Trickery With Delta V

[Tank Buddy]

I am working on a science fair project where I would find under what circumstances the LV-909 and 48-7S produce the same delta V, when the only thing that changes is the amount of fuel. I wondered the effect of adding or removing the mass of the capsule for the calculations, and I found the following:

• LV-909
  W/ capsule-3169.41 m/s
  W/o capsule- 4969.23 m/s
  
• 48-7S
  W/ capsule- 3469.23
  W/o capsule- 6535.97!
  

All calculations were done by hand with a calculator and checked with an online Delta V calculator.

What surprised me was that there was not a proportional chance in the Delta V. The 48-7S had a significant increase in Delta V.

This means you could not do a single set of calculations to determine when you burn X amount of fuel which engine is better. You would have to do calculations for each and every time you added non-fuel related mass.

I do not understand this. Is it just that I'm not making sense of the Logarithm?

Note: To find which engine produces more Delta V, I have set two equal to each other.

Edited November 28, 2013 by Tank Buddy

[zekes]

This is because the 48-7s are have higher ISP and better thrust to weight ratio. That means when you take the capsule off, the engine takes up less of the remaining weight than the 909 and so increases the ratio of full weight to empty weight. I hope that made sense.

[Tank Buddy]

Thrust to weight ratio doesn't effect Delta V. I think you made sense in that last part where you said something about the ratio of the full weight. The ratios between the full and dry masses are very interesting. It doesn't make perfect sense, but I'll just let it marinate in my brain.

[Specialist290]

I think what zekes meant to say is "mass ratio" rather than "thrust-to-weight ratio," because otherwise his statement makes perfect sense. The 48-7S has a lower mass than the LV-909, so with the 48-7S equipped craft, you're increasing the mass ratio by removing the capsule by a larger margin than with the LV-909-equipped craft, since the former craft already has a higher mass ratio.

You may find this series of charts by ThePseudoMonkey to be enlightening. EDIT: I thought I had a link to another, more comprehensive set, but it turned out not to be what I remembered it to be. EDIT2: Aha, found it! It's not as easy to follow, but it's quite comprehensive.

Hope this helps

Edited November 27, 2013 by Specialist290

[Rusty6899]

The 48-7S is very light (0.1t) so when the Delta-V is calculated (assuming an FL-T200 fuel tank is used) m0/m1 ~= 5 (with no capsule) the LV-909 weighs significantly more (0.5t) so with the same amount of fuel m0/m1 < 3. It makes some of this difference with a higher Isp, but not much.

With the capsule added (assuming a Mk 1 lander can) the 48-7S has m0/m1 ~=2.2 and the LV-909 has m0/m1 ~=1.8, so the difference is much smaller.

[allmhuran]

The higher ISP of the 909 will result in the 909 line crossing the 7s line eventually as you add fuel, and then as Rusty has noted, adding a capsule adds, proportionally, much more "dead weight" to the 7s rocket than the 909 rocket (11x as much vs 3x as much)

[zekes]

I think what zekes meant to say is "mass ratio" rather than "thrust-to-weight ratio," because otherwise his statement makes perfect sense. The 48-7S has a lower mass than the LV-909, so with the 48-7S equipped craft, you're increasing the mass ratio by removing the capsule by a larger margin than with the LV-909-equipped craft, since the former craft already has a higher mass ratio.

That's exactly what i was trying to say Thanks

[DrMonte]

I'm not so good at physics but good at math So if fuel consumption constants at wiki are correct http://wiki.kerbalspaceprogram.com/wiki/Parts

48-7S - 0.0087 t/s

LV-909 - 0.0131 t/s

I have quite simple answer. Compared by power we need 5:3 engines ... ok it's an equation

So you have to burn 9,357t (+1t for 48-7S) of fuel per every 5:3 (48-7S vs LV-909) engines for 48-7S become not effective. That is going to 2.5m fuel tanks scale - More than "Rockomax X200-16 Fuel Tank"

That will be almost 12min burn.

[KvickFlygarn87]

To the spreadsheet! (NANANANAAAA!)

Okay, I found that given the same dry mass of the tank, a Probodobyne OKTO2, the different Isp and weight of the engines, the LV-909 needs 2,137719368 times as much fuel. I believe. That's what I got, anyways.

