Planning your rockets: What is Delta-V?

[Xannari Ferrows]

Hello everyone! Today we're going to answer a question that not many people ask themselves when building rockets. What is Delta-V? What does it do? Is it important? Why should I care about it? Today we will be exploring these questions and letting you know how to plan your rockets.

 

However, I need to crush your expectations first:

This is much more complicated then simply getting a mod to calculate it for you. Mods like Mechjeb and Kerbal Engineer are 2 such examples.

Our objective here is to explore the mathematics.

 

 

Spoiler

 

If you clicked on this, you need a quick rundown on Logarithms. So what are Logarithms exactly? They're basically Exponents in reverse. The overall objective is to find your Exponent. Before I give you any examples, I will say that you WILL need a calculator for this.

There are a few rules to Logarithms you may not know.

1. Logarithms are reversible. They work in the equation Log_bx = y, which can be reversed into b^y = x. This works for any number set and will be important.

2. Logarithms have a constant (e / ln) Whenever you see Log_e Or Ln, The value for b is 2.7182818285, which of course requires a calculator.

3. The Rocket Equation only uses ln, so you technically don't need advanced knowledge on these to use it. However, we will still be talking about different kinds.

4. At any time you see Log without a number taking the place of "b", it is always 10.

This is all you need to know, so let's get to an example: Log(100) Look back to rule 4 if you don't understand it.

So Log(100) is the most basic example of Logarithms. The reference number is 10, and the final result is 100. Remember how to reverse Logarithms? Just use that:

10^x = 100. What fits your x? 2? That is correct, and is our answer: Log(100) = 2.

Of course, that was fairly easy, but let's up the ante. What say... Log₂(64)?

Obviously the reference has changed from 10 to 2, so lets reverse this to make it easy:

2^x = 64. Can you guess the answer? I'll let you think about it........................................................................................................................... Got it? ...Yes! It is 6! You're becoming an expert quickly! Log₂(64) = 6! However, I don't think it's safe to let you go off and work for NASA yet, so let's do another.

Log₇(343)? You know the drill; reverse it out: 7^x = 343. Find your answer either with a calculator or with brain power. In the end, x = 3.

Log₇(343) = 3.

In conclusion, Logarithms are simply how many times "b" is multiplied by itself to get "x", and that result is "y".

By now you should have at least a basic understanding of Logarithms. If you still don't get it, either consult the internet or ask me for further detail.

 

 

If you are here, that means you have everything you need to know.

Here's our formula:

ΔV = (I_{sp •} 9.81_m/s^2) • Ln(M1/M0)

 

See, this is one of those big, scary math equations which is actually very easy once you understand what it means. Since you are here, and therefore probably don't understand what it means, here's a quick rundown on the individual parts:

Isp is the Specific Impulse of your engine. Multiply this by 9.81 [1G] to get your exhaust velocity.

Ln is your natural log

M1 is your full mass [Mass of your rocket structure + fuel]

M0 is your dry mass [Mass of just your rocket structure without the fuel]

Still a little lost?

 

Spoiler

 

If you clicked here, you probably need a little more explanation. Here are the individual parts a bit more explained:

Isp is essentially your fuel efficiency. The faster you accelerate your propellant, the more propulsive energy is transferred into your rocket, which we call thrust. This works because of Newtons' third law.

 

Ln is your natural log, of course. Your B value is equal to 2.7182818285

M1 and M0 are 2 types of masses you should know: M1 is the mass of everything, including your fuel. M0 is the mass of everything as if there was no fuel.

M0 can never be greater than M1.

 

 

By now you should know the individual parts and their functions.

So, let's go ahead and use some examples to help visualize the equations' overall function. Imagine a rocket with 1 ton of structural mass and 1 ton of fuel mass. The engine propelling it all has a 400 Isp.

 

ΔV = (400 • 9.81_m/s^2) • Ln(2/1)

 

 

Spoiler

 

We can go ahead and solve for our Delta-V. Ve = 400 x 9.81_m/s^2 = 3924. Ln(2/1) = Ln(2) = 0.69314718056.

3924 x 0.69314718056 = 2719.9m/s.

