Effect of initial TWR on orbit dV cost

[LethalDose]

LethalDose: you've switched coordinate frames. Your OP's coordinate frame was thrusting at a given angle above the horizon, so you were in a rotating frame of reference. Your latest is using a fixed coordinate frame. In a rotating frame of reference, you need a centrifugal term, as arkie pointed out. In a fixed coordinate frame, you don't have centrifugal terms.

First, if you can provide a resource that discusses this issue, I'll happily read it. Not a partially legible scan of some equations where it's impossible to get context or understand what's being presented.

But if all you have is more "you're wrong", you're wasting your time.

Second, I don't see how you can claim the OP is any kind of rotational frame of reference. Burns described take place over a distance of less than 50 km, and usually much less than that. Traveling even 50 km over the surface at sea level leads to practically no deviation from a linear frame of reference, the diameter of the Mun at sea level is over 1,200 km (400 pi km), in which case there is at most a 15 degree change in horizon during the burn. This is means theres less than a 2,000 m deviation from a straight shot over the burn course (again, over 50 km long). I fail to see how any substantial amount of angular momentum (which I think is the real source of "centripetal lift") can occur over that burn. What I'm presenting is basically just the linear frame of reference, and as you state doesn't require a "centrifugal term".

It's basically just a raw linear prograde burn to reach a particular prograde velocity.

Third, what Arkie presented just shoves this term into the equation, demonstrating no rigor or derivation. We simply have to accept it's presence as fact, which is antithetical to this discussion. On top of that, I can barely read what actually is presented due to the small size and poor resolution. Again, I'd really love to discuss what was presented there, but it's not possible given the information available.

Fourth, from what I can glean from the meager information presented about the rotational inertial frame, you would need to intergrate the entire fight path over dR/dt up to AP to determine altitude, and the initial burn would need be expressed in delta theta (radians/sec)... We don't do anything in these terms. You could back convert from polar cordinates to linear terms, but again, I don't see any kind of discussion about this in what arkie's presented. And, again, maybe it's there and I just can't make it out.

I think I get that you need a centifugal term in for this, since dr/dt = 0 in a circular orbit, and you need something to offset the gravity term, but again, there needs to be way more information given about this method before I can draw any conclusions.

And finally, I've got no problem with this over-estimating cost, and under-estimating efficiency, as this is designed to be lower bound for engineering purposes.

Anyway, I'll to sum up: please provide some worthwhile resources to support your statements.

[GoSlash27]

And finally, I've got no problem with this over-estimating cost, and under-estimating efficiency, as this is designed to be lower bound for engineering purposes.

LD's prediction 893m/sec <-- "lower bound"

Total DV to escape 871m/sec <-- *actual result* under less than ideal circumstances

That's a worthwhile resource...

[Anquietas314]

[some data contradicting LD's model]

Having quickly read through LD's paper, there is something that it appears hasn't been accounted for: g (acceleration due to gravity) is not constant with altitude (I may have simply missed it). For the purposes of the model (constant altitude during burn) that's not really an issue, but for testing it empirically (at least in KSP) it is: g drops pretty rapidly with altitude in KSP. It could be enough to explain your data, though I haven't run the numbers (out of laziness, if I'm honest).

EDIT: I just looked at your screenshots; you didn't actually reach true escape velocity going by that orbit curve; you only reached the speed necessary to leave Mun's SOI (this is considerably smaller due to Kerbin's gravity well). That would also be a factor. The orbit should be a parabola or a hyperbola if you've exactly reached or exceeded escape velocity respectively. Your orbit appears to be a cropped ellipse.

Edited December 25, 2014 by armagheddonsgw

[numerobis]

g is constant if you maintain constant altitude, which is what you do if you offset gravity exactly right.

edit: nevermind, you were talking about the experiment.

[GoSlash27]

Having quickly read through LD's paper, there is something that it appears hasn't been accounted for: g (acceleration due to gravity) is not constant with altitude (I may have simply missed it). For the purposes of the model (constant altitude during burn) that's not really an issue, but for testing it empirically (at least in KSP) it is: g drops pretty rapidly with altitude in KSP. It could be enough to explain your data, though I haven't run the numbers (out of laziness, if I'm honest).

