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Just be careful when calculating staged designs, the dry mass for a stage actually includes the fuel in stages above it (this threw me when I first started calculating by hand). Aside from that, it sounds like you're ready to calculate delta-V for existing ships now.

The next step will be designing for a delta-V budget, so instead of asking "How much delta-V does this design have?" the question becomes "How do I change this design to have x m/s of delta-V?"

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Correct iron crown, now I can just solve for x.

One thing I'm not quite understanding is why we wouldn't use the local gravity of the planet we are orbiting? Is it because the ISP is calculated from kerbins g?

And I have no fears of the kraken, frequent offerings are given to tame the beat.

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One thing I'm not quite understanding is why we wouldn't use the local gravity of the planet we are orbiting? Is it because the ISP is calculated from kerbins g?

It's for historical reasons.

Properly speaking, Isp isn't in the Rocket Equation, it's exhaust velocity that matters. But in the postwar US rocketry program there was a mix of Imperial using Americans and metric using Germans, so there was confusion about the units for any given exhaust velocity. It was decided to make a new term, specific impulse, with units in seconds because seconds are the same in both systems. To make it easy to convert a well known constant, Earth gravity, was used as any rocket scientist would have this number memorized.

Things would be a lot easier to understand if we all just used exhaust velocity instead, but there is now a lot of inertia behind specific impulse and its units.

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Things would be a lot easier to understand if we all just used exhaust velocity instead, but there is now a lot of inertia behind specific impulse and its units.

About this, would it be possible to convert the Isp values IG to exhaust velocities through a mod or something ?

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Ok so I am still struggling here a bit.

I am just not sure how to calculate the "ln" function when solving for x. I feel like I am doing it completely wrong, been awhile since I've really had to put forth my algebra skills. Please help me out here:

DV needed: 7000

Dry mass: 21.7t

Wet mass: "X"

Ve: 3433.5 (poodle)

7000=3433.5*ln(x/21.7)

2.03873598=ln(x/21.7)

44.240571=ln(x)

This is where I am stumped, what is the reverse of the ln function? Or maybe I am just going at this all wrong. Either way, I don't have a number that I am confident of.

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ex is the reverse of ln(x), e being the natural logarithm base (approximately 2.718).

Edit: You've also got some algebra errors, should be this

7000 = 3433.5*ln((1.125x+21.7)/(21.7+0.125x)) (1.125x and 0.125x is to account for tank mass, x is propellant mass.)

2.039 = ln((1.125x+21.7)/(21.7+0.125x)

7.681 = (1.125x+21.7)/(21.7+0.125x)

and so on.

Edited by Red Iron Crown
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Sure:

7000 = 3433.5*ln((1.125x+21.7)/(21.7+0.125x) (1.125x and 0.125x is to account for tank mass, x is propellant mass.)

2.039 = ln((1.125x+21.7)/(21.7+0.125x))

7.681 = (1.125x+21.7)/(21.7+0.125x)

7.681(21.7+0.125x) = 1.125x+21.7

166.68+0.96x = 1.125x+21.7

144.98 = 0.165x

878.7 = x

So, 878.7 tons of propellant (+tank mass) to hit 7000m/s of delta-V in a single stage. You'd be much better off staging for such a dV budget using a Poodle. Or switch to LV-Ns for greater exhaust velocity.

Edited by Red Iron Crown
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Thank you much for the info red. My plan did include drop tanks but this is looking much bigger than originally planned, I am guessing that I would just have to pick a combination of tanks that gives me the total of 878.7t of fuel correct?

Lets say I wanted to add a bit more thrust and added 4 909s. How would that affect the equation having different isp? Would I add them in figuring for Ve, so:

Ve=9.81*(350+345+345+345+345)

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For mixed engine clusters, use a thrust-weighted average of the Isps to compute the net Isp for the cluster (F is the thrust for each engine):

5649bb43490708471f734294d36eb565.png

Then multiply by 9.81* to get Ve.

So for a Poodle and four 909s:

Isp = (250+200)/((250/350)+(200/345)) <-- I've combined the 909s into one for simplicity, you can do this with any engines that have the same Isp.

Isp = 347.76s

*Approximated. Hi NathanKell!

