Effect of initial TWR on orbit dV cost

[Deddly]

I'm completely neutral on this discussion, but I'd like to just say that I'm somewhat disturbed that most (but not all) of you have been using your education to look down your noses at each other. I think you have been very fortunate that this thread hasn't been closed.

OP said right from the start:

I'm also basically opening this up for "peer-review"

So I think it's a shame that it's turned into a "my model is better than yours" discussion.

[arkie87]

He just directly explained the terms, without any 'hand-wavy' statements like 'on the other side of the equation it's momentum, not force'

I didnt think my resposnes were handwaivy at all. I was referring directly to the equation.

When the equation is written like this:

m(r'' - V_theta^2/r) = sum(F)

The m*V_theta^2/r term is on the inertia side of the equation To treat it like a force, we have to move it to the force side. i.e. we add m*V_theta^2/r to both sides, we arrive at:

m*r'' = sum(F) + m*V_theta^2/r

Here, m*V_theta^2/r is positive in sign and is on the force side of the equation. It therefore acts as a positive force i.e. lift.

That is what i was trying to say.

In general, we can say:

sum(F) - m*a = 0

Since, m*a is negative and on the force side of the equation we can say m*a, i.e. the inertia term, acts as a negative force to oppose motion. This is D'Alembert's principle.

Edited December 28, 2014 by Vanamonde
Snip.

[LethalDose]

Well, now that we've sorted out the issue with "centripetal lift", I figured I might put my degree to use and have a crack at equation 17 like OP actually wanted (apologies for the incredibly ugly text notation):

I took the rearranged version from immediately after the (17) label - it seemed easier to work with.
equation 17: exp(deltaVp / (g0 * Isp))(1 + sqrt(1 + TWR0^2))/m0 = (1 + sqrt(1 + (g * m_end)^2)) / m_end
goal: rearrange for m_end.

There's a lot of ugly variables in the way, so let's get rid of most of them to make life way easier:
Let E = exp(deltaVp/(g0*Isp))
    A = E * (1 + sqrt(1 + TWR0^2)) / m0

Substituting: 
A = (1 + sqrt(1 + (g * m_end / F)^2) / m_end
A * m_end - 1 = sqrt(1 + (g * m_end / F)^2)   // multiply both sides by m_end, move the 1 over.
(A * m_end - 1)^2 - 1 = (g * m_end / F)^2    // square both sides, move the 1 over.
A^2 * m_end^2 - 2A * m_end + 1 - 1 = g^2 * m_end^2 / F^2  // expand all the things!
(A^2 - (g/F)^2) * m_end^2 - 2A * m_end = 0  // Behold! we have a pretty quadratic!

But there's still some constants that make it a bit harder to read:
Let B = (A^2 - (g/F)^2)

Substituting:
B * m_end^2 - 2A * m_end = 0
B * m_end - 2A = 0  // we can disregard m_end = 0 since it's not realistic
m_end = 2A / B.
      = 2A / (A^2 - (g/F)^2).  // uh, yeah. you're on your own from here  (it's ugly. very ugly.)

disclaimer: I make no claims as to the correctness of this, and it was all done in notepad using my head.

If however this is indeed correct, WolframAlpha's apparently not very good with big equations with lots of variables

Seems to be spot on, it's giving the same values as using solver, which is great because now I can calculate a lot of values simultaneously. I was having trouble getting it to work intitially, but that was due to a typo in original document: equations 15-17 had TWR_0^2 when it should have been TWR_0^-2. Document's been updated, I'll have a new spreadsheet up soon.

[arkie87]

LD:

I am using your version of spreadsheet without vlookups.

Can you explain this result I obtained?

It is giving me an efficiency > 0, which shouldnt be possible since TWR < 1 given starting and end mass?

[LethalDose]

The only explanation I have to offer any situation where TWR < 1 is that it's invalid.

You can try to convince me otherwise, but I doubt you will...

TWR < 1 means you can't take off, since thrust can't exceed the force of gravity. For this to work, you would have to just sit there and thrust until you've burnt off enough fuel to hit TWR = 1. The starting conditions should simply be adjusted to represent the starting m_0 to reflect that.