No, Sir (and merry Christmas!)

The discrepancy between the two models isn't at an altitude where the reduction in gravity is a significant factor. It's in the early phase of the launch where low t/w penalizes the most.

Best,

-Slashy

[Anquietas314]

No, Sir (and merry Christmas!)

The discrepancy between the two models isn't at an altitude where the reduction in gravity is a significant factor. It's in the early phase of the launch where low t/w penalizes the most.

Best,

-Slashy

Considering your result only falls short of the model by a little under 3%, call it 5% accounting for pilot error etc, it may just be enough, especially on top of the part about not reaching escape velocity (see my edit to the post you quoted)

[Greep]

Pilot error would actually make it worse, not better.

Edit nvm point B :/

Edited December 25, 2014 by Greep

[GoSlash27]

Considering your result only falls short of the model by a little under 3%, call it 5% accounting for pilot error etc, it may just be enough, especially on top of the part about not reaching escape velocity (see my edit to the post you quoted)

Arma,

If the model is correct, I should not be able to "fall short" of it under any circumstances.

It should always take me more DV than the model predicts. never less. Especially with terrain in the way.

As for the difference between escape velocity and velocity sufficient to escape SOI, I'd need a clarification on the difference. I was always under the impression that they were the same thing.

Best,

-Slashy

[GoSlash27]

Well 2 things:

A)Pilot error would actually make it worse, not better.

With such a small altitude change from sea level you could just use the crude estimate of dvloss = avg velocity * g / distance which is only a few dV, not a few dozen.

So I think it's either ksp that's wrong or the equation.

(emphasis mine)

Valid point. It could be KSP that's off.

But since the entire point of this is to provide a useful lower bound for KSP, it's clearly the model at fault.

I've tested Arkie's model everywhere from Gilly to Tylo at Isps ranging from 290 to 4,200 and haven't recorded any errors in it beyond the margin of error.

If it's wrong, it's wrong in a way that accurately predicts the results in KSP.

Best,

-Slashy

[numerobis]

As for the difference between escape velocity and velocity sufficient to escape SOI, I'd need a clarification on the difference. I was always under the impression that they were the same thing.

Escape velocity is the speed you need to go so that gravity never gets your speed down to zero.

Speed to escape SoI is the speed you need so that gravity gets your speed to zero exactly when you get out of the SoI.

For Mun at 0m altitude, 773 m/s gets you out of the SoI, but you need 807 m/s to escape.

[Anquietas314]

Arma,

If the model is correct, I should not be able to "fall short" of it under any circumstances.

It should always take me more DV than the model predicts. never less. Especially with terrain in the way.

It's possible to do it in less deltaV when the circumstances include dropping g (since this is a major factor in the efficiency of your ascent - TWR is derived from it).

As for the difference between escape velocity and velocity sufficient to escape SOI, I'd need a clarification on the difference. I was always under the impression that they were the same thing.

In the real world yes, technically. In theory, escape velocity is defined as the minimum velocity required to make an object "go to infinity", under the assumption there are only 2 objects: the thing with a note-worthy gravity well, and the thing the escape velocity applies to. The wikipedia page on it

[LethalDose]

Arkie,

Truth through empirical testing.

I built a ship as close to spec as I could make it and had a go at 1.5:1 t/w. Any test involving a "sea level" launch on the mun necessitates terrain avoidance, so my results were far from ideal.

Anywho...

Initial mass = 2.01t

initial t/w =1.51

launch altitude 822m

elapsed time 5:40

Total DV to escape 871m/sec

LD's prediction 893m/sec

Arkie's prediction 856 m/sec

LD's model doesn't predict that I can achieve 871m/sec to escape at all, let alone with terrain in the way. Arkie's does, so that oughtta settle that.

Screen caps

Merry Christmas,

-Slashy

A few notes here:

1. You're not welcome here.
   
2. No, seriously, go away.
   
3. I will tolerate you if you play nice. I will report you if you take one step out of line to anyone else participating here.
   
