Optimal TWR with Stock vs. FAR Aerodynamics

[Starman4308]

This is what I'm talking about; you've conflated the terminal velocity of a powered rocket with the terminal velocity of a falling body of fixed area and drag coefficient.

They're not the same thing. The terminal velocity of your ship doesn't mean anything at all in determining the ideal launch profile. What matters is the rate at which atmospheric pressure is dropping.

Besides, you can reach terminal velocity in FAR very easily, even in an aircraft that has nearly zero thrust. Just don't climb, and eventually your speed will top out.

Best,

-Slashy

No. Terminal velocity is, was, and forever will be "the velocity at which this object, traveling in its current orientation, would experience a drag force equal to gravity". Yes, during the course of a launch, this changes. However, in stock, you have a curious situation where, for the first 10 km of ascent, terminal velocity is easy to reach, does not increase very rapidly, and is crucial in determining launch efficiency. In stock, you design for that first 10km of crawling at 100 m/s, whereas in FAR, you design on other factors, mostly independent of terminal velocity.

[GoSlash27]

In FAR, you're usually already several degrees off vertical before you ever crack 1000m, and your path from there to around 30 km is determined mostly by aerodynamics and TWR, because your rocket orientation is constrained to a narrow cone around prograde. This is contrasted with stock, where your pitch profile is determined solely by gravity, TWR, and atmospheric density, because aerodynamic stability/instability is just not a thing in stock.

Your "advice" is shown false by the large number of people who have actually played FAR, and the fact that few real-world rockets have first-stage TWR above 1.6 (and those which do generally lose some SRBs in less than a minute).

You're actually saying the exact same thing I am. In FAR, an efficient launch profile has a gravity kick almost immediately (because more streamlined=less drag=less DV losses) and t/w is still where it has always been; about 2 in theory and lower in practice.

Entirely my point, so maybe I'm not the one you should be arguing with

Best,

-Slashy

[Starman4308]

You're actually saying the exact same thing I am. In FAR, an efficient launch profile has a gravity kick almost immediately (because more streamlined=less drag=less DV losses) and t/w is still where it has always been; about 2 in theory and lower in practice.

Entirely my point, so maybe I'm not the one you should be arguing with

Best,

-Slashy

There is no reason why a maximally-efficient launch in FAR should be of 2.0 TWR. In FAR, ignoring aerodynamic stresses, it is the point where increasing engine mass (and thus TWR) to reduce gravity losses is cancelled out by the loss of dV caused by increased engine mass.

2.0 TWR is entirely an artifact of stock aerodynamics and terminal velocity.

If engines were infinitely light, and we ignored aerodynamic stresses, ideal FAR TWR would be something crazy like 30.0, enough to once again hit terminal velocity. Ideal TWR is somewhere around 1.4 because of the parameters (typical engine TWR, Isp, surface gravity) which go into the calculations of efficiency.

EDIT: In short, ideal ascents go like this: at launch, infinite thrust until you hit terminal velocity, followed by maintaining terminal velocity all the way up (and, as such, infinite TWR again in vacuum). This is the absolute minimum amount of dV loss to the sum of gravity and atmospheric drag. The only reason you do not see this is that engines have finite mass and finite thrust: that TWR costs you dV.

Edited December 14, 2014 by Starman4308

[GoSlash27]

There is no reason why a maximally-efficient launch in FAR should be of 2.0 TWR. In FAR, ignoring aerodynamic stresses, it is the point where increasing engine mass (and thus TWR) to reduce gravity losses is cancelled out by the loss of dV caused by increased engine mass.

2.0 TWR is entirely an artifact of stock aerodynamics and terminal velocity.

If engines were infinitely light, and we ignored aerodynamic stresses, ideal FAR TWR would be something crazy like 30.0, enough to once again hit terminal velocity. Ideal TWR is somewhere around 1.4 because of the parameters (typical engine TWR, Isp, surface gravity) which go into the calculations of efficiency.

None of what you said here conflicts with the reasoning used in stock aero, or my own. The change between stock lumberwagondynamics and FAR has not made a change in the ideal t/w at launch. It was around 1.6 before and it's still 1.6 now.

I personally use something much closer to 1.

Best,

-Slashy

[arkie87]

This is what I'm talking about; you've conflated the terminal velocity of a powered rocket with the terminal velocity of a falling body of fixed area and drag coefficient.

They're not the same thing.