 

Alright! First equation done, but lets take it a step further. Our rocket has 30 structural tons and 15 tons of fuel. The engine is a Nuclear Engine (800 Isp).

Clearly this should be a bit harder, right?

 

 

ΔV = (800 • 9.81_m/s^2) • Ln(45/30)

 

 

Think you can solve this one by yourself? Go ahead, and the answer will be below if you can't figure it out.

 

800 x 9.81_m/s^2 = 7848. Ln(1.5) = 0.40546510811

7848 x 0.40546510811 = 3182.1 m/s.mickee for finding that!

 

Delta-V across stages:

 

You may have asked this question in your head: "What about Delta-V across stages?" To which I say, I'm glad you asked! This formula is simple. Even simpler than the last.

 

 

ΔV_t = ΔV₁ + ΔV₂

 

 

This looks different from the last, but it is deceptive. All it is is 2 rocket equations duct-taped together:

 

ΔV =_Stage1(I_{sp •} 9.81_m/s^2) • Ln(M1/M0) + _Stage2(I_{sp •} 9.81_m/s^2) • Ln(M1/M0)

 

Look familiar? Okay, so example time.

Think of a rocket. 32 structural tons and 12 tons of fuel. 6 tons of structural weight, 500 specific impulse engine and 3 tons of fuel used for the first stage. Decouple the first stage. 800 for the second stage with the rest of the fuel used. Think you've got it? Try it out!

 

 

Basically, at the end of the day, Delta-V is the most important thing you can account for. So obviously you want as much as possible. How do you do that? Here are a few things you can do.

1. Increase your specific impulse.

2. Carry more fuel than structural mass.

3. Stage out your rockets across multiple stages.

Increasing your Specific impulse will increase Delta-V, because the number representing Isp will grow larger, thus multiplying a larger factor by your Ln(M1/M0).

Carrying more fuel increases the M1 to M0 ratio. Higher ratios result in bigger numbers. You want the highest you can get.

Stages are important! By shaving excess mass, you increase your M1 to M0 ratio even more.

 

 

EDIT: Those questions at the beginning? Well, maybe I should give those answers instead of giving them to you to answer.

What is Delta-V? Delta-V is a meaning for "Difference in Velocity". Basically how much you can change your current velocity by. 100 m/s to 300 m/s = 200 m/s Delta-V, for example. However, making burns radial to your trajectory will use the same amount of Delta-V, but not alter your velocity. Making a radial burn changes the trajectory you are on.

What does it do? The more you have, the more you can change your velocity, and/or trajectory.

 

Is it important? Absolutely! It is one of the, if not the most important factor to account for when building a rocket, or any spaceship for that matter.

Why should I care about it? If you built all of your rockets without planning your Delta-V out, you would have no idea of it's potential. Always bring extra just in case.

So ends the guide for now.

I know some of you wont get it right away, and that's fine. Read this as many times as you need to. I'll admit, I was reviewing my work for a while. However, I hope in some small way I've been able to help you with a better understanding of the mathematics of Difference in Velocity. (DV, get it?)

Please, please tell me if I missed anything, or perhaps if I need to throw in some extra information. If you have a specific question, however, I will be more than happy to provide you with an answer whenever I can.

Have a nice day to all!

Xannari

Edited December 23, 2015 by Xannari Ferrows
Update

[mhoram]

9.81

As a small addition, it should be noted that as stated in http://wiki.kerbalspaceprogram.com/wiki/Specific_impulse#Conversion_factor

the conversion factor in KSP is not 9.81 m/s², but can rather be approximated by 9.82 m/s².

[Xannari Ferrows]

As a small addition, it should be noted that as stated in http://wiki.kerbalspaceprogram.com/wiki/Specific_impulse#Conversion_factor

the conversion factor in KSP is not 9.81 m/s², but can rather be approximated by 9.82 m/s².

Hi.

This page seems to provide data based on the specific impulse being a constant. However, with lowest mass possible, the specific impulse reaches a peak variable of 290.2, which changes our equation. However, our end result could still be rounded to 9.82. Interesting to know, I actually didn't know this!

However, the purpose of this thread is to give you a real world approximation of your rockets' Delta-V. I will add the addition to the thread, though. Thanks for telling me!

