Help with calculating delta v and twr

[GoSlash27]

So now that I've established the upper stage, I need to repeat the same process for the lower stage with a couple adjustments because I'm working in atmosphere.

Specifically, the Isp will be the average of vacuum and surface and the thrust will be surface thrust.

Average Isp is (Isp (atm)+ Isp (vac))/2

Sea level thrust is T(Isp(atm)/Isp(vac))

Starting with my upper stage (4.11t), I'll add .05t for a decoupler.

Payload= 4.16t

Lower stage mission requirement: 4.16 tonnes payload, 1,800 m/sec DV, and minimum acceleration of 1.4G. Engine specs LV-T30: Thrust= 201KN Mass= 1.25 t, Isp= 290 sec

Lift capacity is 201KN/(1.4x9.81)=14.64t

Rwd is e^(1800/(290*9.81))= 1.883

Fuel percentage is (1.883-1)/1.883=.469

Fuel mass is .469x14.64t= 6.87t

Tank mass is 6.87t/8= .858t

Payload mass is 14.64t-1.25t-6.87t-.86t= 5.67t

Again, our desired payload is less, so we'll use 1 engine

N= roundup(4.16/5.67),0 = 1

Final tank mass is (1.883-1)(4.16t+1.25t)/(9-1.883) = .671t

Fuel mass is 8x .671t= 5.37t

Total mass= 1.25t+4.16t+5.37t+0.67t= 11.44t

Throttle setting is 11.44/14.64= .78, or 78%

- - - Updated - - -

So that's how the process works.

If you put this in a spreadsheet, it will solve the problem for all engines simultaneously, allowing you to see which engines result in the lightest, cheapest, and least-complicated stages.

An example of a simple spreadsheet implementation:

As you can see, the lightest lower stage would have used an aerospike and a 48-7S would also be lighter, but the Aerospike is expensive (and not available) and the 48-7S would require 10 engines, which would be awkward.

Best,

-Slashy

Edited July 1, 2015 by GoSlash27

[Falkenherz]

@Slashy

I am now that far that I am trying to reproduce your steps in my spreadsheet. Sometimes I have the impression that the main reason which scares me away from Math are the abbreviations, so I try to formulate the data a bit more explicit therein.

However, I do not understand the value "maximum acceleration". It must work different to the TWR which I am used to eyeball? How much should I have in space (I use 0,5 TWR)?

Also, I do not understand step 3. What do you mean with "tankage"? Where you get the figure 9 in your formula from (Rwd-1)(Mp+NMe)/(9-Rwd) and how is this formula deduced?

Edited July 16, 2015 by Falkenherz

[Weywot8]

http://forum.kerbalspaceprogram.com/threads/75669-Staging-Methods-Overview

can help inspire some nice setups but calculating the numbers on a spreadsheet gets more tedious with creative fuel/mass saving staging setups, so there is this:

http://garycourt.github.io/korc/

for later in your KSP career.

[Wemb]

... which uses Euler's Number (2.71828)

Which, if you're not a mathematician may seem like a strange and arbitrary number. It's not - e (Euler's number) is up there with 0, 1, À and i as being amongst the five most important numbers in the universe. Not least because of

Wemb

[Red Iron Crown]

What do you mean with "tankage"? Where you get the figure 9 in your formula from (Rwd-1)(Mp+NMe)/(9-Rwd) and how is this formula deduced?

Not to speak for slashy, but I bet the 9 represents mass ratio; the common LFO tanks in KSP have a wet:dry mass ratio of 9:1 (this is a hugely important number).

[Falkenherz]

Thanks! I would never have been able to deduce a formula like this; no idea why this works out like this! The only thing I understand now is that this formula probably does not apply for the rocket fuel tanks when using an LV-N.

[Wemb]

Thanks! I would never have been able to deduce a formula like this; no idea why this works out like this! The only thing I understand now is that this formula probably does not apply for the rocket fuel tanks when using an LV-N.

The Rocket Equation always applies - what may put a spanner in the works of the spreadsheets when looking at the nuke is that it uses fuel in a different way - or rather, it uses different fuel. Or,to be exact, it only uses fuel, but doesn't consume any oxidiser.

So, if you match it up with the normal rocket tanks, any oxidiser you take with you when you're running the nuke engine has to be counted as a dead weight, rather than fuel. The fuel/dry mass ratio will also therefore be worse.

The calculations all work the same, though.

If you want to use the nuke's efficiently, use the spaceplane parts, some of which accommodate only liquid fuel rather than oxidiser - or install some mod that lets you change the oxidiser for fuel in the normal parts.

Wemb

[GoSlash27]

Not to speak for slashy, but I bet the 9 represents mass ratio; the common LFO tanks in KSP have a wet:dry mass ratio of 9:1 (this is a hugely important number).

All,

Apologies for not checking back in on this thread. RIC is absolutely correct. Full fuel tanks weigh 9 times as much as empty tanks when dealing with LF&O tanks (and the small jet fuel tanks as well). RIC is a KSP greybeard who's been there, done that

AFA how it's derived, it's all algebra. Unfortunately, the forum doesn't support easy-to-follow math formatting.

Basically, you take the basic rocket equation DV= 9.81Isp*ln(Mw/Md) and reverse it to get into the "Mw/Md" part.

Then you define Mw as "Md+Mf". Md would equal the mass of the payload + mass of empty tanks + mass of engines( number of engines). Since the mass of a full tank is 9x the mass of an empty tank, it's easy to make one the function of the other.

Walking through the entire process provides a messy wall of text, but I've outlined it in my tutorial on the reverse rocket equation here:

http://forum.kerbalspaceprogram.com/threads/102809-The-reverse-rocket-equation-explained

Best,

-Slashy