Effect of initial TWR on orbit dV cost

[LethalDose]

I've been working to derive the equations needed to describe the effect of initial TWR on the dV cost to achieve a target orbital AP or escape from an airless body. This came up in a previous thread recently, but the thread was locked before I could present the equations.

The equations account for the non-instantaneous expenditure of fuel, the thrust needed to maintain a level trajectory during the burn to reduce gravity losses, and the change in the vessels mass during the burn, as these were factors that were claimed to invalidate previous statements, since they were not accounted for in other conclusions.

Due to the complexity of the equations, I've presented them in a pdf (available here) produced using LaTeX. The document demonstrates and explains the logic behind the entire derivation, gives a glossary of the symbology, and a list of the

Unfortunately, I can't seem to to solve the resultant equations for final mass alone, even with the help of WolframAlpha. This means numerical methods are required to calculate dV, which leaves us without a clean expression relating TWR to dV, and specific values must be plugged in to equation to produce a graph. MS Excel's solver function can be used for the capacity, and I've created a spreadsheet that uses Solver to find solutions to the equation, and have made it available here.

As an example of this method in action, I've compared the costs of a vessel with the properties listed below to reach the edge of the Mun's SoI for escape (the reference example in the original thread).

Vessel description:

• Initial mass: 2 t
  
• ISP: 350 s
  
• Landed altitude: 0 m (Sea Level)
  

At an PE altitude of 200 km from the center of the Mun, a vessel would have to achieve a prograde velocity of 775.8 m/s to achieve an AP of 2,430 km from the center of the planet (the edge of the SoI). Landed at sea level on the equator, the vessel starts with an 'orbital' velocity of ~ 9.0 m/s, which means the vessel must spent 766.7 to accelerate to escape velocity at this altitude.

The method provides the following estimates of dV required to take off over the following values of TWR:

[table=width: 500]

[tr]

[td]TWR_{0, local}[/td]

[td]TWR_{0, KSL}[/td]

[td]Total Req'd dV (m/s)[/td]

[td]Wasted dV (m/s)[/td]

[td]Efficiency (%)[/td]

[/tr]

[tr]

[td]1.5[/td]

[td]0.25[/td]

[td]884.4[/td]

[td]117.6[/td]

[td]86.7[/td]

[/tr]

[tr]

[td]2.1[/td]

[td]0.36[/td]

[td]829.8[/td]

[td]63.1[/td]

[td]92.4[/td]

[/tr]

[tr]

[td]3.1[/td]

[td]0.51[/td]

[td]798.6[/td]

[td]31.8[/td]

[td]96.0[/td]

[/tr]

[tr]

[td]6.1[/td]

[td]1.02[/td]

[td]774.9[/td]

[td]8.1[/td]

[td]98.9[/td]

[/tr]

[tr]

[td]9.2[/td]

[td]1.53[/td]

[td]770.4[/td]

[td]3.6[/td]

[td]99.5[/td]

[/tr]

[tr]

[td]12.3[/td]

[td]2.04[/td]

[td]768.8[/td]

[td]2.0[/td]

[td]99.7[/td]

[/tr]

[/table]

TWR_{0, KSL} is initial TWR at sea level on Kerbin (provided as a comparison) and efficiency is calculated as target dV (766.7 m/s, in this case) divided by total dV spent. Waste dV is calculated as described in eq 19 in the document provided.

There's clearly an asymptotic relationship between increasing TWR and total dV costs, and there appears to be little benefit in increasing TWR beyond ~ 2-3, but lower values of TWR cause increased losses in dV.

If anyone can find a way to solve equation 17 for m_end, I'd love to update the function to a more useful form. I'm also basically opening this up for "peer-review" to see if I've made any mistakes in the derivation or interpretation.

Addendum: I just wanted to add that the purpose of this derivation is primarily to provide an absolute, but realistic, lower bound on the dV required to reach a target AP given a particular landing situation. No one's going to reach these numbers, but this method can provide useful information when it comes to engineering vehicles to lift-off of airless bodies. Secondarily, it acts to debunk that this is a simple matter to calculate using "this looks good enough" math.

Edited December 25, 2014 by LethalDose

[arkie87]

Eq. 7 has a typo (it is inverted); the equations deriving from it seem to be correct though.

What about centripetal lift? It appears this paper neglects it (and in doing so, arrives an an analytical solution).

Please check out my thread, which accounts for centripetal lift (thereby requiring numerical integration), and non-dimensionalizes equations such that similarity requirements become apparent, and thus, results are applicable for all planets/moons etc: http://forum.kerbalspaceprogram.com/threads/103890-Non-Dimensional-Model-for-Optimal-Launch-Efficiency

[LethalDose]

Eq. 7 has a typo (it is inverted); the equations deriving from it seem to be correct though.