Xannari

[Specialist290]

Impressive work with how you managed to explain the topic so thoroughly in such a relatively small post. I'd say this is one for the Drawing Board

[Xannari Ferrows]

Impressive work with how you managed to explain the topic so thoroughly in such a relatively small post. I'd say this is one for the Drawing Board

Thank you! It'd be an honour to be on the drawing board alongside the other impressive works found there. (Now if you have a physics-less part with fuel and all other physics-less parts, what happens when you accelerate? 0/0 it just breaks the game)

Xannari

[mickee]

ÃŽâ€V = (800 • 9.81_m/s^2) • Ln(45/30)

Think you can solve this one by yourself? Go ahead, and the answer will be below if you can't figure it out.

800 x 9.81_m/s^2 = 7848. Log2.7182818285(1.5) = 0.40546510811

7848 x 0.40546510811 = 3812.1 m/s.

i think it should be 3182.1 m/s , Unless i'm doing something wrong. i'm guessing it was just a typo

calculated with 9.82 m/s^2 the solution should be 3185.3 m/s

[Xannari Ferrows]

i think it should be 3182.1 m/s , Unless i'm doing something wrong. i'm guessing it was just a typo

calculated with 9.82 m/s^2 the solution should be 3185.3 m/s

Oh god I should feel ashamed.

Thank you for pointing this out to me. I must have had the two bass-ackwards or something. You will be credited below for fixing my biggest mistake yet.

Or did I make an approximation? I don't know. Oh god this is embarrassing...

I mean, we could go get an ice cream if it's all the same...

Xannari

[SirThawkz]

This is really useful for calculating Delta-v but your original premise hasn't been answered....What is Delta-V? What does it do? Is it important? Why should I care about it?

Could you answer these, as that would be very useful to newbies such as myself.

[Thunderous Echo]

What is Delta-V? What does it do? Is it important? Why should I care about it?

Delta-v is the amount your rocket can change its velocity. If you have 1 m/s of dv, you can accelerate from 0 m/s to 1 m/s, or 1 m/s to 0 m/s, or 200 m/s to 201 m/s...

Delta-v is important because the more your rocket has of it, the 'farther' it can go, the more planets it can visit, the bigger the moons you can land on. See this: http://mononyk.us/wherecanigo.php

[Xannari Ferrows]

This is really useful for calculating Delta-v but your original premise hasn't been answered....What is Delta-V? What does it do? Is it important? Why should I care about it?

Could you answer these, as that would be very useful to newbies such as myself.

Thunderous echo pretty much hit the nail on the head. I was hoping that by the end of this you'd be able to answer the questions yourself... that's what they're there for...

Although, maybe I shouldn't put things like this in the threads... It makes confuzzled.

Xannari

[CosmicCharlie]

Help please. Can't make this effing equation work. Using Rockomax 64 (initial mass = 36, final mass = 4); Mainsail (mass = 6, Isp = 285)

Number of tanks needed = x, each tank with an engine

dV = 4400

Isp = 285

Payload = 10.8

initial mass = m1 = payload + tanks + fuel + engines = 10.8 + 4x + 32x + 6x = 10.8 + 42x

final mass = m0 = payload + tanks + engines = 10.8 + 4x + 6x = 10.8 + 10x

∇∨ = ln(m₁÷m₀) · 9.81 · I_sp

ln(m₁÷m₀) = ∇∨ ÷ (9.81 · I_sp)

(m₁÷m₀) = e^{(∇V ÷ (I}_sp · 9.81))

(m₁÷m₀) = 4.8

m₁ = 4.8 · m₀

10.8 + 42x = 4.8 · (10.8 + 10x)

10.8 + 42x = 52.1 + 48x

-41.3 = 6x

x = ~ -7

So I need -7 tanks to get into the sun's orbit. Dammit.