4. I'm not going to accept any evidence you provide without video. I've stated this before... though I may have removed it. Regardless, vids or it didn't happen.
   
5. The pictures you presented are useless.
   
6. Your useless pictures show you're on the edge of a crater, so you could have gone more horizontally than predicted, and lost some altitude, which would produce a more efficient burn.
   
7. You're misrepresenting my method, which actually predicts the req'd dV to be 885.5883.9 m/s (solver found end mass to be 1.55422461391276t; Struck previous value because I found an error in the spreadsheet that wasn't accounting for altitude above sea level). Though this is still greater than what you've posted. Still, I literally provided the tools and instructions on how to get this.
   
8. If use m0 = 2.014 t (rounds to 2.01) and m1 = 1.5551 (rounds to 1.56), Tsiolkovsky gives 888.7 m/s of dV, which exceeds my bound.
   
9. Also, leave.
   

So, I don't see this as anything that can't be explained in a way that is still consistent with the method I've produced. Given that it looks like you've hyperedited the vessel to exactly 0 deg 0' 0" N, and you have otherwise useless MP on the vessel, I'm guessing there's mass/balast shenanigans going on to pad your actual numbers. Eitherway, that small of a difference is easily attributable to round-off error. Again, unedited videos or I assume you're lying.

All that being said, the whole model is predicated on the assumption that straight horizontal along the surface is singular optimal method for escape. This may not be as true as we believe. I'd be willing to bet that over very long burns, pitching up just barely from horizontal may lead to minuscule gains in efficiency, but it's really not worth considering, the model is complex enough as it stands.

Having quickly read through LD's paper, there is something that it appears hasn't been accounted for: g (acceleration due to gravity) is not constant with altitude (I may have simply missed it). For the purposes of the model (constant altitude during burn) that's not really an issue, but for testing it empirically (at least in KSP) it is: g drops pretty rapidly with altitude in KSP. It could be enough to explain your data, though I haven't run the numbers (out of laziness, if I'm honest).

It's stated in the opening paragraphs that the burn occurs perfectly horizontally with 0 change in vertical altitude. No change in vertical altitude translates to no change in local gravity. It's not going to happen that way, mind you, it's just what is presumed to be the most efficient burn pattern.

Addendum it also vastly simplifies the model, since I only need to track travel in a single axis, and Phi is a function of a single variable (mass), instead of two (mass and gravity). It also doesn't drop quite as quickly as you may think. g at 20km above the munar surface is about 1.35 m/s², which is still ~ 80% of sea level.

EDIT: I just looked at your screenshots; you didn't actually reach true escape velocity going by that orbit curve; you only reached the speed necessary to leave Mun's SOI (this is considerably smaller due to Kerbin's gravity well). That would also be a factor. The orbit should be a parabola or a hyperbola if you've exactly reached or exceeded escape velocity respectively. Your orbit appears to be a cropped ellipse.

I usually use a target Ap of the SoI edge instead of an actual escape velocity, so what he did was consistent with what I've presented.

Edited December 25, 2014 by LethalDose

[GoSlash27]

Escape velocity is the speed you need to go so that gravity never gets your speed down to zero.

Speed to escape SoI is the speed you need so that gravity gets your speed to zero exactly when you get out of the SoI.

For Mun at 0m altitude, 773 m/s gets you out of the SoI, but you need 807 m/s to escape.

Numerobis,

Thanks for the explanation, but now I'm afraid I'm at a loss.

If 773 gets you to the edge of SOI with zero velocity and 807 causes an escape, what happens in between?

Say, at 790 m/sec?

If you leave the SOI with 17 m/sec velocity, why would you not continue to drift away since there's no gravity to bring you back?

Scratchin' mah head

-Slashy

[Anquietas314]

It's stated in the opening paragraphs that the burn occurs perfectly horizontally with 0 change in vertical altitude.

Oh I did notice that, hence the "(constant altitude during burn)", but for Slashy's test it would actually be relevant.

If you don't mind me asking, what's the deal with you and Slashy? It seems odd that you'd have such a negative response simply to him(?) posting in the thread.

- - - Updated - - -

If you leave the SOI with 17 m/sec velocity, why would you not continue to drift away since there's no gravity to bring you back?

