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FlowingOoze

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About FlowingOoze

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  1. FlowingOoze

    Brittney Uber Alles?

    Quote:Original post by LessBread Ironically, in the early years of the United States the call was for Americans to create American culture rather than import it from Europe. What's really ironic is that there already were strong cultures in North America, before anyone even though of USA, before anyone came from Europe (since the last ice age anyway). Anyways, all of us posting here share much of the same popular culture, but in the end, we all represent only a small fraction of the people of Earth. Times change. Throughout history people have mostly just observed the change, rather than actively participated in shaping it. In many ways, we're living a unique time in history. It is up to us to decide what the future will bring. If there's one thing I'd like to teach, it's that you need to decide for yourself, what's really important.
  2. FlowingOoze

    Strange alcoholic beverages

    Quote:Original post by krikkit Can you tell me the difference between a Drink and a Beverage? I knew someone would ask that. Left at readers/posters discretion.
  3. FlowingOoze

    Strange alcoholic beverages

    Quote:Original post by Sneftel "Milk Stout". 'nuff said. Milk Stout. I've gotta taste that. That's insane.
  4. FlowingOoze

    Strange alcoholic beverages

    What's the strangest (traditional) alcoholic beverage/substance that you've come across? (Drinks don't count, because it's too easy to come up with a wacky drink. (Unless you've a good excuse.) ) For starters, the Finnish Jaloviina (Ädelbrannvin på Svenka), a mix of cognac/brandy and spirits/vodka, produced since 1932 as an affordable alternative to Cognac, still very popular. It works very well as a substitute for brandy/cognac in a drink and it's not too bad straight either.
  5. FlowingOoze

    What is the value of 'f' on this force system?!?

    Quote:Original post by ury Quote: That is the only internal force that keeps the system rigid. Any other force would cause the bodies to separate or come closer to each other. There are internal forces at hand but they are hardly proportional to F. In order for the body to stay rigid, very stiff forces must work between the particles so that their structure stays the same. I wasn't implying that the internal forces were proportional to F. Just that in my example the internal force must be F/2 in order for the system to be rigid. The internal forces in a rigid body are of course always such that the particles maintain their relative distances. It just happened to be F/2 in this case. Quote: Nonono! We are not talking about a particle at the center of mass. There's no reason why we can't place a particle at the center of the mass. In a rigid body it would necessarily stay at the center of the mass. So figuring out the total external force is in effect the same as finding out the total force affecting the particle at the center of mass. (If the center of mass falls outside the body, we can still place a particle of zero mass at CM.) Quote: Quote: Anyways, the answer indeed is f=F. Here's how (reading from a book on classical mechanics): This proof is identical to the one that I gave in my first post... So please remind me what was the whole point of this discussion? Was there something unclear in my post? I'm not entirely sure what the point is. ;) But your proof wasn't the same as mine. I objected to your use of Newton's second law. "dP/dt = sum[ fi ] = F". You're saying that the sum of forces is equal to the derivative of the linear momentum of the whole system. I didn't think it's appropriate to use that. To be honest, I can't seem to remember what I really though was wrong with your post. I guess I was just tired and confused. [Edited by - FlowingOoze on October 17, 2005 6:02:38 AM]
  6. FlowingOoze

    optimizing for/while loops...?

    Quote:Original post by Rasmadrak I'm wondering how expensive a loop is? If I have, say 20000 objects, that each need to be checked against the others and the array is sorted by some value, would I gain performance by doing something like this. What exactly are you trying to do? If you need to perform some operation on each pair of objects, then you need to do a nested iteration: for(i=0;i<objects;i++) for(j=i+1;j<objects;j++) check(object,object[j]); There's no way (that I know of) to speed that up in any way.
  7. FlowingOoze

    What is the value of 'f' on this force system?!?

