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The general approach to such mechanical problems is to set up a set of coupled differential equations, which together describe the motion of each weight (and therefore - of the knot). Many packages for numerical solution of differential equations exist.
However, this isn't as simple a problem as it appears at first. If you take a simple case where exactly one weight is heavier than the others, the knot will start sliding towards the heavy weight. This means that the knot is no longer centred in the circle, and the strings are chords, not radii. This means that the forces acting on the edge of the table (caused by the strings) now have transverse components, as well.
At some point, the transverse force for each string will be enough to overcome the static friction between the edge of the table and the string. At this point, the weight will start sliding around the edge of the table (toward the heavy weight?). This will in turn change the dynamics of the problem even more.
This sliding effect may happen even with equal weights, if the initial position of the knot is sufficiently off-centre.
The analytical solution of this problem would probably be worth a nice paper in a peer-reviewed Physics journal.
If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.
--Winston Churchill
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Modeling wise I would expect that understanding the problem would be easier starting with a simpler model.
Start with a square table.
There is no friction on the table or edges.
There are 3 weights: A, B, C
A and B each weigh half as much as C.
A and B are hanging off the left side of the table. There is a 30 degree angle between the ropes and the knot. C is hanging off the right side.
The ropes are very long and have no weight.
Now double the weight of C.
My expectation would be that the knot would move to the right and the angle would decrease.
Fixing the ropes through rings at the table edge would make the force calculations easier (x and y) because the ropes should move (due to force in one direction.) Then one would need to apply calculus to solve once the rings are moved.
Then add three weights on one side and repeat.
I would expect that one would then 'round' the square table to progressively arrive at the final solution.
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Not quite correct.
If the knot is at a constant position, and a string makes a non-perpendicular angle with the edge of the table, we have:
A transverse force (vector pointing along the edge of the table) will exist at the point where the string goes over the edge. If there is no friction, nothing will prevent the string from sliding until it makes a perpendicular angle with the table.
Note that as it slides, the string picks up velocity in the transverse direction. When it reaches the perpendicular position, the direction of the transverse force will reverse, causing the string's velocity to decrease. You end up with an odd type of pendulum.
If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.
--Winston Churchill
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Interesting thought.
However what is the impact on the rope when conserving momentum given that it has no weight.
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The weight of the rope has little bearing on the result. If you take an ideal pendulum and replace the (weightless) string with a rod, the rod has a moment of inertia which affects the pendulum's movement (and period), but does not change the fundamental form of the movement.
For an example, compare a grandfather clock to an ideal pendulum.
If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.
--Winston Churchill
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How can a string that has no mass have a moment of inertia?
Keep in mind that I am not referring to one with a low mass, but rather one that has none.
The force vector in the Y direction (top/bottom) on the string is reduced as the angle decreases but the string has velocity. (Although perhaps that is wrong since how can the force act on it with no mass.)
Daniel Pfeffer wrote: For an example, compare a grandfather clock to an ideal pendulum.
An ideal pendulum has a rod/string which is is controlled by the fixed dynamics of the weight and the connector. It is the mass moving which moves the rod. The rod doesn't move by itself.
If one removes the weight the pendulum no longer works because the physics require gravity on the mass. No mass, no conservation.
Daniel Pfeffer wrote: but does not change the fundamental form of the movement.
Realistically I suspect that if the rope has no mass then the angle cannot change because the force vector in the Y direction would have nothing to act upon.
If the string does have mass then the pendulum dynamics come into play.
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jschell wrote: How can a string that has no mass have a moment of inertia?
The (massless) string may not have inertia, but it does indicate the direction of the force acting between the knot and the weight. For a real-world string (possessing mass), I maintain that when the angle between the string and the edge of the table is sufficiently non-perpendicular, it will slip. The only thing preventing it from slipping is the friction between it and the edge of the table. For an ideal string (possessing no mass), there can be no friction, so as soon as the angle between the string and the edge of the table changes, the position of the string (and the weight attached to it) will change.
jschell wrote: An ideal pendulum has a rod/string which is is controlled by the fixed dynamics of the weight and the connector. It is the mass moving which moves the rod. The rod doesn't move by itself.
If one removes the weight the pendulum no longer works because the physics require gravity on the mass. No mass, no conservation.
Not quite correct. If the rod is rigid and has mass, you could still construct a pendulum with it. The problem with the string is that the string is non-rigid.