Because Kerbin's gravity would take over.

[LethalDose]

Note on comparison with "empirical data", the actually predicted dV is 883.9 (just over 1 m/s change, but it's still there). The spreadsheet wasn't accounting for altitude above sea level. The updated file is here, and the original page has been updated.

If you don't mind me asking, what's the deal with you and Slashy? It seems odd that you'd have such a negative response simply to him(?) posting in the thread.

I don't mind you asking if you don't mind me not answering. I don't feel like getting this thread locked as well because I linked to the one of the multitude of threads that, generally or specifically, demonstrates the font of my ire.

Edited December 25, 2014 by LethalDose

[GoSlash27]

A few notes here:

1. You're not welcome here.
   
2. No, seriously, go away.
   
3. I will tolerate you if you play nice. I will report you if you take one step out of line to anyone else participating here.
   
4. I'm not going to accept any evidence you provide without video. I've stated this before... though I may have removed it. Regardless, vids or it didn't happen.
   
5. The pictures you presented are useless.
   
6. Your useless pictures show you're on the edge of a crater, so you could have gone more horizontally than predicted, and lost some altitude, which would produce a more efficient burn.
   
7. You're misrepresenting my method, which actually predicts the req'd dV to be 885.5883.9 m/s (solver found end mass to be 1.55422461391276t; Struck previous value because I found an error in the spreadsheet that wasn't accounting for altitude above sea level). Though this is still greater than what you've posted. Still, I literally provided the tools and instructions on how to get this.
   
8. If use m0 = 2.014 t (rounds to 2.01) and m1 = 1.5551 (rounds to 1.56), Tsiolkovsky gives 888.7 m/s of dV, which exceeds my bound.
   
9. Also, leave.
   

So, I don't see this as anything that can't be explained in a way that is still consistent with the method I've produced. Given that it looks like you've hyperedited the vessel to exactly 0 deg 0' 0" N, and you have otherwise useless MP on the vessel, I'm guessing there's mass/balast shenanigans going on to pad your actual numbers. Eitherway, that small of a difference is easily attributable to round-off error. Again, unedited videos or I assume you're lying.

All that being said, the whole model is predicated on the assumption that straight horizontal along the surface is singular optimal method for escape. This may not be as true as we believe. I'd be willing to bet that over very long burns, pitching up just barely from horizontal may lead to minuscule gains in efficiency, but it's really not worth considering, the model is complex enough as it stands.

I read all of this as "wharrgarbl you must be lying".

Of course, anybody can reproduce my results.

The vehicle has exactly 8 parts.

1 radioisotope generator

1 command pod (full RCS propellant)

1 RCS tank loaded to 70% for ballast

1 FL-T100 tank (full)

1 48-7S engine set to 16.5% thrust limit

3 cubic octagonal struts to serve as legs.

Any impartial bystander can see that this craft lacks the means to dispose of or add propellant, as all the parts are tallied in the map view and visible in the craft view. The monopropellant is merely to add mass to total 2 tonnes.

Sorry dude, but I've got better things to do than mislead you. Your model is inaccurate.

I invite anyone who's interested to try it for themselves.

-Slashy

Edited December 25, 2014 by GoSlash27

[Anquietas314]

I don't mind you asking if you don't mind me not answering. I don't feel like getting this thread locked as well because I linked to the one of the multitude of threads that, generally or specifically, demonstrates the font of my ire

No no, that's fine I'm sure if I dug enough I'd find one of said threads, but I'm too lazy for that so I'll chalk it up to "unpleasant history"

It also doesn't drop quite as quickly as you may think. g at 20km above the munar surface is about 1.35 m/s2, which is still ~ 80% of sea level.

Well, that's still pretty darn fast. As a *very* rough approximation, you need to be about 5km above sea level just to be fairly safe that you're not going to smash into the side of a hill, which should put g somewhere in the region of 95% of sea level. That means around 5% more TWR which could be enough to explain ~2-3% improved efficiency.

Edited December 25, 2014 by armagheddonsgw

[GoSlash27]

- - - Updated - - -

Because Kerbin's gravity would take over.

