Solving Burn Time manually

[Blue]

So I'm about to head on my way for a pretty big mission, and I wanted to know if there was a good way to solve not really a problem, but a question.

Is there a way to make or calculate a scale analog of relative weights and thrusts?

Simply put, is there a way I could test the thrust and required burn time for a vessel like the one in the shown picture, but not actually have to find that out by launching such a spacecraft (but instead launching a smaller one)?

EDIT!

So I asked my friend who's good at math but doesn't know much about Physics, and he walked me through working the Tsiolkovsky Rocket Equation backwards!

What I can do, is plug my Desired Total Change in Velocity into the Delta-V value, to solve for the mass final. Once I have the mass final and mass initial, I'll be able to calculate how long it takes for the engine to consume that much fuel mass, as long as it consumes fuel at a uniform rate (that is, I don't change the throttle).

dV = V_eg₀ln(m_i/m_f) The famous rocket equation. Put into other terms for those with calculators without a natural logarithm function, and Kerbin's gravity instead of Earth's,

deltaV = (Isp * 9.81) * ((log( mass initial / mass final minus propellant )) / log( e )))

Say, I'm heading to Jool from Kerbin...

Solve for mass final (specific to chosen DeltaV):

deltaV = 2100 m/s²

Isp = 800s

mass _i = 57.5t

mass _f = ?

2100 = (800*9.81)((log( 57.5 / m_f )) / log( e ))) ----- Our initial equation, without a mass ratio complete.

2100 / (800*9.81) = ((log( 57.5 / m_f )) / log( e ))) ----- Divide the Effective Exhaust Velocity to the other side.

e^(2100 / (800*9.81)) = 57.5 / m_f ----- Take the reciprocal so we can solve for m_f (Remember, 1/x = x^-1 , so therefore 1/e^x = e^-x )

e^(-2100 / (800*9.81)) = m_f / 57.5 ----- Divide the initial mass to solve for the final.

(e^(-2100 / (800*9.81)))*57.5 = m_f ----- Solve for the final mass.

44.000 = m_f

The vehicle will have a mass of 44 tons when it's changed its velocity by 2100m/s².

Solve for Burn Time (amount of time it takes to consume the solved amount of propellant mass)

LV-N Fuel Consumption^[1]:

1.53l/s

Rockomax Jumbo-64 Fuel Tank^[2]:

mass initial = 36t

mass final = 4t

mass of propellant: 36-4 = 32t

volume initial = 2880l (liquid fuel)

litres per mass = 2880/32 = 90l/t

Fuel Consumption:

time to consume 1 ton of liquid fuel: ( 90 l/t)/1.53/s = 58.8 seconds of burn time per ton of fuel consumed

57.5t - 44t = 13.5 tons of propellant mass will be burned in order to accomplish the burn.

13.5t * 58.8s = 789.15s total time required to burn that much fuel.

Total Burn Time: 13 minutes, 9 seconds

QED

Please check me

Edited January 7, 2014 by Blue

[Captain Sierra]

You mean launching a different craft that has the same dry mass/fuel mass and thrust/weight ratios but smaller? I do not know. If you are going for Delta V scaling, I can do some rocket equation algebra and try to get you an answer but that will take time and maths.

[nhnifong]

if you need to get to Jool for example, that's about 2100 m/s of delta v. Let's say your ship's mass is 20 metric tons (Mg) and the LV-N produces 50 kN of thrust. Recall that is a derived unit of force equivalent to one kg*m/s^2. In Newtonian physics, F = ma where F is the force, m is the mass, and a is the acceleration. Plugging our mass and force into the equation and solving for acceleration, we get 2.5 m/s^2, and if we take the desired change in velocity and divide it by the acceleration, we get 840 seconds. But that would only give us the acceleration if the tank remained full. at the end of the burn, the tank will not be full so the mass won't be 20 metric tons any more. let's say it's 10. We can deal with this by using a mass that is an average of the starting and finishing masses. But we don't know the finishing mass unless we know how much fuel will be burned. So how do we figure that out? We have a few givens, the specific impulse of the LV-N: 800 seconds, abbreviated Isp. There is an equation that relates Isp to thrust and propellant mass flow rate: F = Isp * m. * g where m. is the mass flow rate in kg/s, and g is 9.8 m/s^2. If we plug in the max thrust and Isp of the LV-N and solve for the mass flow rate we get 6.37*10^-3 kg/s of propellant. We can totally ignore the fact that it's a bipropellant engine. and the densities of those fuels never come into the equation. To determine the mass of propellant used by the burn, we could take the mass flow rate and multiply it by the burn time, but the thru burn time actually depends on the amount of mass used, so lets just write it all out and hope we can solve it anyways.