[SRV Ron]

If you are pushing less mass in orbit, or with the mass equal, have a more efficient engine in vacuum, the Delta V will be higher.

[Rusty6899]

From my spreadsheet, if you have a Mk1 Lander Can, and use any of the FL-T series fuel tanks (total mass/dry mass = 9), and need 2000m/s Delta-V, you would need to use an FL-T200 with the 48-7S (you only need 0.614t of fuel if total mass/dry mass = 9, but in reality you need to use the next biggest tank). The burn would take 114.5 seconds.

For the same lander and Delta-V requirement, with the LV909, you would also need the FL-T200 (you only need 0.826t of fuel if total mass/dry mass = 9). The burn would take 76.5 seconds.

As with any engineering design, there are benefits and shortcomings of either design.

The design with the 48-7S is lighter, and is therefore better if you consider that you will have to have earlier stages in your rocket to launch it into space, for certain Delta-V requirements it will also require a smaller tank than the LV-909 would, increasing this efficiency.

The LV-909 has a higher mass flow rate, so burns will be shorter for the same amount of fuel, and it will be more adept at taking off and landing, due to the higher thrust (in most cases).

You need to do a separate calculation for every major alteration you make to your rocket (the odd strut or thermometer won't make much difference) this can be set up in a spread sheet, where you just have to input the engine mass, additional mass, Isp, the full/empty mass of fuel tank you plan to use. etc. and it will tell you the minimum fuel requirement.

[Rusty6899]

I forgot to mention that the other consideration, which you will have to take account of eventually, is cost. In this case the 48-7S is cheaper, but you probably don't care about that yet.

Also KvickFlygarn87, I am guessing that you are using the comma as a decimal separator otherwise I would have to question your results.

[KvickFlygarn87]

Also KvickFlygarn87, I am guessing that you are using the comma as a decimal separator otherwise I would have to question your results.

Yes, Swedish number system ftw.

[Yasmy]

What surprised me was that there was not a proportional chance in the Delta V. The 48-7S had a significant increase in Delta V.

This means you could not do a single set of calculations to determine when you burn X amount of fuel which engine is better. You would have to do calculations for each and every time you added non-fuel related mass.

I do not understand this. Is it just that I'm not making sense of the Logarithm?

Hi Tank. This is all correct. You do have to redo the calculations each time you change the payload, since the two engines have different masses.

As you guessed, there isn't a proportional change in dV, when you remove a fixed amount of payload. There is a logarithmic change in dV:

proportional: dV = a * m0/m1

logarithmic: dV = a * ln(m0/m1)

(a = g * Isp)

Remove some payload mass dm: dV = a * ln((m0-dm)/(m1-dm))

m0 and m1 are different for the two rockets with different engines,

thus there is no simple proportional change in dV when changing the two different rockets by removing some payload.

[DrMonte]

Hi Tank. This is all correct. You do have to redo the calculations each time you change the payload, since the two engines have different masses.

As you guessed, there isn't a proportional change in dV, when you remove a fixed amount of payload. There is a logarithmic change in dV:

proportional: dV = a * m0/m1

logarithmic: dV = a * ln(m0/m1)

(a = g * Isp)

Remove some payload mass dm: dV = a * ln((m0-dm)/(m1-dm))

m0 and m1 are different for the two rockets with different engines,

thus there is no simple proportional change in dV when changing the two different rockets by removing some payload.

So why you are not making them equal?

[capi3101]

Okay, so you're trying to find situations where the delta-V for an LV-909 equals that available from a 48-7S, right?

First thing's first then: a 48-7S has a mass of .1 tonnes and a vacuum I_sp of 350. For the LV-909, it's .5 tonnes and a 390 I_sp.