For a non-rigid (but fixed-length) pendulum string, each infinitesimal element dl of the string at distance l from the fulcrum would attempt to move at its natural frequency, sqrt(g/l)/2pi. This would cause the string to change its shape from a line to some more complex, time-dependent curve (the exact details are too hairy for me to describe off-hand). The movement is no longer periodic, which you interpret as the pendulum "not working".
If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.
--Winston Churchill
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Daniel Pfeffer wrote: For a real-world string (possessing mass), I maintain that when the angle between the string and the edge of the table is sufficiently non-perpendicular, it will slip.
Of course.
Daniel Pfeffer wrote: If the rod is rigid and has mass, you could still construct a pendulum with it.
However that is not what I said nor what was started with. If the rod has a mass then the mass is added to the system, but the rod itself does not define the system because it is consider to be significantly less than the system especially in a "ideal pendulum" thus what I was referring to with "No mass".
Daniel Pfeffer wrote: For a non-rigid (but fixed-length) pendulum string,... which you interpret as the pendulum "not working"
Nothing to do with what I said. You will need to take your own meaning from "ideal pendulum" but I meant that it has no mass. Once you remove the weight in that system the string no longer functions in any way.
In the table problem I am rather certain, now, that even with no friction on the table, if the string had no mass then the angle would never change because there is nothing for the Y force to act upon. With a very small mass and no table friction the string would act as a pendulum but would not conserve the swing because the knot would be moving. That would reduce the angle (conceptual moment to moment) and thus reduce the force vector in the Y direction over time.
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Would this not be just a problem of solving for equilibrium among the various line tensions? Resolve the tensions into X-Y components, sum them , then find that value of X and Y that renders the sum as zero. Interesting problem, Bill. I've always enjoyed such challenges, but I'm many years out of practice.
Will Rogers never met me.
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As I have mentioned above, in the general case this is a dynamic problem, which may have no equilibrium position (i.e. the knot eventually slides off the table).
If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.
--Winston Churchill
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Conjecture: if the knot moves at all, it has to fall off eventually.
I can't really formally prove it. But if the knot slides somewhere, the weights will redistribute such that the weights in the direction the knot moved into can be closer together (they will attempt to all move to the place such that their length of string on the table is minimized, that will be the same location for all weights now because the knot isn't in the middle any more). So if there was an imbalance to begin with, it gets worse. So whatever weird things happen with weights swinging on their strings like pendula and weird outward forces from being swung around the table, there is no equilibrium to reach.
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Intuitively, I agree with you, but for different reasons.
The weights will move so that the total energy of the system is minimized.
At any given point in time, the force on the knot is the vector sum of the tensions in all of the strings. If this vector sum is non-zero, the knot will accelerate. Once it starts accelerating in a given direction, I can see no way that a returning force would develop, so the knot would eventually slide off the table.
If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.
--Winston Churchill
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Thank you all for a most interesting, and evocative, discussion !
My current thinking is that what I really want to model is some kind of parallel to a Bezier curve, but with #n control points, instead of two.
An analogy might be consider, for a 2 control-point Bezier curve, given a movement in 2-d space of either control-point: what is the x,y displacement of the center of the total length of the curve ?
«In art as in science there is no delight without the detail ... Let me repeat that unless these are thoroughly understood and remembered, all “general ideas” (so easily acquired, so profitably resold) must necessarily remain but worn passports allowing their bearers short cuts from one area of ignorance to another.» Vladimir Nabokov, commentary on translation of “Eugene Onegin.”
modified 2-Feb-16 8:21am.
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Not sure if it is the right place to ask this question. Is there anyone know where I can download the lecture videos for MIT subject "Distributed Algorithms" (course number : 6.852J / 18.437J) and "Advanced Algorithms" (course number 6.854J / 18.415J). I could not find the download link in its official website. Thank you.
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May be you should attend to the course ?
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
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MIT Open Courseware has these subjects, but without the Video Lectures.
Advanced Algorithms: Here[^] and here[^].
Distributed Algorithms: Here[^].
Till they release the video lectures over there, it's a good idea to go over the material available here.
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http://cgp.wikidot.com/circle-to-circle-collision-detection
radius of both circle cannot longer than the distance apart
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When posting links to external sites, please make them clickable.
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Circles are trivial. The interesting part is ellipses. I've found this in Graphics Gems V.