What about centripetal lift? It appears this paper neglects it (and in doing so, arrives an an analytical solution).

Please check out my thread, which accounts for centripetal lift (thereby requiring numerical integration), and non-dimensionalizes equations such that similarity requirements become apparent, and thus, results are applicable for all planets/moons etc: http://forum.kerbalspaceprogram.com/threads/103890-Non-Dimensional-Model-for-Optimal-Launch-Efficiency

First: corrected the typo (that was freaking embarrassing).

Second: I presume you when you say "centripetal lift" you mean the increase in altitude that occurs as a result of the orbital velocity. If this is the case, I explicitly state that there is no vertical acceleration during the burn. This assumption was made for a variety or reasons.

• Through most of the burn, the centripetal lift is negative, and only hits 0 when you attain orbital velocity (in this case, when you get to about 563 dV out of the 766 req'd, so 75% of the burn) and is basically, in my mind, just part of weight being offset with the downward thrust.
  
• g₀ is conveniently constant
  
• The location where the burn is taking place is also essentially at the AP of the new orbit until you reach orbital velocities, when it becomes the PE. At the AP and PE, there is basically no change vertical altitude over time (i.e. no centripedal lift occurs at these locations).
  
• These burns are usually only taking 1-2 minutes, the the lift gained during the burn is, in my opinion negligible.
  
• Finally, this is an analytic solution intended to give a lower bound for dV requirements to guide vessel design.
  

In the documents you linked, it appears you're locking phi (which you refer to as theta) at certain value, where my solution varies phi as a function of instaneous mass, because, as mentioned, no accounting for how that changed during the burn caused someone to krap their pants, even though they ignored that problem in their own equations later.

Edited December 24, 2014 by LethalDose

[arkie87]

First: corrected the typo (that was freaking embarrassing).

Dont sweat it!

Second: I presume you when you say "centripetal lift" you mean the increase in altitude that occurs as a result of the orbital velocity. If this is the case, I explicitly state that there is no vertical acceleration during the burn.

No, by "centripetal lift" i mean decrease in apparent gravity force due to horizontal velocity around a circular/spherical body i.e. V^2/R term, before the craft reaches orbital velocity. This occurs even if no change in height occurs during burn. At orbital velocity, centripetal lift = gravity. When accelerating up to orbital velocity, this term is very important; after you've already achieved orbital velocity, i agree, we can assume planet/moon is large enough such that very little height is gained during acceleration.

In the documents you linked, it appears you're locking phi (which you refer to as theta) at certain value, where my solution varies phi as a function of instaneous mass, because, as mentioned, no accounting for how that changed during the burn caused someone to krap their pants, even though they ignored that problem in their own equations later.

I am locking phi (theta) to the angle it needs to be to cancel gravity (minus centripetal lift). The angle is, however, NOT constant during the burn for two reasons:

(1) Mass is decreasing, so instantaneous TWR is as well

(2) Centripetal lift is increasing, so instantaneous TWR is as well

The only main difference between our two models is centripetal lift term (that, and you define efficiency using escape velocity whereas i define it as just getting into orbit). I have compared our two methods, correcting for the fact that my model only accelerates up to orbital velocity, rather than escape (so i've added in the difference). Here are the results:

So you can see, my model predicts a higher possible efficiency because it includes effect of centripetal lift, which benefits the craft as it approaches orbital velocity.

Edited December 24, 2014 by arkie87

[Greep]

So I skimmed the pdf, and I'm wondering how practical this research is given two of the assumptions:

1) no drag

2)Single stage

These are pretty big assumptions. In a normal launch would these results have much benefit? After an even modest TWR you're going to hit terminal velocity anyways.

This would be excellent research if you're talking about landers on tylo, though.

Edit: But even then, you kind of need a high TWR to land efficiently anyways, right? Efficient landings require burning at the last possible moment, and that requires a relatively high TWR in the first place.

Edited December 24, 2014 by Greep

[arkie87]

So I skimmed the pdf, and I'm wondering how practical this research is given two of the assumptions:

1) no drag

2)Single stage

These are pretty big assumptions. In a normal launch would these results have much benefit? After an even modest TWR you're going to hit terminal velocity anyways.

This would be excellent research if you're talking about landers on tylo or mun, though.

No drag isn't an assumption; it's a requirement for the model i.e. in only applies to atmospherless bodies. That said, terminal velocity under realistic aerodynamics models, such as NEAR or FAR, is unreachable.

