Hand calculating with multiple motors

[zeppelinmage]

I have a lander in Munar orbit that has 4x nuke motors on. I need to calculate delta-V and burn time (specifically, the fuel mass required for a certain delta-V) for it.

My question relates to how multiple motors work together. Do I add the I_sp together? Average? Other? Fuel consumption would add, I would think?

Edited January 9, 2014 by zeppelinmage

[Silverchain]

If the specific impulse is the same for all motors, you use it exactly as is.

If you have multiple motors with different specific impulse, the combined impulse is ( Thrust1 + Thrust2 + (...) + ThrustN ) / (Thrust1/isp1 + Thrust2/isp2 + (...) + ThrustN/ispN) if I remember right.

e.g. an LV-N and an LV-T30 the combined isp would be:

60 + 215 / (60/800) + (215/370)

= about 420

[wasmic]

I think you'd take the average thrust-adjusted Isp. So if you have two engines with the same thrust, you'd just take the average Isp.

[zeppelinmage]

EDIT: Crap, I chose the wrong numbers... stand by.

[smunisto]

In other news - why would you need 4 NERVAs for a munar lander 0_o 6 weight of each.

[zeppelinmage]

Balance, and fuel efficiency. I used to use 909s, but thought I'd try nukes this time around. And I can't fly a 2-motor ship stably, so I have to use 4. (In hindsight, I could have probably used 3...)

But, I can't get the equation to behave. Basically, I'm trying to figure out how much fuel I need to keep in my tanks to be able to take off from the Mun and retrn to Kerbin. My math:

ÃŽâ€V = I_sp * g * ln(m₀/m_f)

1000 = 800 * 1.63 * ln(m₀/16.36)

0.767 = ln(m₀/16.36)

e^0.767 = m₀/16.36

2.153 = m₀/16.36

m₀ = 16.36*2.153

m₀ = 35.223

My ship is 22.36t and only has 6t of fuel on it. Kerbal Engineer says I currently have on the order of 2500m/s of ÃŽâ€V, so that doesn't match at all.

If I add all the I_sp together, that provides 3200s:

ÃŽâ€V = I_sp * g * ln(m₀/m_f)

1000 = 3200 * 1.63 * ln(m₀/16.36)

0.192 = ln(m₀/16.36)

e^0.192 = m₀/16.36

1.211 = m₀/16.36

m₀ = 16.36*1.211

m₀ = 19.817

Which comes out to 3.457t of fuel, or 691.467L (311L of liquid fuel). That makes a little more sense. Is that right?

[Rhomphaia]

g in this case is a constant, not the gravity of the body in question., so 9.82 rather than 1.63.

[capi3101]

My ship is 22.36t and only has 6t of fuel on it.

Alright - to use your terms: ÃŽâ€V = I_sp * g * ln(m₀/m_f)

m₀ = 22.36

m_f = 16.36

I_sp = 800

g = 9.81 (always use standard gravity here, regardless of the reference body)

ÃŽâ€V = 800 * 9.81 * ln(22.36/16.36) = 2451.984 m/s

Now, what you're trying to figure out is how much fuel you'll need for the 1000 m/s mark, right? So...

1000 = 800 * 9.81 * ln(x/16.36]), where x=m₀

ln(x/16.36) = .12742

x = 16.36 * e^.12742 = 18.58 tonnes

18.58 - 16.36 = 2.22 tonnes

So when you're down to 18.58 tonnes, you have 1,000 m/s of delta-V remaining.

Actually, let's try it this way instead:

1000 = 800 * 9.81 * ln(22.36/x) where x=m_f

ln(22.36/x) = .12742

x = 22.36/e^.12742 = 19.68 tonnes

22.36 - 19.68 = 2.68 tonnes

So you'll need 2.68 tonnes of fuel for the first 1,000 m/s of delta-V.

Edited January 9, 2014 by capi3101

[Smidge204]

g in this case is a constant, not the gravity of the body in question., so 9.82 rather than 1.63.

This. An easy mistake to make.

The actual equation is dV = V_e * ln(m_o/m_f), where V_e is the exhaust velocity in m/s. Specific impulse is this value divided by gravity, in m/s², giving a final value in units of seconds.

