Vertical Ascent vs. To LXO First

[Superfluous J]

I'm getting sick of repeating myself. This is a theoretical question.

Sorry. Then yes. By burning horizontal instead of vertical you would in a theroretical perfect universe save fuel. In our actual, practical universe the act of turning your ship on the surface before firing your engines imparts enough thrust on your ship to introduce significant errors in your results, enough that they will easily hide any gain.

It's not? I thought it was...

From the wiki: "Gilly's odd shape and fast rotation give the surface near the peaks an apparent velocity near 5 m/s." Also:

Sidereal orbital period 4 d 11 h 56 m 27.4 s

Sidereal rotation period 7 h 50 m 55 s

Edited December 19, 2014 by 5thHorseman

[arkie87]

Aye, but the cliff has to actually *exist*. The Mun has no 10km cliffs (or 10km terrain anywhere, for that matter). Gilly, while lumpier, also has no cliffs anywhere that's a going concern.

You could pick a spot near the pole of nearly any body and find significant cliffs. You won't launch into any important direction, but you can at least test. Otherwise, moholes or an Eeloo canyon. I don't recommend landing in moholes

Cool. I do want to pick your brain about something real quick if you have a moment...

-Slashy

No, the cliff doesnt have to exist. It's theoretical.

Sure. pick away...

[GoSlash27]

Thanks,

Do you have a function that relates DV losses to achieve orbit in terms of escape velocity as a function of initial t/w neglecting drag?

I'm thinking that

1) the theoretical cost to achieve orbit in DV is the energy of the orbit itself/mass minus sidereal rotation

2) the minimum and maximum cost are an inverse function of t/w with escape velocity as a coefficient

and

3) Isp doesn't affect it terribly much in practical ranges.

My thinking is that if I'm looking at the DV loss as an additional DV requirement that I must pay for with fuel and tankage, that the Isp will tend to cancel.

No biggie if you don't, I was planning on testing.

Thanks,

-Slashy

Edited December 19, 2014 by GoSlash27

[GoSlash27]

No, the cliff doesnt have to exist. It's theoretical.

It does if you intend to test

Or... maybe it doesn't. You don't need the cliff to actually exist in order to test it's impact. Just *pretend* it's there.

In that sense, you can test it anywhere you choose.

-Slashy

[arkie87]

It does if you intend to test

Or... maybe it doesn't. You don't need the cliff to actually exist in order to test it's impact. Just *pretend* it's there.

In that sense, you can test it anywhere you choose.

-Slashy

That is very true-- good point. Though, given my track record, i feel like if i did that test, people would tell me that I didnt need to climb up to 10 km, misunderstanding that i'my pretending there was a cliff in the way...

But like i said. I'm sick of debating by now. I wan't to just play. And kill Kerbals. The way the game was meant to be played.

Maybe after i play through everything 0.9 has to offer, i'll revisit

Thanks,

Do you have a function that relates DV losses to achieve orbit in terms of escape velocity as a function of initial t/w neglecting drag?

I'm thinking that

1) the theoretical cost to achieve orbit in DV is the energy of the orbit itself/mass minus sidereal rotation

2) the minimum and maximum cost are an inverse function of t/w with escape velocity as a coefficient

and

3) Isp doesn't affect it terribly much in practical ranges.

My thinking is that if I'm looking at the DV loss as an additional DV requirement that I must pay for with fuel and tankage, that the Isp will tend to cancel.

No biggie if you don't, I was planning on testing.

Thanks,

-Slashy

Hmm, Idk if i can give it to you in that form.... but,

Did you see this: http://forum.kerbalspaceprogram.com/threads/102947-Vertical-Ascent-vs-To-LXO-First?p=1612130&viewfull=1#post1612130

The final form of the dimensionless equation is very useful. And the contour plots it can provide are applicable to ANY craft taking off from that planet with that TWR, FMR, and ISP (though i think the contour plot i provided with it is wrong... im still working on the objective function a bit). I can definitely use the model to provide you with contour plots of extra (dimensionless) velocity requirements. btw: the normalizing factor for velocity isnt escape, it's orbital, for convenience (and since one you are in orbit, TWR doesnt matter...)

Incidentally, the dimensionless form indicates the need to define FMR (fuel to mass ratio i.e. fuel mass to total mass), which Rocketeer used.

I'll probably start a new thread once I am sure about the results. I have asked numerobis to verify my work... but like i said, i'm pretty sure my objective function for fuel is wrong, since TWR * dimensionless time is not fuel, but dimensionless fuel. I need to multiply by t0 to turn it into actual fuel... and when i do that, the graphs all change...

Hopefully i'll post soon, but probably in the science labs, not here.

Edited December 19, 2014 by arkie87

[GoSlash27]

Interesting!

I'll have to spend some time on this.

