Water Rockets

[CPL.Luke]

like I said before though, the water will be flowing equally in both directions, thus conserving momentum. no need for the hopper to move.

[D H]

like I said before though, the water will be flowing equally in both directions, thus conserving momentum. no need for the hopper to move.

 

The hopper proper, the water in the tank on top of the hopper, the water flowing between the tank and the drain hole, and the water that has left the hopper together form an isolated system. The total linear momentum of this isolated system is constant.

 

EDIT. Something is seriously wrong with scienceforums.net. LaTeX is not working. I tried posting the math, but all that came out was

 

[ LaTeX Error: One or more directories do not exist ]

 

Is this related to the performance problems described in this post:http://www.scienceforums.net/forum/showthread.php?p=313047&highlight=performance#post313047 ?

 

Bottom line: If the water that has left the vehicle has a non-zero momentum, the hopper must end up with a non-zero velocity to compensate. This is exactly what happens.

 

CPL.Luke,

Show the math that exhibits how the entire system starts at rest, the hopper ends at rest, but the exhaust water has a non-zero velocity.

[Cap'n Refsmmat]

Sorry about that. LaTeX should work now.

[CPL.Luke]

hmm still doesn't work for me.

 

 

but as for the hopper, lets isolate the system such that were only looking at the exhaust water and can ignore everything else.

 

to do this lets consider a hopper filled with someamount of water, and someone pokes a hole in the bottom such that the water falls directly downward into a perfectly level channel.

 

the water will hit the channel with some downward momentum, which will of course be absorbed by the channel and the ground in turn.

 

the water now has zero momentum, but as more water is still coming down out of the hopper, the water has to go somewhere, so it goes out in both the right and left directions, gaining some amount of momentum in each direction but because the rate of flow and velocity in each direction is equal, momentum is conserved.

 

you seem to be assuming that more water will flow in one direction than the other, I don't see why your doing that, perhaps were both looking at different diagrams ie. on yours the channel is inclined?

[Cap'n Refsmmat]

Fixed again.

[CPL.Luke]

I see the latex in the previous posts but the latex error message is still up on post 27

[D H]

Allrighty, then.

 

The momentum of the hopper proper is

[math]p_v(t) = m_v \dot r_v(t)[/math]

 

The tank is rigidly attached to the hopper. Ignoring sloshing in the tank,

[math]p_f(t) = m_f(t) \dot r_v(t)[/math]

 

The fuel flowing between the tank and the hole in the hopper has momentum

[math]p_{flow}(t) = \dot m_f(t) r_{v \rightarrow f}[/math]

 

The momentum of the water that has left the hopper is a bit trickier. It is the cumulative momenta of the differential quantities of water ejected from the hopper from the time the valve was opened to the current time:

[math]p_{exhaust}(t) = - \int_{t_0}^t \dot m_f(\tau) r_v(\tau) d\tau[/math]

 

Summing and differentiating with respect to time yields the time rate of change in momentum of the isolated system hopper+water system:

[math]\dot p_{sys}(t) \equiv 0 =

m_v \ddot r_v(t) +

\dot m_f(t) \dot r_v(t) + m_f(t) \ddot r_v(t) +

\ddot m_f(t) r_{v \rightarrow f} -

\dot m_f(t) r_v(t)[/math]

 

Simplifying yields the equation of motion for this system:

[math](m_v + m_f(t)) \ddot r_v(t) + \ddot m_f(t) r_{v \rightarrow f} = 0[/math]

 

Now assume that shortly after opening the valve the flow rate reaches a constant, steady-state rate [math]\dot m_{f_s}[/math]. The flow rate stays at this rate until the valve is closed, at which time the flow rate quickly drops to zero. The changes in the hopper velocity due to opening and closing the valves at times [math]t_o[/math] and [math]t_c[/math] are, to first order,

[math]

\Delta \dot r_v(t_o) \approx

- \frac {\dot m_{f_s}}{m_v + m_f(t_o)} r_{v \rightarrow f}

[/math]

[math]

\Delta \dot r_v(t_c) \approx

+ \frac {\dot m_{f_s}}{m_v + m_f(t_c)} r_{v \rightarrow f}

[/math]

 

Nothing changes if the mass of water left in the tank after the valve is closed is considered to be a part of the hopper itself. Denoting [math]\Delta m_f[/math] as the total amount of water ejected from the fuel tank while the valve is open,

[math]

\Delta \dot r_v(t_o) \approx

- \frac {\dot m_{f_s}}{m_v + \Delta m_f} r_{v \rightarrow f}

[/math]

