Effect of initial TWR on orbit dV cost

[Superfluous J]

GoSlash,

Escape Velocity is the speed you need to go so that, ASSUMING MUN WAS THE ONLY BODY IN EXISTENCE, you would fly away from it and never return.

The only body in KSP that you can actually see this with is Sun. Even though its reach goes on forever, there is a speed you can go so that your orbit goes from an ellipse to hyperbolic. At that point, you are at escape velocity.

[LethalDose]

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.

First off, still waiting for any kind of resource that supports your statements. Would love to read it. Seriously.

Second, yes, if you follow my algorithm, you'll gain altititude. In fact, you'll get as high as the SoI boundary (or whatever you set your target AP to). That's the point of the exercise.

Finally, as I said before, during the very short period of thrust to reach target velocity in the vast majority of instances, no meaningful horizontal velocity will be gained.

Phi is not dependent on speed. If you're not accelerating upward, then you're not losing weight due to decreased gravity. It's that simple. It has nothing to do with speed. A car with a full tank at sea level going 10 mph will weigh the same amount as a car going 100 mph with the same amount of gas at sea level.

If you want to provide any supporting evidence for the math behind your magical lift, I'm willing to read it.

[GoSlash27]

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.

^ "Grudging"

So there we have it. Grudging concurrence.

The model doesn't predict the outcome, and is therefore flawed. It doesn't serve as a lower bound if empirical results wind up under it. Arkie's model has not been outperformed by empirical results (at least not within the margin of error) and is therefore superior.

Really all that needs to be said on the subject.

-Slashy

[GoSlash27]

Phi is not dependent on speed. If you're not accelerating upward, then you're not losing weight due to decreased gravity. It's that simple. It has nothing to do with speed. A car with a full tank at sea level going 10 mph will weigh the same amount as a car going 100 mph with the same amount of gas at sea level.

False, and this is the crux of the matter.

When going orbital velocity, you "weigh" nothing. With zero angular velocity, you "weigh" appropriately to the gravitational field. Between the two, you "weigh" less in proportion to the angular velocity.

If what LD states is true, we'd have lots of pictures of astronauts walking around on the ISS instead of floating.

The misconception is a core part of his model, and is why it doesn't work.

[LethalDose]

^ "Grudging"

So there we have it. Grudging concurrence.

The model doesn't predict the outcome, and is therefore flawed. It doesn't serve as a lower bound if empirical results wind up under it. Arkie's model has not been outperformed by empirical results (at least not within the margin of error) and is therefore superior.

Really all that needs to be said on the subject.

-Slashy

Had you provided any credible evidence that couldn't also be easily explained as round-off error, you'd have a point.

You've still failed on both points though. The model stands.

Also, if you had actually read any of what I've written in the last hour, you'd see I've acknowledged there may be more efficient methods. You're sure as crap not doing anything to help, though, since you won't produce any credible evidence of what actually happened.

[Greep]

Soo... anyways, I tried looking up the part masses of GoSlashy's ship and... couldn't find an RCS tank that has 100 capacity? I'm just a little confused, wondering if this is a modded part or if I'm just missing something obvious.

[GoSlash27]

Had you provided any credible evidence that couldn't also be easily explained as round-off error, you'd have a point.

You've still failed on both points though. The model stands.

Also, if you had actually read any of what I've written in the last hour, you'd see I've acknowledged there may be more efficient methods. You're sure as crap not doing anything to help, though, since you won't produce any credible evidence of what actually happened.

What LD will accept <> "credible". "Repeatable" is the scientific standard.

All anyone has to do to confirm or reject my results is repeat the experiment.

[GoSlash27]

Soo... anyways, I tried looking up the part masses of GoSlashy's ship and... couldn't find an RCS tank that has 100 capacity? I'm just a little confused, wondering if this is a modded part or if I'm just missing something obvious.

FL-R25.

*edit* Hmm... now that you mention it, I've got 2 different RCS tanks in my list with that name.

One has a capacity of 100 (.55t), while the other has a capacity of 250 (1.15t)

Must've been a cleanup error during an update?

Shouldn't matter so long as you have a 2t ship with a 48-7S set to 16.5% throttle.

Best,

-Slashy

Edited December 25, 2014 by GoSlash27

[Greep]

Yeah mine and wiki shows 250 and tweaking the fuel still has it show 260 for "full" capacity in the ship information :/ Is yours edited to have 100 or something?

[LethalDose]

If what LD states is true, we'd have lots of pictures of astronauts walking around on the ISS instead of floating.