And that gravity is Earth's (Kerbin's) gravity... so you need to use that value in your equation to restore the proper value of V_e.

=Smidge=

[capi3101]

^^^

V_e ≡ I_sp * G_o; don't want to confuse you - people tend to expand that one out.

[draeath]

I see people using both 9.82 and 9.81 for g around here.

Which one is it actually? It would be so much nicer had Squad used the real g value instead :/

[capi3101]

I've seen both; 9.81 is Kerbin's surface gravity so I use it. Don't know if it's actually right or not and to be honest it doesn't throw things off by that much either way.

[Soda Popinski]

According to Wikipedia, 9.8066. So that rounds up to 9.81. Unless you want to go with ft/sec, then you use 32.16

[Smidge204]

It actually varies by altitude in real life, so 9.81 is good enough for game purposes and is slightly more conservative, even if the difference is negligible.

=Smidge=

[UmbralRaptor]

I see people using both 9.82 and 9.81 for g around here.

Which one is it actually? It would be so much nicer had Squad used the real g value instead :/

Within KSP (but not in real life), 9.82. This becomes obvious when you test it. eg:

g₀ == thrust/(I_sp * mass_flow)

g₀ == 1500/(330 * 0.46288) == 9.81994...

[draeath]

It does not vary, we are talking about Standard Gravity. 9.80665 m/s² in real life.

Squad either rounded/truncated or mistook it for another value, but as stated I've seen 9.81 and 9.82 - but I would like to know what is actually being used.

EDIT: ninja'd by UmbralRaptor - thank you for the confirmation!

EDIT2: I should point out this whole time I've been asking about g₀ not g - UmbralRaptor understood what I meant apparently.

Edited January 9, 2014 by draeath

[UmbralRaptor]

EDIT2: I should point out this whole time I've been asking about g₀ not g - UmbralRaptor understood what I meant apparently.

\o/

As best I can tell, g (not g₀!) for Kerbin's surface is 9.81 m/s². The discrepancy showed up several versions ago.

[zeppelinmage]

Alright - to use your terms: ÃŽâ€V = I_sp * g * ln(m₀/m_f)

m₀ = 22.36

m_f = 16.36

I_sp = 800

g = 9.81 (always use standard gravity here, regardless of the reference body)

ÃŽâ€V = 800 * 9.81 * ln(22.36/16.36) = 2451.984 m/s

Now, what you're trying to figure out is how much fuel you'll need for the 1000 m/s mark, right? So...

1000 = 800 * 9.81 * ln(x/16.36]), where x=m₀

ln(x/16.36) = .12742

x = 16.36 * e^.12742 = 18.58 tonnes

18.58 - 16.36 = 2.22 tonnes

So when you're down to 18.58 tonnes, you have 1,000 m/s of delta-V remaining.

Actually, let's try it this way instead:

1000 = 800 * 9.81 * ln(22.36/x) where x=m_f

ln(22.36/x) = .12742

x = 22.36/e^.12742 = 19.68 tonnes

22.36 - 19.68 = 2.68 tonnes

So you'll need 2.68 tonnes of fuel for the first 1,000 m/s of delta-V.

Excellent, thanks! The first one is what I'm going for - at what point in my tank will I have exactly 1000m/s left. So, by conversion, that wold equate to 444L of total fuel, or 199.8L of the liquid portion.

Yeah, the difference between "local" g and g₀ is what confused me. I couldn't figure out if you used the g of your local body, or just 9.82ms^-2. Local body made more sense (at the time) as I thought the force on the craft would change depending on what gravity you're fighting against. Rather than simply the force of the propellant out the back of the motor.

[Borogove]

Yeah, the difference between "local" g and g₀ is what confused me. I couldn't figure out if you used the g of your local body, or just 9.82ms^-2. Local body made more sense (at the time) as I thought the force on the craft would change depending on what gravity you're fighting against. Rather than simply the force of the propellant out the back of the motor.

Nope -- all the delta-V calculations ignore "local" gravity. The use of g in the equations is because of the conventional use of the unit "seconds" for Isp; if you factor out g, Isp is in different units (m/s) -- the effective velocity of the rocket exhaust -- and has larger numeric values. See http://en.wikipedia.org/wiki/Specific_impulse.