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


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In my model, theta is always w.r.t. to horizontal....

Then this statement

At higher velocities, pointing horizontally results in a negative θ WRT prograde.

Is wrong. If horizontal in your model means θ = 0, then horizontal can't also result in θ < 0.

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Sure, but there'll be some way to express it as an integral or differential equation or whatever.

A lot of times, there's no ready way to solve for the integral of a function in weird forms. If there's a way to do 1/(C+e^(yadayada)), I'm not familiar with it.

Best,

-Slashy

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A lot of times, there's no ready way to solve for the integral of a function in weird forms. If there's a way to do 1/(C+e^(yadayada)), I'm not familiar with it.

I believe you're talking about this, and that's a very special case that really doesn't come up that often unless you're doing something related to Gaussians

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Speaking of the devil, guys, guess what I have been working on???

hmmm?

An excel version of my Model :sticktongue:

Get it while it's hot! https://www.dropbox.com/s/sddikc3qzv37zyx/Horizontal%20Orbital%20Insertion%20Simulator%20v1.0.xlsx?dl=0

Booyah!

Now to find a way to convert it to .ods...

Minor point: Go= 9.82.

Sweet! This is gonna save me so much time!!

-Slashy

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Then this statement

Is wrong. If horizontal in your model means θ = 0, then horizontal can't also result in θ < 0.

At higher velocities, means V > V_orbital, which is out of the range of my model. So he clearly wasnt talking about my model's theta, but rather, your model's theta/phi, since mine stops at V=V_orbital.

If you are going above V_orbital such that you have an elliptical orbit, after you pass periapsis, "horizontal" will be below "prograde".

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I think you can surmise that it would be, simply from your personal experience launching from airless bodies. Any V above Vo is going to exhibit vertical acceleration. It's unavoidable.

The argument seems to be whether or not this is important at V< Vo.

I think the consensus here (please correct me if I'm mistaken) is that it is.

Best,

-Slashy

No, there isn't consensus, and no one has presented any reliable evidence that there is.

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No, there isn't consensus, and no one has presented any reliable evidence that there is.

I have shown the practical difference between our two models on the first page of LD's thread:

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.

As you can see, the difference is only significant for low TWR. LD has said that his model assumes a small distance traveled/angle changed during the burn, which automatically means his model requires a high TWR for accuracy, or else it violates its own assumptions (how high a TWR it requires is unclear).

Our models are both accurate at high TWR. At low TWR, only my model is accurate.

Edited by arkie87
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At higher velocities, means V > V_orbital, which is out of the range of my model. So he clearly wasnt talking about my model's theta, but rather, your model's theta/phi, since mine stops at V=V_orbital.

If you are going above V_orbital such that you have an elliptical orbit, after you pass periapsis, "horizontal" will be below "prograde".

In my model, phi is also defined as horizontal.

Further, the sentence that describes figure 4.8 states "[the flight path angle] is positive when the velocity vector is directed away from the primary". This is the situation when V > V_orb, so if anything, pointing prograde when V > V_orb should give a positive flight path angle.

But in either case, pointing prograde isn't theta equal zero as you stated.

I think horizontal is relevant reference direction when below orbital velocity; and prograde is relevant reference direction at or exceeding orbital velocity.
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In my model, phi is also defined as horizontal.

Further, the sentence that describes figure 4.8 states "[the flight path angle] is positive when the velocity vector is directed away from the primary". This is the situation when V > V_orb, so if anything, pointing prograde when V > V_orb should give a positive flight path angle.

But in either case, pointing prograde isn't theta equal zero as you stated.

I think the confusion is that it is well know that burning prograde, when above orbital velocity, is most efficient due to Oberth effect. You do not want to burn horizontal, since we established that horizontal and prograde are two different directions, unless you are in a circular orbit or at apoapsis/periapsis in an elliptical one.

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The evidence you cite is here:

wfe7Mti.png

shows, at TWR = 1.5, estimated dV costs for your model and mine were 856.3 m/s and 893.4 m/s, respectively. The difference is 37.1 m/s, and less than 5%...

