

I think the way matlab plots it, the beginning of the contour line is the value (you can tell this from the fact that the colors end at 0.9, but clearly, efficiency increases above 0.9 in the far right). Thus, your estimate from the graph is incorrect. The orange contour begins at 0.7 and ends at 0.8 not begins at 0.65 and ends at 0.75.

I have also run the numbers for TWR = 1.1 and TVR = 4.98, and arrived at eta = 0.7326.

arkie,

Oh, wait... I think I screwed that up. I shouldn't be using Vo, huh? *facepalm*

You never posted the minimum DV for your final orbit, but it's got to be more than your final Vo, right?

Confusin' myself again,

-Slashy

Why not??? V0 is correct... ?

[Anquietas314]

The orange contour begins at 0.7 and ends at 0.8 not begins at 0.65 and ends at 0.75.

I think the bit in question is 0.6-0.7 going by the scale on the right (it's the third highest grouping), but then of course I am colourblind. EDIT: Unless the colours are based on the lowest value in the range...

Edited December 26, 2014 by armagheddonsgw

[arkie87]

I think the bit in question is 0.6-0.7 going by the scale on the right (it's the third highest grouping), but then of course I am colourblind.

Maybe my post was vague...

If you were to interpolate for TWR = 2 TVR = anything, what would efficiency be? Less than 0.9 or greater?

The way matlab plots, it is greater. The line of efficiency equals 0.9 is exactly on the border between colors, not in the center of its color...

[arkie87]

What i am trying to say is this:

[Anquietas314]

Maybe my post was vague...

If you were to interpolate for TWR = 2 TVR = anything, what would efficiency be? Less than 0.9 or greater?

The way matlab plots, it is greater. The line of efficiency equals 0.9 is exactly on the border between colors, not in the center of its color...

That makes sense I guess. As I said, colourblind. I can distinguish the borders on the graph, but I can't very accurately compare the colours on the graph with those on the scale.

[arkie87]

That makes sense I guess. As I said, colourblind. I can distinguish the borders on the graph, but I can't very accurately compare the colours on the graph with those on the scale.

Oh, given our previous exchanges, i wasnt sure if you were being sarcastic...

[GoSlash27]

What i am trying to say is this: http://i.imgur.com/hDVHeHr.jpg

Oh, okay! That explains it. So at the moment, I have exceeded the predicted DV and you haven't.

And of course LD won't accept my results.

I propose a test:

1.0 t/w launch from Eeloo using whatever engine.

If my suspicion is correct, I'm thinking everybody's results will exceed the predictions of both models by a wide margin.

Best,

-Slashy

[arkie87]

Oh, okay! That explains it. So at the moment, I have exceeded the predicted DV and you haven't.

And of course LD won't accept my results.

I propose a test:

1.0 t/w launch from Eeloo using whatever engine.

If my suspicion is correct, I'm thinking everybody's results will exceed the predictions of both models by a wide margin.

Best,

-Slashy

First, can you tell me what altitude your Mun tests were conducted at? I can adjust my model to see what prediction is for higher altitude (since effectively TWR increases)

[GoSlash27]

arkie,

I can check and see if my original test launcher is still in the same spot. I never recorded that during the first batch of tests.

If it's not, no worries. I'm gonna rerun the test.

*edit* update: Sorry, the original lander is gone. But as I said, I'm gonna rerun the test.

Best,

-Slashy

Edited December 26, 2014 by GoSlash27

[GoSlash27]

Okay, my Eeloo results are in, and it blows my theory out of the water. Looking at the contour plot, it *precisely* matched what Arkie's model predicts.

Initial state: 2.35t launch vehicle with 4kN/ 290s Isp engine. t/w=1.01

alt=1,971m time= 04:00

End state:

M=1.71t V=568.7m/sec alt=20,048m time=31.48

DV consumed= 905m/sec

efficiency= .629

et= 27:48

If I'm reading the table correctly, arkie's model predicts 63% efficiency, so right down the middle. Whatever the deal is, it doesn't seem to be the sidereal rotation.

Best,

-Slashy

[arkie87]

Okay, my Eeloo results are in, and it blows my theory out of the water. Looking at the contour plot, it *precisely* matched what Arkie's model predicts.

Initial state: 2.35t launch vehicle with 4kN/ 290s Isp engine. t/w=1.01

alt=1,971m time= 04:00

End state:

M=1.71t V=568.7m/sec alt=20,048m time=31.48

DV consumed= 905m/sec

efficiency= .629

et= 27:48

If I'm reading the table correctly, arkie's model predicts 63% efficiency, so right down the middle. Whatever the deal is, it doesn't seem to be the sidereal rotation.