burn_time = delta_v / acceleration

acceleration = thrust / average_mass

average_mass = (m_start + m_end) / 2

m_end = m_start - mass_flow_rate * burn_time

putting it together...

burn_time = delta_v / (thrust / ((m_start + m_start - mass_flow_rate * burn_time) / 2))

burn_time = delta_v * (2 * m_start - mass_flow_rate * burn_time) / 2 * thrust

burn_time * 2 * thrust / delta_v = 2 * m_start - mass_flow_rate * burn_time

2 * thrust / delta_v = (2 * m_start) / burn_time - mass_flow_rate

2 * thrust / delta_v + mass_flow_rate = (2 * m_start) / burn_time

(2 * thrust / delta_v + mass_flow_rate) / (2 * m_start) = 1 / burn_time

(2 * m_start) / (2 * thrust / delta_v + mass_flow_rate) = burn_time

m_start / (thrust / delta_v + mass_flow_rate / 2) = burn_time

Don't assume any of this is right I'm just working it out in my head and I haven't really done that much algebra in about 5 years.

[Blue]

I solved it out for myself with a friend's assistance. Please check my calculation.

[Kasuha]

Your equation looks ok.

I'm not sure what does your solving for final mass have to do with scale models.

And I'm not sure why do you need to solve it for final mass. You have your full mass, you have the amont of propellant/oxidizer -> you have dry mass (1 unit = 5 kg) -> you can calculate the delta v forward and check your delta v is higher than what you need.

Because you need to check your dry mass anyway, you won't be able to pull any more fuel out of it.

As long as almost all our fuel tank use constant ratio of dry mas and full mass, interesting would be to solve amount of fuel for given dead weight (engines+command pods+whatever else that isn't a fuel tank). That's of course also possible.

Edit: ok, I solved it for you.

Assuming:

v = delta v you want to achieve

I = your ship's ISP

g = g0 coefficient

D = dead weight (anything but fuel tanks)

F = amount of fuel to take (we're solving for it)

Note that the weight of fuel tanks you're carrying will be 9F/8 and their dry weight will be F/8

then the equation comes as:

v = I * g * ln((D + 9F/8)/(D + F/8))

v/(I * g) = ln((D + 9F/8)/(D + F/8))

e^(v/(I * g)) = (D + 9F/8)/(D + F/8)

Now the thing in the left is a constant value for us, let's give it a name

K = e^(v/(I * g))

and our equation turns to

K = (D + 9F/8)/(D + F/8)

K * (D + F/8) = (D + 9F/8)

K * D + K * F/8 = D + 9F / 8

(K - 1) * D = F * (9 - K) / 8

D * 8 * (K - 1) / (9 - K) = F

And there you are. Any more fuel will do the job, too. And clearly there are such delta-v values which cannot be achieved this way (K >= 9).

Edited January 7, 2014 by Kasuha

[Blue]

Your equation looks ok.

I'm not sure what does your solving for final mass have to do with scale models.

Well another thing I wanted to test was the orbital time-- for my mission I'm going to be doing the departure burn split into two burns: one will take me to a highly eccentric Kerbin orbit with the AP beyond Minmus, and then when the vessel reaches PE again, I'll burn a second time for escape and trajectorizing to Jool. I didn't know how long the orbit would take if it was such a large one, but I remembered that I could just calculate it from a Semi-Major Axis.

[draeath]

K = e^(v/(I * g))

What is "e^(x)" here? Are you raising Euler's constant by the power of x? Eg, if x was to be 4, would this be equal to approx 54.598? Plain text is not an optimal method for this kind of thing.

(this forum really could use MathML support)

Edited January 7, 2014 by draeath

[Kasuha]

What is "e^(x)" here? Are you raising Euler's constant by the power of x? Eg, if x was to be 4, would this be equal to approx 54.598? Plain text is not an optimal method for this kind of thing.

(this forum really could use MathML support)

Yes, it's the exponential function.

[draeath]

Ah! It's been far too long.