studiot

The equation says The modulus of the expression = 1. The modulus is a real number, what is the expression for it in terms of x and y? What is the locus of all point in the x - y plane whose modulus is 1? You should be able to plot expressions ax+by = 1 on the xy plane; What does that plot look like? Let the complex quotient in the expression be equal to [math]\alpha x + \beta iy[/math] Then reduce the quotient to this form and equate coefficients; Do you know how to reduce a complex quotient to a single complex number (ie bring the complex part from the bottom to the top) ?

So the purpose of the collection was to look at zet's post83 'experiment'. It is suprising what can be deduced from a simple static situation in the bent tube. Zet is right to try to simplify, but it takes experience to know what can be left out and what must be included and the idea is to examine this as we go along. I have drawn a square tube of side 'a', since flat surfaces are easy to work with and section area calcuations are simple. The initial setup is shown in Fig1 with elevation (z) reckoned upwards from the base of the horizontal part of the tube. The tube is shown as constant cross section,a x a where a is small compared to the height of the fluid level in the vertical column. This allows a simple representation of the fluid pressure in the lower horizontal part of the tube as the pressure at z=0. Talking of pressure it is worth noting what is meant by 'fluid pressure' and I have inset Fig1a to show this. Consider a small cube within the fluid, shown solid. This cube has a face ABCD on which the cube exerts a fluid pressure PR as shown. The fluid in the adjacent cube, shown dashed, must push back with an equal and opposite pressure PL, or the cube would move. This must, of course be true of the other five faces of the cube. If the face of the cube is against a container, not more fluid, then the container, must supply this pressure instead. These can also be cast in terms of Newton's 3rd law, using force instead of pressure. I will return to these comments later. Now that we have our tube set up, but capped off at section 6 so the fluid can't flow out. To start with let us examine the flow without the airfoil and introduce it later at section 5. Our control volume is from section 1 to section 2, as shown in the green box in Fig2. Since the column is vertical these sections are horizontal. Section 1 is the top free surface of the fluid. The pressure here is atmospheric. In the inset in Fig2a I have returned to my earlier point about the pressure on the walls of the container. The forces to the left (FL) must be equal and opposite to the forces to the right (FR) or again the whole container will move. Again since the fluid is pressing on the container, the container pushes back equally. This may be stating the obvious, but it will become important, along with momentum later. This now leads to Fig3 and a Bernoulli analysis at sections 1 and 2. This is where working in terms of head comes into its own. There are three terms in the simple Bernoulli equation I called equation 1 in post146. At section1 the elevation head is simply z1 Since the velocity is zero the velocity head is zero And the pressure is the ambient (atmospheric) pressure at the free surface. At section 2 The velocity head is still zero The elevation head is now zero, since z2=0 and the pressure head has therefore increased. I have shown this in the equation although I have used vertical section 4 to save a drawing (Remember I said keep 'a' small so we can consider the pressure the same over the whole of vertical sections) This enables us to calculate the pressure along the whole of the horizontal section of fluid. If we now suddenly remove the end cap at section 6 so the fluid flows out, as shown in Fig4. Fig 4 is also about the law of conservation of mass and continuity. Since our fluid is incompressible the same quantity of fluid that flows aout through section 6 must also flow through each other section as shown in red. But since all the sections have the same cross section this means that the velocity has constant magnitude throughout the fluid. If we change thearea at section 5 by introducing the airfoil we must change the velocity to compensate. Unless we know the volumetric flow rate we cannot calculate this changed velocity at section 5 to put into Bernoulli. Fig 5 returns to the static situation before endcap removal for a bit more analysis. Note clearly that the velocity here is zero throughout the fluid. Therefore the momentum is zero throughout the fluid. This comment is important to the airfloil lift, which see the same mechanism about to be discussed, in relation to momentum. I showed in inset Fig 2a that the pressures and forces on the walls of the container balance, and that the wall supply a reaction to the fluid. If we suddenly remove part of those walls in the form of the cap at section 6 we must also remove that reaction. We replace that reaction with the ambient (atmospheric) pressure Pa This explains why the fluid starts to move. The fluid is still at pressure head of (z1 + Pa/rhog) whilst this is opposed by Pa/rhog alone. So the driving head is the difference, z1, as shown inf Fig 6, which allows us to calculate the starting exit velocity of the fluid. Before the cap is removed there is zero momentum. After removal the fluid possesses momentum to the right. So there must be some equal and opposite momentum to the left, because momentum is a conserved quantity. The rate of change of momentum constitutes a force and this force acts on vertical section 3 of the container, pushing it back of the left. This is, of course, how a rocket works. The changes that occur as the level falls require calculus to address and I will leave to a possible later post, along with the application of the Work equation, I have called equation 3, in post 146 to estimating the head loss due to the airfoil. Hopefully this is enough to convince you that because of the interaction with the container walls the analysis of contained fluids is different from that of a semi infinite fluid like an atmosphere. However it is always encouraging when Newton's laws, conservation laws and any specialist laws applicable to the subject in hand, all offer the same answer and fit together nicely.

