Trickery With Delta V

[thereaverofdarkness]

1. If your empty mass was ZERO the Delta-V is undefined (not infinite) due to divide by zero error in the function. However, (to be more precise), you can say that Delta-V approaches infinity as empty mass approaches zero. But, the logarithmic function demonstrates the diminishing return of this effect. You quickly reach a point where you have to decrease empty mass a lot to get a small change in Delta-V.

It's only undefined in the equation you chose to use to determine the value, as that equation makes use of division in which you get a denominator of zero. In actuality, having 100% fuel mass would lead to an actually infinite delta-v, assuming a theoretical fuel fluid that is non-particulate, and no speed limit c. In real life, it would never happen because of a variety of constraints, but the actual physical equation not only approaches infinity, it reaches it.

It is not a logarithmic function but an exponential function. It may appear logarithmic due to physical constraints such as dry mass, c, or "imperfect" real fluid dynamics, but if you take away all components other than the thrust vs rate of consumption (unnecessary but can be included), fuel mass, and ISP (also unnecessary but can be included), you get an exponential function.

[Tank Buddy]

Yes, that is a more clear explanation, thank you.

[capi3101]

So in my original derivation, this was the last step I had correct:

ln(M_48-7S/M_{o, 48-7S}) = 1.11429*ln(M_LV-909/M_{o, LV-909})

So if I take e of both sides and do it correctly, I should now have:

M_48-7S/M_{o, 48-7S} = (M_{, LV-909}/M_{o, LV-909})^1.11429

If you then sub in the M = 9M_o relationship, you have:

9M_{o, 48-7S}/M_{o, 48-7S} = (9M_{o, LV-909}/M_{o, LV-909})^1.11429

And now take out the deadmass in each part of the equation:

(9M_{o, 48-7S} + M_{dead, 48-7S})/(M_{o, 48-7S} + M_{dead, 48-7S}) = ((9M_{o, LV-909} + M_{dead, LV-909})/(M_{o, LV-909} + M_{dead, LV-909}))^1.11429

Set M_{o, 48-7S} = x and M_{o, LV-909} = y to simplify and you have

(9x + M_{dead, 48-7S})/(x + M_{dead, 48-7S}) = ((9y + M_{dead, LV-909})/(y + M_{dead, LV-909}))^1.11429

Now that should be as far as that can go without putting in specific terms.

Okay. So now for the case where M_{dead, 48-7S} = 0.7, M_{dead, LV-909} is 1.1 and y = 1:

(9x + 0.7)/(x + 0.7) = ((9 + 1.1)/(1 + 1.1))^1.11429

(9x + 0.7)/(x + 0.7) = ((10.1)/(2.1))^1.11429

(9x + 0.7)/(x + 0.7) = 5.75519

(9x + 0.7) = 5.75519(x + 0.7)

(9x + 0.7) = 5.75519x + 4.02863

3.24481x = 3.32863

x = 1.02583

Let's verify:

ln(10.1/2.1) * 9.81 * 390 = 6008.95 m/s

ln (((9*1.02583) + 0.7)/ (1.02583 + 0.7)) * 9.81 * 350 = dV

ln (9.93249/1.72583) * 9.81 * 350 = dV

dV = 6008.98 m/s

The difference can be attributed to rounding errors, so this derivation is correct. Hallelujah.

For the case where y = 0.5 tonnes, x = .42875 tonnes, so the "tipping point" is somewhere between 0.5 tonnes and 1 tonne.

Now I suppose the trick is going to be to figure out exactly where the "tipping point" is - the point at which it becomes better to use one engine over another given cases of x amount of dead mass. Should be a matter of setting x=y, right? Let's try it for the same case:

(9x + 0.7)/(x + 0.7) = ((9x + 1.1)/(x + 1.1))^1.11429

Oof. Who wants to tackle that equation?

Edited November 29, 2013 by capi3101

[numerobis]

Just solve it numerically.

There was a similar thread in general discussion at about the same time. I threw up my hands and said it was too complicated to solve everything, so I computed the break-even amount of propellant for equal *momentum* using two different engines (T30 vs Poodle), adding fuel to the T30 to make up for its reduced mass. While not exactly what we actually want I don't think, it was a good approximation, at least for the KSP space.

[capi3101]

So brute force it, eh? Very well. Set up a spreadsheet to solve the equation for x given values of y, where y is incremented in values of 0.0625 tonnes (the dry mass of an FL-T100 fuel tank, the smallest tank for which the 9:1 assumption holds - I tend to think of fuel requirements in FL-T100 equivalents; e.g. an FL-T400 is 4 FL-T100 equivalents, an FL-T800 is 8 FL-T100s, etc.). And I'll just go ahead and report the results rather than post screenies of the spreadsheet.

Okay...for the case where the non-engine dead mass is 0.6 tonnes, the tipping point is somewhere between 14-15 FL-T100 equivalents (so with 14 FL-T100s, you're better with the 48-7S; with 15, you're better with the LV-909).

14 FL-T100s equivalents = a dry mass 0.875 tonnes.

15 FL-T100 equivalents = 0.9375 tonnes.

16 FL-T100 equivalents = 1 tonne, our initial test case.

For the case where there is no non-engine dead mass, the tipping point is somewhere between 20-21 FL-T100 equivalents.

For the case where the non-engine dead mass is 1.25 tonnes (a Mk1 Cockpit), the tipping point is somewhere between 9-10 FL-T100 equivalents.

For the case where the non-engine dead mass is 2.5 tonnes (a Mk2 Lander Can), the tipping point is somewhere between 3-4 FL-T100 equivalents.

For the case where the non-engine dead mass is 4 tonnes (a Mk1-2 Command Pod), it's always better to go with the LV-909.

So it looks like the higher the non-engine dead mass, the more likely it is that the LV-909 is the better choice. Not a surprise really.

The highest non-engine dead mass for which I'm detecting any delta-V advantage whatsoever with the 48-7S is 3.098639 tonnes (3.098640 tonnes favors the LV-909).

21 FL-T100 equivalents is 1.3125 tonnes, 11.8125 tonnes fuel, and 7,330.07 m/s for the LV-909.

I should so totally go through this whole exercise with nukes. Anybody want to see that, and what engine should I compare it against?

Edited November 29, 2013 by capi3101