Ellipses can be represented by a characteristic 3x3 matrix S, such that the points on the ellipse satisfy x̄Sx̄T = 0, x is a 3-vector representing the coordinate of the point with a homogeneous coordinate for translation. This is probably not too surprising yet, it's reminiscent of a circle with radius zero (lol, but the point is the form) x*x+y*y=0 => x̄x̄T=0 (I'll omit bars on vectors from now on, you can figure it out from context) but with a matrix in between to potentially do some scaling/rotating/translation, and an ellipse is clearly just a transformed circle.
The point(s) of intersection satisfy the equations of both ellipses. So xS1xT = 0 and xS2xT, those points must therefore satisfy axS1xT + bxS2xT = 0. Because linear algebra is, well, linear, you can write that as x(S1 + qS2)xT = 0 where q = b/a. This test is not as strict, it will allow some spurious points, but they can simply be tested against the original equations.
H = S1 + qS2 will again be the characteristic matrix of some conic, not necessarily an ellipse, indeed it could be a bunch of lines. That would be useful, so solve for q (recall that the points we want will work for any q, so we can pick it) such that det(H) = 0. Since H is a 3x3 matrix, this is just a cubic equation which can be solved directly (though slightly annoyingly). The three roots defined three H's.
For each H found that way (most H's will work, but some solutions may not have a useful form), determine which type it is (because of round-off, do the usual floating point "is nearly zero" thing instead of the exact math thing), say the matrix has the form
A B D
B C E
D E F (it is, because a linear combination of two symmetric matrices is also symmetric, this just names the elements)
- A=B=C=0, now vHvT=0 is just Dx+Ey+F=0, the familiar line equation. Because we will be applying linear transformations to the line(s), represent it as two points. Just pick any two points on the line and remember them.
- B2-AC=0, some number of parallel lines. Make a rotation matrix for the angle -0.5 atan(2B / (A-C)) and apply it to H. The quadrant doesn't matter so atan2 isn't necessary. The lines will be either horizontal or vertical, if C=0 then they're vertical with the equation Ax²+2Dx+F=0 (beware typos in the book, if you're reading it alongside this), just solve that the regular way giving you up to 2 vertical lines, again represent them by pairs of points. Since we had rotated, apply the reverse rotation to those points. If C!=0, then they were horizontal lines, with the equation Cy²+2Ey+F=0, do the same thing with that.
- B²-AC > 0, crossing lines, it's a bit tricky to actually find them. Rotate the same way as above, then translate with (using the rotated H) xoffset=-(CD-BE)/(B²-AC), yoffset=-(AE-BD)/(B²-AC), finally in the rotated and translated H, 0,0 is a trivial point on both lines (their intersection is now in the origin) and we also have 1/sqrt(|A|),1/sqrt(|B|) and 1/sqrt(|A|),-1/sqrt(|B|). Translate them back, then rotate them back.
- Some other form, pick a different H.
Now you have a bunch of lines. Transform them and the first ellipse such that the ellipse becomes a circle, test all the transformed lines against the circle, then inverse-transform the intersections you found. Testing points against the second ellipse is easy, but keep in mind you will have plenty of rounding error by now.
There's C code implementing this here: http://www.realtimerendering.com/resources/GraphicsGems/gemsv/ch2-6/conmat.c[^] and you should probably read the relevant chapter from the book as well.
modified 6-Jan-16 6:12am.
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Thanks for reply.
I am studying your comments and links.
It will help me definitely.
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Hi, I am learning probability and need a good working algorithm for selecting between two options successfully. Any help will be appreciated. Please post the algorithm in equation form.
Thanks.
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- Remove coin from pocket
- Flip coin
- Examine coin. If it fell "heads" up, choose option 1. If it fell "tails" up, choose option 2.
- Repeat as necessary.
The case of the coin falling on the side is left to the student
If you have an important point to make, don't try to be subtle or clever. Use a pile driver. Hit the point once. Then come back and hit it again. Then hit it a third time - a tremendous whack.
--Winston Churchill
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(Apart from question being unclear) Probability is a "chance", it does not guarantee 100% "successful" answer. So even if you could get an algorithm it won't guarantee to be correct.
The accuracy of answer is 100%, only when there is one answer. In two options, there is a probability of 50/50 (50%) and so on.
answer := 1/n
Where n is the number of answers; options. The example of "coin flip" is a good way to demonstrate this. You flip a coin and your friend selects one. That is the probability. It can be correct or it can be false.
The sh*t I complain about
It's like there ain't a cloud in the sky and it's raining out - Eminem
~! Firewall !~
modified 5-Jan-16 6:47am.
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