Best,

-Slashy

I have computed for this case:

TVR: 9.81*290/sqrt(1.69*210000) = 4.775

Running model for TWR = 1.01, TVR = 4.775 yields: efficiency = 0.6455

So you are slightly below, as expected, since let's be honest, you are no Jeb

[Anquietas314]

let's be honest, you are no Jeb

In fairness, Jeb's famous more for enjoying a wild ride than being an expert pilot (mechjeb aside)

[GoSlash27]

I ran a new mun test, and this time I was *way* under the model's prediction (which is good).

At launch:

M=2.43t T=4kN Isp=290s t/w=1.01 alt=2,361m t=30.00

At circularization:

M=1.70t Vo=547.5 m/sec alt=10,009 t=52:31

DV expended= 1,017 m/sec

efficiency= .538

st= 22:31

According to the contour plot I would've expected 63%, and the terrain wasn't even awful this time.

Hmm...

-Slashy

[GoSlash27]

I have computed for this case:

TVR: 9.81*290/sqrt(1.69*210000) = 4.775

Running model for TWR = 1.01, TVR = 4.775 yields: efficiency = 0.6455

So you are slightly below, as expected, since let's be honest, you are no Jeb

Don't I know it! *rueful grin*...

How did the elapsed time stack up?

-Slashy

[LethalDose]

Well, now that we've sorted out the issue with "centripetal lift", I figured I might put my degree to use and have a crack at equation 17 like OP actually wanted (apologies for the incredibly ugly text notation):

I took the rearranged version from immediately after the (17) label - it seemed easier to work with.
equation 17: exp(deltaVp / (g0 * Isp))(1 + sqrt(1 + TWR0^2))/m0 = (1 + sqrt(1 + (g * m_end)^2)) / m_end
goal: rearrange for m_end.

There's a lot of ugly variables in the way, so let's get rid of most of them to make life way easier:
Let E = exp(deltaVp/(g0*Isp))
    A = E * (1 + sqrt(1 + TWR0^2)) / m0

Substituting: 
A = (1 + sqrt(1 + (g * m_end / F)^2) / m_end
A * m_end - 1 = sqrt(1 + (g * m_end / F)^2)   // multiply both sides by m_end, move the 1 over.
(A * m_end - 1)^2 - 1 = (g * m_end / F)^2    // square both sides, move the 1 over.
A^2 * m_end^2 - 2A * m_end + 1 - 1 = g^2 * m_end^2 / F^2  // expand all the things!
(A^2 - (g/F)^2) * m_end^2 - 2A * m_end = 0  // Behold! we have a pretty quadratic!

But there's still some constants that make it a bit harder to read:
Let B = (A^2 - (g/F)^2)

Substituting:
B * m_end^2 - 2A * m_end = 0
B * m_end - 2A = 0  // we can disregard m_end = 0 since it's not realistic
m_end = 2A / B.
      = 2A / (A^2 - (g/F)^2).  // uh, yeah. you're on your own from here  (it's ugly. very ugly.)

disclaimer: I make no claims as to the correctness of this, and it was all done in notepad using my head.

If however this is indeed correct, WolframAlpha's apparently not very good with big equations with lots of variables

This is awesome, I'll try applying it when I get back home (traveling currently). If this works, it'll make life much easier.

I'm glad the way Iskierka explained it to you made sense. I was trying my best :-(

He just directly explained the terms, without any 'hand-wavy' statements like 'on the other side of the equation it's momentum, not force'

Oh, okay! That explains it. So at the moment, I have exceeded the predicted DV and you haven't.

And of course LD won't accept my results.

I'll accept your results when you provide video. I've made this very clear.

Edited December 28, 2014 by Vanamonde
Snip.

[LethalDose]

Overall, I'm not sure why people here are insisting on using extremely low TWR vessel to test this model. The model shows TWR ~ 1 will take nearly 50% more total dV to liftoff than TWR ~ 3. That's the point.

The low TWR vessels are also clearly violating the underlying assumptions of the model as stated in the write-up, since the vessel are achieving orbital velocities over extremely long distances.

Also, if you're going to try to prove Arkie's model, please do it on his thread.

[GoSlash27]

Overall, I'm not sure why people here are insisting on using extremely low TWR vessel to test this model.

The reason for this is because incorrect models tend to diverge from empirical results the most at low t/w ratios. If it's incorrect at the bottom end, then the slopes will be incorrect at the higher end, and anything inferred from them (such as ideal t/w for a given engine) will yield incorrect answers.

The model shows TWR ~ 1 will take nearly 50% more total dV to liftoff than TWR ~ 3. That's the point.

This would only hold true for a medium-low Isp engine on the Mun. For a low Isp engine on Tylo the DV efficiency at 1.0 t/w can exceed 75%, while a high Isp engine on Pol might only achieve 32%.

Regards,

-Slashy