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Everything posted by baxtrom

  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 random coordinate point in x = -1..1, y = -1..1) x = 2*rand-1 y = 2*rand-1 R = sqrt(x^2 + y^2) (checks if [x,y] is within unit circle, if so -> increase n) if R <= 1 then n = n + 1 end end (n/N is approximately pi/4) pi_approx = 4*n/N; Disclaimer, code may be wrong, didn't test it. Even if it works, it will be a seriously slow method for getting correct decimals of pi
  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 ~ 3.14160046787255 (correct digits in bold) with 512 subintervals and Richardson extrapolation: pi ~ 3.14159265358979 In comparison, using the area method (computing the area of a unit circle), i.e. [math]\pi = 4 \int_{0}^{1} \sqrt { 1 + x^2 } \mathrm{d}x [/math] ..one arrives at pi ~ 3.14159246129562 after 16384 subintervals + Richardson extrapolation. It looks like the 1/8 arc length method is converging faster than the area method.
  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, allthough I don't have experience from fatigue testing, just from fatigue calculation.. Good luck with your report!
  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) methods.
  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 in Matlab (or manually using some appropriate iterative algorithm). You could also choose a different cost function instead of the least squares approach, for example minimizing the sum of the absolute values of the error. There are advantages (and disadvantages) with such an approach, google robust curve fitting. Good luck.
  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 guess to use with some smart numeric method, i.e. perhaps some form of power method or similar. Good luck! PS. I hope my reply above does not infuriate anyone. Recent experience indicates it is easy to step on academic toes here on scienceforums. We'll see what happens. DS.
  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 you need to get your hoop stress checked. Yeah, you saved "milliions of dollars, and likley a life or two", that doesn't sound like a load of manure at all. Myself I saved thousands of orphans and got Nelson Mandela out of prison. Top that if you can. Funny this, how my two line reply on friday could start this chain reaction of indignation. It's like that small pertubation of a thread seemingly in equilibrium caused DrRocket's ego to suddenly reach critical mass. Wonder what the man is like IRL.
  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, a stone wall, an asteroid, a midget or anything in between. Right. So, those of you who are still steaming from indignation could use the "ignored users list" feature, it works beautifully, I tried. Ciao
  16. Since everbody's getting very excited about this, I'm going to try to explain why I posted that by now infamous 3rd reply to the topic. Yes, it was based on assumption, and I did believe it to be obvious but perhaps I should have spelled it out. The idea was, assuming the distance traveled by the COG of the car during the negative acceleration imposed by the contact condition could be approximated, a time average of the force acting on the car could be approximated (yes, that's that word again). I understand this might be a very upsetting thought and if so, please take a moment to catch your breath before you hit the "fast reply" button. Btw, I found this on a website hosted by Georgia State University's Department of Physics, with pictures and all. But hey, what were they thinking not doing a full dynamic elasto-plastic analysis using an explicit FEA solver to obtain the time history of all 12 stress and strain components in every inch of the car? Really, Georgia State..
  17. I don't know which "energy transfer" you mean but the energy absorbed during impact is known according to the TS. The distance traveled by the car during impact is not stated in the original post but I believe I was quite clear in pointing out that the time average of the force could be approximated from a reasonable guess of this parameter. Yes, that is an assumption, and I thought it was evident from my post. Stay away from civilization, that's my best tip honeybunny!
  18. Oh, my. Scienceforum's own Professor Snape is at it again. Not that I didn't see it coming, just fun to watch the fireworks. No, the approach is not nonsense and deep inside you know it, but you have to defend your position and try to be the big bad bully of the forum. Really, I don't mind, it's sort of entertaining. A secret tip: don't fly, or even go by car or train, because design methods and codes are based on empirical observations and engineering judgement. Better stay in your cave scribling those integral equations on the walls.
  19. Which can be assumed by means of engineering judgement. For example, a typical car is 4-5 m long. The COG is going to be close to the middle of the car, i.e. perhaps at 2 m from the front. It is therefore a reasonable assumption that the distance traveled by the COG will not be greater than 2 m even for high energy crashes. Sure, you could do it the hard way. Explicit FEA solvers like LS-DYNA are available for complex impact analysis. Nonlinear material models and contact conditions put into a million-or-so DOF model would do the trick. I do not believe however that is the answer the TS is looking for.
  20. ..or you put on your engineering goggles, approximate the distance traveled by car center of mass during impact and then compute the average force during the impact event as [math]F = \frac{\textrm{kinetic energy}}{\textrm{distance}}[/math].
  21. Like in Matlab, that command will create a vector t with values from zero to 5 with stepsize 0.1, i.e. t = [0, 0.1, 0.2, 0.3, .... 4.8, 4.9, 5.0].
  22. Good point. Didn't think about flow induced flutter. Perhaps it doesn't matter if the actuator is located in front of or behind the hinge. However, damping out flutter using added mass or dissipative dampers seems intuitively simpler than controlling an aircraft with a control surface jammed hard into an extreme position. As far as I understand, the crew on the Alaska flight tried to compensate for the jammed trim panel by pulling hard on the control column, i.e. using the elevators to pull the nose up. When the jackscrew finally disintegrated the trim panel went all the way due to the airflow pressure and the pilots lost all control of the aircraft. Edit: what I find hard to understand is that the pilots did not immediately declare emergency and land the aircraft after the first signs of problems with one of the control surfaces. Instead they tried to fix it while airborne. Then again, it's easy to be a keyboard pilot..
  23. Thanks for the link. A tragic example of Murphy's law. The Alaska 262 flight is, on the other hand, mostly the result of criminal negligence regarding the maintenance of the aircraft. Apparantly the company stretched the maintenance interval to save money. Still the design with the actuator in front of the hinge puzzles me. I believe the design is similar on for example the Boeing 737 and perhaps on most commercial jet aircraft. Check the photo below - if the actuator breaks the control surface is likely to snap into one of the extremes, and the plane would go either full nose down, or up.
  24. Yes, it's me with an airplane disaster-related question again.. Still recovering from shivering induced by watching a scary documentary about the Alaska airlines flight 261 crash, I ask myself why the trim tabs (is that correct terminology?) are designed in the way they are (at least on the MD-80). As far as I understand, control of the aircraft was irrevocably lost when a jammed jackscrew finally disintegrated and the control surface pitched violently up around its hinge due to the airflow. To me, it seems intuitively safer to reverse the design and have the jackscrew located behind the hinge. In case of total jackscrew failure, which I believe is less unlikely than total hinge failure, I imagine the airflow would cause the control surface to flatten out, i.e. become a passive follower instead of being pushed into an extreme position. What is the reason for this design? Practical considerations? (Would have posted this in airliners.net if I wasn't too cheap to pay $25 for membership..)
  25. Resistance to bending greatly depends on the cross-sectional geometry of the pizza. Compare bending of a plastic ruler - if oriented in its weak direction, it is very flexible. On the other hand it is very stiff if rotated 90 degrees. This is because bending is produced by a moment carried as tension in the upper part of the pizza cross-section and compression in the lower part. If the cross-section is made bigger in the vertical direction, the distance between these opposing forces becomes larger (forces get better leverage) and thus the pizza can carry a greater bending moment.
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