How can that be used as evidence to qualify:

I have shown the practical difference between our two models ...

:confused:

Addendum:

Further, if you push the TWR even lower, then both models very clearly indicate that you're wasting nearly 20% or more the dV used to get off the ground.

If you believe either model, why would you build a lander with TWR <1.5!? Both models clearlydemonstrate increasing TWR is a dV saving strategy. Remember this was the reason both models were created in the first place!

Edited by LethalDose
Addendum
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The evidence you cite is here:

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

shows, at TWR = 1.5, estimated dV costs for your model and mine were 856.3 m/s and 893.4 m/s, respectively. The difference is 37.1 m/s, and less than 5%...

How can that be used as evidence to qualify:

:confused:

I have shown the practical difference... In practice, difference is small... (though it might get a lot larger for TWR ~= 1)

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Our models are both accurate at high TWR. At low TWR, only my model is accurate.
I have shown the practical difference... In practice, difference is small... (though it might get a lot larger for TWR ~= 1)

That's a pretty freaking small margin to use to stake your claim that you're the only one to have an "accurate" model.

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What is your model's result for TWR = 1, 1.1, 1.2, 1.3, 1.4?

Besides, i have shown/am willing to bet that for high TVR (i.e. even smaller moons), the models will diverge even further, since very little fuel mass is burned during ascent, so TWR @ burnout will be approximately equal to TWR at liftoff.

If you dont believe me, give me some data points to compare, and I will gladly do it. :D

(though now anyone can since i have released an excel version)

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Addendum:

Further, if you push the TWR even lower, then both models very clearly indicate that you're wasting nearly 20% or more the dV used to get off the ground.

If you believe either model, why would you build a lander with TWR <1.5!? Both models clearlydemonstrate increasing TWR is a dV saving strategy. Remember this was the reason both models were created in the first place!

This is true -- the overall gist -- increasing TWR is good for efficiency-- is still there, but the specific value of TWR needed for, say, 90% efficiency, will vary between our models, and the difference might change for different planets or engines. You have only provided data for one body so far, and i cannot get your excel sheet to work (perhaps due to vlookup?)

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What is your model's result for TWR = 1, 1.1, 1.2, 1.3, 1.4?

Under the following conditions for Mun escape (repeated from OP for clarity):

  • m_0 = 2 t
  • ISP = 350 s
  • Starting alt = Sea Level (0 m)

[table=width: 250]

[tr]

[td]TWR[/td]

[td]Total dV[/td]

[td]Eff[/td]

[/tr]

[tr]

[td]1[/td]

[td]1013.8[/td]

[td]75.6%[/td]

[/tr]

[tr]

[td]1.1[/td]

[td]977.5[/td]

[td]78.4%[/td]

[/tr]

[tr]

[td]1.2[/td]

[td]948.5[/td]

[td]80.8%[/td]

[/tr]

[tr]

[td]1.3[/td]

[td]924.9[/td]

[td]82.9%[/td]

[/tr]

[tr]

[td]1.4[/td]

[td]905.5[/td]

[td]84.7%[/td]

[/tr]

[/table]

Same information to a 10 km AP instead:

[table=width: 250]

[tr]

[td]TWR[/td]

[td]Total dV[/td]

[td]Eff[/td]

[/tr]

[tr]

[td]1[/td]

[td]763.5[/td]

[td]74.5%[/td]

[/tr]

[tr]

[td]1.1[/td]

[td]734.6[/td]

[td]77.4%[/td]

[/tr]

[tr]

[td]1.2[/td]

[td]711.6[/td]

[td]79.9%[/td]

[/tr]

[tr]

[td]1.3[/td]

[td]692.9[/td]

[td]82.1%[/td]

[/tr]

[tr]

[td]1.4[/td]

[td]677.6[/td]

[td]83.9%[/td]

[/tr]

[/table]

Edited by LethalDose
Corrected table values.
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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.

Edited by LethalDose
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If you believe either model, why would you build a lander with TWR <1.5!? Both models clearlydemonstrate increasing TWR is a dV saving strategy. Remember this was the reason both models were created in the first place!

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

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