Specific impulse & Velocity?

[GeeKay]

The quote below is the most succinct description I've found on the net explaining the relationship between spacecraft velocity and performance:

 

"The speed of a rocket is directly affected by 2 factors. The first is
mass ratio, which is a ratio the weight of the rocket at lift-off
compared to the weight of the rocket at engine shutdown. The second
factor is specific impulse, which is the amount of thrust produced from
each pound of propellant per second. The higher the mass ratio and the
higher the specific impulse, the faster the rocket can go."

 

Mass ratio I understand. Likewise, I can get my head around specific impulse - although I still have to strain it a bit. What I don't understand is how all this translates into spacecraft velocity. It's all very well to talk about exhaust velocity, but at risk of sounding fatuous, were I a petrolhead wanting to buy a sports car, I might be interested in fuel consumption (possibly) and knowing how efficiently the exhaust functions (well, maybe) but really, I'd love to know how fast the car goes. Unfortunately, concerning specific impulse and 'forward' velocity as it apples to spacecraft, I can find nothing anywhere that can join up these two dots. Many thanks.

 

PS. Again, I apologise for the crassness of the question. But how can one find out if you don't ask?

 

 

[swansont]

For a car, the analogous performance is the acceleration, which is a matter of the torque the engine can deliver at some value of RPMs, and the car's mass, because a = F/m. One huge difference between the rocket and the car is that the rocket will accelerate the whole time it has fuel. Then it coasts. If a car did that, the top speed would be dependent on acceleration and fuel capacity.

 

So the specific impulse and mass ratio are combinations of the acceleration you can deliver and the size of the gas tank.

[Fuzzwood]

What you calculate with this (the rocket equation) is called the ΔV or change in velocity. Do not confuse this with acceleration as the latter is a change in velocity per time unit. I can directly add any quantity of ΔV to the velocity I already have, turning this in an addition of vectors.

 

For example: I have a rocket with a ΔV of 1000 m/s. If I start at 0 m/s and burn all my fuel in one go in a single direction, I will end up going 1000 m/s.

 

I can also burn my rocket for 500 m/s in one direction. Starting at 750 m/s, I will be going at 1250 m/s after this burn, then 500 m/s in the opposite direction to end back up traveling the same initial speed of 750 m/s.

 

I can also make this a bit more complicated by burning at a certain angle from my initial velocity vector. If I am going at 500 m/s in the x direction, then burn 500 m/s in the y-direction, I will end up going 707 m/s in a direction 45 ° from my initial velocity vector.

Edited March 9, 2015 by Fuzzwood

[Skeptic134]

The quote below is the most succinct description I've found on the net explaining the relationship between spacecraft velocity and performance:

 

"The speed of a rocket is directly affected by 2 factors. The first is

mass ratio, which is a ratio the weight of the rocket at lift-off

compared to the weight of the rocket at engine shutdown. The second

factor is specific impulse, which is the amount of thrust produced from

each pound of propellant per second. The higher the mass ratio and the

higher the specific impulse, the faster the rocket can go."

 

What I don't understand is how all this translates into spacecraft velocity.

 

 

The mass ratio tells you the amount of fuel available to be burned by the rocket which correlates to how long the rocket can produce an accelerating force. The specific impulse provides information on the energy density of the fuel and thus the amount of force that can be produced will the fuel is burning.

 

With those two pieces of information, force and time, you can calculate the final velocity of the rocket.

[GeeKay]

 

The mass ratio tells you the amount of fuel available to be burned by the rocket which correlates to how long the rocket can produce an accelerating force. The specific impulse provides information on the energy density of the fuel and thus the amount of force that can be produced will the fuel is burning.

 

With those two pieces of information, force and time, you can calculate the final velocity of the rocket.

 

 

Regarding these "two pieces of information, force and time", I'd appreciate it very much if you could provide me with an example how a given final velocity may be calculated. Many thanks.

[Janus]

Regarding these "two pieces of information, force and time", I'd appreciate it very much if you could provide me with an example how a given final velocity may be calculated. Many thanks.

The "rocket equation" is:

 

[math]\Delta v = v_e \ln(MR)[/math]

 

Where Ve is the exhaust velocity and MR the mass ratio.

 

The exhaust velocity is dependent on the specific impulse of what you are burning as fuel. The more energy released by burning the fuel, the hotter and faster the exhaust.