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Ugh this just keeps getting more and more complex, but I'll give it a shot:

Here are my actual numbers:

M1=X

M2=17.6

Ve=3433.5

DV=6800 m/s

6800=3433.5*ln((1.125x+17.6)/(.125x+17.6))

1.98=ln((1.125x+17.6)/(.125x+17.6))

7.246=(1.125x+17.6)/(.125x+17.6)

7.246(.125x+17.6)=1.125x+17.6

.90575x+127.5296=1.125x+17.6

109.9296=.21925x

501.34=x

And if I went with the 1 poodle and 4 909's:

Ve=9.81((250+60+60+60+60)/((250/350)+(240/345))

Ve=9.81(490/(.7143+.1739+.1739+.1739+.1739)

Ve=9.81(347.542)

Ve=3409.387

6800=3409.387*ln((1.125x+20.1)/(.125x+20.1))

1.9945=ln((1.125x+20.1)/(.125x+20.1))

7.3485=((1.125x+20.1)/(.125x+20.1))

.9186x+147.7049=1.125x+20.1

127.605=.2064x

618.24=x

I think a different setup would be best. But I hope I have the concept down now. And it looks like you got the 909 thrust wrong or I am still missing something. Aren't they at 60KN.

If I went with multiples of the same engine like 2 poodles. The Ve would still be 3433.5 correct?

Edited by ForScience6686
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Er whoops, that's what I get for using old numbers from memory, the 909 used to be 50kN.

Another important thing is that a single stage has a finite maximum amount of delta-V (based on Isp and tank wet:dry ratio), and the propellant mass goes up exponentially the closer you approach that maximum. Two 3500m/s stages will mass much less combined than a single 7000m/s stage.

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So one last question and I should be set.

4 lvn would equate to 46.43t fuel required

while 1 lvn would equate to 31.45t

Ve would be constant in this correct? With the only difference being mass.

and TWR of the 4 lvn setup would be:

TWR=240/(74.33*9.81)

TWR=.3:1?

74.33 I don't think is quite correct as it only equates the mass of the vehicle and engines plus the result of x=46.43t fuel required, so no tank mass is in that number correct?

I guess that's more than one question...

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Adding more of the same engines will increase the TWR but lower dV available (as engines are dry mass), so your numbers look ok at first glance.

You need to multiply the propellant mass by 1.125 to account for the tanks' dry mass. And when you go to build the thing you won't be able to hit those numbers exactly because tanks are only available in discrete sizes.

Finally, if using LV-Ns be sure to use LF-only tanks, like the Mk1 or Mk3 LF fuselages. Otherwise the tanks will have more dry mass per ton of propellant and throw off your calculations.

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Allow me to make a suggestion. I have seen many math and physics students insert numbers into equations first, then solve for the desired quantity. This is more work and less instructive than solving the equation first, and then inserting numbers. The solution is meaningful before you put in numbers.

It is more work because you have to resolve the equation if you want to change any of the numbers. It is also more to write down, since numbers are longer than symbols, usually. It is less instructive because in the end all you have is number soup. You can't easily look at a pile of arithmetic and see the meaning or the effect of each term.

Let a = 1/8. The fuel mass is x. The dry mass is a*x. The full tank + fuel mass is (1+a)*x. Now solve for x:

dv = ve ln((m + (1+a)*x)/(m + ax))

e^(dv/ve) = (m + (1+a)*x)/(m + ax)

x = m (e^(dv/ve) - 1) / (1 - a (e^(dv/ve) - 1))

x = m / ( 1/(e^(dv/ve) - 1) - a)

I skipped a couple simplification steps along the way, but nothing difficult. Now you can reuse this equation, changing ve or m without resolving from the rocket equation. You could plot the fuel mass versus dv. You can think and reason about a mathematical function in ways you can't to something like x = 500 tons.

Edit: To elaborate, let's look at the effect of the dry tank fraction, a. I said we could reason about an equation, so let's do so. If a gets bigger, the denominator gets smaller, so x gets larger. If your tanks weigh more per unit fuel, you need more fuel to push the thicker tanks. We already knew this had to be true from our understanding of the rocket equation, but we now know exactly how the dry fuel fraction enters into the wet fuel mass.

Edited by Yasmy
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(To be fair, developing a core understanding of the rocket equation is pretty key to this game)

Exactly. I'm no math genius, but I forced myself to at least grasp the concepts and even manually calculate my own dV values a few times before I installed KER. Having a good understanding of how it's calculated makes designing ships much quicker and easier even though I now use KER on all my saves (even my "stock" install has KER with all the my HUDs pre-configured as the only mod so they're the same every time I copy the directory for a new install)

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