If you do start up this discussion, I'm game to participate and share my results.

My goal is to pin down an equation that accounts for this in terms of fuel.

Thanks and have fun!

Best,

-Slashy

[arkie87]

Interesting!

I'll have to spend some time on this.

If you do start up this discussion, I'm game to participate and share my results.

My goal is to pin down an equation that accounts for this in terms of fuel.

Thanks and have fun!

Best,

-Slashy

Ok. so there were a lot of bugs/errors in my program. Do not look at that first graph.

Almost positive it is correct now after reviewing and checking everything. I also came up with a definition of "efficiency" which is basically deltaV_min/deltaV_spent, so, if you have an efficiecny of 90% with your craft, an orbital velocity is 900 m/s, then you need 1000 m/s. Pretty straightforward calculation, and dimensionless (meaning it applies to all craft/planets etc... with the same dimensionless or relative parameters).

LOTS of interesting things from new graphs...

Will start a new thread in Science Lab, and paste a link here so you can find it

[arkie87]

http://forum.kerbalspaceprogram.com/threads/103890-Non-Dimensional-Model-for-Optimal-Launch-Efficiency?p=1613079#post1613079

[arkie87]

Actually, i dont know how we missed this: Minmus actually seems like a perfect candidate.

It is not tidally locked (from what i can tell), so you can wait until you are on trailing edge. Its peak elevation is 5725 m out of 60km radius, and it has sharp hills/cliffs, such that you might have to travel vertically before you can go horizontally at all. For impulse burns, i calculated 25% savings from vertical approach. For practical TWR, I will need to see what LethalDose's model predicts (or ill have to make my own), and/or test it in KSP.

[GoSlash27]

As an aside, I think I've got it modeled as a function, and it's ridiculously simple:

Gravity loss(m/sec) ≈ ÃŽâ€v/2Rtw^4

where ÃŽâ€V= theoretically "perfect" ÃŽâ€V to orbit (I estimate it at approx. 612 m/sec for the mun)

and Rtw= t/w ratio at launch.

Could you do me a favor and see how this compares?

I ran a test launcher using o-10 engines in selectable banks and checked the DV expenditure at various t.w ratios. Here's the results as compared to the model:

Initial t-w/predicted DV/actual DV

0.91 / 1,058 / 1,103

1.84 / 639 / 657

3.68 / 614 / 630

7.37 / 612 / 625

Best,

-Slashy

*edit* Formula had a typo. Corrected now.

Edited December 21, 2014 by GoSlash27
Typo

[arkie87]

As an aside, I think I've got it modeled as a function, and it's ridiculously simple:

Gravity loss(m/sec)= ÃŽâ€v/2Rtw^2

where ÃŽâ€V= theoretically "perfect" ÃŽâ€V to orbit (I estimate it at approx. 612 m/sec for the mun)

and Rtw= t/w ratio at launch.

Could you do me a favor and see how this compares?

I ran a test launcher using o-10 engines in selectable banks and checked the DV expenditure at various t.w ratios. Here's the results as compared to the model:

Initial t-w/predicted DV/actual DV

0.91 / 1,058 / 1,103

1.84 / 639 / 657

3.68 / 614 / 630

7.37 / 612 / 625

Best,

-Slashy

It's close; red 'o's is your correlation.

[GoSlash27]

It's close; red 'o's is your correlation.

http://i.imgur.com/Bb1wNQb.jpg

The red dots aren't located quite correctly for my formula. Here's the corrected graph for my formula.

efficiency ≈ 1-[1/2R_tw⁴]

My equation is completely independent of all factors other than Rtw (sorry "TWR" by your standard), so it'll read the same on Gilly as it would on Tylo.

Are there any airless bodies where it's out of the ballpark?

Thanks,

-Slashy

Edited December 20, 2014 by GoSlash27

[GoSlash27]

^ we could be using different different definitions of "efficiency", tho'.

How do you define "efficiency" in your model? By mine, it's the ratio of achieved ÃŽâ€V to ideal ÃŽâ€V.

[LethalDose]

Alright, this is written as non-confrontationally as possible.

I think it's massively unfair, and frankly hypocritical, for you to present any kind of equation, model, or conclusion without a substantial amount proof and/or derivation behind what you're presenting.

After posting this:

As for LD's analysis, the holes in his thinking are threefold:

1) fuel spent fighting gravity decreases as horizontal velocity approaches orbital due to centripetal lift

2) The t/w of the vehicle increases over time as fuel is consumed.

and

3) The pilot is going to vary the pitch of the vehicle over the duration of the launch, not keep it at a fixed angle.

Best,

-Slashy

I don't see how you can post this:

As an aside, I think I've got it modeled as a function, and it's ridiculously simple:

Gravity loss(m/sec)= ÃŽâ€v/2Rtw^2

This accounts for none of what you claim invalidated my models, statements, and conclusions.