[math]

\Delta \dot r_v(t_c) \approx

+ \frac {\dot m_{f_s}}{m_v} r_{v \rightarrow f}

[/math]

 

Summing the two terms yields the total change in the hopper velocity:

[math]

\Delta \dot r_v \approx

\frac {\dot m_{f_s} \Delta m_f}{m_v(m_v + \Delta m_f)}

r_{v \rightarrow f}

[/math]

 

CPL.Luke,

To avoid having you berate me for ignoring the time it takes for the water to flow from the tank to the hole and from the hole to the hopper to the ground, I am doing the following:

• Placing the valve at the end of the line (i.e., at the hole) rather than at the tank. This eliminates the lag between opening the valve and water falling into the hopper.
  
• Making the vehicle negligibly thin. This eliminates the lag between the water falling from the top to the bottom of the the hopper.
  

 

You said in some previous post,

In your first two setups your essentially trying to put a boat engine in a bathtub and make the bathtub move, outside of wiley cyote this doesn't work.

 

Look at what you said about the moving water that has left the vehicle and the stationary hopper after the valve is closed in light of this statement.

[D H]

the water will hit the channel with some downward momentum, which will of course be absorbed by the channel and the ground in turn.

 

Ignore all that. Assume that the hole at the bottom of the hopper turns the exhaust water into dark matter that has momentum but does not interact with anything, including itself. The end result is the same.

 

 

BTW, the

[ LaTeX Error: One or more directories do not exist ]

 

in post #27 is not a math error. I was reproducing in text the error message I received when I posted LaTeX. There is no math in that post.

[CPL.Luke]

ah, I was just skimming for latex mistakes and saw that.

 

I think were thinking of two entirely different problems here. and we need a common diagram.

 

because right now I'm saying that the water drops out the bottom of the hopper do to gravitational acceleration (therefore no momentum was borrowed against the hopper) instead momentum is borrowed against the earth, and the momentum is returned when the water hits the ground.

 

 

and like you said if the water turned into dark matter and didn't interact with anything but instead was only attracted to the earth, then the water still merely accelerates towards the earth and the earth towards it. no reason why the hopper should move.

 

now, if the hopper has a hole punched in it from any direction other than straight down, then its a different story and the hopper will gain momentum and accelerate.

[CPL.Luke]

also i'm using the term borrowed to mean that the sum of object A and object B's momentum is zero.

 

so as an asteroid plummets towards the earth it borrows momentum from the earth, and accelerates. Likewise the earth borrows momentum from the asteroid and accelerates towards it.

[D H]

because right now I'm saying that the water drops out the bottom of the hopper do to gravitational acceleration (therefore no momentum was borrowed against the hopper) instead momentum is borrowed against the earth, and the momentum is returned when the water hits the ground.

 

Ignore the vertical motion. Treat this as a one dimensional problem. The key things your are not seeing are

• The water leaves the hopper with zero velocity relative to the hopper. If the hopper has some horizontal velocity, then so does the water falling out of it. Simple physics. Think of what happens when you toss a ball inside a car traveling at 60 mph. The ball doesn't lose that 60 mph horizontal speed when you let it go.
  
• The water flowing from the tank to the hole in the center of the hopper has momentum. This momentum cannot be ignored. Simple fluid dynamics concepts yield the flow momentum as [math]p_{flow} = \dot m_f r_{v \rightarrow f}[/math].
  
• It doesn't really matter what happens to the water once it leaves the hopper. There is indeed no interaction between the water and the hopper once the water leaves the hopper. However, if we make a horizontally isolated system, it is easier to see what happens to the hopper by looking at what happens to the water ejected from the vehicle. I put a frictionless channel between the tracks to ensure a horizontally closed system for this reason.
  

 

You agreed earlier that the hopper does move when the hole is plugged (i.e., when the center of mass of the water moves from the off-center tank to the center of the hopper). The hopper+water forms a horizontally closed system. The hopper gains momentum to exactly balance the fluid flow momentum of the water running from the water tank to the center of the hopper. It has to in order to keep the center of mass of the entire system stationary.

