studiot

My description was rather short. You can find all the gory detail of the method in chapter IV (miscellaneous theorems and methods) of Applied Differential Equations, by Relton Formally the substitution of the first derivative (Let d&/dt = p) and working in terms of the substituted variable, p has many applications. The usual reason for using this method is when one or both of the variables are missing from the equation in direct form. The method reduces the order of the differential equation by one and in your case allows a solution by separation of variables. Your equation is discussed on page 109.

Wouldn't it be nice if every molecule and atom had the same mass and weight? But they don't, any more the every person has the same size of show or hat. The equation of the chemical reaction tells you how many molecules of one substance react with how many molecules of another to produce how many molecules of product. So your equation says (get used to saying it to yourself) One molecule of aluminium hydroxide reacts with three molecules of hydrogen chloride to produce one molecule of aluminium chloride and three molecules of water. Now all you need are the molecular weights. You get these by adding the atomic weights of the consitiuent atoms. You get the atomic weights by looking them up in a table, so:- Edit as fuzzwood pointed out I looked in the wrong table (atomic numbers) A serious mistake I have now corrected. Hydrogen is 1 Aluminium is 13 27 Chlorine is is 17 35.5 Oxygen is 16 Now you are motoring. The molecular weights are Aluminium hydroxide is one aluminium plus 3 oxygens plus 3 hydrogens = 13 + (3x16) + (3x1) = 13 + 48 + 3 = 64 27 + (3x16) + (3x1) = 27 + 48 + 3 = 78 Hydrogen chloride is 1 + 17 = 18 1 + 35.5 = 36.5 Aluminium chloride is 13 + (3x 17) = 64 27 + (3x35.5) = 133.5 Water is 2 + 16 = 18 Now I have not said what the units of molecular weights are. That is because we can work in any weight units we like so long as we use the same ones throughout. Common metric units of weight are micrograms, milligrams, grams, kilograms, tonnes Any any will do - different folks use different ones so doctors use micro and milligrams, physicists use grams, bakers use kilograms and cement manufacturers use tonnes. Chemists like to work in grams and even coined the term gram-molecule whichis just the molecular weight in grams. This was shortened to mole. But kilogram-molecules or microgram-molecules could also be used just as well. So back to your stochiometric equation, and using grams this says one gram-molecule of aluminium hydroxide reacts with three gram-molecules of hydrogen chloride to produce one gram-molecule of aluminium chloride and three gram-molecules of water. or as pavel had it One mole of aluminium hydroxide reacts with three moles of hydrogen chloride to produce one mole of aluminium chloride and three moles of water. In numbers (1x78) grams of aluminium hydroxide reacts with (3x36.5)grams of hydrogen chloride to produce (1x133.5) grams moles of aluminium chloride and (3x18)grams of water. 78grams of aluminium hydroxide reacts with 109.5grams of hydrogen chloride to produce 133.5grams of aluminium chloride and 54grams of water. Notice as a check that the total weight on one side of the equation = the total weight on the other side 78 + 109.5 = 187.5 = 133.5 + 54. This is always true of weights, but the number of molecules may be different. So to complete your question (78/78)grams of aluminium hydroxide reacts with (109.5/78)grams of hydrogen chloride to produce (133.5/78)grams of aluminium chloride and (54/78)grams of water. 1gram of aluminium hydroxide reacts with (109.5/78)grams of hydrogen chloride to produce (133.5/78)grams of aluminium chloride and (54/78)grams of water. 10grams of aluminium hydroxide reacts with (10x48/64)grams of hydrogen chloride to produce (10x133.5/78)grams of aluminium chloride and (10x54/78)grams of water. Finally we usually talk about molecular masses, rather than molecular weights these days. They are the same thing to all intents and purposes. Does this help?

Janus, doesn't all this assume a homogenous density function? As we know the actual density/depth function is concave down starting from the surface. It rises quite sharply just down from the surface, with the rate of increase steadily decreasing with depth, finally becoming asymptotic to some max value near the middle. Have you any working to show the effect of this?

A quart of milk? Don't you buy flagons? Interestingly I grew up in a dual world UK education system. Metric and Imperial. When I first went to work at the end of the sixties, civil engineering had converted to nearly fully metric. Cement came in hundredweight bags or metric tonnes. Concrete was just being converted from cubic yard mixers to cubic metre mixers. So concrete was ordered in multiples of 4.8 cubic metres. So imagine my suprise when I went to work for the americans in the late seventies and found myself working in Imperial, for the first time. Bigger suprise, the company was working in grads not degrees.

In the 1950s and 1960s (valve) car radios were operated by vibrator based power supplies that ran continuously at 100 to 200 Hz. You may like to watch these as they contain much pertinent wisdom, most particularly about the switching voltage spikes. Watch this one first from a service engineer https://www.youtube.com/watch?v=Fp6PkRTmb8U Then this one from a modern experimenter. https://www.youtube.com/watch?v=nHq6vArLVc8 There is considerably more circuit theory you need to know to use these things (electromechanical devices) successfully in continuous mode, good luck in your search since you don't want help here.

