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To say Gauss-Jordan elimination is the only way is a bit strong, to say the least. I rarely use Guass-Jordan elimination, and I deal with matrices a lot. I frequently use various decomposition techniques: LU, Cholesky, QR, SVD, ... Guass-Jordan elimination is the only way to do it by hand, perhaps.

Strictly speaking, 12 noon is 12 noon, period. Designating 12 noon as AM or PM is a contradiction in terms. AM means before noon, PM means after noon. We say 12 noon is 12:00 PM because even a tiny fraction of a second after noon is after noon: Such a time gets the PM designator when using a 12-hour clock. It makes more sense for the AM/PM designator to switch in sync with the hour than some infinitesimal time after the hour, does it not?

It is a lot less humid in Arizona than in Georgia. Lower humidity => less scattering of light. Twilight is short in the desert.

Wikipedia has a fairly good article on pseudo-random numbers. Tree discussed linear congruent generators and pointed to a Wiki article on this subject. Tree, did you notice the disclaimer in this article Unfortunately, it is easy to write a bad pseudo-random number generator. For example, Microsoft's basic random number generator 'RAND' remains notoriously bad (surprise, surprise). It repeats after generating 32,768 values. I do a lot of Monte-Carlo analyses, sometimes running tens of thousands of cases, each of which involves many thousands of calls to the random number generator. I need a very good pseudo-random number generator. I also need repeatable random numbers, which is exactly what a pseudo-random number generator gives. Suppose test case #89623 (for example) displayed some very wierd results. If I used a true random number generator, how could I possibly re-create that one specific test? With a pseudo-random number generator, I can re-create it and analyze it to death.

It is fundamental because it relates integration and differentiation. What could be more fundamental (in realm of the calculus) than a theorem that relates the two core concepts of the calculus? A proof is here. Google is your friend.

Yet another example of the utility of percussive reparation techniques.

The obvious critical factor is the vapor pressure, which is much lower at -40C (0.13 mmHg) than at 40C (55.35 mmHg). Raising water-saturated air from -40C to 40C without adding any moisture results in 0.24% relative humidity. The humidity is low in the desert, but not that low (relative humidity is in the low teens or single digits). Perhaps the instructor was meaning relative humidity and used some words that made you think absolute humidity.

Is zero a positive integer?

Many mathematicians define 0^0 as 1. Even with that definition in hand, x=0,y=0 is not a solution because 0 is not a positive integer. If this is homework, the deadline is undoubtedly long past. The only solution I could find was x=2,y=5. I cannot prove this is the only solution.

Neither is a solution. [math]0^0[/math] is undefined, but more importantly, This is your stinking problem, and you have violated your own constraints.

Neither x=0, y=0 nor x=1, y=1 are solutions. The OP asked for positive integral solutions. Define [math]f(x,y) = x^y - (y^x + x + y)[/math]. Then [math]f(1,y) = -2y\ \forall y \ne 0[/math] [math]f(y,y) = -2y \ \forall y \ne 0[/math] [math]f(x,y) < 0 \ \forall (x,y)\in (\mathbb I \times \mathbb I), x > y >= 2[/math] [math]f(y-1,y) > 0 \ \forall y\in \mathbb I, y >= 4[/math] There are a two positive real solutions [math]x[/math] to [math]f(x,y)=0[/math] for each integer [math]y\in(4,5,...)[/math]. (Actually, there are two positive real solutions [math]x[/math] to [math]f(x,y)=0[/math] for each [math]y>3.6475[/math] or so.) There is at least one integral solution (i.e., I found one).

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)?

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.

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

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.

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.

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.

Is what? I see two things wrong with saying "0.999... is equal to one". Nothing wrong with saying 'equals'. The two problems: Use of the passive voice. One should say "0.999... equals one" instead. Not strong enough. Even better stated as [math]0.999... \equiv 1[/math], or in active English, "0.999... equals one, identically". Saying "0.999 is one" still uses the passive voice and uses a word I can't find in my mathematical dictionary (i.e., "is").

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

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.

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.

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. 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). Exactly.

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.

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 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.

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, 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.