I think the confusion is the definition of "terminal velocity". The accepted definition of terminal velocity is not the velocity where acceleration is zero (despite it's name) i.e. when a craft under thrust stops accelerating or a gliding craft stops accelerating. It is accepted to mean the velocity where a falling craft stops accelerating due to weight perfectly balancing drag force from air resistance.

We are not interested in the "terminal velocity" of a craft under thrust since it has no bearing on optimum launch, and likewise, we are not interested in "terminal velocity" of a gliding craft.

What DOES have a bearing on optimal launch is falling terminal velocity i.e. the accepted definition of terminal velocity.

It can be shown mathematically that the terminal velocity of a falling object (no thrust) happens to be the optimum ascent velocity to minimize fuel consumption. This is the result of a balance between drag force and gravitational losses (any slower and you are spending too long climbing; any faster, and drag losses are too great). If you dont believe me, look it up in a textbook, or ask me to derive it for you, and i will.

The terminal velocity of your ship doesn't mean anything at all in determining the ideal launch profile. What matters is the rate at which atmospheric pressure is dropping.

The rate at which atmospheric pressure is dropping does matter, and is accounted for in the variation of terminal velocity with altitude.

Besides, you can reach terminal velocity in FAR very easily, even in an aircraft that has nearly zero thrust. Just don't climb, and eventually your speed will top out.

This is not what is meant by terminal velocity. See above.

Edited December 15, 2014 by arkie87

[GoSlash27]

I think the confusion is the definition of "terminal velocity". Terminal velocity does not usually mean the point where acceleration is zero (despite it's name) i.e. when a craft under thrust stops acceleration or a gliding craft stops accelerating. It almost always means the point where a falling craft stops accelerating due to weight perfectly balancing drag force from air resistance.

We are not interested in the "terminal velocity" of a craft under thrust since it has no bearing on optimum launch, and likewise, we are not interested in "terminal velocity" of a gliding craft.

What DOES have a bearing on optimal launch is falling terminal velocity i.e. terminal velocity as it is usually used.

It can be shown mathematically that the terminal velocity of a falling object (no thrust) happens to be the optimum ascent velocity to minimize fuel consumption. This is the result of a balance between drag force and gravitational losses (any slower and you are spending too long climbing; any faster, and drag losses are too great).

DINGDINGDING!! We have a winner!! This about a million.

So now given that *Kerbin's* atmospheric pressure and gradient haven't actually changed between stock and FAR, what does this imply about your previous analysis of the terminal velocity and how it relates to an optimal path?

Hint: The readout you've been using was of your rocket's terminal velocity under acceleration

Expectantly,

-Slashy

[arkie87]

DINGDINGDING!! We have a winner!! This about a million.

So now given that *Kerbin's* atmospheric pressure and gradient haven't actually changed between stock and FAR, what does this imply about your previous analysis of the terminal velocity and how it relates to an optimal path?

I never said FAR changed anything about the atmosphere, only the interaction between the craft and the atmosphere i.e. drag. My Previous analysis makes no assumptions regarding the planet's atmosphere, or the drag model used-- stock or FAR-- only the relationship between drag force and velocity i.e. quadtratic. This quadratic relationship holds for both FAR and stock, and is the reason that optimal ascent velocity equals (falling) terminal velocity.

Hint: The readout you've been using was of your rocket's terminal velocity under acceleration

Expectantly,

-Slashy

I assume by "acceleration" you mean "thrust", i.e. m*a = 0 = T - W - D; T = W + D, where D is calculated at V_terminal.

Anyone who knows anything about terminal velocity will know this suggestion is preposterous.

A quick googling of "terminal velocity" results in the wikipedia definition, which refutes your claim:

"The terminal velocity of an object is the velocity of the object when the sum of the drag force (Fd) and buoyancy equals the downward force of gravity (FG) acting on the object. Since the net force on the object is zero, the object has zero acceleration."

However, since I know now that you will not listen to anything i say, I have asked Ferram himself (the maker of FAR)... Stay tuned.

Edited December 15, 2014 by arkie87

[Wanderfound]

However, since I know now that you will not listen to me, I have asked Ferram himself (the maker of FAR)... Stay tuned.

(Wanderfound opens popcorn bag)

This should be fun...

[GoSlash27]

arkie,

You're right on the edge of confirming what I personally confirmed earlier.

Please understand that none of this is an attack on you personally. I'm just trying to explain why your analysis was incorrect and where it went wrong.