 

Suppose we pull the plug in the bottom of the hopper some time after the valve is opened but well before the valve is closed. Up until the time the plug is pulled, the hopper has the momentum that exactly counters the fluid flow momentum. The only thing that makes the hopper stop in this closed configuration is for the water flow to stop. By assumption, that hasn't happened yet. Now pull the plug. Pulling the plug doesn't make the hopper stop, does it? The collected water in the hopper drains to the ground, but it has horizontal velocity equal to that of the hopper. The ground doesn't absorb that horizontal momentum; the water is spilling onto a frictionless track. It doesn't matter whether the water piles up, sloshes around, whatever. The aggregate horizontal momentum remains unchanged. Now close the valve. The only possible moving objects are the water that has left the hopper and the hopper itself. The water is still moving. Claiming the hopper comes to a rest violates the conservation of linear momentum.

[CPL.Luke]

ah see now in the system that your describing the water does not leave the hopper with zero velocity.

 

If I understand you correctly your placing your hole to the side of the hopper, in which case the fluid will be propelled by the pressure etc. inside of the hopper and generate some thrust.

 

I thought we were placing the hole directly beneath the hopper in which case the hopper won't experience any thrust.

 

The point however is that the final momentum will ony depend on the total momentum ejected from the hopper. (note this is the only the final momentum, the momentum before the fuel flow is stopped will depend on the internal fluid movements)

 

but like I said I thought we were just looking at two different mental diagrams and that appears to be what occored.

[D H]

ah see now in the system that your describing the water does not leave the hopper with zero velocity.

 

The water leaves the hopper with zero velocity --- relative to the hopper. This is just like a baseball tossed straight up inside a speeding car. The ball leaves the hand with zero horizontal velocity --- relative to the hand. That ball is moving 60 mph with respect to a stationary observer. The water leaving the hopper is similarly moving with respect to a stationary observer.

 

There is no hole in the side of the hopper, there is no ramp. The hole is straight down. I don't need pressure. I don't need thrust. I said in my first post on this subject that this is a "zero-thrust rocket". I just need conservation of momentum.

 

I thought we were placing the hole directly beneath the hopper in which case the hopper won't experience any thrust.

 

Yes! This is a "zero-thrust rocket". The hopper moves in the first place because of the fluid flow on top of the hopper. The hopper continues to move after the flow ends because, even with no thrust, momentum is ejected from the hopper. That 'ejected momentum' is the stream of ejected water that follows the hopper (until the flow inside the hopper stops).

 

The point however is that the final momentum will ony depend on the total momentum ejected from the hopper.

 

Exactly.

[CPL.Luke]

if the water exits the hopper with zero velocity relative to it, then there is no ejected momentum.

[D H]

You are not looking at this the right way.

 

Here is a different scenario with the same equations of motion. Connect a vehicle to a frictionless rod in space. The rod is infinitely long and infinitely massive. The sole purpose of the rod is to constrain the vehicle to one dimensional motion.

 

Put a water tank and squirt gun on the vehicle. The squirt gun will be used to propel the vehicle. For some reason, the vehicle designers mounted the squirt gun so that it ejects water normal to the rod.

 

If we co-locate the water tank with the squirt gun the vehicle cannot change its velocity. Suppose the vehicle has some longitudinal velocity (longitudinal = along the rod) with respect to some inertial observer. In this case, firing the squirt gun will merely create an outgoing stream of water that moves along with the vehicle. The vehicle's velocity does not change.

 

Things become much more interesting when the water tank and squirt gun are separated longitudinally.

 

Suppose the vehicle is at rest with respect to some inertial observer prior to firing the squirt gun. The vehicle fires the squirt gun for some length of time and then stops firing. The flow rate out of the gun is constant except for short intervals of time immediate following start up and shut down. What is the state of the of vehicle with respect to this same observer after firing terminates?

 

Firing the squirt gun creates a flow of water from the tank to the gun. Conservation of momentum requires that the vehicle move in a manner that counteracts this flow momentum. The vehicle transfers momentum to the flow during the start-up phase: The vehicle undergoes some start-up acceleration. The vehicle stops accelerating once the flow reaches a constant rate. Now the vehicle is moving at a constant, non-zero velocity with respect to the inertial observer. The water is being ejected normal to the rod. From the perspective of the vehicle, this exhaust flow has no longitudinal velocity. It is moving in a straight line normal to the rod. From the perspective of the inertial observer, the vehicle and the exhaust water are moving with the a common longitudinal velocity. The exhaust water has a transverse velocity as well (the same transverse velocity seen from an observer on the vehicle).

 

Suppose the vehicle continues firing long enough to make an observable change in the mass of the vehicle+tank water system before firing terminates. When firing terminates, the vehicle transfers momentum to the flow inside the vehicle that exactly counterbalances the momentum transfered to start the flow. However, the vehicle's mass has changed considerably. To achieve that equal but opposite change in momentum, the change in the vehicle's velocity at flow termination must be greater in magnitude and opposite in sign compared to the change in velocity at flow initiation.

