Tsiolkovsky Rocket Equation

[Hurricane]

A friend of mine gave me a derivative of the rocket equation to find the amount of fuel needed to achieve a certain Delta-v:
Mf=Me*e((Dv/Ve)-1)
Where:
Mf is the fuel mass
Me is the empty mass
Dv is the the Delta-v
Ve is the exhaust velocity
However, when comparing it with the original equation, the values are different.
Can anyone spot a faulty in this formula?

~Thanks in advance, Oz.

 

[Urwumpe]

A friend of mine gave me a derivative of the rocket equation to find the amount of fuel needed to achieve a certain Delta-v:
Mf=Me*e((Dv/Ve)-1)
Where:
Mf is the fuel mass
Me is the empty mass
Dv is the the Delta-v
Ve is the exhaust velocity
However, when comparing it with the original equation, the values are different.
Can anyone spot a faulty in this formula?

~Thanks in advance, Oz.

The -1 in the exponent there looks strange. The german engineering notation version of the simple rocket equation is like that:

[math]\frac{m_1}{m_0} = r = e^{\frac{-\Delta v}{w}}[/math]

With [math]m_1[/math] being the post-burn mass and [math]m_0[/math] being the pre-burn mass and w being the specific impulse. r is the mass ratio of that stage or rocket.

The inverse would be:

[math]\Delta v = -w \cdot \ln{r} = -w \cdot \ln{\frac{m_1}{m_0}}[/math]

 

[Hurricane]

But I'm searching for the amount of fuel needed to attain a known Dv.

 

[Urwumpe]

But I'm searching for the amount of fuel needed to attain a known Dv.

That is [math]m_0 - m_1[/math] - the amount of mass ejected at the average exhaust velocity or specific impulse of the engines.

 

[dgatsoulis]

I was wondering where that -1 in the solution for the fuelmass [math](m_0 - m_1)[/math] came from, so I tried to solve it myself step by step.
[math]\Delta V[/math]= Delta Velocity [math]u_e[/math]= exhaust velocity [math]m_0[/math]= initial mass [math]m_1[/math]= final mass

[math]\Delta V = u_e \cdot\ln{\frac{m_0}{m_1}}[/math]

[math]
\ln{\frac{m_0}{m_1}} = \frac{\Delta V}{u_e}
[/math]

[math]
\frac{m_0}{m_1} = e^{\frac{\Delta V}{u_e}}
[/math]

[math]
{m_0} = {m_1} \cdot\ e^{\frac{\Delta V}{u_e}}
[/math]

[math]
{m_1} = \frac{m_0}{e^{\frac{\Delta V}{u_e}}} = {m_0} \cdot\ e^{-\frac{\Delta V}{u_e}}
[/math]

[math]
{m_0}-{m_1} = ({m_1} \cdot\ e^{\frac{\Delta V}{u_e}}) - ({m_0} \cdot\ e^{-\frac{\Delta V}{u_e}}) = ({m_1} \cdot\ e^{\frac{\Delta V}{u_e}}) - ({m_1} \cdot\ e^{\frac{\Delta V}{u_e}} \cdot\ e^{-\frac{\Delta V}{u_e}})
[/math]

[math]
e^{\frac{\Delta V}{u_e}} \cdot\ e^{-\frac{\Delta V}{u_e}} = 1
[/math]

[math]
{m_0}-{m_1} = ({m_1} \cdot\ e^{\frac{\Delta V}{u_e}}) - ({m_1} \cdot\ 1) = {m_1} \cdot\ (e^{\frac{\Delta V}{u_e}} - 1)
[/math]

This seems to be correct, but it's been ages since I last did something like this. Would someone mind checking it?

 

[Urwumpe]

Would someone mind checking it?

That looks correct to me. You also get very similar formulas, if you calculate for payload mass fraction or with structural mass ratio.