And of course we have the rocket equation: dV = ln(M/M_o) * I_sp * G_o

If we're looking for situations where the delta-V from the two engines is equal, we can just set two rocket equations equal to one another for the same case. Thus:

dV_48-7S = ln(M_48-7S/M_{o, 48-7S}) * I_{sp, 48-7S} * G_o

dV_LV-909 = ln(M_LV-909/M_{o, LV-909}) * I_{sp, LV-909} * G_o

Where dV_48-7S = dV_LV-909, ln(M_48-7S/M_{o, 48-7S}) * I_{sp, 48-7S} * G_o = ln(M_LV-909/M_{o, LV-909}) * I_{sp, LV-909} * G_o

G_o cancels out, so we're left with

ln(M_48-7S/M_{o, 48-7S}) * I_{sp, 48-7S} = ln(M_LV-909/M_{o, LV-909}) * I_{sp, LV-909}

We have the I_sp of our engines, so we can plug that data straight in - we're left with:

350*ln(M_48-7S/M_{o, 48-7S}) = 390*ln(M_LV-909/M_{o, LV-909})

Since we have a constant on both sides of the equation, we can simplify - we get:

ln(M_48-7S/M_{o, 48-7S}) = 1.11429*ln(M_LV-909/M_{o, LV-909})

Take the inverse natural logarithm (e) of both sides! We get:

M_48-7S/M_{o, 48-7S} = 3.04739 *(M_LV-909/M_{o, LV-909})

Now, the key thing about the mass factors in the rocket equation is that there are actually two components to it - a part that handles the mass that changes (accounting for what's changing it mass - fuel) and a part that handles the mass that doesn't (everything else); that's called dead mass, of which the engine's mass is a part. We need to split off those bits from one another; we get:

(M_{fuel, 48-7S} + M_{dead, 48-7S})/(M_{dry tank, 48-7S} + M_{dead, 48-7S}) = 3.04739 *((M_{fuel, LV-909} + M_{dead, LV-909})/(M_{dry tank, LV-909} + M_{dead, LV-909}))

You can then use the assumption that M_fuel = 9M_{dry tank} (which is true for all liquid fuel tanks in KSP except the Round-8 and Oscar- and carry through the constant on the right hand side. You get:

(9M_{dry tank, 48-7S} + M_{dead, 48-7S})/(M_{dry tank, 48-7S} + M_{dead, 48-7S}) = (27.42651M_{dry tank, LV-909} + 3.04739M_{dead, LV-909})/(M_{dry tank, LV-909} + M_{dead, LV-909}))

That's about as simple as you can get it without putting in specific terms.

I'm out of time right now, but I'll see if we can't apply this equation to a specific case in the near future.

[Rusty6899]

Yes, Swedish number system ftw.

Let's not start an argument over number convention. It could get pretty heated.

[KvickFlygarn87]

Let's not start an argument over number convention. It could get pretty heated.

Oh, didn't mean that it was better or anything, just saying it's the one I used.

[EtherDragon]

Thrust to weight ratio doesn't effect Delta V. I think you made sense in that last part where you said something about the ratio of the full weight. The ratios between the full and dry masses are very interesting. It doesn't make perfect sense, but I'll just let it marinate in my brain.

I have the PERFECT video for you...

This video breaks down the math and why certain things affect Delta-V and others don't.

[capi3101]

Alright - time for a specific case. Let's just go with a basic setup - a Mk1 Lander Can (0.6 tonnes), the fuel tank and the engine.

As a reminder, the 48-7S has a mass of 0.1 tonnes and the LV-909 has a mass of 0.5 tonnes.

That means M_{dead, 48-7S} = 0.1 + 0.6 = 0.7 tonnes

M_{dead, LV-909} = 0.5 + 0.6 = 1.1 tonnes

Let's set M_{dry tank, 48-7S} = X and M_{dry tank, LV-909} = Y, and substitute in our masses. We have:

(9X + 0.7)/(X + 0.7) = (27.42651Y + 3.04739*1.1)/(Y + 1.1)

Alright. Let's multiply through by the divisors:

(9X + 0.7)(Y + 1.1) = (27.42651Y + 3.35213)(X + 0.7)

Multiply through:

9XY + 9.9X + .7Y + .77 = 27.42651XY + 19.19856Y + 3.5213X + 2.34649

Now, let's get all the terms involving X on one side of the equation and everything else on the other side:

6.3787X -18.42651XY = 18.49856Y + 1.57649

For the case where Y = 1 tonne

6.3787X -18.42651X = 18.49856 + 1.57649

-12.04781X = 20.07509

X = -1.66629

I've got a math error somewhere...let's do some verification.

