Effect of TWR, TVR, and DVR on Orbital Launch Efficiency (deltaV)

[LethalDose]

Because saving DV <> saving weight or cost, and these are both engineering priorities.

A low t/w launcher may come out lighter overall than a high t/w launcher when the mass penalty of fuel & tankage is less than the mass penalty of adding engines to achieve higher t/w.

Likewise, fuel and tankage is cheaper than an equivalent mass of engine, so it can be economically advantageous to design a low t/w launcher.

I'm working this problem, and I need accurate numbers to find the solutions. It makes a huge difference if the model is overestimating the "minimum bound" by several percentage points when there's less than 0.1% change in overall vehicle mass between t/w=1.3 and t/w=2.6 (which is, incidentally, the case when using LV-1 engines on Gilly according to arkie's model).

Best,

-Slashy

Then don't use my model. Ever. Please don't.

Obviously, if you hold to notions like this:

A model either accurately predicts empirical results or it does not.

There's no use in trying to discuss anything with you. George Box, an exceptional statistician stated in 1976:

"Since all models are wrong the scientist cannot obtain a "correct" one by excessive elaboration. On the contrary following William of Occam he should seek an economical description of natural phenomena. Just as the ability to devise simple but evocative models is the signature of the great scientist so overelaboration and overparameterization is often the mark of mediocrity."

Summarized in his book later, and used frequently since:

"Essentially, all models are wrong, but some are useful"

He even included the ideal gas law in this generalization of models (source). You can argue a "black-white" PoV regarding the simplicity of models with that, I'm done.

To everyone: If what I've presented isn't useful for your work, then qualify it is such, and stop trying to discredit the entirety of the model over < 50 m/s.

Edited December 28, 2014 by LethalDose
Added source for quotations.

[sal_vager]

This thread is getting overheated and from what I've seen so far it really shouldn't.

You're arguing over a predictive model, there's many models for many kinds of things and they all have some utility, some are just easier to use or understand, others may be more accurate but all are intended to find a ballpark figure for what they attempt to model.

Just look at weather modelling, it's very mature and capable but still gets it wrong when you were hoping for a sunny day

LethalDose's model looks to be just as useful as arkie87's or any other model proposed, they give the user an idea of what kind of craft will preform better in a given situation, but like any model, the proof is in the actual flight itself, not in the graphs.

I think there's little more that can be said by any of the parties involved in this thread, you're just going to end up going round in circles, I'll leave this open a bit longer though and well see.

[GoSlash27]

Then don't use my model. Ever. Please don't.

Fair 'nuff.

Best,

-Slashy

[arkie87]

Uh, arkie, plugging in different values for the dry mass is changing the results of numeric integration/efficiency results from your spreadsheet... Not by much, but it is changing. That should not be happening...

The excel file with the VLOOKUPs removed is available here.

Yes, i noticed that too.

First off, I never said the slope equals 0, just that slope is approximately 0 (for all intents and purposes). Besides, there is no reason off the bat to assume that if slope doesnt equal zero, there is something wrong... after all, it was this model itself that actually "discovered" that the slope is approximately zero...

Second, the small differences can be due to numerical integration. I used ~1000 time steps (maximum) of first order, forward Euler integration. Errors are inherent in any numerical method used, so it should be no surprise if values change slightly.

Edited December 29, 2014 by arkie87

[arkie87]

LD: Any chance you can simplify your model to only integrate up to orbital velocity, to make it more easy to compare to my model? Pretty please

[arkie87]

Yes, i noticed that too.

First off, I never said the slope equals 0, just that slope is approximately 0 (for all intents and purposes). Besides, there is no reason off the bat to assume that if slope doesnt equal zero, there is something wrong... after all, it was this model itself that actually "discovered" that the slope is approximately zero...

EDIT: i take this first point back. I think i agree that slope should be zero in theory. The reason its not is because of the numerical integration i.e. if DVR >> 1, then it wont end up using all 1000 integration steps, and so, results wont be as accurate. I am revising my model to always use 1000 steps, and to calculate necessary DVR and/or FMR (sort of what you are doing). This is better for accuracy. Thanks!

[arkie87]

Okay. I have updated model to always run at DVR = 1/eta such that it is using approximately the same number of integration steps every time

I have also added a "physics assumptions" section, where you can enable and disable "centripetal lift" to see how this term effects results.

