2 Rocket equation questions

[Kyrt Malthorn]

Hello! I have some math questions.  So if you don't like math, well, I do.  Kinda.

I consider myself an intermediate experience player, currently getting on just fine with the numbers kerbal engineer gives me.  I hate actually doing math so I'm not likely to give that up any time soon, but I am comfortable with mathematical concepts.  In fact, I have some interest in getting a grasp on themathematical science behind simple rockets....But I never learned logarithms.  And the rocket equation involves a natural logarithm.

Question 1: what are logarithm and natural logarithm, and how to do them? 

 

For the second question: the rocket equation traditionally appears using exhaust velocity.  I have read it can also use isp.  But I haven't seen the version of the rocket equation that uses isp, that I could plug in the stats we get in KSP.  Also, I know isp is counted in seconds and relates to the engine efficiency. As I understand Newtonian physics, exhaust velocity IS your efficiency, so.... how is isp calculated and why is it more popular than exhaust velocity (other than because that's what SQUAD gave us).

[Rhomphaia]

I Logarithm is the inverse on an exponent.

So is Log₁₀ X = Y then X = 10^Y 

For a natural logarithm we use replace the 10 in the above with the constant e

Log_e is usually written as ln.

To do logarithms, use your calculator

 

Isp is a conversion factor, to avoid confusion when exhaust velocity is given in different units, and uses the acceleration due to gravity at earths surface as a constant for conversion. this gives out an answer in seconds which is a unit common to both metric and imperial.

[OhioBob]

Exhaust velocity in the rocket equation is actual the effective exhaust velocity, denoted C, where

C = I_sp*g_o

In KSP, I_sp is obtained from the engine stats visible in the VAB, while g_o = 9.80665 m/s².

 

Edited March 25, 2016 by OhioBob

[Crown]

Answer 1 (Logarithms):

I would describe the logarithm as the inverse operation of the power. Like subtracting is the inverse operation of adding, and dividing is the inverse operation of multiplying. I will start at the very low so it's hopefully a good guide to any level

The power (mathematically) is a short way of writing how many times one wants to multiply a number with itself. So if we have

2 * 2 * 2

we can also write it as

2³

This saves time and paper (because maths people are lazy too ). When the line would go even longer, we save even more time. No-one wants to write or read or count a very long formula like below

3 * 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3 * 3

It's shorter, easier, and more pleasing to write and read

3¹⁵

A shout-out to the copy&paste function here. Using powers is very useful in this case. The "3" is called the base, the "15" is called the power (please correct me if I'm wrong).

Now let's to the reverse:
We have 8 as the answer and want to know the power. What has to be the "high" number to get eight?

2^x = 8

This is where the logarithm comes into play. Like inverting an addition my subtracting. The logarithm gives us the power (literally) and it's written like this:

log₂(8) = x

So here we have the "logarithm to the base two of eight". I think that's how it's pronounced, I very rarely "speak" math .
But the cool thing with this is that it can be done with any number. Almost. More below.

Most of the time when we talk about logarithms, it's the logarithm to the base of 10. I don't know why but that's what we do. Most calculators don't even give the option to enter a base for the logarithm. So the log button on the calculator is a log₁₀ but with some maths magic it's possible to use a log₁₀ to calculate the log for any base.

Onwards to the natural logarithm.
Or in short ln. Minor L, minor N. This one's is very special in many ways. It's the inverse operation to e^x. That's the "natural" part in the "natural" logarithm.

e is is about 2.72 and is used to describe - or more-so describes - e.g. the natural decay of something. I fell like that's one of the very few occasions where maths meets the real world, waving through a thick window. So the natural logarithm is just the inverse operation of

e^x

And because we need it quite often, it has his own name.

ln(16) = 2.773

The connection to KSP is almost anywhere where physics come in. In the rocket equation and in the atmosphere thickness. And even more but these are the ones I use quite often.
 

