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April 13

Submarine depth gauge

In the movie "Das Boot," about a Uboat, or in submarines in general, when the depth gauge shows, say 200 meters, what is the reference level in the sub? The bottom or top of the pressure hull, the bottommost point of the nonpressurized fuel tanks under the pressure hull or any projection from the bottom , the center of the sub, the topmost point on the superstructure (sail, radar mast?), or what? On the one hand, the topmost point would make sense, since you want it to be submerged far enough a passing ship doesn't see it or hit it. On the other hand, if you know the water is shallow, you don't want the bottom to hit a rock. From an engineering standpoint, greater depth matters more to the pressurized central portion than to fuel tanks around it which in some cases are at the surrounding pressure and under no particular increased stress at extreme depth. Neither Submarine nor Submarine depth ratings gives an answer to this. If the gauge is calibrated to show how far the bottommost part is below the surface, then how high above that point was the topmost projection of U-96, the basis of Das Boot, when the periscope was down? Edison (talk) 03:39, 13 April 2014 (UTC)[reply]

The gauge shows depth to keel -- so, when the sub is surfaced, it shows the sub's draft. 24.5.122.13 (talk) 04:22, 13 April 2014 (UTC)[reply]

Yes. Scroll down for picture of a depth gauge labelled DEPTH TO KEEL with zero off the scale. 84.209.89.214 (talk) 11:43, 13 April 2014 (UTC)[reply]

It seems the U-96 was a VIIC and here is its specification manual (secret under 1934 Reich Penal Code § 88). The top of the conning tower was 9397 mm above the keel and the periscopes could be extended to 14663 mm at most above the keel. Thincat (talk) 15:16, 13 April 2014 (UTC)[reply]

Canola consumption

From the article intro: Consumption of the oil is not believed to cause harm in humans. Since we know that oils and fats are typically harmful when consumed in large quantities, what is this supposed to mean? Two sources are presented: the second basically says that it doesn't have anything that's actively toxic, and the first is a 404 error. Neither one says that you can't get harmed by consuming canola oil. Nyttend (talk) 03:40, 13 April 2014 (UTC)[reply]

In large quantities, EVERYTHING is harmful, even sugar -- so what this statement means is that it doesn't contain anything actively toxic. 24.5.122.13 (talk) 04:24, 13 April 2014 (UTC)[reply]

Also, there's acute toxicity, meaning you get sick in short order, versus chronic diseases, which is what too much fat can cause, eventually. StuRat (talk) 06:33, 13 April 2014 (UTC)[reply]

Canola is a specially bred variety of rapeseed to be low in erucic acid which some believe is dangerous to humans. Hence the claim that canola is safe. Rmhermen (talk) 12:49, 13 April 2014 (UTC)[reply]

Calculus in engineering

Is it true that you don't necessarily need to be good at calculus to do engineering at college? I've heard people say it's not as important as it is in a maths degree. — Preceding unsigned comment added by 2.218.65.60 (talk) 10:41, 13 April 2014 (UTC)[reply]

There are many branches of engineering; can you be more specific?--Shantavira|^{feed me} 11:03, 13 April 2014 (UTC)[reply]

Off the cuff, I would say that calculus is probably more important for many kinds of engineers (and indeed most applied science professions) than for pure mathematicians. You probably need less algebra, though. But in general, if you go to a decent university, you will encounter math far beyond everything you imagined in high school for any hard science or technical degree (says the guy who once thought that with sine, cosine, Pythagoras, and integrating single-variable polynomials, all real-world problems could be solved ;-). --Stephan Schulz (talk) 11:47, 13 April 2014 (UTC)[reply]

I am an engineer and I would say that it is essential to have a broad understanding of the principles of calculus, and to be confident in applying those principles in practical situations. If, by writing "be good at calculus", you mean in the top ten percent etc., I would say no. The most important thing is not where the engineer lies in percentiles in the results of an examination; it is how confident the engineer is to apply the principles of calculus without being prompted by a teacher or fellow engineer.

Another consideration is that most subjects that are included in any engineering course will make use of calculus some of the time. If the engineering student isn't confident in application of calculus, he or she will have difficulty assimilating the content of most subjects in the course.

If you don't feel comfortable with calculus, try focusing more of your time and effort on the subject. If you still don't feel comfortable with it, perhaps engineering is not for you. Dolphin (t) 12:09, 13 April 2014 (UTC)[reply]

While at work many engineers will more often use formulas derived with calculus than calculus itself, engineering school is another matter. There they use calculus quite a bit. I'm a computer programmer with an engineering degree, and have needed calculus exactly once in 20 years. Interestingly, it was my inability to find the formula I needed that prompted me to ask my first Q here, and, when I didn't get an answer, derive the answer myself with calculus. The Q was how to find the CG point of a pipe elbow. StuRat (talk) 13:01, 13 April 2014 (UTC)[reply]

Calculus is used heavily in engineering education, both in classes taught by the engineering departments, and in other required classes, especially physics. These classes often cover idealized cases that can be solved in closed form with calculus.