One of the properties of infinity is adding any real number to it still makes infinity, or if you like subtracting any real number still makes infinity. Neither of these are properties of (9recurring).

Of course not. Is infinity a constant? What does adding 100 to infinity make? is (9recurring ) + 100 = (9recurring) ? Infinity is not a real number or an integer, 9 recurring is and obey all the rules of integers and ral numbers. There are number systems which include an infinity, but not the real numbers or the integers.

I asked my [insert] favourite politician [/insert] this one, The reply was Rename them Walter currents.

I think I originally misunderstood you. Why will taking the log(77.960603) and subtracting log(37.960603) and then taking the antilog not give your answer?

Have you ever tried telling the corporate IT environment that you want permission to install a particular program, not on their list?

What is 20 the log of? and what is 3.873 the log of? And what is the log of A added to the log of B the log of?

Here is the result of (left) clicking once near the top of the "big empty box". In this case it opens with the font size selector selected, but it could be any of the random options available.

Thank you, Larry, for picking that up. Unfortunately I can't now correct post142. The energy equation should, of course contain velocity squared terms. [math]\frac{{V_1^2}}{2} + {P_1}{v_1} + g{z_1} + {U_1} + Q - W = \frac{{V_2^2}}{2} + {P_2}{v_2} + g{z_2} + {U_2}[/math]...........3 I started by using all capitals for quantities, then realised I had two Vs, one for velocity and one for volume and things wnet awry when I sorted that. The work equation does not apply to distributed mass such as fluids, it applies to identifable bodies such as airfoils. Edit I see the error carried right through so the proper Bernoulli equations are without losses/inputs [math]\frac{{V_1^2}}{{2g}} + \frac{{{P_1}}}{{\rho g}} + {z_1} = \frac{{V_2^2}}{{2g}} + \frac{{{P_2}}}{{\rho g}} + {z_2}[/math].............1 and with losses/inputs [math]\frac{{V_1^2}}{{2g}} + \frac{{{P_1}}}{{\rho g}} + {z_1} \pm h = \frac{{V_2^2}}{{2g}} + \frac{{{P_2}}}{{\rho g}} + {z_2}[/math]...............2 I have also now numbered the equations for future reference.

One of the most annoying traits of the input editor is that , depending upon the position of my mouse, the editor also starts off whatever icon my mouse pointer is headed, be it underline, strikeout, lists, links overlays Wouldn't it be just fantastic if it was fixed so this did not happen.

Of course depending upon the nature of your subject, here is the classic mathematical study poster http://prefrontal.org/files/posters/Bennett-Salmon-2009.pdf

Larry, I didn't notice the density appearing in my equation, so how could it affect it? Please bear with me I haven't labelled my variables etc, as I hope that all will become clear in the next post(s). But your post has encouraged me that the interest is still there to complete this discussion.