 

[Hurricane]

I was wondering were that -1 in the solution for the fuelmass [math](m_0 - m_1)[/math] came from, so I tried to solve it myself step by step.
[math]\Delta V[/math]= Delta Velocity [math]u_e[/math]= exchaust velocity [math]m_0[/math]= initial mass [math]m_1[/math]= final mass
[math]
{m_0}-{m_1} = ({m_1} \cdot\ e^{\frac{\Delta V}{u_e}}) - ({m_1} \cdot\ 1) = {m_1} \cdot\ (e^{\frac{\Delta V}{u_e}} - 1)
[/math]

This seems to be correct, but it's been ages since I last did something like this. Would someone mind checking it?

So, the mass of fuel needed to attain a known Dv will be Me*e((Dv/Ve)-1)?
(Where Me= empty mass, Dv=Delta-v and Ve is the exhaust velocity)
Or am I missing something? :shifty:

 

[Urwumpe]

So, the mass of fuel needed to attain a known Dv will be Me*e((Dv/Ve)-1)?
(Where Me= empty mass, Dv=Delta-v and Ve is the exhaust velocity)
Or am I missing something? :shifty:

yes, you are missing some mathematics. No surprise you get such results as in the first post, the order of operations and the existence of most fun stuff in mathematics seems to be unknown to you.

 

[Hurricane]

yes, you are missing some mathematics. No surprise you get such results as in the first post, the order of operations and the existence of most fun stuff in mathematics seems to be unknown to you.

Huh?
Well, sorry for being an idiot! :tiphat::lol:
Also, it wasn't me who made the equation in the first post, it's a friend of mine. I don't know all this stuff yet.

 

[dgatsoulis]

So, the mass of fuel needed to attain a known Dv will be Me*e((Dv/Ve)-1)?
(Where Me= empty mass, Dv=Delta-v and Ve is the exhaust velocity)
Or am I missing something? :shifty:

Place Euler's number e inside the first parenthesis and keep in mind that e is NOT multiplied with dV/Ve but raised to the power of that.

First you need to have an understanding of the order of operations in math and then what ln and e are.
Do a wiki search for e mathematical constant, logarithm and natural logarithm.
Just by reading through the intros and the layman's definitions will help you get a grasp of the concepts, then look at Urwumpe's equations and the solution I posted again.

 

[Urwumpe]

Huh?
Well, sorry for being an idiot! :tiphat::lol:
Also, it wasn't me who made the equation in the first pot, it's a friend of mine. I don't know all this stuff yet.

So you better get aware, that exponents are not decorative.

[math]a^b \not= ab[/math]

And maybe, you should look for the laTex math notation to also describe formulas in this forum, it makes things easier to read since it gives you some kind of standard.

 

[Hurricane]

So you better get aware, that exponents are not decorative.

[math]a^b \not= ab[/math]

And maybe, you should look for the laTex math notation to also describe formulas in this forum, it makes things easier to read since it gives you some kind of standard.

well, I used the notation my friend used, whereby e(a) is e^a... That's the way he described it to me. And laTex math notation? Is that what you all use?

 

[Urwumpe]

well, I used the notation my friend used, whereby e(a) is e^a... That's the way he described it to me. And laTex math notation? Is that what you all use?

Yes, with [math]...[/math] tags around it.

and your friend should be aware that most programming languages use "exp(a)" for expressing [math]e^a[/math]

 

[Hurricane]

and your friend should be aware that most programming languages use "exp(a)" for expressing [math]e^a[/math]

Well, probably that's what he said... I don't quite remember.
~Thanks again, Oz.!

 

[Urwumpe]

Still:

Mf=Me*exp((Dv/Ve)-1)

is

[math]m_f = m_e \cdot e^{\frac{\Delta v}{v_e} - 1}[/math]

not

[math]m_f = m_e \cdot \left (e^{\frac{\Delta v}{v_e}}-1 \right )[/math]

 

[Hurricane]

Yup, sorry for everything. I REALLY shouldn't have done all that at 1 a.m.