After leaving university and entering the profession, it will probably be found that the problems of practical interest do not lend themselves to closed-form solutions, so they will be solved with numerical methods, including numerical differentiation and integration. These methods may be embedded into computer software tailored to the particular branch of the profession (see "SPICE" for example). It will still be necessary to understand the principles of calculus to deal with situations where the software fails or produces nonsensical results. Jc3s5h (talk) 14:45, 13 April 2014 (UTC)[reply]

Note that calculus is usually not taught in a rigorous way to engineering and physics students. In theoretical physics, you do have to study some math courses for math students. A good example is this textbook. An engineering student may study from this book, but only after he/she has already learned enough from earlier less rigorous math courses. A math student may learn about complex analysis for the first time straight from this book.

A first year engineering may e.g. learn from his book that you can solve differential equations using the following trick. Suppose that we want to find a particular solution of y'' + y' - y = x^2, then we write this as (D^2 + D - 1 ) y = x^2, we then formally write (note that what comes after "formally" is heresy in a rigorous math class) that y = 1/(D^2 + D - 1) x^2, if we then expand in powers of the differential operator, we get y = - (1 + D + 2 D^2 + ...) x^2 = -(4 + 2x + x^2). Count Iblis (talk) 15:25, 13 April 2014 (UTC)[reply]

I nowikied your '' to fix the wiki markup issue. Wnt (talk) 17:46, 13 April 2014 (UTC)[reply]

The original question, with emphasis, asked whether calculus is needed "to do engineering at college." Almost categorically, the answer is yes - to study engineering at most reputable universities that grant engineering degrees, you will have to repeatedly demonstrate your mastery of advanced topics in calculus, both in the formal setting of a mandatory mathematics course that is prerequisite to matriculation into the engineering degree-program; and in the more informal setting as a part of your mathematical toolkit that will be used to complete work for other engineering courses. More bluntly: your engineering department will eventually expel you if you cannot pass your mandatory calculus class(es).

With a bit of cynicism, those of us who practice engineering outside of college might wryly belittle the amount of mathematics that we actually conduct. Truthfully, it is pretty rare for an engineer to write and solve a closed-form equation. Nonetheless, we are performing analysis whenever we engineer a solution to a problem - in the formal mathematical sense of the word, and in the more informal sense of engineering analysis. This process requires an understanding of relationships - the concept that mathematicians might call functional analysis; and it requires understanding of change - the relationship between a system and its sub-components as they change. This is the concept that mathematicians call calculus. And once in a while, we actually do need to write and formally derive or solve a mathematical expression in order to decide how many bricks to use, or which steel alloy provides the required thermal characteristics, or whether the aircraft will wobble safely, or whether the software program will function correctly when ten million users are entered in to the database.

From a psychometrics perspective, individuals who excel at calculus tend to perform better at these other tasks: broad categories like "planning" and "quantitative reasoning" and "spatial awareness" and "risk analysis;" and narrow categories, like statics, circuit analysis, and computer programming - even when the actual work involved in solving such problems uses no formal method of calculus. Some organizations, universities, and job-recruiters will therefore use your measured performance in calculus as a benchmark of your potential skill at other tasks. A lot of people do not like this fact; but it is a scientific conclusion that has been repeatedly demonstrated. For example, here's a full book-length review on the subject: Proceedings of the National STEM Assessment Conference (1996), put forth by the NSF, which demonstrated that student assessment of calculus concepts (assessed by any method) correlated with ability to perform well in Physics, Engineering... even Biology. Some psychologists call this the "g factor," and they find that it is well-correlated to an individual's performance in formal mathematics. The theme of this year's 2014 conference was scaling technical education to the lowest-quartile income families; a fascinating topic in its own right; but 1996 was the most recent year I found a strong theme on the topic of assessment - so there for your consideration is the present state of the art).

Every statistical study like the ones I linked above will footnote its conclusions with numerous caveats. There are many confounding factors; each student is unique, and has unique backgrounds, socioeconomic constraints, and so forth. So, we can obviously concede that some student, somewhere, might be an engineering prodigy yet fail to grasp elementary concepts of calculus. But as a statistical aggregate, it has to be acknowledged that calculus is a requirement for a proper engineering education at the university level. Outside the academic environment, perhaps the mathematics takes a less rigorous form; it may appear in more informal ways; but these mathematical concepts underlie the daily theory and practice of engineering.