Strange offered another gem of perceptual insight. That is if it was ever rotating in the first place. I offered a condition whereby this might happen before. It will only start rotating if there is a net influx of mass and momentum from a particular direction. If, on the other hand, it is hit many times at random from all directions then the net turning effect will be zero.

Yes indeed. There is a big difference between a constrained (by the bucket) low viscoscity fluid with next to ne shear stress and an unconstraned (except by surface tension) ultra high viscoscity fluo-solid.

The material is likely borne in suspension in the water. Since it is black after dry heat it could be carbon from particles of poorly deburred plastic pipes and fittings. (you said they were relatively new) Manganese (as robittybob noted but I would hope it isn't) would indeed leave a black residue, as would copper oxide. If the area is fed from old iron pipes then iron oxide could be transported into your new pipes. Can you not complain to the water quality department of your water supplier? Finally do you have any tanks in the system where sediment could collect/have collected?

Is it my aftershave or what? I offer you a formula containing all the variables inherent in determining cycle time The response is total silence. I tell you that cycle can be asymmetric and that the current you can draw depends partly on one of the variables in my first equation. Again total silence. I find you a fast switching relay (the fastest type available) And you are rude about it. I can't see why you persist with your post#6 relay, it is clearly not suitable for your needs.

Good points, strange. Particularly the last one. +1

gpe = mass x g x separation distance. Mass is constant, although g is smaller distance is larger. My objection to MigL's sentence is the same as someone who earlier said "cart before the horse" An object whether solid or liquid or some combination tends to a spherical ball in the absence of other disturbing forces to minimise its free surface surface energy (surface tension). I think you will find that there are other configurations that minimise gravity by balancing opposing pulls on opposite sides of the COG. Anyway having taken up the minimum free surface configuration, or even before, some disturbing mass impacts the ball and adds its mass and momentum to the pot. This in general will add moment of momentum to the ball and if enough such impacts happen, and also if the impacts arrive from a preferential direction and so are not cancelled out by randomness, will set the ball rotating. If, once the ball is rotating, parts of it are capable of movement relative to other parts, yes a belly will develop. The equatorial material that is lifted to form the belly will actually gain rotational energy compared to its slower polar counterpart. Alternatively the polar material will loose gpe by subsidence (flattening). Most likely some combination of both will occur. So horses and carts, which is it?

I didn't realise that was your original contention. That bit is most definitely wrong.

If you don't want to engage in discussion, why ask the question? I linked to a large matrix switching device, but Farnell sell spare parts, which you require. You didn't respond to a polite offer to discuss a subject in which you are clearly non expert and I thought were seeking help. Sorry I wasted my time and effort.

It depends upon the current drawn. Frequency is not an appropriate parameter to be using Did you not manage to follow my post #3 ? What needs explaining further? http://www.farnell.com/datasheets/1884830.pdf

Reed relays get you down to 0.5ms switching time which make kHz frequencies quite feasible. CPC offer reeds with amps capability.

Doesn't this give the general solution? Boundary conditions are applied to eliminate the arbitrary constant for an ODE, or arbitrary function for PDE.

Relays are usually specified by switching time. Cheap relays offer something like 5 - 15 millisecond times. You have the following sum Switch on time + on time + switch off time + off time = time for one cycle = period. With 5 milliseconds each you have Period = 5+5+5+5 = 20 milliseconds. This translates to a frequency of 50 cycles/second. As regards current, this only flows during the on time so in the example the duty cycle is 5/20 = 0.25. The average current = current during one time x duty cycle. The on current is determined by the resistance of the external components. So you are looking for a 24volt relay with better than 5ms switching time. CPC offer some cheap ones at 8ms But they have several hundred in their catalog so try their technical helpline.

Shape of Earth, nonspherical or what? Well I suppose all this ballyhoo really depends on what you take the shape of. This the prime decision to make before debating the shape. The second one is what do you mean by centre? Geologically speaking the most abundant rock at the surface is water, mostly liquid. The surface of this rock is decidedly non spherical. So perhaps you mean the geoid? The average density of the non hydrous rock at the surface is just under 3 times that of water. But we can deduce from astronomic studies that the average density of the equivalent globe is around 5.5 times the density of water. So we can say for certain that the variation of gravity with radius deviates significanly from the inverse square law since the density is not constant.

Is this a typo for 9.8 metres per second2 ?

Here is a good pdf on the 3-moment method for continuous beams like yours. http://people.duke.edu/~hpgavin/cee201/three-moment.pdf and here is a simpler worked hand calculated example. http://www.roymech.co.uk/Useful_Tables/Beams/Continuous_Beams.html

You have already described the correct interpretation in the first part. Think about what happens before you start and during the 'crank up'. Which way do you direct the bob when you start it spinning and which way do you jiggle it to keep it spinning? That is the key and equivalent to Mig and his toy car.

Why no smiley at the end? Gooseberry wine Mmmmm. Reminds me of the nicest drink I have ever tasted.

What do you mean by 'continuous beam theory' ? I don't see any sign of the three moment equation and the resulting simultaneous equations in your working pdf. It would also be nice to see the reactions identified and labelled.

Or the string is pulling Bob inwards. Much more satisfactory.

Nah, it's the flattening effect of the vodka, stops things and people being upright.