I did say I would post more detail. This will take several posts and I have been trying to get some stuff together for this. This post will be an introductory post to collect together some preliminaries, for ease of referral. First dimensional analysis is important in Physics, but particularly important in fluids. Some useful dimensions are Now it can clearly be seen that neither pressure nor velocity is energy. So in order to write an equation we need to bring all our terms to a common denominator, preferably a simple one. Engineers have traditionally used 'head' as this common denominator, which has the dimensions of length. Energy is just too complicated. Here are some quantities that have the dimensions of length or head. [math]{\rm{Head}} = \frac{{{\rm{Energy}}}}{{{\rm{Force}}}}{\rm{ = }}\frac{{{\rm{Energy}}}}{{{\rm{Weight}}}}{\rm{ = }}\frac{{{\rm{Pressure x Volume}}}}{{{\rm{Force}}}} = \frac{{{{\left( {{\rm{Velocity}}} \right)}^{\rm{2}}}}}{{{\rm{Acceleration}}}}{\rm{ = }}\frac{{{\rm{Pressure x Volume}}}}{{{\rm{Mass x Acceleration}}}}{\rm{ = }}\left\{ {\frac{{{\rm{Pressure}}}}{{{\rm{Acceleration}}}}{\rm{x}}\frac{{\rm{1}}}{{{\rm{Density}}}}} \right\} = {\rm{Length}}[/math] Now we need some equations of motion to employ these quantities. The simplest form of Bernoulli's equation (in terms of head) as applied to a pipe or duct as in post 83 is [math]\frac{{V_1^1}}{{2g}} + \frac{{{P_1}}}{{\rho g}} + {z_1} = \frac{{V_2^1}}{{2g}} + \frac{{{P_2}}}{{\rho g}} + {z_2} = H = {\rm{A constant}}[/math] Note that all these terms have the dimensions of length. This equation is applied to a pipe running full, it does not apply to a pipe only partly full. Further the fluid must act in an incompressible way. However it can be used (with modifications) for a pipe with or without friction, The fluid can have viscoscity or be inviscid And we can introduce machines such as Zet's airfoil in the stream as some sections of the pipe. Each of these adds a term to the equation, increasing or reducing the head at any point. So friction and fluid driven machinery (eg airfoils) introduce a negative head, whilst pumps introduce a positive head, section 1 (subscripts) is taken before the extra and section 2 after it. [math]\frac{{V_1^1}}{{2g}} + \frac{{{P_1}}}{{\rho g}} + {z_1} \pm h = \frac{{V_2^1}}{{2g}} + \frac{{{P_2}}}{{\rho g}} + {z_2}[/math] Where +h is the work added by a pump or -h taken out by a turbine. Compressible fluids, principly gases, have several more terms and are best dealt with by what is known as the enrgy equation. [math]\frac{{V_1^1}}{2} + {P_1}{v_1} + g{z_1} + {U_1} + Q - W = \frac{{V_2^1}}{2} + {P_2}{v_2} + g{z_2} + {U_2}[/math] Finally we will need an estimate of the work done by the airfoil machine against gravity. The work equation for any object lefted in a gravitational field is [math]W = \left( {\frac{1}{2}mV_2^2 - \frac{1}{2}mV_1^2} \right) + \left( {mg{z_2} - mg{z_1}} \right)[/math] We can convert this to a head loss by dividing W by the airfoil weight, which will produce some interesting results when applied to the post83 setup.

Thank you again, imatfaal, but I'm not sure of your point(s) here. I am just testing a few things out using MathType, which I seem to have working for me for once. Eventually I want use these items in a particular thread.

Thank's imatfaal, that's now created a bookmark trail for some useful other threads. Others may also find these useful. However I didn't see anything about the subject of this thread, posting a table. test [math]{\rm{Head}} = \frac{{{\rm{Energy}}}}{{{\rm{Force}}}}{\rm{ = }}\frac{{{\rm{Energy}}}}{{{\rm{Weight}}}}{\rm{ = }}\frac{{{\rm{Pressure x Volume}}}}{{{\rm{Force}}}} = \frac{{{{\left( {{\rm{Velocity}}} \right)}^{\rm{2}}}}}{{{\rm{Acceleration}}}}{\rm{ = }}\frac{{{\rm{PressurexVolume}}}}{{{\rm{MassxAcceleration}}}}{\rm{ = }}\left\{ {\frac{{{\rm{Pressure}}}}{{{\rm{acceleration}}}}{\rm{x}}\frac{{\rm{1}}}{{{\rm{density}}}}} \right\} = {\rm{Length}}[/math]

Maths and English education is allegedly plummeting in the US and the UK Does Donald Clark's wry comment have any bearing? http://donaldclarkplanb.blogspot.co.uk/2015/02/the-problem-with-maths-is-english-20.html

As a matter of interest, conventional geologists are not insensitive to other mechanical ideas. Indeed these days they like to try them out practically in models. Here are a couple of extracts from Twiss and Moores. The first shows mountain building as a result of tension. The second shows mountain building as a result of upwelling. I think the experimenter's ingenuity is to be commended. While you are a way the final chapters of Twiss and Moores Structural Geology and Ramberg's Gravity, deformation and the earth's crust would be worth a punt.