Single-stage i'll give you though, so clearly it applies only if you are single stage, which might realistic if your lander is single stage to ground and back up to orbit, where it docks with the main craft.

[LethalDose]

I think you're misunderstanding the physics of the situation.

And going back to the equations you've presented in that link, I'm really clear that's the situation.

What you're describing as "centripetal lift" seems to be as much of an illusion as centrifugal force (in fact I think they're the same thing). You describe this as "'apparent' decrease in gravity". It might "appear" that lifting force is there, but it's not really.

Centripetal force is only exerted downward, towards the inside of the circular trajectory (in this case, an orbit). It can't produce an upwards force (lift) in any circumstance. In our case, the force is exerted by gravity, instead of a tether or roller coaster track (classic examples).

If you want to convince me of this, you're going to have to draw a free-body diagram that demonstrates this "lift".

Another issue with what you're presenting is that the equation you're using: a_c = V/r² or F = mV/r² at all points.

This relationship is only true for a circular track. The centripetal force/acceleration experienced in an elliptical track wouldn't be described by this relationship at all.

As a counter example, lets look at a vessel orbiting the Mun at 10 km above sea level. It's orbital velocity is 556.9 m/s, and it's local gravity is 1.48 m/s². If there was some "centripetal lift" acting on this vessel, it would also by equal to 1.48 m/s². If this acceleration were real, then the vessel would have 0 net acceleration, and by Newton's first law the vessel would be traveling in a straight line.

But it's not traveling in a straight line, it's traveling in a circle.

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).

Edited December 24, 2014 by LethalDose

[LethalDose]

So I skimmed the pdf, and I'm wondering how practical this research is given two of the assumptions:

1) no drag

2)Single stage

These are pretty big assumptions. In a normal launch would these results have much benefit? After an even modest TWR you're going to hit terminal velocity anyways.

This would be excellent research if you're talking about landers on tylo, though.

Edit: But even then, you kind of need a high TWR to land efficiently anyways, right? Efficient landings require burning at the last possible moment, and that requires a relatively high TWR in the first place.

IMO, I find these are safe assumptions for vessels on the surface of the following bodies:

• Moho
  
• Gilly
  
• Mun
  
• Minmus
  
• Ike
  
• Dres
  
• Vall
  
• Bop
  
• Pol
  
• Eeloo
  

Give me TWRs and engine ISPs and I can give you necessary mass ratios required for single stage to orbits. They're pretty reasonable. Even the mass ratio for Tylo is only about 2 with an engine ISP of 370 (an LVT-45), which is doable.

As far as "You need a High TWR to land efficiently"... I would counter that "High TWR" and "efficiently" are pretty subjective. And it really depends on your approach: If you're in a grazing orbit, you actually have a preeeeety good amount of time to slow down very near to the planet, and don't need a really potent TWR.

Addendum: I included the "single stage to orbit" assumption (Technically, it's really single stage to AP, but, whatever) because I figured someone would "gotcha" me on it if I didn't. I've been thinking about it, and the method described can accommodate multiple stages to orbit, but requires separate integrations of equation 14 for each stage. For example, for a two stage ascent, you would integrate the equation for the first stage values over the interval of the initial mass of the entire first stage to the dry mass of the first stage, then integrate over the mass of the second stage to the end mass. When staging, you're effectively making m(t) and phi(t) discontinuous functions over the entire interval of the burn, and you have to use this approach anytime your functions aren't continuous over the domain of the integral.

Edited December 24, 2014 by LethalDose

[arkie87]

As a counter example, lets look at a vessel orbiting the Mun at 10 km above sea level. It's orbital velocity is 556.9 m/s, and it's local gravity is 1.48 m/s². If there was some "centripetal lift" acting on this vessel, it would also by equal to 1.48 m/s². If this acceleration were real, then the vessel would have 0 net acceleration, and by Newton's first law the vessel would be traveling in a straight line.

But it's not traveling in a straight line, it's traveling in a circle.

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).

Rather than respond point by point, i think the example you provided is poignant:

The "centripetal lift" term results from transformation of coordinates. Either we have to treat the craft moving in a circle (in which case, centripetal lift term is not necessary) but we must calculate gravity vector in both x- and y-coordinates; or, we rotate x-coordinate (theta coordinate) and y-coordinate (radial coordinate) around with the craft, such that x =0 always (i.e. stationary) and gravity is always pointed down in the y-coordinate (in which case, centripetal lift is necessary).

