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baxtrom

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About baxtrom

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    Meson

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  • College Major/Degree
    Dr.Tech, Lic.Tech, M.Sc
  • Favorite Area of Science
    Solid mechanics, strength of materials, structural dynamics
  • Occupation
    Structural analysis specialist, vehicle engineering industry
  1. From the department for cumbersome ways of estimating pi [math]\pi = \Gamma^2(1/2)[/math] ..where Gamma is the Gamma function, [math]\Gamma(z) = \int_0^\infty e^{-t} t^{z-1} \mathrm{d}t[/math] I was just going to suggest to use a Stirling approximation* to estimate the gamma function instead of computing the integral, but of course all those approximations contain pi.. *) Yes, these approximations are accurate only for large arguments, but Gamma(0.5) can be rewritten 2 Gamma(1.5) et c
  2. True, but then it would require some additional explanation so I skipped it. Same goes with the random coordinates, "2*rand-1" could be replaced with just "rand" since the ratio of areas of a quarter unit circle to a quarter square is also pi/4. Oh, by the way: it's important that the random numbers are uniformly distributed. Typically the rand functions are UD but some generators produce normally distributed random numbers which would screw up the algorithm and produce pi_approx > pi.
  3. Did anyone mention trial and error? There's a fancy name for this type of analysis: Monte carlo algorithms! Imagine you are throwing a dart onto a 2x2 square (area = 4). You are good enough to always hit the square, but them darts land all over the place. Now, a ratio of those darts will fall inside a unit circle concentric with the square (area = pi). That ratio is the same as the ratio of areas: pi / 4. I assume you have a random number generator in your c64. Then you can write a program which does something like this: randomize n = 0 N = big number! for i = 1 to N (creates
  4. The quarter circle arc length method, although being intuitive, is problematic since any numerical integration algorithm will converge slowly due to nasty integrand when approaching singularity at x = 1. However, one way of getting around this is to compute the arc length of a 1/8 unit circle, which should be pi/4. Hence, replacing the upper integration limit 1 -> sqrt(0.5) and doubling the constant in front of the integral, the identity becomes [math]\pi = 4 \int_{0}^{\sqrt{1/2}} \sqrt { 1 + \frac{x^2}{1-x^2} } \mathrm{d}x [/math] Using Simpson's rule with 16 subintervals gives pi
  5. Perhaps an indicator on the hinge, being pushed by the pendulum to its extreme position but not pulled back when the pendulum returns? The friction needed to keep the indicator needle in place would theoretically influence the test, but could probably be neglected in the real world..
  6. I guess a problem with that formula is that it requires implicit knowledge of pi beforehand. This is more clear if you rewrite the limit in terms of radians instead. Your formula would then become [math]\pi = \lim_{n \to \infty} \frac{n \sin(\frac{2 \pi}{n})}{2} = \text{(sine expansion for small arguments)} = \lim_{n \to \infty} \frac{\frac{2 n \pi}{n}}{2} \ldots[/math]
  7. It's an algorithm for approximating pi using numerical integration (midpoint rule) to obtain the arc length of a quarter unit circle (which equals pi/2). For a given function the arc length can be computed using integral calculus. In this case, however, the integrand is not very well behaved close to x = 1 so accuracy may suffer. For a more practical method using integral calculus I would suggest instead computing the area of a quarter unit circle = pi/4.
  8. Also Richardson extrapolation is worth mentioning here. If [math]I_n[/math] is the result using the trapezoidal method for n steps, then [math]I_{2n} + \frac{I_{2n}-I_n}{3}[/math] ..will give super-duper accuracy compared to [math]I_{2n}[/math] alone. Good way to improve results without too much added coding.
  9. Fatigue testing typically involves applying a known stress loading until the specimen breaks after N cycles. Counting the cycles would'nt present much of a problem, however knowing exactly the stress level could be. Presumably, the stress level in a test involving for example rotation bending is easier to keep track of than the membrane stress in a pure axial test. Rotation bending tests are typically performed applying a known static load to the end of a rotating shaft. Knowing the RPM of the shaft gives you the cycles. An axial fatigue test requires a more complicated setup I guess
  10. Hi, E-N curves refer to strain life (from strain = epsilon = "e"), and is mostly used for low-cycle fatigue, i.e. for high loads that lead to more rapid failure. S-N curves are applicable for high cycle fatigue, when nr of cycles exceeds 100,000 or so. Check out this website.
  11. You could check out the gradient descent method, which is perhaps the simplest minimization algorithm. Basically it means taking downhill steps in the direction of "greatest slope" until the bottom (miminum) is reached. So, you need the gradient of the cost function, i.e. the derivatives wrt a, b and c. Those you could approximate using for example central difference approximations. The best way to learn about minimization is probably to pick up an introductory textbook on numerical analysis. There you will find examples on how to implement this and other more advanced (and more effective
  12. Without the c constant in your equation that log operation would have been slightly more justified since it would turn your nonlinear regression problem into a linear one in variables log x and log y. However, it's possible you are stuck with having to do the full nonlinear thing. With scilab or matlab that shouldn't be so hard if you are familiar with numerical methods. Define a cost function, for example [math]cost(a, b, c) = \Sigma_i (y_i - a x_i^b - c)^2[/math], where [math](x_i, y_i)[/math] are your measured data points and minimize that wrt a, b and c using for example fminsearch
  13. Hi there, perhaps Gershgorin circle theorem could help you, especially if the diagonal entries of the matrix are dominant. I don't know if you are familiar with Gershgorin's theorem but it states that the eigenvalues of a matrix will be located within special circles in the complex plane. The location of the circles will be given by the diagonal elements, and the radius by the off-diagonal elements in the same row (sum of absolute values). In your case, if the off-diagonal entries are small then the eigenvalues should be close to the diagonal entries. That could provide you with an initial g
  14. My numbering above. 1) Please consult an introductory textbook on physics, page 1. Energy is a scalar quantity and is obtained by means of the dot product if working with force and distance vectors. I'm sure you are aware of the properties of the dot product, otherwise, consult an introductory textbook on linear algebra, page 1. 2) We are of course referring to the energy dissipated by predominantly plastic work on the car during impact, which btw is assumed not to be made of your "indeal" rigid bricks. 3) You are yourself quite an example. Really, you are so full of yourself yo
  15. Thank you for adding some sense to the topic. I got some support from John too, appreciate that, and some critical but civil remarks from swansont, respect that. Had I known however that I would arouse the wrath of Dr Evil and his Bride I wouldnt have replied in the first case, however, alas, I naively tried to help with some humble ideas. Do I understand that the problem is not fully defined in the original post? Yes, I'm not an idiot in spite of mad drooling ramblings by Dr Evil above. The "car" could be a radio controlled toy or a 3 ton pickup, and the impact could involve another vehicle,
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