Here are some of the definition / rules about powers and logarithms:
n⁰ = 1
n¹ = n
log_n(0) = undefined
log_n(b) is only defined for b greater 0.

 

I'll come back after breakfast. Back from breakfast.

Onwards to the Specific Impulse, I_sp

Like @OhioBob already said, Specific Impulse is the engine efficiency and the exhaust velocity. Both. Together. Kinda. By mathematical magic.

I believe that the I_sp is used in seconds because (besides of being shorter) it's unit - the second - is the only thing that American and German rocket engineers had in common. But I don't know if that's really true. But using seconds makes Russian and American rocket engines very easily comparable.

For myself, it was quite a challenge to wrap my head around the Isp but once it's in the head, it won't get out anymore.

On Wikipedia one can come across this neat (some would call it ugly) formula:

with
as the burn time,
as the fuel mass,
mean thrust,
thrust the the time ,
as the standard earth gravity.

You can ignore most of it.

A specific impulse of 1'000 m/s = 1'000 Ns/kg ≙ 102 s means that 1 kg of fuel can change the impulse by 1'000 Ns.

This means an engine that has a thrust of 1'000N and burns for 1 second uses 1 kg of fuel. Or a smaller engine with 10N of thrust that burns 10s and uses 0.1 kg of fuel.
1'000m/s is the change in speed the mass (1 kg) would experience if it would fall for 102s in nominal earth gravity.

Maybe it's easier to demonstrate the formula on an example. Let's take the beloved LV-T45 "Swivel" liquid fuel engine:

It has a vacuum thrust of 200kN of 200'000N and an vacuum I_sp of 320 seconds.

When we put this in our neat formula, we get

With the units in bracket, it's not that confusing. But still, the newton is annoying, so ...

because 

 

Now the bad memories from 8th grade come back. But be can now kill almost any unit. the two s² are eliminated, the meter m and the kg from the the newton and the mass. All what is left is the lonely second.

That's how we get the unit for the specific impulse. By killing units and leaving nothing more that the second alive.

 

But wait, there's more:

To get the exhaust velocity of the fuel multiply the Isp with the standard earth gravity. Use either Imperial or Metric:

 

Sometimes I want to know if I have enough fuel to perform a maneuver node burn. That's why I want to know the fuel consumption of the engine.

Let's take the formula and change it to our needs like Play-Do.

becomes

with the units        or   

With a burn time of 1 second, aka. fuel consumption per second

gives us 63.71 kg fuel per second. Divided by 5kg/unit we have 12.74 units fuel per second.
The fuel is mixed in a ratio (I hope it still is) of 11 Ox to 9 RP1, or 1.22. At this point I divide by 20 and multiply by 9 to get the RP1 consumption. That gives me 5.733 units RP1 per second. A FL-T400 with 180 units (liters?) of RP1 will last 31.39 seconds in vacuum with a LV45.

I use RP1 and Ox to have a linguistic difference between a tank's content ("fuel") and the resource ("fuel").

We could test our calculation by inserting the mass of 2'000kg and burn time of 31.30 seconds into the equation above. If we get an I_sp of 320s, we are correct.

For my burn time calculations I usually ignore a few things. But I keep them in mind and in case something is odd I know where to look at. These things are the t_b in the calculation above because it's 1 anyways. When it comes to the burn time for a delta-v, I ignore the whole basic rocket equation (e.g how long does it take to do a 850m/s burn and does my tank last this long?). This makes the burn a few seconds shorter and is good for my estimations.

I hope that was helpful in both mathematical and engineering ways. In case you have more questions, just ask ;-)

Lunch time!

Edited March 25, 2016 by Crown

[GoSlash27]

Kyrt Malthorn,

 The formula using KSP stats:

DV = g₀*I_sp*ln(M_w/M_d)

where

DV= change in velocity in m/sec

g₀ = approx. 9.81 m/sec²

I_sp = specific impulse in seconds

M_w = Mass in tonnes at the beginning of the burn

M_d = Mass in tonnes at the end of the burn.