 

The basics behind rocket propulsion is this:

 

Imagine that you are standing on a wagon full of bricks. You start throwing bricks backwards. Assuming that friction is negligible, when you throw the first brick, the wagon moves forward in response. How fast it moves depends on the speed you throw the brick and its mass compared to total mass of the wagon(including any remaining bricks).

Every brick you throw after that adds to the wagon's forward momentum.

How fast you throw the brick is the exhaust velocity, and with rocket fuel, the energy to do so is contained within the brick itself. (the amount of energy in each brick is the specific impulse.)

 

 

As you throw bricks, the number of remaining bricks and the thus the total mass of the wagon goes down. This means that each succeeding brick adds more velocity to the wagon than the previous one did.(the thrust from throwing the brick remains the same, but the mass it is propelling is decreasing).

This why the rocket equation takes the form it does, it has to account for the decreasing mass of the rocket as you burn fuel.

 

There are two ways to increase the maximum velocity of a rocket. One is to add more fuel, and the other is to use fuel that has more energy per kg.

 

Increasing the amount of fuel is not as effective as increasing the exhaust velocity, Doubling the exhaust velocity will double the final speed.

 

Doubling the fuel is not as effective. For example, If we start with an fuel mass that equals the empty mass of the rocket, this gives a mass ratio of 2 and a final rocket velocity of 0.693 that of the exhaust velocity. If we double the fuel mass, we get a mass ratio of 3, and a final speed of 1.099 times the exhaust velocity which is only an increase by a factor of 1.56

 

Doubling the exhaust velocity again, doubles the final velocity again. Doubling the fuel mass gives a mass ratio of 5 and only increases the final velocity by a factor of 1.46 (doubling the fuel the second time was even less effective than doubling it the first time was.)

 

Theoretically, there is no limit to a top speed of a rocket(except that imposed by Relativity, and if you have a rocket capable of reaching such speeds you need to use the Relativistic form of the rocket equation) given enough fuel.

[Skeptic134]

An overly simplistic way of looking at it is using Newton's second law and understanding the relationship between force, time and velocity. If a certain force is applied over a specific amount of time and the net force is calculated, the acceleration of a mass and subsequently the change in velocity is straightforward.

 

In rocketry the mass of the system isn't static however and that complicates the calculation so for your question the correct way to perform the calculation is using the Tsiolkovsky rocket equation.

 

[math]\Delta V = Isp * g * ln (MR)[/math]

 

Where Isp is the specific impulse and MR is the mass ratio.

[pavelcherepan]

As others have pointed out in previous posts, you can calculate your delta-v from the rocket equation but not all of your available delta-v translates into the velocity of the rocket.

 

Say, you start from the ground and want to get to the Low Earth Orbit. While the velocity you of the craft on LEO is about 7800 m/s, the actual delta-v required to get there is somewhere between 9500 and 11000 m/s depending on several factors, for example whether you're heading East or West after the initial vertical part of trajectory. That extra delta-v required is lost to atmospheric drag and gravitational losses.

 

If, on the other hand you start from geostationary orbit, where both the drag and the gravity losses are negligible, pretty much all of the delta-v you have available will translate directly into your velocity.

Edited March 11, 2015 by pavelcherepan

[Skeptic134]

As others have pointed out in previous posts, you can calculate your delta-v from the rocket equation but not all of your available delta-v translates into the velocity of the rocket.

 

Say, you start from the ground and want to get to the Low Earth Orbit. While the velocity you of the craft on LEO is about 7800 m/s, the actual delta-v required to get there is somewhere between 9500 and 11000 m/s depending on several factors, for example whether you're heading East or West after the initial vertical part of trajectory. That extra delta-v required is lost to atmospheric drag and gravitational losses.

 

If, on the other hand you start from geostationary orbit, where both the drag and the gravity losses are negligible, pretty much all of the delta-v you have available will translate directly into your velocity.

 

Wouldn't the gravitational losses be taken into account by the specific impulse portion of the rocket equation?

[GeeKay]

Many thanks for the explanations - brick analogies and all Yes, I think I finally understand the equation now. On a personal note, I suffered badly on the maths front at school. Partly the fault was my own; but I do contend that a substantial chunk of it was due to the teaching practices typical during those times, especially when it came to rote-learning techniques. This method has its uses, but for me back then, especially when it came to the teaching of maths, it was heavy on the 'how' and light on the 'why'. It wasn't until I did GCSE Maths as a mature student that I began to understand that maths is a language - and a wonderful language at that. At least I know that now, even if I remain a poor reader of numbers. I just wanted to say that. End of personal note.