The misconception is a core part of his model, and is why it doesn't work.

... Dude, now you're just embarrassing yourself.

Let's look at what I've said, in this thread:

The "illusion" of weightlessness in spaceflight is caused by the fact that everything around you is also plummeting towards the earth at the same speed you are. You're just moving fast enough that you're eternally missing the earth (or, in this case, Kerbin).

Okay, here's a resource for middle school students...

and a

...

and a Wired article.

One of those has got to be your speed.

The all make it very clear that the weightlessness is due to everything around you is falling with you. You orbit because of your speed/angular velocity, but that has crap all to do with weightlessness. If what you were saying is true, then "Vomit comet" wouldn't be able to induce weightlessness unless you were traveling at orbital velocities!

[Greep]

FL-R25.

*edit* Hmm... now that you mention it, I've got 2 different RCS tanks in my list with that name.

One has a capacity of 100 (.55t), while the other has a capacity of 250 (1.15t)

Must've been a cleanup error during an update?

Shouldn't matter so long as you have a 2t ship with a 48-7S set to 16.5% throttle.

Best,

-Slashy

Ah okay, Well I'll check it out in the morning. Right now I was just going to quick check the masses for tsiolkovsky, but kinda hard to do that if I don't have 100% the same parts as you

[LethalDose]

What LD will accept <> "credible". "Repeatable" is the scientific standard.

All anyone has to do to confirm or reject my results is repeat the experiment.

I've been very clear about what's credible: Video. It's what I've done to demonstrate results. It's not that high of a bar. OBS is free, YouTube's free.

And it sounds like that whole "reproducible" thing ain't going so well for ya.

[LethalDose]

Ah okay, Well I'll check it out in the morning. Right now I was just going to quick check the masses for tsiolkovsky, but kinda hard to do that if I don't have 100% the same parts as you

You can still report your dV findings in addition to initial TWR, ISP, alt, etc.

[Greep]

Ah yeah but I've never used hyper edit and it's getting late. So I will try this in the morning, though

[GoSlash27]

Ah yeah but I've never used hyper edit and it's getting late. So I will try this in the morning, though

Outstanding!

I look forward to seeing the results!

Merry Christmas,

-Slashy

[arkie87]

I'm sorry arkie, but I feel compelled to educate you on basic mathematics: Let's assume a perfect circular orbit, then r is a positive constant, which means r'' is 0. Now, v_theta is a real number (i.e. not a complex one), which means v_theta^2 is positive (or 0 - but this is a very odd case), which means a_r is negative (or 0). The only explanation for this is gravity; the fact constants are not mentioned is irrelevant, because these are absorbed into the v_theta value.

I wouldnt recommend being so rude on these forums, especially since nothing you've said "corrected" anything I've said. I don't understand why people on these forums are so rude and condescending so quickly.

a_r is negative, due to the radial unit vector changing direction (in the Cartesian plane). But even so, this means you accept the form of the equation I've provided, which includes the V^2/R term i.e. centripetal lift...

yeah, well, neither can I. It's simple: The only two forces acting the body in this situation is thrust and gravity.

There's no such thing as 'centripetal lift'. Please provide an independent source that claims otherwise.

You seem to be very confused about your math. In polar coordinates, r isn't a "direction", its a magnitude. It's the magnitude of the position vector that is Theta radians away from 0.

Using polar coordinates, any derivative of r over time is zero for circular orbit. But it's direction of travel is obviously changing, which is what's given by dTheta/dt. I'm trying to figure out how to explain this with polar coordinates, but it's been a long time since I had to take the derivative of polar coordinates.

In fact, I'm going to bail out and just use Cartesian coords like I planned to originally.

- - - Updated - - -

Get out get out get out get out.

There are only two forces acting on the body -- thrust and gravity-- but from Newton's Law: sum(F) = m*a, and it is the m*a term that changes when switching to polar coordinates i.e. the V^2/R term.

I'm not sure how the reference i supplied isnt an "independent source" -- it is a textbook on dynamics... what else could i possibly provide?

I'm not confused about polar coordinates. From your own admission, it's been a long time since you've dealt with them, so i would like to humbly recommend refreshing yourself.

You can use either polar or cartersian coordinates to solve this problem. Cartesian does not require a V^2/R term, but polar does. And i'm willing to bet that both approaches will yield identical results...

[arkie87]

I have taken pics (no scanner, sorry) of the derivation of time derivatives in polar coordinates. Sorry for the quality but it will have to do. Please review and get back to me.