For the LV-909 case with one tonne of dry mass:

dV = ln ((9+1.1)/(1+1.1)) * 9.81 * 390 = 5160.83

5160.83 = ln ((9M + 0.7) / (M + 0.7) * 9.81 * 350

4.49552 = (9M + 0.7) / (M + 0.7)

4.49552(M + 0.7) = 9M + 0.7

4.49552M + 3.14686 = 9M + 0.7

4.50448M = 2.44686

M = .543206

So full mass is 9M = 4.88885

Okay - definite error there. I think I see where...over did my first equation.

[thereaverofdarkness]

This is because the 48-7s are have higher ISP and better thrust to weight ratio. That means when you take the capsule off, the engine takes up less of the remaining weight than the 909 and so increases the ratio of full weight to empty weight. I hope that made sense.

Thrust to weight ratio doesn't effect Delta V. I think you made sense in that last part where you said something about the ratio of the full weight. The ratios between the full and dry masses are very interesting. It doesn't make perfect sense, but I'll just let it marinate in my brain.

Actually, the ISP has nothing to do with the proportion change, while the thrust to weight ratio has everything to do with it. The percentage of non-fuel mass you have vs your ISP directly determines your exact delta-v. If ISP remains the same, then cutting the non-fuel mass in half is the same as adding the total delta-v of a rocket with the same mass as your old dry mass, that is half fuel and has the same ISP. Basically, all unspent fuel is effectively a payload fraction. As your effective payload fraction decreases (as your rocket gets lighter by burning fuel), your thrust-to-mass ratio increases, and you get more velocity change for the same volume of fuel burn.

If your dry mass were zero, your delta-v would be infinite.

[numerobis]

capi3101 -- you're assuming e^(ab) = e^a e^b, rather than (e^a)^b. That's the error. From e^(1.x ln(m1/m0)) you should get (m1/m0)^1.x

[EtherDragon]

Actually, the ISP has nothing to do with the proportion change, while the thrust to weight ratio has everything to do with it. The percentage of non-fuel mass you have vs your ISP directly determines your exact delta-v. If ISP remains the same, then cutting the non-fuel mass in half is the same as adding the total delta-v of a rocket with the same mass as your old dry mass, that is half fuel and has the same ISP. Basically, all unspent fuel is effectively a payload fraction. As your effective payload fraction decreases (as your rocket gets lighter by burning fuel), your thrust-to-mass ratio increases, and you get more velocity change for the same volume of fuel burn.

If your dry mass were zero, your delta-v would be infinite.

The math from the Rocket Equation doesn't bear this out, in any way though. Couple things to consider:

1. If your empty mass was ZERO the Delta-V is undefined (not infinite) due to divide by zero error in the function. However, (to be more precise), you can say that Delta-V approaches infinity as empty mass approaches zero. But, the logarithmic function demonstrates the diminishing return of this effect. You quickly reach a point where you have to decrease empty mass a lot to get a small change in Delta-V. Lastly, on this point, there is a functional limit to the Delta-V for any stage because the propellant tank always has empty mass - so the absolute best full to empty mass ratio is limited to 9/1 (for a given stage).

2. The Rocket Equation doesn't include thrust-to-weight ratio in any way - which means the TWR has nothing to do with the rocket's total Delta-V. In most cases, increasing TWR decreases the total Delta-V because you have to add empty mass to do it. Although it is true that higher TWR means you can bring more empty mass for the same Delta-V (given equal Full / Empty mass ratios) it is not true that you get more Delta-V from it. You have more thrust, but you also consume more propellant in the same time; double the thrust - double the rate of propellant consumption. As your TWR approaches infinity, your rate of propellant consumption also approaches infinity equally.

Edited November 27, 2013 by EtherDragon
typos fixed

[capi3101]

capi3101 -- you're assuming e^(ab) = e^a e^b, rather than (e^a)^b. That's the error. From e^(1.x ln(m1/m0)) you should get (m1/m0)^1.x

Ah....thank you. I've been blowing my mind for the last hour or so trying to figure out why nothing was working out right.

Gonna take a break and try again later.

[Rusty6899]

Oh, didn't mean that it was better or anything, just saying it's the one I used.

I was only kidding. I haven't quite gotten to grasp with the Kerbal Smileys yet, which probably would have made that more clear.