During the design process, I reorganize the equation to solve for M_w/M_d; the "wet-to-dry ratio", or R_wd

R_wd= e^{(DV/g₀I_sp)}

 

Best,
-Slashy

[Kyrt Malthorn]

2 hours ago, OhioBob said:

Exhaust velocity in the rocket equation is actual the effective exhaust velocity, denoted C, where

C = I_sp*g_o

In KSP, I_sp is obtained from the engine stats visible in the VAB, while g_o = 9.80665 m/s².

 

Ooh....  so isp is an expression in seconds because gravity is expressed in acceleration per second as well.  That explains a few things.

 

47 minutes ago, Crown said:

Answer 1 (Logarithms):

*snip*

I'll come back after breakfast.

Super helpful.  I get logarithms much better now and see how they fit in - I'm quite familiar with powers.

 

And @GoSlash27 these questions have largely been raised in my effort to understand your Reversing the Rocket Equation thread.  =D  You're inspiring, sir.

[Plusck]

6 hours ago, Kyrt Malthorn said:

.. how is isp calculated and why is it more popular than exhaust velocity (other than because that's what SQUAD gave us).

The "why" bit is historical. US rocket scientists back in the early days were a mix of home-grown "foot-pound per bushel" sort, and formerly German "newtons per litre" sort. Isp measured in seconds depends only on the constant for gravity to be turned into either an imperial measurement or a metric one, so it avoids confusion.

Isp is what it is because of the cancelling-out of units, leaving just seconds. The physical interpretation of those "seconds" is actually "how long can this engine hover above the Earth's surface per unit of fuel" (discounting engine weight, and with that "unit" of fuel being equal to the weight that the engine can support while hovering).

On a side note, on the question of cancelling out units from equations, take a look at xkcd for a couple of interesting examples (such as the fact that gas mileage is a measure of distance (miles = 63360 inches) divided by distance cubed (gallons = 15.199*15.199 inches) and therefore resolves down to "per distance squared" (1 mpg = 274.3 per square inch), and is therefore the fraction (1/274.3 square inches/mpg = 2.32 mm^2/mpg) of the cross-section of the fuel line that your car would have to scoop from while running along its length).

Edited March 25, 2016 by Plusck

[GoSlash27]

Kyrt Malthorn,

 Thanks, that's very kind of you!

Be sure to check out this thread also:

This has practical examples of using the reverse rocket equation, as well as incorporating thrust to weight ratio in the design process.

Good luck!

-Slashy

[soulsource]

If I may add some trivia to Crown's excellent explanation: The natural logarithm is called natural, because its base, Euler's constant, has some very interesting mathematical properties. First and foremost, the exponential function f(x)=e^x is its own rate of change (mathematically speaking: its own derivative: ). This furthermore means, that the area under the curve e^x (its integral) is given by e^x + a constant as well. Exponential functions of any other base get additional factors when calculating the rate of change or the area below. The general formula for the rate of change of an exponential function to arbitrary base a is given by , with ln(a) being the natural logarithm of a.

Another remarkable property of the function e^x is that it's related to the trigonometric functions sin(x) and cos(x). To fully understand this, you'd need to know about complex numbers, which are a 2D generalization of the real numbers, and still form a field. They have, as a basis, the real unit (1), and the imaginary unit (typically called i, but sometimes also j), which is given by . You can imagine them as being a 2D plane, where the real-axis is horizontal, and the imaginary axis vertical. So, any point in this plane is a complex number. These numbers can (obviously) be rewritten with their magnitude (distance from the 0-point), and the angle with respect to the real axis. And this is, where e^x becomes interesting again. Euler showed that . By this, complex numbers, typically written as (where a and b are real numbers) can just as well be written as . So, although we are talking about an exponential function here, which basically boils down to how often one has to multiply a number by itself, we can do trigonometry with it. Isn't that amazing? It's also a "trick" commonly found in science, to use complex numbers to be able to combine these two remarkable properties, so that one can use the simple rule that the derivative of e^x is e^x again to calculate the derivative of a more complicated formula containing trigonometric functions, or to combine two coupled differential equations in one single equation (for instance the Schrödinger equation).