 

Your concept of 'borrowing' momentum is essentially correct. The flow 'borrows' momentum from the vehicle at flow initiation and returns that 'borrowed' momentum to the vehicle at flow termination. The change in vehicle mass means the vehicle doesn't stop. Instead, it reverses direction.

[Severian]

The key to understanding this is that it is the centre-of-mass which obeys the Newtonian equations.

 

So for example, if you have an ensemble of n particles with position vectors [math]\vec{r}_i[/math] and mass [math]m_i[/math] each, then the centre-of-mass is

[math]\vec{r}_{\rm cm} = \frac{\sum_{i=1}^n m_i \vec{r}_i}{\sum_{i=1}^n m_i}[/math]

 

Differentiating this with respect to time twice:

[math]\vec{a}_{\rm cm} = \frac{\sum_{i=1}^n m_i \vec{a}_i}{\sum_{i=1}^n m_i}[/math]

Notice the masses [math]m_i[/math] do not depend on time because they are fixed (I am not allowing mass->energy conversion). This is also true in a rocket - the mass of rocket+fuel doesn't change - the mass only appears to change because we stop counting ejected fuel as part of the rocket.

 

But if each particle feels a force [math]\vec{F}_i[/math], then by Newton's second law, [math]m_i \vec{a}_i = \vec{F}_i[/math], so

[math]\vec{a}_{\rm cm} = \frac{\sum_{i=1}^n \vec{F}_i}{\sum_{i=1}^n m_i}[/math]

 

So the acceleration of the centre-of-mass frame can only come from a total net force on the constituent particles. Since Newtons first law tells us that internal force will be balanced, any net force must be external. The centre-of-mass does not accelerate unless there is an external force.

 

Note, that this is not saying that the rocket can't accelerate - it does, because it is not the entire ensemble. One can also make the rocket wiggle or turn by transfering angular momentum between the rocket and the ejected fuel.

 

However, if we consider that we only fire the fuel straight out the back, then the rocket (and its centre-of-mass) must exactly balance the momentum loss from the fuel. The 'external force' in this case is just the rate of change of momenta lost in the ejected fuel. If you want to consider the remaining fuel as part of the rocket (and you should) then it doesn't matter how much it moves about, it will not change the motion of the centre-of-mass of the rocket, because it is itself included in the centre-of-mass calculation.

[CPL.Luke]

So I take it you think the hopper will move?, or the rocket of post 40?

[D H]

However, if we consider that we only fire the fuel straight out the back, then the rocket (and its centre-of-mass) must exactly balance the momentum loss from the fuel.

 

As you noted, the fuel that remains in the tank is essentially a part of the rocket. Unless the fuel tank is located at the center of mass of the dry rocket, the center of mass of the rocket with respect some fixed point on the rocket changes as the rocket expels fuel. This 'motion' of the center of mass means that the "rocket (and its centre-of-mass) exactly balance(s) the momentum loss from the fuel" is not quite correct.

[Severian]

As you noted, the fuel that remains in the tank is essentially a part of the rocket. Unless the fuel tank is located at the center of mass of the dry rocket, the center of mass of the rocket with respect some fixed point on the rocket changes as the rocket expels fuel.

 

Yes, I agree with that. The position of the centre-of-mass with respect to (say) the nose tip will change.

 

But it is a bit of a semantic dispute. I would say that the rocket's position is defined by its centre-of-mass, so from this viewpoint the internal movement of fuel does not change the trajectory of the rocket. If you define the position of the rocket by its outer shell, then it can, simply by moving around its centre-of-mass within the structure.

 

Similarly, the centre-of-mass of the hopper (including the fluid) does not move, but the hopper shell may move because of the shift of the centre-of-mass.

[D H]

I would say that the rocket's position is defined by its centre-of-mass, so from this viewpoint the internal movement of fuel does not change the trajectory of the rocket. If you define the position of the rocket by its outer shell, then it can, simply by moving around its centre-of-mass within the structure.

 

The movement of fuel does change the trajectory of the rocket because the vector from the center of mass of the rocket to the thruster is not constant. The time derivative of that vector is a part of the equations of motion for the rocket.

 

Similarly, the centre-of-mass of the hopper (including the fluid) does not move, but the hopper shell may move because of the shift of the centre-of-mass.

 

Find the math error in this post. The hopper shell and the center of mass move.

[D H]

if the water exits the hopper with zero velocity relative to it, then there is no ejected momentum.