[arkie87]

I agree with LethalDose that my approach isnt "rigorous" w.r.t. to a_theta, but i think if d(theta)/dt is small (d(theta)/dt = V_theta/R << 1), then my approach is accurate enough.

[Greep]

Hey, so I tried GoSlashy's approach and got essentially the exact same results xD

Initial conditions:

altitude: 3k. High, but flat, so I could just go straight horizontal, had to aim for 4k apoapsis after hitting about 200 m/s as there was a hill somewhere around that altitude far out

intial mass: 2.0025 (big parachute, 2 radial chutes, full fl-100, capsule untweaked, engine)

thrust limit of 16.5 with rockomax

dV 875 to SoI, slight overshoot, though, since I didn't throttle limit towards the end. So I think it literally may have been exactly his answer of 871 :/

Admittedly, since this is "SoI" and not "escape velocity" I have no idea if this is a confirmation of LD's or a refutation lol, I'll just leave that up to you guys.

If you want screenshots I could upload them, but if the results are identical I don't see much point :/

Edited December 25, 2014 by Greep

[arkie87]

Hey, so I tried GoSlashy's approach and got essentially the exact same results xD

Initial conditions:

altitude: 3k. High, but flat, so I could just go straight horizontal, had to aim for 4k apoapsis after hitting about 200 m/s as there was a hill somewhere around that altitude far out

intial mass: 2.0025 (big parachute, 2 radial chutes, full fl-100, capsule untweaked, engine)

thrust limit of 16.5 with rockomax

dV 875 to SoI, slight overshoot, though, since I didn't throttle limit towards the end. So I think it literally may have been exactly his answer of 871 :/

Admittedly, since this is "SoI" and not "escape velocity" I have no idea if this is a confirmation of LD's or a refutation lol, I'll just leave that up to you guys.

If you want screenshots I could upload them, but if the results are identical I don't see much point :/

If you get deltaV values below LD's prediction, it means that either KSP is wrong, or LD is wrong. Obviously, I will suggest LD's approach is too conservative because it neglects centripetal lift, which becomes significant once you approach orbital velocity. This, in turn, means that you dont have to aim as high above the horizontal, which makes the burn slightly more efficient.

[LethalDose]

First, arkie, I want to be just crystal clear that I want to understand what you're talking about. If I didn't, I wouldn't be talking to you.

a_r is negative, due to the radial unit vector changing direction (in the Cartesian plane). But even so, this means you accept the form of the equation I've provided, which includes the V^2/R term i.e. centripetal lift...

First off, let's not put words in anyones mouth, like "you accept the equation I've provided". I have no trouble seeing the equation is in a book, but that doesn't automatically explain it's meaning.

When you posted that, I didn't have a clue what those dots meant, and no explanation was provided. Apparently those dots indicate derivatives in with respect to time? That's a heck of a lot more apparent now that I can actually read the accompanying text! If I've never seen this notation before, how is a shot of equations using it supposed to explain anything?

Further, the equation you provide is: a_r = r.. - rTheta.^2

If you're saying that "rTheta.^2" = V^2/R, again, that term is negative, hence that's not lift, and it's not clear where you claim the lift term is! The only thing left in equation is "r..". Is "r.." your V^2/R?

And why are you stating a_r is negative? If you're maintaining a constant altitude (either in a circular orbit, or the assumptions made in this exercise), wouldn't a_r be zero?

I'm not sure how the reference i supplied isnt an "independent source" -- it is a textbook on dynamics... what else could i possibly provide?

See above. It's an independent source, but explained nothing.

I can't take a shot of equations out of my epidemiology texts and expect you to understand logistic regression from that alone, can I? But I can send you to an online source like KhanAcademy and actually make it useful...

I'm not confused about polar coordinates. From your own admission, it's been a long time since you've dealt with them, so i would like to humbly recommend refreshing yourself.

Again, why do you think I'm requesting resources to do that? Your explanations have not made any sense in my frame of reference, so I'm trying to go back and find some source that can explain it.

You can use either polar or cartersian coordinates to solve this problem. Cartesian does not require a V^2/R term, but polar does. And i'm willing to bet that both approaches will yield identical results...

So, I'm going back to your equations here.

For all intents and purposes, it looks like you're using Cartesian coords... Your equation line starts with "Sum Fy ...". That's not consistent with your statements that you're using polar coordinates for this, based on your source. If you're using Cartesian coords, use y and x, if you're using polar, use theta and r.