Because the condition is invalid, I haven't included any code to handle TWR < 1, beyond highlighting it in red, indicating that it's invalid.

[arkie87]

In your paper, you say phi_min(t) = arctan(1/TWR) (eq. 8)

Why are you using arctan... shouldnt it be arcsin? Or do i have an old version of the paper?

T sin(phi) = m*g, no?

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If we use arcsin instead of arctan, then my numerical model and your analytical solution match perfectly, when i disable centripetal lift.

For convenience, the solution of the ODE:

dv/dt = TVR*FMR/(1-FMR*t) cos(arcsin((1-FMR*t)/TWR))

is (via Wolfram Alpha, but after manual simplification... which Wolfram Alpha couldnt do for some reason?):

v(t) = c1 - TVR/TWR*X(t) - TVR*ln(m(t)) + TVR*ln(TWR*(X(t)+TWR))

where:

X(t) = sqrt(TWR^2 - m(t)^2)

m(t) = 1 - FMR*t

And c1 is the constant of integration which must be found by enforcing v(t=0) = 0

m, t, and v are dimensionless

I dont think t can be solved for @ V = 1 i.e. V_orbital, given the non-linear nature of the equation.

Edited December 29, 2014 by arkie87

[LethalDose]

In your paper, you say phi_min(t) = arctan(1/TWR) (eq. 8)

Why are you using arctan... shouldnt it be arcsin? Or do i have an old version of the paper?

T sin(phi) = m*g, no?

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If we use arcsin instead of arctan, then my numerical model and your analytical solution match perfectly, when i disable centripetal lift.

For convenience, the solution of the ODE:

dv/dt = TVR*FMR/(1-FMR*t) cos(arcsin((1-FMR*t)/TWR))

is (via Wolfram Alpha, but after manual simplification... which Wolfram Alpha couldnt do for some reason?):

v(t) = c1 - TVR/TWR*X(t) - TVR*ln(m(t)) + TVR*ln(TWR*(X(t)+TWR))

where:

X(t) = sqrt(TWR^2 - m(t)^2)

m(t) = 1 - FMR*t

And c1 is the constant of integration which must be found by enforcing v(t=0) = 0

m, t, and v are dimensionless

I dont think t can be solved for @ V = 1 i.e. V_orbital, given the non-linear nature of the equation.

Nope, you're right as far as the arctan goes... I can't remember why I was using that. This is why I opened this to peer review, that looks like a mistake...

Let me go over it again and see what comes up. I don't see an easy published substitution for acos(sin(x)), but on [0, pi/2], it appears to be just pi - x. I'll probably have to revise the document. nope that wrong, too.

[arkie87]

Nope, you're right as far as the arctan goes... I can't remember why I was using that. This is why I opened this to peer review, that looks like a mistake...

Let me go over it again and see what comes up. I don't see an easy published substitution for acos(sin(x)), but on [0, pi/2], it appears to be just pi - x. I'll probably have to revise the document. nope that wrong, too.

Use Wolfram Alpha to integrate it. It gave me an analytical solution (though matlab gave up)

[Anquietas314]

Use Wolfram Alpha to integrate it. It gave me an analytical solution (though matlab gave up)

It's worth mentioning that even if wolfram fails to find an analytical solution, doesn't mean there isn't one

Also, what exactly are you guys trying to integrate or solve? Between the slightly disjointed posts and the PDF I can't make heads or tails of it

[LethalDose]

So cos(arcsin(x)) is basically just a circle. Specifically, for x[0,], its a quarter circle in the first quadrant.

I got a integral result from wolfram that looks very similar to the original (i.e. roots and logs). Curious if there's a sol'n for this too. Can't work on this now though.

[arkie87]

So cos(arcsin(x)) is basically just a circle. Specifically, for x[0,], its a quarter circle in the first quadrant.

I got a integral result from wolfram that looks very similar to the original (i.e. roots and logs). Curious if there's a sol'n for this too. Can't work on this now though.

There is an analytical solution, since the non-dimensional case got a solution, and Wolfram Alpha can provide it

Let me know what result you get...

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An Excel Spreadsheet with a Macro to solve for the burn time can be found here: here