Happy simulating!

https://www.dropbox.com/s/9o3vqf9hlzyzpmf/Horizontal%20Orbital%20Insertion%20Simulator%20v2.0.xlsx?dl=0

[GoSlash27]

Okay. I have updated model to always run at DVR = 1/eta such that it is using approximately the same number of integration steps every time

I have also added a "physics assumptions" section, where you can enable and disable "centripetal lift" to see how this term effects results.

Happy simulating!

https://www.dropbox.com/s/9o3vqf9hlzyzpmf/Horizontal%20Orbital%20Insertion%20Simulator%20v2.0.xlsx?dl=0

I've downloaded the new version, but since I've already begun my work by doing a hatchet job on v1.0, I think I'll stick with that.

So far, all of my t/w recommendations for Gilly are in the 2-3 t/w range for minimum mass and 1-2 range for minimum cost.

I'll be checking all engines on all airless bodies to see how they affect the results.

Thanks again!

-Slashy

[LethalDose]

LD: Any chance you can simplify your model to only integrate up to orbital velocity, to make it more easy to compare to my model? Pretty please

If I understand your model correctly, no simplification should be needed. Simply set the target altitude to be equivalent to starting altitude. e.g. If starting at sea level, enter "0" into the target altitude field.

My model also removes the rotational velocity from the planet when calculating target dV.

Can you change your sheet to provide dV spent, instead of only efficiency? or at least end mass?

Edited December 29, 2014 by LethalDose
added examples

[arkie87]

If I understand your model correctly, no simplification should be needed. Simply set the target altitude to be equivalent to starting altitude.

Ok, i've tried doing this in your model, but wasnt sure if it was equivalent or not. Thanks.

Can you change your sheet to provide dV spent, instead of only efficiency?

just divide V_orbital by efficiency, as per the formula/definition given: dV_expended = V_orbital / efficiency

[arkie87]

I have created a third (and final) version that uses Macros available here: https://www.dropbox.com/s/16lgb6o9ey4g0xs/Horizontal%20Orbital%20Insertion%20Simulator%20v3%20with%20Macros.xlsm?dl=0

It uses a Macro to perform the numerical integration. The macro can be called as a normal Excel function i.e. type in "=Efficiency(TWR,TVR,?CL)" into any cell, and it will give you the efficiency for that TWR, TVR, and centripetal lift assumption.

It also includes total deltaV (which is also the same as expended deltaV), as per LD's request.

[LethalDose]

So, I've tried using your model arkie to compare values, but there seems to be some kind of odd inversion, where my model is now predicting a lower expended dV than yours, which seems contrary to what you've presented.

Values being entered into the model:

• T= 3.257
  
• Mwet = 2
  
• Mdry = 1
  
• g = 1.628
  
• Isp = 350
  
• R = 200
  

Results e = 0.6069 & dV exp = 940.2 m/s

done using v1.

Also, I provided numbers, as requested, for 1 - 1.4. Can you provide the same for your model? I obviously seem to be using your model wrong.

Edit: Consistent results with v3

Edited December 29, 2014 by LethalDose

[arkie87]

So, I've tried using your model arkie to compare values, but there seems to be some kind of odd inversion, where my model is now predicting a lower expended dV than yours, which seems contrary to what you've presented.

Values being entered into the model:

• T= 3.257
  
• Mwet = 2
  
• Mdry = 1
  
• g = 1.628
  
• Isp = 350
  
• R = 200
  

Results e = 0.6069 & dV exp = 940.2 m/s

done using v1.

Also, I provided numbers, as requested, for 1 - 1.4. Can you provide the same for your model? I obviously seem to be using your model wrong.

Edit: Consistent results with v3

You are using my model correctly. I noticed that your model predicts less deltaV under those conditions, and got confused as well, since this shouldnt be possible, and it's what led me to post this question on your thead.

However, that might be a trivial problem since your model might not be set up to handle the case when TWR < 1 (after all, mine wasnt until Slashy tested it under those conditions which led me to modify the code to handle it).

Nevertheless, i think a better test case is this one:

• T= 3.2560
  
• Mwet = 2
  
• g = 1.628
  
• Isp = 1,000,000
  
• R = 200
  

i.e. what does your model predict for ISP --> infinity @ TWR = 1?