Now back to your example:

In the case of the spacecraft at 10 km perfectly circular orbit, either we track its position in x-y plane using gravity vector (which has components in both x- and y-coordinates), which changes with time, and will cause the craft traveling at orbital velocity to travel in a circle

or

We transform coordinates, such that x- and y-coordinates follow craft around 10 km orbit-- with this method, it is obvious that x-coordinate remains zero always (since it is moving with the craft); however, the y-position will not remain at zero due to gravity always pointing downward in the y-direction -- to counter this, we need centripetal lift.

As a side note, in an ellipse, centripetal acceleration does equal V^2/R; however, R is not constant, so instantaneous R must be used.

[arkie87]

Edit: But even then, you kind of need a high TWR to land efficiently anyways, right? Efficient landings require burning at the last possible moment, and that requires a relatively high TWR in the first place.

This is called a suicide burn. What's more efficient is a horizontal one (albeit much harder to pull off and not always possible due to terrain).

What's more practical is to slow down horizontally over your target and then fall (from as low as possible) onto your target, performing a suicide burn from a low altitude.

[Anquietas314]

What's more practical is to slow down horizontally over your target and then fall (from as low as possible) onto your target, performing a suicide burn from a low altitude.

Technically you don't have to be falling nearly-vertical to do a suicide burn. A tiny burn to bring your periapsis just below the surface (near the landing site) will let you do a suicide burn almost horizontally. The tricky part of course is getting from that orientation into a vertical one before you hit the ground

[arkie87]

Technically you don't have to be falling nearly-vertical to do a suicide burn. A tiny burn to bring your periapsis just below the surface (near the landing site) will let you do a suicide burn almost horizontally. The tricky part of course is getting from that orientation into a vertical one before you hit the ground

This is the "horizontal" approach described. It is more efficient, but more dangerous.

I recommend buying life insurance for your Kerbals before you attempt it

[LethalDose]

Rather than respond point by point, i think the example you provided is poignant:

The "centripetal lift" term results from transformation of coordinates. Either we have to treat the craft moving in a circle (in which case, centripetal lift term is not necessary) but we must calculate gravity vector in both x- and y-coordinates; or, we rotate x-coordinate (theta coordinate) and y-coordinate (radial coordinate) around with the craft, such that x =0 always (i.e. stationary) and gravity is always pointed down in the y-coordinate (in which case, centripetal lift is necessary).

Now back to your example:

In the case of the spacecraft at 10 km perfectly circular orbit, either we track its position in x-y plane using gravity vector (which has components in both x- and y-coordinates), which changes with time, and will cause the craft traveling at orbital velocity to travel in a circle

or

We transform coordinates, such that x- and y-coordinates follow craft around 10 km orbit-- with this method, it is obvious that x-coordinate remains zero always (since it is moving with the craft); however, the y-position will not remain at zero due to gravity always pointing downward in the y-direction -- to counter this, we need centripetal lift.

As a side note, in an ellipse, centripetal acceleration does equal V^2/R; however, R is not constant, so instantaneous R must be used.

Dude... you're so wrong... The only force acting on an orbiting object is gravity, there's no lift.

First, draw the free body diagram for yourself, then share it with the class. Show all the vectors you think an object in orbit experiences.

I'm going to try to put together a new document that explains this you in a Cartesian plane.

[Anquietas314]

This is the "horizontal" approach described. It is more efficient, but more dangerous.

I recommend buying life insurance for your Kerbals before you attempt it

Having landed on Tylo back when it was first added, purely stock, I think I can handle a nice easy - if risky - Mun landing

However, it isn't quite the same thing as the horizontal approach; there you generally have to start aiming above the horizon to stop yourself from falling. The idea with a suicide burn (intended from mission start) is to go nuts on TWR and come to a stop as fast as possible, which if done right means you don't need to aim above the horizon (of course, you do need to leave enough space underneath you ) because you won't start falling quickly enough for it to matter. It's a subtle difference, but an important one

[arkie87]

Dude... you're so wrong... The only force acting on an orbiting object is gravity, there's no lift.

First, draw the free body diagram for yourself, then share it with the class. Show all the vectors you think an object in orbit experiences.

I'm going to try to put together a new document that explains this you in a Cartesian plane.

I unburied my dynamics textbook:

Turn your attetion to polar coordinates (which are the coordinates i am using). r-coordinate is y-direction and theta-coordinate is x direction.

a_r = r'' - r*(theta')^2

v_theta = r*theta'

So:

a_r = r'' - v_theta^2/r

the V_theta^2/r term is the "centripetal lift"

[LethalDose]

a_r = r'' - v_theta^2/r

the V_theta^2/r term is the "centripetal lift"

That's not lift, it's gravity.