[Kyrt Malthorn]

@Crown, did you have to add that to the same post?  Now I can't give you more rep.  Thanks, man, you've shown a lot of patience putting all that together.  I expected leads to the kind of articles I needed to advance, you assembled a topical math lesson with the why's, how's, and what's-it-for's.  That's how I learn.  Thank you.   Definitely pushed my envelope of following mathematical notation a little wider.

@GoSlash27, I'll be tackling your methods again when I've gotten some sleep.  Thanks for the responses!

And more thanks to @Plusck, @OhioBob, and @Rhomphaia for your contributions!

 

[Zhetaan]

On a related-but-unnecessary note, I can add a little bit to @Crown's post and explain why most logarithms are to the base 10, otherwise called common logs.

Before scientific calculators, there were two common ways to calculate logs:  you could use a slide rule, or you could use the log tables (pages of pre-calculated logs that you would use as a sort of directory to look up the one you wanted).  The problem with either of these is that they are physical objects that take up space.  Adding more capability makes them take up more space, and since numbers go to infinity, it's not a solvable problem.

Or is it?  Mathematical trickery comes to the rescue!  The thing about common logs is that because they are taken to the base ten, and because our numbering system is also base ten, there are some helpful patterns that appear.

For example:

log₁₀ 5.501 = 0.7404 (approximately, as all of these will be).

log₁₀ 55.01 = 1.7404

log₁₀ 550.1 = 2.7404

log₁₀ 5501 = 3.7404

Common logs make calculation easier because they are a special case:  for larger numbers, move the decimal place to just after the first digit, and take the log of that number.  Then count up how many places you moved the decimal, and put that answer in front of the log.  This means that, so long as you can shift decimal points around, you can get logs of any number.  If you're using a slide rule, you can usually get three-place accuracy.  If you're using tables, you can get four-place accuracy (good enough for almost anything) if you pre-calculate the first ten thousand logs.  It doesn't take up much space (and you can make copies for everyone, so they only have to be calculated once).  You can take care of the rest on your own easily.

[OhioBob]

As long as we're getting really heavy into the math, I might as well add a couple more things to the discussion.

First, let me say something about the I_sp equation given by Crown,

I_sp = (F * t_b) / (m * g_o)

where t_b is the burn time and m is the fuel mass (technically the "propellant mass," which includes both the fuel and the oxidizer).  In this case, m is the mass of propellant burned in time t_b.  In many cases, the variable m is replaced by the mass flow rate, denoted ṁ (pronounced "m dot"), where ṁ is measured in mass units per second.  (It is common practice that a dot placed over a variable indicates that it is a rate.)  Since ṁ is mass units per 1 second, t_b is always equal to 1 s, therefore we can drop t_b from the equation.  The specific impulse equation is more commonly written as,

I_sp = F / (ṁ * g_o)

My second point is to elaborate on something I said earlier.  I mentioned that Tsiolkovsky’s rocket equation uses effective exhaust gas velocity, so let me explain a little about that.

Computing the velocity at which exhaust gases are expelled from a rocket nozzle involves a lot of complex thermodynamics that I don't want to get into.  However, the equation used is

V_e = SQRT[ (2ϒ/(ϒ-1))*(RT_c/M)*(1-(P_e/P_c)^(ϒ-1)/ϒ) ]

where,
V_e = exhaust gas velocity (note, not "effective")
ϒ = specific heat ratio
R = universal gas constant
T_c = combustion chamber temperature
M = exhaust gas molecular weight
P_e = pressure at nozzle exit
P_c = combustion chamber pressure

Once we have the exhaust velocity, we compute an engine's thrust from the equation

F = ṁV_e + (P_e – P_a)*A_e

where,
F = thrust
ṁ = propellant mass flow rate
V_e = exhaust gas velocity
P_e = pressure at nozzle exit
P_a = ambient air pressure
A_e = cross-sectional area of nozzle exit