I can think of another tech forum that has lost many long standing members because it wasted a lot of money trading 'up' and moving from Vbulletin to XenForo. [math]\begin{array}{*{20}{c}}{{\bf{Acceleration}}} \hfill & {{\bf{L}}{{\bf{T}}^{{\bf{ - 2}}}}} \hfill \\{{\bf{Force}}} \hfill & {{\bf{ML}}{{\bf{T}}^{{\bf{ - 2}}}}} \hfill \\ {{\bf{Energy }}} \hfill & {{\bf{M}}{{\bf{L}}^{\bf{2}}}{{\bf{T}}^{{\bf{ - 2}}}}} \hfill \\ {{\bf{Pressure}}} \hfill & {{\bf{ M}}{{\bf{L}}^{{\bf{ - 1}}}}{{\bf{T}}^{{\bf{ - 2}}}}} \hfill \\{{\bf{Velocity}}} \hfill & {{\bf{L}}{{\bf{T}}^{{\bf{ - 1}}}}} \hfill \\ {{{\left( {{\bf{Velocity}}} \right)}^{\bf{2}}}} \hfill & {{{\bf{L}}^{\bf{2}}}{{\bf{T}}^{{\bf{ - 2}}}}} \hfill \\ {{\bf{Head}}} \hfill & {\bf{L}} \hfill \\\end{array}{\rm{ }}[/math] Yippee, I'm getting there. Now all I need is a way to see the code in a few weeks time, ie the next time I want to do a table. Edit I just discovered that if you click on the above table the image overlay opens and includes the code. Someone really ought to write a help section for the forum, explaining the functions, 'cause some are really cool.

b has the units of volume so the formula per mole is [math]P = \frac{{RT}}{{\left( {V - b} \right)}}[/math] It is a cut down Van der Waal's equation.

Thank you Sensei for providing that background +1 It is no reflection on you personally, but a sad statement of 'progress' that I would expect an 8 year old with a paper and pencil to be able to draw a better table than that. I certainly could at that age. In Word or Frontpage I can make a simple table that where I can make the rows and columns of sensible size for the job, with or without separating lines, and then type the data into the spaces. I particularly wanted to use this site's excellent superscript facility to display a table of quantities and their dimensions for a dimensional analysis in another thread so I might try your suggestions. That was one drawback to endy's link (and also Word/Frontpage) - the table generator doe not allow this.

Not exactly my point. 2 millionyears v 200 million years is a big difference. If you look closely at my four maps you can see that two hotspots, roughly the presentday Azores and Tristan-Walvis hotspots , (Nos 1 and 42 here) http://en.wikipedia.org/wiki/File:Hotspots.jpg Doming up under the landmass of the 200MY BP starting the North Atlantic and South Atlantic oceans respectively. At this time there was no oceanic basin , ridge or floor. The spreading of the ocean floor, the building of the ridge and the eventual joining of the two oceans can be followed over the first three maps for 150 MY. Note that the northen end to this ridge between Greenland and Canada becomes a side spur and the ridge branches and extends further north in th last (50MY) map with the upwelling of the modern Icelandic hotspot. You have not fully appreciated the implications of the regular increase in age of the ocean floor rocks from as distance increases both east and west of the ridgeline. If, as you contend, the ridge is the result of pre-existing being thrust up under pressure from both sides (you cannot have compression from one side only) then, by definition of pre-existing, the ridge would be older rock. However some of the ridge rock is so young as to be dated 0 Y. If, you prefer to contend that compression led to the original upwellings then that upwelling cannot have been the present day ridge material because even the space it now occupies did not exist 200 MY BP, let alone the material to occupy it. Let me take this opportunity to wish you wwell in your break and to thank you for focusing attention on an interesting subject. I do not endorse this site, but have you come across the continually expanding theory? (Not expansion/contraction cycles like yours) http://www.expansiontectonics.com/page3.html Edit today was pancake day (Shrove Tuesday). I don't know if you have ever cooked pancakes, but you can see the doming produced by hot upwelling beautifully illustrated in the pancake pan as the hot air bubbles up from underneath.

Greenwood & Earnshaw refer to "poorly formed graphite" for the structure of coke. Wells gives more detail It should be noted that there are two full forms of the graphitic structure, with slightly different interplanar separations. So the structure would appear to consist of crystallites, being small pieces of containing the regular hexagonal graphitic planes, but not fully aligned and linked in the third dimension; the crystallites being randomly oriented with respect to each other, like the grains in a metal.

Good morning, Billy and welcome to ScienceForums. So just how long is a piece of string? The card you link to looks good and solid in itself, but needs additional circuitry to make it do anything. I particularly like the high current connection provision. But no one can really make any more detailed comments without some idea of your application and source of power.