Water Rockets

[CPL.Luke]

D H

 

Here's sevarian's point and after looking at your post I belive that we are all in agreement, its just that were talking about different problems.

 

 

If you lock the problem into one dimension so that you have a rocket of length L a fuel tank and a thruster all perfectly balanced along at the origin along any set of axes made perpendicular to the the line from nosecone to thruster. and assume that at the beginning the center of mass along the L axis is located in the fuel tank.

 

the thruster begins to fire and ejects mass out the back end of the rocket, after some time the center of mass of the rocket (not including the ejected mass) has shifted towards the back end of the rocket slightly. and after the fuel tank has been completely emptied the center of mass will lie at the the center of the rocket along the L axis.

 

now in this problem the total momentum of the rocket (just the shell, again not the expended fuel) is going to be exactly equal to the negative of the momentum carried by the fuel in the opposite direction.

 

it does not matter what path the fuel took along its one dimensional trip (whether it zigzaged back and forth between the fuel tank the nose cone and the thruster several times before being shot out of the thruster doesn't matter) to get the final momentum all that you need to look at is the ejected mass.

 

 

however if your looking at the total discplacement of the rocket from where it started its burn, your going to need to take into account the center of mass change.

 

The problem becomes a bit more complex when you add in the possibility of the center of mass change along multiple axis, thus allowing the possibilty that the thruster will cause the rocket to change direction and possibly spin do to the change in the vector between where the force is applied and the center of mass.

 

but the final momentum of the rocket will be exactly equal to the momentum carried away by the fuel in the opposite direction of the rocket.

 

 

one other point however, in your zero thrust rockets, you kept insisting that the velocity of the ejected mass relative to the rocket was 0 this would mean that the mass would not move away from the rocket and would infact stay with the rocket.

[Severian]

I am not spouting a crackpot notion here.

 

Well, it is not terribly clear what you are trying to say. You seem to be saying that momentum is not conserved, which really would be a crackpot notion, but I suspect you are being deliberately obtuse. Conservation of momentum is a very simple concept and is required for space-translation-invariance of the laws of physics. You can't just throw it away.

 

As I said before, the only effect that you could have of the kind you describe would be from the internal motion of the fuel changing the centre-of-mass such that the fuel is ejected in a different direction. Since I had this in my previous post and you ignored it I can only assume that you disagree. So please tell me why you disagree with conservation of momentum?

[D H]

Well, it is not terribly clear what you are trying to say.

 

To address this problem, one of our team members came up with the idea of a one-dimensional vehicle with two opposing thrusters, one on each end of the vehicle. The net effect is zero thrust. Standard treatments say the rocket doesn't move. There is no thrust afterall. This can be further reduced to the hopper (exact same equations of motion), or the squirt gun-on-a-wire (post #40, again, the exact same equations of motion). I like the squirt gun-on-a-wire model. No gravity to confuse the issue, no puddling of the exhaust. Nice and simple.

 

Look at post #40.

 

now in this problem the total momentum of the rocket (just the shell, again not the expended fuel) is going to be exactly equal to the negative of the momentum carried by the fuel in the opposite direction.

 

Correct. However, how much momentum has been transferred to the exhaust depends on the behavior of the center of mass. The difference between the simple model that yields the rocket equation and a model that accounts for the motion of the center of mass is small: The rocket equation remains a good approximation for the velocity.

 

However, the rocket velocity is always a bit smaller than that predicted using the simplified equations of motion that yield the rocket equation. This is a bias. Integrating biased estimates can result in significant errors, even when the bias is small.

 

 

The problem becomes a bit more complex when you add in the possibility of the center of mass change along multiple axis, thus allowing the possibilty that the thruster will cause the rocket to change direction and possibly spin do to the change in the vector between where the force is applied and the center of mass.

 

Not just a bit. The problem becomes a lot more complex when you add in the two dimensions that have been ignored in all of the previous posts. I don't want to go there in this forum. Multiple pages of LaTex.

 

 

one other point however, in your zero thrust rockets, you kept insisting that the velocity of the ejected mass relative to the rocket was 0 this would mean that the mass would not move away from the rocket and would infact stay with the rocket.

 

Exactly. Until the fuel flow terminates, that is. Look at what you said earlier:

 

now in this problem the total momentum of the rocket is going to be exactly equal to the negative of the momentum carried by the fuel in the opposite direction.

 

If (a) the entire system started at rest, and (b) the exhaust has some aggregate non-zero momentum when thrusting terminates then the vehicle must have non-zero momentum that counterbalances the exhaust momentum. Since (a) is true by definition, the only question that remains is, Does the exhaust have some non-zero linear momentum? It does. Look at post #40.

[D H]

I turned this into an interview question. Simple counterintuitive problems like this lets me get insight into how interviewees think.

 

A space vehicle with two thrusters is depicted below.

+=====v=========+

|.....|.........|

|.....+-----O...|

|.....|.........|

+=====^=========+

 

The thrusters (the 'v' and '^' in the diagram) direct their exhaust normal to the long axis of the vehicle. The two thrusters have the same Isp and generate identical force magnitude. The vehicle is perfectly symmetric about the long axis. A single fuel tank (the 'O' in the diagram) feeds the two thrusters and is located some distance from the thrusters along the long axis of the vehicle.

 

The vehicle starts in a quiescent state at rest wrt some inertial observer. The two thrusters are activated simultaneously. The thrust from each thruster quickly ramps up to a constant value; both thrusters ramp up in an identical manner. After firing for some time, both thrusters are shut down simultaneously. Both thrusters ramp down to null firing quickly and in the same manner.

 

Describe the behavior of the vehicle.

[Spyman]

Describe the behavior of the vehicle.

OK, I have not read the thread from the beginning so I might come in from a wrong angle here...

 

During the quickly ramp up, the fuel tank will act as an internal thruster and accelerate the vehicle in opposite direction of the fuel flow.

 

When the ramp up is completed the acceleration stops and the vehicle moves with constant speed. (As long as the fuel flow remains constant.)

 

After the firing time and the ramp down starts the fuel flow in the pipe will deccelerate the vehicle by terminating in the T-connection.

 

At the time of null firing and zero fuel flow the vehicle will have slowed down to the same initial speed as before the ramp up.

 

The vehicle are now at rest in a new position with respect to the inertial observer.

[D H]

You got the first part right (state during thrusting), but not the final state. What is the net momentum of the exhaust gas, and what does that say about the final state of the vehicle (use conservation of momentum)?

[Spyman]

Point taken, net momentum of the exhaust gas will be in the direction of movement so by use of conservation law the vehicle must be moving in opposite direction.

 

During the acceleration the vehicle contains more fuel, (more mass), than when deccelerating, and since the force from the thrust of the fuel tank also must equal the termination in the T-split, (conservation law), the difference in mass will cause the ship to deccelerate more than accelerate.

 

You are correct, the final state of the vehicle will be a constant speed with respect to the inertial observer.