Your description of terminal velocity and how it relates to the optimal flight path was completely spot-on (and well- written, I might add). It's your reliance on the KER readout of terminal velocity that has led you astray.

It's talking about your ship's terminal velocity under acceleration, *not* the terminal velocity of a free- falling body at your altitude.

That's why it varies with your Mach number and seems infinite after you go supersonic.

Your pressure gradient, pressure, and gravity are the same as they've always been and therefore the optimal ascent path is the same as it has always been.

The path your ship takes deviates from that due to changes in the ship, not the atmosphere itself.

Stock lumberwagons go higher before starting their turns because stock lumberwagons are draggier than FAR missiles. FAR missiles should begin their gravity turns earlier to take advantage of their low drag coefficient.

Best,

-Slashy

[arkie87]

arkie,

You're right on the edge of confirming what I personally confirmed earlier.

Please understand that none of this is an attack on you personally. I'm just trying to explain why your analysis was incorrect and where it went wrong.

Your description of terminal velocity and how it relates to the optimal flight path was completely spot-on (and well- written, I might add). It's your reliance on the KER readout of terminal velocity that has led you astray.

It's talking about your ship's terminal velocity under acceleration, *not* the terminal velocity of a free- falling body at your altitude.

That's why it varies with your Mach number and seems infinite after you go supersonic.

Your pressure gradient, pressure, and gravity are the same as they've always been and therefore the optimal ascent path is the same as it has always been.

The path your ship takes deviates from that due to changes in the ship, not the atmosphere itself.

Stock lumberwagons go higher before starting their turns because stock lumberwagons are draggier than FAR missiles. FAR missiles should begin their gravity turns earlier to take advantage of their low drag coefficient.

Best,

-Slashy

KER does not have a terminal velocity readout. Terminal velocity readout comes from FAR directly.

Pressure, pressure gradient and gravity are the same; but the interaction of pressure on the craft is completely different in FAR. Furthermore, we have already established in another thread that the rockets in my youtube videos are outperforming the nominal FAR values, so increased drag forces are clearly not a problem, even if i was "reading the KER readout wrong".

And once again, there is no such thing as terminal velocity under acceleration from thrust. It varies with Mach number because that is what happens physically, and is accounted for in FAR. It goes to infinity after Mach 1 because drag coefficient is decreasing and air density is approaching zero. Terminal velocity in space above Kerbin is infinity, since air density (and therefore drag) is zero.

Anyway, let's wait for Ferram...

Edited December 15, 2014 by arkie87

[arkie87]

arkie,

You're right on the edge of confirming what I personally confirmed earlier.

Please understand that none of this is an attack on you personally. I'm just trying to explain why your analysis was incorrect and where it went wrong.

Your description of terminal velocity and how it relates to the optimal flight path was completely spot-on (and well- written, I might add). It's your reliance on the KER readout of terminal velocity that has led you astray.

It's talking about your ship's terminal velocity under acceleration, *not* the terminal velocity of a free- falling body at your altitude.

That's why it varies with your Mach number and seems infinite after you go supersonic.

Your pressure gradient, pressure, and gravity are the same as they've always been and therefore the optimal ascent path is the same as it has always been.

The path your ship takes deviates from that due to changes in the ship, not the atmosphere itself.

Stock lumberwagons go higher before starting their turns because stock lumberwagons are draggier than FAR missiles. FAR missiles should begin their gravity turns earlier to take advantage of their low drag coefficient.

Best,

-Slashy

What did you "personally" confirm earlier and how/by asking whom?

[GoSlash27]

KER does not have a terminal velocity readout.

Terminal velocity readout comes from FAR directly.

And once again, there is no such thing as terminal velocity under acceleration from thrust. Let's wait for Ferram...

Whatever that is in the lower right hand corner of your screen. The readout that said "Terminal Velocity" that you kept referring to in your video.

If it's from FAR, then so be it.

You *know* that the atmospheric gradient and pressure aren't affected by your mach number, yet you referred to it specifically when describing "terminal velocity" and how it increases.

But you just said that the optimal flight path isn't dependent on your ship's terminal velocity. That's clearly what it's referring to if it's changing with thrust and drag.

You accidentally conflated the two.

-Slashy

[arkie87]

KER does not have a terminal velocity readout. Terminal velocity readout comes from FAR directly.

Pressure, pressure gradient and gravity are the same; but the interaction of pressure on the craft is completely different in FAR. Furthermore, we have already established in another thread that the rockets in my youtube videos are outperforming the nominal FAR values, so increased drag forces are clearly not a problem, even if i was "reading the KER readout wrong".