The OP's second question is a request for homework advice. I would solve this problem by setting up a constrained linear system, which would not require the use of calculus. But were I to recreationally engage myself in the solution of that problem, I might program a piece of software to solve the general-case using the linear simplex method, which is an application of the calculus of variations to solve numerical optimization problems. In this case, I think the demonstration proves that you do not need calculus to solve many problems in engineering statics; but those who choose to use it anyway tend to found successful companies and then retire to the Swiss Alps to build home-made torsion-balance apparatuses to measure the gravitational constant G_c.

Nimur (talk) 17:50, 13 April 2014 (UTC)[reply]

I think you probably meant something other than the calculus of variations. Personally I think a bit of calculus should be taught at a much earlier age or at least get children to try using finite differences for some simple things like distance and velocity. Dmcq (talk) 19:57, 13 April 2014 (UTC)[reply]

Many methods associated with optimization theory, such as the generalized conjugate gradient method, are numerical implementations of calculus of variations. The purpose is to iteratively compute the gradient of the objective function until you find a location at which the gradient is zero. In this case, the parameters of the function are varied; the gradient is the composite of the partial derivative of the function with respect to every argument, evaluated at the present functional value. Using calculus of variations to analyze the gradient can be faster than direct iterative calculation at all points. This is why conjugate gradient method converges faster than the method of steepest descent; it uses analytical calculus to project the gradient into an orthonormal space, ensuring that each iteration is linearly independent from all previous iterations. To do so, you have to muck with the basis set for the derivative - in other words, to use a functional instead of a function - which is a method of the calculus of variations. Nimur (talk) 21:26, 13 April 2014 (UTC)[reply]

Thanks for that. It's always nice to hear about a connection one hasn't dreamed existed, I must have a look into that. 07:05, 14 April 2014 (UTC)

Even if you don't necessarily need calculus for a discipline, such as perhaps computer engineering, the thinking you obtain when learning it is indispensible. For me, calculus was how I began loving mathematics again after being bored by algebra classes.--Jasper Deng (talk) 21:20, 13 April 2014 (UTC)[reply]

Beam analysis

I have a main beam with a fixed support to the left and a pin at the other end, 5 m long. 3.6m from the left there is a cantilever beam sticking out at a 29 degrees angle. I've used stiffness analysis to determine that the bending moment at the right of the main beam is 123.23kN. The bending moment at the side of the cantilever fixed to the main beam is 160kN. How do I calculate the bending moments for the rest of the structure through equilibrium so that I can draw a bending moment diagram for this structure. — Preceding unsigned comment added by 82.132.232.35 (talk) 15:55, 13 April 2014 (UTC)[reply]

It's been awhile since I've done this, but I'm confused by the information you provided. You say that the bending moment "at the right of the main beam" is 123.23kN: but you said this end is pinned, so the bending moment must be zero. I am also unsure about the cantilever: if the cantilever is loaded, then it will twist the main beam in addition to bending it. You don't say if/how the cantilever is loaded. Are the moments you have calculated "using stiffness analysis" bending moments or axial torque (twist)?--Dreamahighway (talk) 20:54, 15 April 2014 (UTC)[reply]

Okay, after thinking about this some more: Assuming the cantilever is in the same plane as the main beam, there is no torque. The load on the cantilever can be resolved onto a concentrated moment, vertical load, and axial load on the main beam at the 3.6m point. The axial load doesn't matter for the bending moment. Assuming the stiffness analysis that you did shows the bending moment in the beam on each side of the concentrated load -- caveat: I don't remember how that part is done. Then I think that the bending moment diagram along the main beam to the right of the concentrated load has to be triangular from 123.23 to zero. Because the shear diagram for this part is rectangular. I am assuming there are no other loads on the main beam. That means the vertical reaction at the pinned end is 123.23/1.4. That should allow you to determine the remainder of the diagram--yes? — Preceding unsigned comment added by Dreamahighway (talk • contribs) 23:01, 15 April 2014 (UTC)[reply]

protein and boiling

Does boiling soya meal and pulses etc... decrease the protein content ?? — Preceding unsigned comment added by 119.235.54.187 (talk) 18:07, 13 April 2014 (UTC)[reply]

Cooking denatures the protein, meaning it changes their structural bending, and may break (or form) bonds between the constituent amino acids. But without charring this doesn't destroy the amino acids, which are the ultimate building blocks your body wants. μηδείς (talk) 20:19, 13 April 2014 (UTC)[reply]