 

Momentum is transfered. The general equation for the momentum of the exhaust produced by a rocket is

 

[math]p_g = \int_{t_0}^t \dot m_g(\tau)(\dot r_t(\tau) + v_e)d\tau[/math]

 

where [math]\dot m_g(\tau)[/math] is the rate at which the exhaust mass increases, [math]\dot r_t(\tau)[/math] is the velocity of the thruster, and

[math]v_e[/math] is the exhaust velocity relative to the thruster (assumed constant). Differentiating with respect to time,

 

[math]\dot p_g = \dot m_g(t)(\dot r_t(t) + v_e)[/math]

 

In other words, momentum is transfered to the gas even in the case of a zero-thrust rocket. This had better be the case or the rocket equation would not be valid at all. The term [math]\dot m_g(t)\dot r_t(t)[/math] is essential.

 

Ignoring fuel flow, the rocket's momentum is

 

[math]\dot p_b = m_b(t) v_b(t)[/math]

 

Where the subscript [math]b[/math] denotes the body in question (i.e., the rocket which includes the fuel in the tank). Differentiating wrt time,

 

[math]\dot p_b = \dot m_b(t) v_b(t) + m_b(t) \dot v_b(t)[/math]

 

By conservation of mass,

 

[math]\dot m_b(t) = \phantom{-}\dot m_f(t)[/math]

[math]\dot m_g(t) = -\dot m_f(t)[/math]

 

If we model the rocket as a single point, then [math]\dot r_t(t) = v_b(t)[/math]. The time derivative of the exhaust gas momentum becomes

 

[math]\dot p_g = -\dot m_f(t)(v_b(t) + v_e)[/math]

 

By conservation of momentum,

 

[math]\dot p_b(t) + \dot p_g(t) = 0[/math]

 

Substituting and simplifying,

 

[math]m_b(t) \dot v_b(t) + \dot m_f(t) v_e = 0[/math]

 

Integrating this leads directly to the rocket equation. Note that the zero-thrust momentum term [math]\dot m_g(t)\dot r_t(t)[/math] cancels the term [math]\dot m_b(t) v_b(t)[/math] with the assumption of a point mass rocket.

[Severian]

The movement of fuel does change the trajectory of the rocket because the vector from the center of mass of the rocket to the thruster is not constant. The time derivative of that vector is a part of the equations of motion for the rocket.

 

It makes no difference where the back of the rocket is compared to the centre-of-mass. As long as the fuel leaves the rocket in a constant direction, there is no change, since the only important thing is the change in momentum.

[D H]

It makes no difference where the back of the rocket is compared to the centre-of-mass. As long as the fuel leaves the rocket in a constant direction, there is no change, since the only important thing is the change in momentum.

 

Try again. It does matter. Do the math.

 

I work in the field of spacecraft guidance, navigation, and control. We recently had to chase down a 30 meter discrepancy in our simulated orbit insertion position. One source of that error was how we propagated state. We formerly ignored what center of mass motion does to the equations of motion. We instead accounted for the change in center of mass after each propagation step.

 

What I have recently posted is an offshoot of our investigations into the source of that error. The motion of the center of mass in the vehicle structure does impact the equations of motion.

[Severian]

Try again. It does matter. Do the math.

 

*sigh* You don't seem to be understanding this, do you? Which bit aren't you following? The rate of change of momentum is given by the net force. If the momentum of the fuel is in one direction the rocket goes in the other, period. You can't change that - as Scotty would say 'you cannie change the laws of physics'.

 

The effect that you are claiming is probably due to you changing the direction of the fuel flow by changing where the exhaust is in relation to the centre-of-mass of the rocket and remaining fuel. Obviously if you change the direction that you eject the fuel, the rocket will change directions, but that is hardly... rocket science... and is completely not the point of the discussion here.

[D H]

*sigh* You don't seem to be understanding this, do you? Which bit aren't you following? The rate of change of momentum is given by the net force. If the momentum of the fuel is in one direction the rocket goes in the other, period. You can't change that - as Scotty would say 'you cannie change the laws of physics'.

 

Jeez. DO THE F***ING MATH.

 

I am not spouting a crackpot notion here. I work at NASA; NASA regards me as an expert in this exact field. I live and breath the rocket equation; these results were quite surprising to me as well. We had to chase down an error. Surprise, surprise. Sometimes mathematics yields rather counterintuitive results. But it doesn't lie (unless one resorts to statistics: liars, damn liars, and statisticians). Do the f***ing math.