I agree with LethalDose that my approach isnt "rigorous" w.r.t. to a_theta, but i think if d(theta)/dt is small (d(theta)/dt = V_theta/R << 1), then my approach is accurate enough.

You're approach isn't rigorous, which leaves a pile of unsupported assumptions and confusing notation. Proving these statements requires rigor, and until you produce a document that rigorously demonstrates what you're presenting in with clear defined notation, I can't guarantee you I'll be convinced.

Also, please make your images larger in the future. What you've posted on that other thread is near illegible due to size and resolution. I'm getting eye strain trying to decipher it.

[LethalDose]

Hey, so I tried GoSlashy's approach and got essentially the exact same results xD

Initial conditions:

altitude: 3k. High, but flat, so I could just go straight horizontal, had to aim for 4k apoapsis after hitting about 200 m/s as there was a hill somewhere around that altitude far out

intial mass: 2.0025 (big parachute, 2 radial chutes, full fl-100, capsule untweaked, engine)

thrust limit of 16.5 with rockomax

dV 875 to SoI, slight overshoot, though, since I didn't throttle limit towards the end. So I think it literally may have been exactly his answer of 871 :/

Admittedly, since this is "SoI" and not "escape velocity" I have no idea if this is a confirmation of LD's or a refutation lol, I'll just leave that up to you guys.

If you want screenshots I could upload them, but if the results are identical I don't see much point :/

/Sigh So I keep asking for video because I want to see flight paths, and I just don't trust some one in particular. But you're not him, so...

My method, given a landing at 3000 m altitude and those vessel parameters, predicts a total delta V cost of 873.4, observed was 875. I don't see a problem.

As always, questions:

• How are you calculating dV? KER? End mass?
  
• If end mass, what level of precision are you getting (round-off errors are important)
  
• Can you be more precise about your landing altitude?
  
• Any record of your burn time?
  

If you want to fly it again from a different location, please get a video of it, and r-click on your parts before take off so we can see settings for the trial.

If you get deltaV values below LD's prediction, it means that either KSP is wrong, or LD is wrong. Obviously, I will suggest LD's approach is too conservative because it neglects centripetal lift, which becomes significant once you approach orbital velocity. This, in turn, means that you dont have to aim as high above the horizontal, which makes the burn slightly more efficient.

Even if previous empirical evidence is to believed to be accurate, then my estimate is <2% too conservative.

Given it's purpose, I'm pretty freaking okay with that.

[Greep]

So anyways, in practical ksp terms, think I get what arkie is saying. When I was doing the experiment, with a TWR that is very low, like 1.5, you actually burn for a very long time. If you go straight horizontal for, say, 10 seconds, you can maintain altitude. However if you keep in that direction, you're altitude rises significantly over the course of the burn, so I had to go from pointing at 40 degrees to 5 (eventually 0, but the stupid hill :/).

At a high TWR of like 12, I would've just started at 7 degrees and stayed there. Is this the "lift" arkie is trying to explain?

Edit: Can't do video as I'm using a laptop and KSP is a RAM beast, but I think you can get my flight path from this post. Basically I just minimized apoapsis as much as I could for 90% of the burn.

Precision of dV calculation is pretty much perfect, I added up the masses by hand for the initial mass, and for end mass I calculated total fuel mass (.5 tons), and percentage used, and used 3 decimal points (since display only shows two, no point in even going that precise), then subtracted the amount.

Landing altitude was about 3400 if you need more precise I can check, I didn't realize it was that important to be very precise there.

I will try again in a minute if you want a start and final screenshot. Same location? And this is the correct flight path right described here?

Edit: wait this is silly heh, it is pretty easy to test. If you want improved accuracy and to know the flight path is correct, why don't you try it yourself?

Edited December 25, 2014 by Greep

[arkie87]

First, arkie, I want to be just crystal clear that I want to understand what you're talking about. If I didn't, I wouldn't be talking to you.

First off, let's not put words in anyones mouth, like "you accept the equation I've provided". I have no trouble seeing the equation is in a book, but that doesn't automatically explain it's meaning.

When you posted that, I didn't have a clue what those dots meant, and no explanation was provided. Apparently those dots indicate derivatives in with respect to time? That's a heck of a lot more apparent now that I can actually read the accompanying text! If I've never seen this notation before, how is a shot of equations using it supposed to explain anything?

Sorry, I didnt post the derivation just the result. I have since posted the derivation as well. I hope that helped explain the notation and nomenclature.