I believe μηδείς is correct. Note however that although the protein content may not significantly change, if it's intended for the protein to be absorbed (after consumption), cooking may affect this via a variety of means (the denatured proteins may be digested and absorbed better or worse, enzymes or other substances which would inhibit or aide absorption may be damaged etc). See [1] which relating to eggs not soya meal or pulses gives some idea of the complexity. (The rate of absorption will likely also change which could affect other things.) Nil Einne (talk) 20:51, 13 April 2014 (UTC)[reply]

Soybeans are a bit peculiar, because they contain trypsin inhibitors, which interfere with the digestion of protein. However, to the best of my knowledge boiling inactivates the trypsin inhibitors, so it should actually increase the usability of soy protein. Looie496 (talk) 01:17, 14 April 2014 (UTC)[reply]

Even and odd wavefunctions

Hi,

I was recently doing an exercise in Griffith's Introduction to Quantum Mechanics(2nd) ed. that was as follows:

Show that if $V(x)$ is an even function then $\psi (x)$ can always taken to be either even or odd.

(Here $\psi (x)$ is a solution to the time-independent Schrodinger equation ${\hat {H}}\psi =E\psi$ )

The hint said to show that if $\psi (x)$ satisfies the equation above then so too does $\psi (-x)$ . This part was not too tricky. The hint then said to consider the odd an even combinations $\psi (x)\pm \psi (-x)$ . This was where I got stuck.

Firstly for any real map $f(x)$ the combinations $f(x)\pm f(-x)$ will always be odd and even, and as it's absurd to claim that by examining these one can deduce that any real map is odd or even so I assumed the question relied on some defining property of the wavefunction in order for the claim to be true.

When I looked at the solution however, it was approximately as follows:

Letting $\psi _{\pm }(x)$ denote $\psi (x)\pm \psi (-x)$ , we have that $\psi (x)={\frac {1}{2}}(\psi _{+}(x)+\psi _{-}(x))$ so any solution can be expressed as a linear combination of even and odd solutions. QED.

The solution seems to have answered a different question, namely whether every wavefunction that is a solution to the time-independent Schrodinger equation (with the $V(x)$ condition) can be expressed as a linear combination of even and odd solutions, NOT that every wavefunction is purely odd or even. Or am I missing something? Certainly an arbitrary linear combination of odd and even functions is not necessarily odd or even, (take $f=x^{3}$ and $g=1,f+g$ is not odd or even.)

I checked the errata for both the textbook and the solution manual but to my knowledge nothing was corrected.

Neuroxic (talk) 23:32, 13 April 2014 (UTC)[reply]

Can you clarify the wording? You said "

\psi (x)

can always taken to be either even or odd." So that wouldn't seem to mean that

\psi (x)

is necessarily even or odd, merely that you can cover all (basic) solutions using even or odd functions, and linear combinations of these. I don't know the area very well, although I studied a bit of it at home for interest. So can you check what exactly is meant by "can always (be) taken to be"? IBE (talk) 00:45, 14 April 2014 (UTC)[reply]

In particular, if two solutions differ only by their phase, that would mean to me that they are the same solution. But I don't know if that would apply here. IBE (talk) 03:24, 14 April 2014 (UTC)[reply]

One slight thing, the function g=1 is not a linear function. At least by the mathematical definition. g=0 is a linear function. That just caught my eye and I haven't looked at the rest. --DHeyward (talk) 04:24, 14 April 2014 (UTC)[reply]

It sounds like it's just a case of a sloppily-worded problem. You're interpreting the vaguely-worded phrase "ψ can always be taken to be either even or odd" to mean "for any solution ψ, either ψ is even, or ψ is odd". But Griffiths intended the phrase to mean "there always exists a solution ψ which is even, and there always exists a solution ψ' which is odd". Red Act (talk) 07:39, 14 April 2014 (UTC)[reply]

Even or odd, you can only have even and odd if the energy level is degenerate. Count Iblis (talk) 17:58, 14 April 2014 (UTC)[reply]

The point here is that the solution can be even or odd to start with in which case the assertion is valid. Then you look at the case where the wavefunction is't either even or odd. It is in this case that the fact that psi(-x) is also a solution yields something non-trivial, as you find a new linearly independent solution. But then you can then construct two eigenstates, an even one and an odd one. If you were to perturb the Hamiltonian in a way that doesn't break parity symmetry, then that would cause the degenerate energy level to split into two different levels and the two wavefunctions with the definite parities would be the eigenstates to a first (zeroth order) approximation. Count Iblis (talk) 17:58, 14 April 2014 (UTC)[reply]