    Quote:Original post by ury I know that my answer seems strange. But it's true. You see, the only "real" test must be performed in total vacuum and because your wasn't, the two attempts looked different because of the air friction. I doubt that air friction plays a significant part in the pen experiment, but that's not really relevant. Quote: A fairer test would be this: Drop a pen and simultaneously tap it in two places equally distant from the center of mass. If you do that right, there won't be any angular movement at all. Of course, since there won't be any torque. Quote: Quote: The equal and opposite reactions don't cancel out. What does that mean? It means that the equal and opposite forces affect different particles. The overall sum of the force vectors may be zero, but the forces don't necessarily cancel out, because they're applied to different points of the system. Quote: Quote: Consider a two particle system with a force affecting the right particle along the axis: ---> O<-->O Both masses are equal (m), the overall acceleration is F/(2*m). Acceleration of what? Of the whole system of course and consequently both of the particles. Quote: Quote: The internal forces are equal to F/2. What makes you say that? That is the only internal force that keeps the system rigid. Any other force would cause the bodies to separate or come closer to each other. Quote: Quote: The situation is completely different if the external force is perpendicular to the axis. Why? Because the directions of the force vectors are different. The magnitude of the internal force is likely to be different too. Quote: Quote: The original problem is asking what the total internal force affecting the particle at the center of mass is. You mean external force, right? I think that we were asked to find the external force acting on the center of mass. If you think of the body in the original problem as a mesh of particles, then the force F was not applied to particle at the center of the mass, thus there's no external force being applied to CM. There may be a force that affects the particle at CM, but it's an internal force. Quote: Please read my previous post more carefully. It's very basic rigid body mechanics and I don't see any point arguing about it unless you have a solid and a well defined argument. It no doubt is a simple problem, but saying that f=F just because the vector sum of internal forces affecting different particles is zero is not entirely correct argument. Quote:Original post by Sneftel The key to the answer, and why it's difficult to accept, is that conservation of energy has nothing to do with this problem. ... But that's an incorrect understanding of the nature of force. That's true. But I don't see how that's the key to the answer. Quote:Original post by Sneftel In the case of the "tapping a stick" experiment that was suggested, here's the key: In each case (tapping in the middle, tapping at the end) you are tapping with the same force, but you are doing a different amount of work. Could be. However, isn't work defined as the product of the force and distance? If the force in the tap is the same and moves the point of the stick that it affect the same distance, then isn't the work the same too? Of course it's even more likely that the forces aren't the same at all. Anyways, the answer indeed is f=F. Here's how (reading from a book on classical mechanics): Newton's second law written for the ith particle: Σj Fij + F(e)i = dpi/dt F(e)i is the external force on the ith particle. Fij is the internal force on the ith particle due to jth particle. pi is the linear momentum of the particle. Summing over all particles: d2/dt2Σimiri = Σi F(e)i + Σi,j Fij If the internal forces are subject to Newtons third law (not all forces are), then each pair Fij + Fji is zero and the second term on the right vanishes. R = (Σ miri) / (Σ mi) = (Σ miri) / M R is of course the center of mass. M is total mass. Now, M d2/dt2 R = Σi F(e)i That is, the center of the mass of a system moves as if a force equal to the sum of external forces is acting on it.
  8. FlowingOoze

    Linked Lists, Binary Trees

    Quote:Original post by silverphyre673 Here is a site that has descriptions of STL standard containers and how to use them. That's a pretty good site, but some of the information is outdated. This is a much better reference to STL.
  9. FlowingOoze

    What is the value of 'f' on this force system?!?

    Quote:Original post by ury The answer is f=F. A simple empirical test appears to contradict your answer: Drop a pen and simultaneously tap it with a finger in the middle of the pen. It'll fly away from you at some velocity. Tap it at the end and it'll spin, but won't fly very far.
  10. FlowingOoze

    What is the value of 'f' on this force system?!?