And once again, there is no such thing as terminal velocity under acceleration from thrust. It varies with Mach number because that is what happens physically, and is accounted for in FAR. It goes to infinity after Mach 1 because drag coefficient is decreasing and air density is approaching zero. Terminal velocity in space above Kerbin is infinity, since air density (and therefore drag) is zero.

Anyway, let's wait for Ferram...

Watch this video if you think FAR reports "terminal velocity under acceleration". I repeatedly floor the engine and then cut it and terminal velocity does not increase by a factor of TWR, as it would if your definition was correct.

[GoSlash27]

What did you "personally" confirm earlier and how/by asking whom?

Sorry,

I got the word from NathanKell back around 2:00 local time.

He confirmed that FAR doesn't alter atmospheric values, but rather alters Cd and surface area based on local parameters. These are, of course, functions of the part itself, not the air.

Best,

-Slashy

[arkie87]

You *know* that the atmospheric gradient and pressure aren't affected by your mach number, yet you referred to it specifically when describing "terminal velocity" and how it increases.

But you just said that the optimal flight path isn't dependent on your ship's terminal velocity. That's clearly what it's referring to if it's changing with thrust and drag.

You accidentally conflated the two.

-Slashy

Terminal velocity is supposed to depend on drag (i.e. velocity), but not thrust. I have just posted a video, above, showing this to be true.

Edited December 15, 2014 by arkie87

[arkie87]

Sorry,

I got the word from NathanKell back around 2:00 local time.

He confirmed that FAR doesn't alter atmospheric values, but rather alters Cd and surface area based on local parameters. These are, of course, functions of the part itself, not the air.

Best,

-Slashy

No one was arguing about that...

Admittedly, i wasnt sure if FAR changed atmospheric parameters, but it is entirely irrelevant to this argument.

[GoSlash27]

Watch this video if you think FAR reports "terminal velocity under acceleration". I repeatedly floor the engine and then cut it and terminal velocity does not increase by a factor of TWR, as it would if your definition was correct.

Negatory.

You said yourself (and we all watched) that the terminal velocity decreased as you approached Mach 1, then increased dramatically "no way we can catch it" once supersonic.

Does any of that affect the terminal velocity of a free- falling body that happens to be at your altitude? You know it doesn't.

It's not talking about the air, it's talking about your ship.

-Slashy

[GoSlash27]

No one was arguing about that...

Admittedly, i wasnt sure if FAR changed atmospheric parameters, but it is entirely irrelevant to this argument.

Au contraire, it is *everything*.

You explained it perfectly earlier; the terminal velocity of a rocket under power has absolutely nothing to do with defining an ideal launch profile. What matters is the terminal velocity of a falling body at that altitude because it demonstrates a balance between gravity and pressure gradient.

FAR does not alter the gravity or the pressure gradient, so the ideal profile remains the same.

Edited December 15, 2014 by GoSlash27

[arkie87]

Negatory.

You said yourself (and we all watched) that the terminal velocity decreased as you approached Mach 1, then increased dramatically "no way we can catch it" once supersonic.

Does any of that affect the terminal velocity of a free- falling body that happens to be at your altitude? You know it doesn't.

It's not talking about the air, it's talking about your ship.

-Slashy

The trend i was describing was the result of this trend, which occurs (qualitatively) for every object, whether in thrust or free fall, whether aiming up, horizontal, or down: http://upload.wikimedia.org/wikipedia/commons/0/0e/Qualitive_variation_of_cd_with_mach_number.png

As Mach 1 is approached, Cd increases, thus, terminal velocity decreases. Above Mach 1, Cd decreases, and so, terminal velocity increases again.

So yes, it will happen to a free-falling object as well. Ferram can confirm this as well.

[arkie87]

Au contraire, it is *everything*.

You explained it perfectly earlier; the terminal velocity of a rocket under power has absolutely nothing to do with defining an ideal launch profile. What matters is the terminal velocity of a falling body at that altitude because it demonstrates a balance between gravity and pressure gradient.

FAR does not alter the pressure gradient, so the ideal profile remains the same.

Terminal velocity demonstrates the balance between gravity force and drag force, not gravity and pressure gradient. FAR alters the drag force (drastically).

The good news is, i think i understand where you are coming from and what your misunderstanding is. Did this help clear it up?