[Thunderous Echo]

Help please. Can't make this effing equation work. Using Rockomax 64 (initial mass = 36, final mass = 4); Mainsail (mass = 6, Isp = 285)

Number of tanks needed = x, each tank with an engine

dV = 4400

Isp = 285

Payload = 10.8

initial mass = m1 = payload + tanks + fuel + engines = 10.8 + 4x + 32x + 6x = 10.8 + 42x

final mass = m0 = payload + tanks + engines = 10.8 + 4x + 6x = 10.8 + 10x

∇∨ = ln(m₁÷m₀) · 9.81 · I_sp

ln(m₁÷m₀) = ∇∨ ÷ (9.81 · I_sp)

(m₁÷m₀) = e^{(∇V ÷ (I}_sp · 9.81))

(m₁÷m₀) = 4.8

m₁ = 4.8 · m₀

10.8 + 42x = 4.8 · (10.8 + 10x)

10.8 + 42x = 52.1 + 48x

-41.3 = 6x

x = ~ -7

So I need -7 tanks to get into the sun's orbit. Dammit.

Figured it out.

No matter how many tanks you add, you will never have enough delta-v to get to the sun. A tank with a mainsail alone would not have enough delta-v.

Try reducing the number of mainsails per tank.

[Xannari Ferrows]

Figured it out.

No matter how many tanks you add, you will never have enough delta-v to get to the sun. A tank with a mainsail alone would not have enough delta-v.

Try reducing the number of mainsails per tank.

This isn't actually true. However, I'm pretty sure 69086785 tanks worth of fuel, even with 1 mainsail, would not be worth it, especially since it really shouldn't work in the first place.

Edited November 2, 2015 by Xannari Ferrows

[FancyMouse]

Because of tank mass ratio, you'll never get more than g0*Isp*ln(9) dV for a single stage, since the full/empty ratio of KSP LFO tanks is 9. Even for KR-2L you're not going to exceed 8km/s. That's why empty fuel tank needs to be dropped via stages.

[Thunderous Echo]

This isn't actually true.

What do you mean?

Because of tank mass ratio, you'll never get more than g0*Isp*ln(9) dV for a single stage, since the full/empty ratio of KSP LFO tanks is 9. Even for KR-2L you're not going to exceed 8km/s. That's why empty fuel tank needs to be dropped via stages.

Yes, that's another good solution.

[Xannari Ferrows]

What do you mean?

The equivalent amount of fuel in any given amount of tanks can in theory get you unlimited Delta-V.

[Sampa]

Delta-V refers to the amount of velocity change.

[Thunderous Echo]

The equivalent amount of fuel in any given amount of tanks can in theory get you unlimited Delta-V.

Actually, as the mass of fuel increases, the mass of tanks increases proportionally. Instead of approaching zero, the dry mass approaches a certain part of the wet mass, resulting in it approaching a certain real amount of delta-v, not infinite.

[Xannari Ferrows]

Actually, as the mass of fuel increases, the mass of tanks increases proportionally. Instead of approaching zero, the dry mass approaches a certain part of the wet mass, resulting in it approaching a certain real amount of delta-v, not infinite.

Well with the tank mass yes, but not with just the fuel itself.

[GoSlash27]

Xennari,

Correction to the earlier correction:

g0 *used to be* approximated as 9.82 prior to 1.0. That has since been rectified. It is now 9.81 like it should be.

Proof:

Best,

-Slashy

- - - Updated - - -

Well with the tank mass yes, but not with just the fuel itself.

Aye, but the tank mass matters as a practical concern since you can't add fuel without adding tank mass to hold it.

Since the LF&O tanks weigh 1/8 the fuel they contain, you will never be able to build a rocket with a Rwd higher than 9.

This sets an absolute maximum DV of Isp*9.81* ln(9), or approx. 21.6*Isp. And this is with an infinite number of fuel tanks.

Best,

-Slashy

Edited November 2, 2015 by GoSlash27

[FancyMouse]

Since the LF&O tanks weigh 1/8 the fuel they contain, you will never be able to build a rocket with a Rwd higher than 9.

This sets an absolute maximum DV of Isp*9.81* ln(9), or approx. 21.6*Isp. And this is with an infinite number of fuel tanks.

Just to clarify so that other people don't get confused or misunderstand anything - this absolute maximum only applies to a single stage.

[CosmicCharlie]

Figured it out.

No matter how many tanks you add, you will never have enough delta-v to get to the sun. A tank with a mainsail alone would not have enough delta-v.

Try reducing the number of mainsails per tank.

Thanks Echo. It ws driving me bat.....

Also, I'm glad I sparked a conversation I can't follow.