Ah, but Kerbin's gravity will take over regardless. Does it take over in such a way as to return you to the mun?

Actually, nevermind. LD has already confirmed that my method was in line with his model.

Best,

-Slashy

[Anquietas314]

Ah, but Kerbin's gravity will take over regardless. Does it take over in such a way as to return you to the mun?

Just to wrap this up: this post provided a link to the technical definition of "escape velocity". It seems like you might have missed it. As to your question, there are escape paths which can return you to Mun's SoI shortly after leaving it (in particular, ones that go further from Kerbin and towards Mun's prograde vector). I've had a few unexpected surprises with that back when I wasn't very good at plotting my route home

[numerobis]

Escape velocity is a theoretical concept; doesn't have much relevance in KSP except for Kerbol escape.

[GoSlash27]

As a *very* rough approximation, you need to be about 5km above sea level just to be fairly safe that you're not going to smash into the side of a hill, which should put g somewhere in the region of 95% of sea level. That means around 5% more TWR which could be enough to explain ~2-3% improved efficiency.

Arma,

My burn was done long before I had reached 5kM altitude.

As I said, the inefficiency of low t/w happens at launch, not at altitude. LD will (grudgingly) concur on this point.

The error is in how the craft behaves in the early stages, not the latter.

The problem is that his model doesn't take into account the centrifugal force generated by angular velocity. Arkie's does, and that's what makes the difference.

Arkie's model doesn't take into account the sea level radius of the body or effects of launching above/ below that altitude, and neither model takes into account the body's axial rotation. (unless I missed something in the math)

But that centrifugal force is a pretty big deal. That's why arkie's model works and LD's doesn't.

Best,

-Slashy

Edited December 25, 2014 by GoSlash27

[LethalDose]

Well, that's still pretty darn fast. As a *very* rough approximation, you need to be about 5km above sea level just to be fairly safe that you're not going to smash into the side of a hill, which should put g somewhere in the region of 95% of sea level. That means around 5% more TWR which could be enough to explain ~2-3% improved efficiency.

Well, it's not even 2-3% efficiency, it's closer to 1.5%. If we assume that the numbers he reported were perfectly accurate and there's no roundoff, he burned 871.1 m/s, and my method predicted 883.9 m/s, that's < 1.5%.

But let's not quibble over that, it's freaking close.

In theory It would be more efficient to burn higher. It's well established dV for escape drops for circular orbits as altitude increases. No debate there. However, you have to get to that higher altitude, first, and if you're circularizing up there, you are, in theory "wasting" a very small amount of dV during the circularization. And if you're burning at AP, you're burning where your orbital velocity is lowest, which means it's the least efficient place to transfer chemical potential energy (fuel) into kinetic energy (speed).

Hence, it's thought that it's just more efficient to burn from your starting altitude to "maximize the Oberth effect" by burning low in a gravity well when possible. There's some merit there, but overall, I think the changes are really incredibly minor.

While I'm willing to chalk the differences posted above up to round-off error, it's still evidence enough that there may be a more efficient way. Since us philistines aren't worthy of seeing video evidence of what he did... /shrug we can't base it on anything more than conjecture, and I'm more than happy with my method for what it's intended to be: a lower bound for engineering guidance in the VAB.

[Anquietas314]

My burn was done long before I had reached 5kM altitude.

Given your screenshots only show you on the surface and at 12,665m (map view info tab), some proof of that would be good. (Also, it's 5:30am here I should probably be off to bed >_>)

[numerobis]

The theoretical error was pointed out right away at the start. If you follow LethalDose's algorithm you'll gain altitude, because the thrust angle doesn't take account of horizontal speed.

[LethalDose]

Arma,

My burn was done long before I had reached 5kM altitude.

As I said, the inefficiency of low t/w happens at launch, not at altitude. LD will (grudgingly) concur on this point.

The error is in how the craft behaves in the early stages, not the latter.

Best,

-Slashy

Are you still here?

Don't put words or tone in my mouth. There's nothing "grudging" about what is established fact. I've never denied the inefficiency was early in the flight, or stated anything of the kind.