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PDF Editor FAQ

What if I throw a cube of pure uranium in the Mariana Trench, will the fission reaction start under that high pressure?

The pressure experienced at the bottom of the Mariana Trench while great1,086 bars (15,750 psi) or 1071.72 atmospheres [standard] would probably not be enough to cause a cube of pure U235 or pure U233 to become supercritical.Your intuition regarding high pressures reducing the critical mass of pure fissile like U235 is however a good one.Powerful laser or heavy ion beams can produce extremely high pressures on small samples. Large lasers have been predicted to compress a small fissionable sphere which is on the order of a mm to the point it could become a highly supercritical assembly. Pressures up to 10^12 atm, comparable to the pressure in the center of the sun, are achievable using advanced giant lasers like NIF which can expose a small pellet simultaneously from all sides with intense laser light.(Picture of a RAMP pressure target used at NIF to produce momentary extreme pressure on small samples).If you are using pure fissile materials like U233, U235, or Pu239 and you expose them to ultra-high pressure, you can greatly reduce the mass of fissile required to form a critical mass. Irradiating the small pellet simultaneously from all sides using a intense fast pulse laser like NIF can cause the fissile material to be compressed to about 250 times normal density.The critical mass of an unreflected sphere will vary inversely as the square of the densityMc ~ sigma^-2whereMc = critical masssigma = density of fissileIf for example two systems, differing only in density, were critical (keff = unity) but the density of the first was 100 times that of the second, its critical mass would be only 1/10,000 that of the second.Further details can be found in the following unclassified documentPNNL-19176, Anomalies of Nuclear Criticality - by E. D. Clayton

What are the differences between average RMS and peak voltages?

In electronics we have to measure waveforms. The most common ones are DC and sine waves, that most people with any knowledge of electronics know.With a constant voltage or DC, the peak is the DC value and the RMS is also the DC value.With AC sine waves, the peak value is the highest point in time of the sine and the peak to peak is twice that being that the signal is symmetric around zero. The RMS represents the heating value that an equivalent DC would create when either is driving a pure resistor. In this case of a sine the RMS value is 0.707 (1/square root of 2) times the peak value.One way of thinking of this is if you flip the negative half up (rectifying it) since positive and negative values heat resistors the same, and then bulldoze the peaks down into the valleys created... then the RMS will be the resulting level height created.For a square wave the peak is the height of the square wave and the RMS is the height as well.For a triangle wave the Peak is the tip of the height of the triangle and the RMS is 0.577 of the peak. That’s 1/ sqrt(3).More complex waveforms can be determined by the integration operation but most computers calculate it from a large number of samples of a cycle of the waveform. Each sample is squared, all the squares are added together and divided by the number of samples, and then take the square root of the sum of the squares/N. This is called the Root Mean Square or RMS for short.The beauty of RMS voltage is that is relates all the different waveforms together, The RMS value of a waveform is the "heating" value, what an equivalent amount of DC voltage into a resistor would heat that resistor.Thus a 100V DC and 100V RMS applied to a resistor would use the same power and generate the same heat.Peak is useful because it describes the biggest momentary voltage. This is needed for amplifer compliance and electrical breakdown calculations.

Can anyone briefly explain the differences between the mealy and moore state machines..?

The most general model of a sequential circuit has inputs, outputs, and internal states. It is customary to distinguish between two models of sequential circuits: the Mealy model and the Moore model. They differ only in the way the output is generated. In the Mealy model, the output is a function of both the present state and the input. In the Moore model, the output is a function of only the present state.A circuit may have both types of outputs. The two models of a sequential circuit are commonly referred to as a finite state machine, abbreviated FSM. The Mealy model of a sequential circuit is referred to as a Mealy FSM or Mealy machine. The Moore model is referred to as a Moore FSM or Moore machine.The above circuit is an example of mealy machine. Output y is a function of both input x and the present state of A and B . The corresponding state diagram given below shows both the input and output values, separated by a slash along the directed lines between the states.An example of a Moore model is given below.Here, the output is a function of the present state only. The corresponding state diagram isIt has only inputs marked along the directed lines. The outputs are the flip-flop states marked inside the circles.In a Moore model, the outputs of the sequential circuit are synchronized with the clock, because they depend only on flip-flop outputs that are synchronized with the clock. In a Mealy model, the outputs may change if the inputs change during the clock cycle. Moreover, the outputs may have momentary false values because of the delay encountered from the time that the inputs change and the time that the flip-flop outputs change. In order to synchronize a Mealy-type circuit, the inputs of the sequential circuit must be synchronized with the clock and the outputs must be sampled immediately before the clock edge. The inputs are changed at the inactive edge of the clock to ensure that the inputs to the flip-flops stabilize before the active edge of the clock occurs. Thus, the output of the Mealy machine is the value that is present immediately before the active edge of the clock.Thanks for A2A!

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