Even though I provided complete transparency in regards to the math behind my position, you rejected them all as flawed without providing contrary evidence. Here, you barely make any effort to support novel equations.

The equation is wrong. It does not accurately represent the relationship between TWR and "wasted" dV, even if it does fit 4 data points.

I don't see why you expect the community to accept this level of evidence for anything you say after your condescension and incorrect rejection of vastly better evidence in this thread.

I've put in the effort to actually derive the relationship between TWR and lost dV, and it's nowhere near as simple as you present. As soon as I get LaTex figured out, I'll publish the method.

[Bryce Ring]

Not sure if this would help or not.

but...

Does anyone know how to stop Kerbin from spinning. (not sure if it is required for your tests, see Pictures)

Bryce.

Edited December 20, 2014 by Bryce Ring

[GoSlash27]

LD,

Frankly, I'm surprised that you would expect me to respond to anything you have to say at this point, even if you *are* trying to be "non-confrontational".

There's only so much rude behavior people will tolerate before they say "you know what... I'm really better off just not interacting with this guy"

Nevertheless, I will *this one time* to say that:

1) You actually do have a point here which I wouldn't mind discussing... with somebody else.

2) There was a breakdown in communication that caused your model to either be misrepresentedby arkie or misinterpreted by myself and several others.

3) Do not expect any further responses from me. I have absolutely zero interest in interacting with you. I will happily glean all the knowledge I can from your posts (you are, after all, a very brainy guy), comment on points that you raise, and report posts that violate forum policy. I will not respond to posts addressed to me or address any further posts to you.

Apologies,

-Slashy

[GoSlash27]

Arkie,

Since LD misunderstood my intent here, you may well have misunderstood it also due to a failure on my end.

I'm not trying to attack your model as incorrect or claim that mine is correct. In fact, it's the opposite; I have no reason to think that your model isn't correct, which is why I'm asking you to compare them for me.

I'm only trying to verify that my equation produces a reasonable estimate of DV losses as a function of t/w for calculation purposes. My formula really should be using the "approximately" symbol. I'll see if I can update it...

You're seeking "exact accuracy from a model", while I'm just looking for "good enough for government work from a formula".

Good morning!

-Slashy

Edited December 20, 2014 by GoSlash27

[arkie87]

The red dots aren't located quite correctly for my formula. Here's the corrected graph for my formula.

http://i52.photobucket.com/albums/g13/GoSlash27/graphcorrect1_zpsaa8cc119.jpg

efficiency ≈ 1-[1/2R_tw⁴]

My equation is completely independent of all factors other than Rtw (sorry "TWR" by your standard), so it'll read the same on Gilly as it would on Tylo.

Are there any airless bodies where it's out of the ballpark?

Thanks,

-Slashy

The exponent in your graph changed from 2 to 4... i graphed with a 2... you graphed with a 4...

That said, 4 looks better

Arkie,

Since LD misunderstood my intent here, you may well have misunderstood it also due to a failure on my end.

I'm not trying to attack your model as incorrect or claim that mine is correct. In fact, it's the opposite; I have no reason to think that your model isn't correct, which is why I'm asking you to compare them for me.

I'm only trying to verify that my equation produces a reasonable estimate of DV losses as a function of t/w for calculation purposes. My formula really should be using the "approximately" symbol. I'll see if I can update it...

You're seeking "exact accuracy from a model", while I'm just looking for "good enough for government work from a formula".

Good morning!

-Slashy

It's definitely a reasonable approximation. But my model predicts it varies with FMR and TVR as well. We should probably continue this exclusively in the science lab post before people get confused.

^ we could be using different different definitions of "efficiency", tho'.

How do you define "efficiency" in your model? By mine, it's the ratio of achieved ÃŽâ€V to ideal ÃŽâ€V.

I define it the same way: ideal i.e. minimum deltaV/used i.e. required deltaV

I think the way you wrote it is backwards since ideal deltaV <= achieved deltaV always.

Edited December 20, 2014 by arkie87

[GoSlash27]

The exponent in your graph changed from 2 to 4... i graphed with a 2... you graphed with a 4...

That said, 4 looks better

What the... what?? *D'OH!* Typo. I'll fix it.

It's definitely a reasonable approximation. But my model predicts it varies with FMR and TVR as well. We should probably continue this exclusively in the science lab post before people get confused.

I define it the same way: ideal i.e. minimum deltaV/used i.e. required deltaV

I think the way you wrote it is backwards since ideal deltaV <= achieved deltaV always.

[Vanamonde]

The discussion in this thread had become and stayed rather personal, and it's time to close things down and allow tempers to cool. Those of you who would like to discuss the physics behind the question, and can do so without attacking each other personally, can start a new thread in a day or two. But for now, please turn your attention to other aspects of the game.