If you work out the direction of the acceleration vector, you'll see it's in the opposite direction of the location vector (proportionate magnitude, and Theta(acceleration) = Theta(position) - pi) . That nomenclature is sloppy, but it works.

That's also not how polar coordinates work. Theta is not the "x coordinate", Theta is arctan(y/x)

Your book is even telling you draw a free body diagram of the situation. DO THAT.

Edit:Actually, what I'm saying is even better supported by this because the sign on the term you're citing is negative, indicating a downward force, not an upward force. That term is saying you're losing r!

Edited December 24, 2014 by LethalDose

[arkie87]

That's not lift, it's gravity.

Gravity has nothing to do with the page i pasted from my textbook... my textbook derives the phantom forces experienced due to transformation of coordinates...

Where do you see any value related to gravitational constant? or the planet? Very confused....

If you work out the direction of the acceleration vector, you'll see it's in the opposite direction of the location vector (proportionate magnitude, and Theta(acceleration) = Theta(position) - pi) . That nomenclature is sloppy, but it works.

Absolutely no idea what you are trying to convey here. The acceleration vector is given in the theta and r directions. If the term on the right-hand-side is positive, so is the acceleration.

That's also not how polar coordinates work. Theta is not the "x coordinate", Theta is arctan(y/x)

You misunderstood. In my model, what i call "x" is really "theta" and what i call "y" is really "r" i.e. even though my model uses "x" and "y" coordinates, it is not Cartesian, but rather, polar.

Your book is even telling you draw a free body diagram of the situation. DO THAT.

Free body diagrams wont help you, since we are discussing transformation of coordinates. The book derives what happens to conservation of momentum in a rotating reference frame.

I really cant believe we are even debating this.

Edited December 24, 2014 by arkie87

[Anquietas314]

Absolutely no idea what you are trying to convey here. The acceleration vector is given in the theta and r directions. If the term on the right-hand-side is positive, so is the acceleration..

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.

Edited December 25, 2014 by armagheddonsgw

[LethalDose]

I'm sorry arkie, but I feel compelled to educate you on basic mathematics

Hey, be nice. This isn't basic. I'm frustrated, too, but I'm not in the mood to have this thread get locked, too.

[Anquietas314]

Hey, be nice. This isn't basic. I'm frustrated, too, but I'm not in the mood to have this thread get locked, too.

Orbital mechanics are not basic, no, however the error was.

[GoSlash27]

Edit: But even then, you kind of need a high TWR to land efficiently anyways, right? Efficient landings require burning at the last possible moment, and that requires a relatively high TWR in the first place.

Greep,

This is a whole 'nother subject, but actually... no. You don't need a high t/w ratio to land, and a suicide burn is actually the least efficient way to do it.

Scope out my tutorial here for a technique that's more efficient, easier, more precise and safer than the suicide burn and only requires 1.2:1 t/w ratio.

Best,

-Slashy

[LethalDose]

I really cant believe we are even debating this.

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 - - -

Greep,

This is a whole 'nother subject, but actually... no. You don't need a high t/w ratio to land, and a suicide burn is actually the least efficient way to do it.

Scope out my tutorial here for a technique that's more efficient, easier, more precise and safer than the suicide burn and only requires 1.2:1 t/w ratio.

Best,

-Slashy

Get out get out get out get out.

[LethalDose]

Ugh, it's sloppy as [expletive deleted] but some derivations of the direction of the accelerations in orbit are here.

[numerobis]

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.

[GoSlash27]

Dont sweat it!

No, by "centripetal lift" i mean decrease in apparent gravity force due to horizontal velocity around a circular/spherical body i.e. V^2/R term, before the craft reaches orbital velocity. This occurs even if no change in height occurs during burn. At orbital velocity, centripetal lift = gravity. When accelerating up to orbital velocity, this term is very important; after you've already achieved orbital velocity, i agree, we can assume planet/moon is large enough such that very little height is gained during acceleration.

I am locking phi (theta) to the angle it needs to be to cancel gravity (minus centripetal lift). The angle is, however, NOT constant during the burn for two reasons:

(1) Mass is decreasing, so instantaneous TWR is as well

(2) Centripetal lift is increasing, so instantaneous TWR is as well

The only main difference between our two models is centripetal lift term (that, and you define efficiency using escape velocity whereas i define it as just getting into orbit). I have compared our two methods, correcting for the fact that my model only accelerates up to orbital velocity, rather than escape (so i've added in the difference). Here are the results:

http://i.imgur.com/wfe7Mti.png

So you can see, my model predicts a higher possible efficiency because it includes effect of centripetal lift, which benefits the craft as it approaches orbital velocity.

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

Edited December 25, 2014 by GoSlash27