Further, the equation you provide is: a_r = r.. - rTheta.^2

Yes, r*theta_dot^2 = V_theta^2/R, since V_theta = R*theta_dot

If you're saying that "rTheta.^2" = V^2/R, again, that term is negative, hence that's not lift, and it's not clear where you claim the lift term is! The only thing left in equation is "r..". Is "r.." your V^2/R?

This term is negative, but it is on the same side as m*a (i.e. inertia side); when we bring it over to the sum(F) side, it becomes a positive force

And why are you stating a_r is negative? If you're maintaining a constant altitude (either in a circular orbit, or the assumptions made in this exercise), wouldn't a_r be zero?

r_dot and r_dot_dot are zero since radius is not changing. Acceleration in the radial direction (a_r) is not zero since particle is following a curved path.

See above. It's an independent source, but explained nothing.

I can't take a shot of equations out of my epidemiology texts and expect you to understand logistic regression from that alone, can I? But I can send you to an online source like KhanAcademy and actually make it useful...

Again, why do you think I'm requesting resources to do that? Your explanations have not made any sense in my frame of reference, so I'm trying to go back and find some source that can explain it.

Sorry for my part in this confusion. I should have posted the derivation instead of the result. My bad.

So, I'm going back to your equations here.

For all intents and purposes, it looks like you're using Cartesian coords... Your equation line starts with "Sum Fy ...". That's not consistent with your statements that you're using polar coordinates for this, based on your source. If you're using Cartesian coords, use y and x, if you're using polar, use theta and r.

You're approach isn't rigorous, which leaves a pile of unsupported assumptions and confusing notation. Proving these statements requires rigor, and until you produce a document that rigorously demonstrates what you're presenting in with clear defined notation, I can't guarantee you I'll be convinced.

I agree my approach isnt rigorous (though i think the equations in my textbook justify my approach) and my use of x- and y-coordinates is confusing since i'm really using polar coordinates. I will change the document to reflect this (though i think it might be obvious to some people).

Also, please make your images larger in the future. What you've posted on that other thread is near illegible due to size and resolution. I'm getting eye strain trying to decipher it.

The images I posted can be zoomed in on/they are plenty large on my screen...

[LethalDose]

So anyways, in practical ksp terms, think I get what arkie is saying. When I was doing the experiment, with a TWR that is very low, like 1.5, you actually burn for a very long time. If you go straight horizontal for, say, 10 seconds, you can maintain altitude. However if you keep in that direction, you're altitude rises significantly over the course of the burn, so I had to go from pointing at 40 degrees to 5 (eventually 0, but the stupid hill :/).

At a high TWR of like 12, I would've just started at 7 degrees and stayed there. Is this the "lift" arkie is trying to explain?

Edit: Can't do video as I'm using a laptop and KSP is a RAM beast, but I think you can get my flight path from this post. Basically I just minimized apoapsis as much as I could for 90% of the burn.

Thanks for posting this, Greep, because it's a perfecting jumping off point for some other points I kinda wanted to make.

First off, I believe that the "lift" arkie is trying to explain is increase in altitude along an ellipse as you move from Pe to Ap. The problem is that there is no force being applied to move you upward, it's just gravity pulling you down, so there's no "lift". You're moving upward because you have velocity in that direction from earlier in the burn that gravity hasn't canceled out yet.

I really think the source of deviations from minimum predicted values are due to violations of the assumptions. Specifically, the first one: altitude gains due traveling around the planet. As you pointed out, these low TWR burns are very long. If my math is correct, your burn lasted > 5 minutes, and you traveled ~ 120 km during the flight. That's more than 30 degrees longitude change on the Mun. Yeah, that's going to violate this assumption.

And even with that rather egregiously violated assumption, the error is <2% at worst, even if that example is believed to be accurate.

If that invalidates the model in your, mind, fine. I'm done trying to convince people otherwise.

The other point is exactly yours, Greep, RE: TWR. If the thrust limiting was removed from the engine, my model predicts the req'd dV drops by ~ 100 dV! That's the one of the primary points of this model: Increasing TWR saves dV to a point, then levels out. You stop seeing really improved returns after TWR ~ 2.5-3.

Keeping the thrust limited does not seem like a "practical" change. Someone in this thread, who may or may not have designed a vessel similar to yours, has previously gone off on a tear trying to say this model would only show benefits for completely impractical vessel that no one would design, but is now using vessels designed to increase dV costs, which sure seem impractical, to invalidate that same model...

Edited December 25, 2014 by LethalDose