Efficiency should approach zero, since TWR will hover around 1 for a long-ass time until enough mass is burned to get it moving. When I use my model with centripetal lift disabled, I get 3.72% efficiency since angle is almost 90 degrees during the entire burn. When i enable centripetal lift, i get 23.04% efficiency since upon approaching orbital velocity, angle approaches zero.

Moreover, it is my understanding that the only difference between our models is my inclusion of the centripetal lift term. However, i have added a mechanism to turn this term off, and our answers still differ. Thus, i think i need to investigate whether our models really are the same except for this one difference. I will get back to you.

Edited December 29, 2014 by arkie87
Values Changed in Table

[LethalDose]

First, I answered your question on the other thread. Basically, until you convince me otherwise, I view TWR < 1 is an invalid starting condition. I'm not going to spend much, if any, energy dealing with it.

You are using my model correctly. I noticed that your model predicts less deltaV under those conditions, and got confused as well, since this shouldnt be possible, and it's what led me to post this question on your thead.

However, that might be a trivial problem since your model might not be set up to handle the case when TWR < 1 (after all, mine wasnt until Slashy tested it under those conditions which led me to modify the code to handle it).

Nevertheless, i think a better test case is this one:

• T= 3.2560
  
• Mwet = 2
  
• g = 1.628
  
• Isp = 1,000,000
  
• R = 200
  

i.e. what does your model predict for ISP --> infinity @ TWR = 1?

Oddly, the results of my model seem to give an asymptotic result of around 70.7%. I'm looking into why it's doing that. I'd still like to see your results from 1-1.4, since I provided mine. I feel like that's good faith here...

Efficiency should approach zero, since TWR will hover around 1 for a long-ass time until enough mass is burned to get it moving. When I use my model with centripetal lift disabled, I get 3.72% efficiency since angle is almost 90 degrees during the entire burn. When i enable centripetal lift, i get 23.04% efficiency since upon approaching orbital velocity, angle approaches zero.

I wonder if this has something to do with the fact that my model ultimately integrates over mass instead of time, though I feel like it shouldn't matter since one is a linear transformation of the other. I see where you're going with this, but if ISP is infinite and TWR is 1, the vessel still isn't going to go anywhere.

Moreover, it is my understanding that the only difference between our models is my inclusion of the centripetal lift term. However, i have added a mechanism to turn this term off, and our answers still differ. Thus, i think i need to investigate whether our models really are the same except for this one difference. I will get back to you.

This is my understanding as well.

[arkie87]

First, I answered your question on the other thread. Basically, until you convince me otherwise, I view TWR < 1 is an invalid starting condition. I'm not going to spend much, if any, energy dealing with it.

Oddly, the results of my model seem to give an asymptotic result of around 70.7%. I'm looking into why it's doing that. I'd still like to see your results from 1-1.4, since I provided mine. I feel like that's good faith here...

I wonder if this has something to do with the fact that my model ultimately integrates over mass instead of time, though I feel like it shouldn't matter since one is a linear transformation of the other. I see where you're going with this, but if ISP is infinite and TWR is 1, the vessel still isn't going to go anywhere.

This is my understanding as well.

I think the problem is the use of arctan instead of arcsin. See this

Though it looks like you found that already...

I dont think its possible to solve for t without using Excel's solver or a macro.

I can make a macro button which will solve for the t needed. How does that sound?

- - - Updated - - -

In good faith, here are my results:

[GoSlash27]

arkie,

Results are up for my study, and they're surprising in one sense, and not surprising in others:

http://forum.kerbalspaceprogram.com/threads/105422-Ideal-T-W-ratios-for-airless-body-launch-%28KSP-90%29

For all engines on Gilly, lowest mass is achieved in the 2-3 t/w range, while for lowest cost they're in the 1-2 range.

On Tylo, all these numbers drop considerably, with lowest mass in the 1.5-2.5 range and lowest cost in the 1-1.8 range.

I didn't expect the numbers to work out so low, but the kicker is efficiency holds steady for optimal solutions on all airless bodies.

An engine that yields the lowest stage mass at 92% efficiency on Gilly will also yield the lowest stage mass at 92% on Tylo.

Just thought I'd pass that on and thank you again for all your work on this model!

Best,

-Slashy

Edited December 30, 2014 by GoSlash27