    Quote:Original post by ury Quote:Original post by FlowingOoze Quote:Original post by ury The Newton's second law states: (6) dP/dt = sum[ fi ] = F where: fi the force applied on each particle. Not true. The fis are the forces affecting the particle/body. You've missed rotation and angular momentum entirely. If you want to use Newtons laws of motion and treat bodies as consisting of particles without moments of inertia, you'll have to consider forces between the particles too. In a rigid body, these forces are such that the relative positions of the particles never change. I am afraid that I don't understand entirely what you were trying to say there so I'll reply as best as I can. Since we treat the rigid body as a closed system, the forces acting between the particles (the internal forces) cancel each other out. This is derived from the Newton's third law that states that for every action there's an equal reaction. So if fi = fint,i + fext,i Then because: sum[ fint,i ] = 0, we get: sum[ fi ] = sum[ fint,i ] + sum[ fext,i ] = Fext. The equal and opposite reactions don't cancel out. The overall sum of internal forces in a rigid body is zero, but the internal forces affecting a single particle aren't necessarily zero. Consider a two particle system with a force affecting the right particle along the axis: ---> O<-->O Both masses are equal (m), the overall acceleration is F/(2*m). The internal forces are equal to F/2. Forces affecting the particle on the left are F and the reactive internal force of -F/2, sums up to F/2, causing an acceleration of F/2/m=F/(2*m). The only force affecting the right particle is the equal but opposite internal F/2. The sum of internal forces is F/2-F/2=0. The situation is completely different if the external force is perpendicular to the axis. The original problem is asking what the total internal force affecting the particle at the center of mass is. Quote: I have no idea why I should use moments of inertia. You shouldn't. I was just being clear that we're dealing with imaginary particles that, unlike real particles, don't have a moment of inertia. [Edited by - FlowingOoze on October 16, 2005 5:06:53 PM]
  11. FlowingOoze

    Linked Lists, Binary Trees

    Quote:Original post by Gaj Thanks for all the replies! But, flowingooze, I don't even know where to begin to understand what you wrote. :( What's a .hpp file? It looks like you are using commands I never saw before; where are they from? But that's also a lot less than what the book used as an example, so it looks cool - what is it? :) Since you're still very much a beginner, I won't scare you with too much details. Boost is a collection of helpful libraries for C++. .hpp extension is often used to indicate that the file is a C++ header, as opposed to a C header. The compiler doesn't care what extensions are used, or even if you don't use any at all (as the standard library does). boost::shared_ptr is a smart pointer class that counts references to objects. The boost documentation is a bit technical, but shared_ptr is basically really easy to use and by using it, you don't need to worry about deleting objects so much. Binary trees are usually used as a data structure to make searches quicker. Here's an animated applet. I'm sure you'll find plenty of information on trees using Google. If you wish to use binary search trees in C++, you should probably use std::set as it's (usually) implemented as a balanced binary tree. (See red-black tree)
  12. FlowingOoze

    What is the value of 'f' on this force system?!?

    Quote:Original post by ury The Newton's second law states: (6) dP/dt = sum[ fi ] = F where: fi the force applied on each particle. Not true. The fis are the forces affecting the particle/body. You've missed rotation and angular momentum entirely. If you want to use Newtons laws of motion and treat bodies as consisting of particles without moments of inertia, you'll have to consider forces between the particles too. In a rigid body, these forces are such that the relative positions of the particles never change.
  13. FlowingOoze

    Linked Lists, Binary Trees

    How about an organisation like this: #include <boost/shared_ptr.hpp> struct Army { std::string name; //Name of the Army std::list< boost::shared_ptr<Unit> > units; }; std::list< boost::shared_ptr<Army> > armies; You can use boost::shared_ptr to handle deletion of units and armies automagically and avoid unnecessary copying of Units. Army contains the name of the army to show that you can place extra data in there. Of course you should do some data encapsulating in your game.
  14. FlowingOoze

    newbie question about newton-euler equations

    Quote:Original post by dsecrieru One more thing, though. Is the inertia tensor calculation dependent on the shape of the body only? I was under the impression that it also depends on how the mass is distributed in the object (ie the engine might be heavier, some cargo bays lighter and so on). Are you saying that I can't (or at least is very difficult to) calculate a pretty accurate inertia tensor, knowing the shape of the body and its mass distribution? I've been thinking about this myself lately. Calculating the moment of inertia (ie. inertial tensor) involves integrating over the body in question. It can be very difficult for arbitrary shapes and density distributions. However, Monte Carlo integration or some other similar method can be used. Of course, for games you can probably settle for a combination of simple shapes, (like a car motor could be a box with constant density), so solving the inertial tensor is not that hard (because you can integrate the elements separately). (A bigger problem for me is figuring how I should generate the data.) Incidently, what do you mean by Newton-Euler equations? I've a hard time finding anything about them using Google.
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