As you can see, thrust is broken down into two parts:  momentum thrust, ṁV_e, and pressure thrust, (P_e–P_a)*A_e.  As long as ṁ remains constant, V_e and P_e are also constant.  In this case we see that momentum thrust is constant; however, pressure thrust varies as a function of ambient air pressure, P_a.  This is why rocket engines have different performance at sea level versus in a vacuum.  We can isolate the variable part by writing,

F = ṁV_e + P_eA_e – P_aA_e

where we can see that the difference in thrust between vacuum and sea level is simply A_e times the sea level air pressure.  The different values for vacuum and sea level I_sp are simply computed using the different vacuum and sea level values of F.

In Tsiolkovsky’s rocket equation we don't want to use V_e by itself because that would ignore the contribution of pressure thrust.  To combine the effects of both momentum and pressure thrust, we define effective exhaust gas velocity, C, where

C = V_e + ((P_e – P_a)*A_e) / ṁ

This simplifies the thrust equation by reducing it to

F = ṁC

Substituting ṁC for F in the I_sp equation gives

I_sp = ṁC / (ṁg_o) = C/g_o

which we rearrange to give

C = I_sp*g_o

 

Edited March 26, 2016 by OhioBob

[Padishar]

7 hours ago, Zhetaan said:

The thing about common logs

It is only the fact that the number base is the same that makes this simple.  It uses this fact:

log_b(x*y) = log_b(x) + log_b(y)

This is true for any base but when it matches the number base you use (b) you get:

log_b((b^n)*x) = log_b(b^n) + log_b(x) = n + log_b(x)

 

5 hours ago, OhioBob said:

T_c = combustion chamber pressure

I presume this should be temperature...

Edited March 25, 2016 by Padishar

[OhioBob]

1 hour ago, Padishar said:

I presume this should be temperature...

Fixed it.  Thanks.

[Crown]

On Donnerstag, 24. März 2016 at 3:33 PM, Kyrt Malthorn said:

@Crown, did you have to add that to the same post?  Now I can't give you more rep.  Thanks, man, you've shown a lot of patience putting all that together.  I expected leads to the kind of articles I needed to advance, you assembled a topical math lesson with the why's, how's, and what's-it-for's.  That's how I learn.  Thank you.   Definitely pushed my envelope of following mathematical notation a little wider.

Thank you much. I'm happy this helped.

Maths is very fun and useful when one knows how. Too sad it's spoiled by maths in school. That stuff is boring and has no connection to real life.

Since KSP I love maths. And I am happy I was able to help you.

[Wemb]

On 24/03/2016 at 7:26 AM, Crown said:

Answer 1 (Logarithms):....

This is what I love about this game and this community - you just don't get lessons on arithmetic like this on Call of Duty forums.

Wemb

Edited May 6, 2016 by Wemb

[Wemb]

One thing which might help illuminate why logarithms are a thing at all is that as well as being incredibly fundamental in calculations involving rates of change growth - they are also tremendously useful in simplifying manual arithmetic - and this was why they were first developed and investigated.

Imagine you want to multiple two large numbers togther - 4324924823 x 6575676575  - this is a farily horrible and time consuming job if you don't have a calculator or computer (or, if you existed before they were invented).

However, one of the rules that govern how logarithms work is that

log (x) + log(y) = log (xy)

So that if you had a compiled table of logs and their inverses, you could easily perform otherwise very long-winded mulitplications by simply looking up a couple of numbers from a table, adding them together and looking up the result in a second table. And, if you're using logs to the base 10, rather than the natural log, you don't have to worry about the decimal places till to the end, and that's also a trivial exercise in adding. Slide rules allow you do to the simialr, but in a somewhat more flexible way.

This facility is why the  exams I did in the late1980s and early 1990's still had printed log tables in the back of the formula book for those without calculators, or where calculators were not allowed. 

Wemb.