I think your assumption is that drag force is only a function of altitude i.e. pressure gradient, and not velocity. If drag force was only a function of altitude, then, you would be right: since atmosphere is not changed, then optimal flight path shouldnt be either.

Edited December 15, 2014 by arkie87

[GoSlash27]

It demonstrates the balance between gravity force and drag force. FAR alters the drag force.

Pressure gradient is only relevant through effect of pressure on air density, and therefore drag force.

No, you had it right the first time.

You're killin' me, Smalls...

The correlation between G and Vt was valid in stock *only* because Cd and area don't change in stock.

The changes you're seeing in FAR aren't because the air is different or the gravity is different, but because the CD and area are different.

It's still the same planet, so the optimal path off of the planet is the same.

Your readout of Vt at mach 1 and 5KM is invalid because an object doesn't free-fall at mach 1 at 5KM.

You're looking at data about your ship and mistaking it as data about the air.

Your Vt readout is worthless in FAR. It was sorta- relevant in stock, but not now.

[arkie87]

No, you had it right the first time.

You're killin' me, Smalls...

The correlation between G and Vt was valid in stock *only* because Cd and area don't change in stock.

The changes you're seeing in FAR aren't because the air is different or the gravity is different, but because the CD and area are different.

It's still the same planet, so the optimal path off of the planet is the same.

Your readout of Vt at mach 1 and 5KM is invalid because an object doesn't free-fall at mach 1 at 5KM.

You're looking at data about your ship and mistaking it as data about the air.

Your Vt readout is worthless in FAR. It was sorta- relevant in stock, but not now.

I think your assumption is that drag force is only a function of altitude i.e. pressure gradient, and not velocity. If drag force was only a function of altitude, then, you would be right: since atmosphere is not changed, then optimal flight path shouldnt be either.

[arkie87]

The correlation between G and Vt was valid in stock *only* because Cd and area don't change in stock.

The changes you're seeing in FAR aren't because the air is different or the gravity is different, but because the CD and area are different.

It's still the same planet, so the optimal path off of the planet is the same.

This just isnt true. I would recommend you read up on drag coefficient, terminal velocity, and the way drag force is treated in KSP.

Your readout of Vt at mach 1 and 5KM is invalid because an object doesn't free-fall at mach 1 at 5KM.

Who says objects dont free-fall at Mach 1 at 5 km? If its dense and aerodynamic enough, of course it will!

You're looking at data about your ship and mistaking it as data about the air.

Data about the air? The air has terminal velocity? What? (incidentally, terminal velocity of air is zero, since air is neutrally buoyant).

Your Vt readout is worthless in FAR. It was sorta- relevant in stock, but not now.

Vt readout is worthless in FAR because you will never reach it, not because it isnt optimal.

[GoSlash27]

No, I think you're confusing yourself, and it's way past my bedtime, so last try for this evening.

In your video, you state (and the video shows) that terminal velocity *decreases* with altitude for the first 5KM or so.

Is this true for a free- falling body?

No, of course not. that's absurd.

So your readout of "terminal velocity" isn't talking about an object in free- fall at your atmosphere, but rather your ship under acceleration.

The only reason terminal velocity matters is because it maps the pressure gradient of the atmosphere vs. gravity, which defines your balance between drag losses and pressure losses for an optimal ascent.

That display has nothing to do with it unless you're arguing that objects fall slower when higher above the ground.

That's it for me. I need sleep.

Best,

-Slashy

[NathanKell]

Quick note, GoSlash27, arkie87 explained why you're misinterpreting the info I gave you. What I said (which is true) is that in FAR, pressure is the same as in stock Kerbin, but density becomes a function of pressure and temperature, rather than just pressure. This means, first, that drag (which is a function of density, not pressure) will not scale linearly with pressure.

Second, you are missing the significance of the fact that Cd varies with mach. If an object has a high enough ballistic coefficient, it will accelerate falling past Mach 1 and you will watch terminal velocity go down more slowly than if it had a lower BC (because the decrease in Cd past Mach 1 somewhat offsets the increase in density).

It is *absurd* to claim there is One True Ascent Path, since performance at various mach numbers will vary with rocket shape--again, unlike stock, Cd is not fixed and it doesn't just vary with mach, but the *amount* and *direction* it varies with mach depends on craft geometry.

If you want, you can literally read the code that creates the FAR terminal velocity readout. It does nothing with vessel acceleration; it is based on getting the force of gravity at the position the vessel is.