I have been working on this all day but I haven't figured it out yet. So I thought I may as well ask on here and see if someone can help.
The problem is as follow:
----------
F(input)(t) --> | | --> F(output)(t)
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Given a sample with a known length, density, and spring constant (or young's modulus), find the 'output' force against time when a known variable force is applied at the 'input'.
My current solution can already discretise the sample into finite elements, however I am struggling to figure out how the force should transmit given that the change in transmission speed in the material changes itself with respect to the force (using equation c = sqrt(force*area/density)).
If someone could point me to a solution or any other helpful resources, it would be highly appreciated.
A method for applying damping to the system would also be helpful but I should be able to figure out that part myself. (losses to the environment via sound or internal heating)
I will remodel the porbem in the following way:
___ ___
F_input(t) --> |___|--/\/\/\/\/\/\/\/\--|___|
At time t=0 the system is in equilibrium, the distance between the two objects is L, the mass of the left one (object 1) is m1 and the mass of the right one (object 2) is m2.
___ ___
F_input(t) --> |<-x->|___|--/\/\/\/\/\/-|<-y->|___|
During the application of the force F_input(t), at time t > 0, denote by x the oriented distance of the position of object 1 from its original position at time t=0. Similarly, at time t > 0, denote by y the oriented distance of the position of object 2 from its original position at time t=0 (see the diagram above). Then the system is subject to the following system of ordinary differential equations:
x'' = -(k/m1) * x + (k/m2) * y + F_input(t)/m2
y'' = (k/m2) * x - (k/m2) * y
When you solve it, you get the change of x and y with time, i.e. you get two functions x = x(t), y = y(t). Then, the output force is
F_output(t) = m2 * y''(t)
The problem isn't well defined at all. For starters for F_out to exist, there must be some constraint it must obey. Otherwise, the system will have more unknowns than equations.
The discretization will lead you to a system like
M*xpp = -K*x + F
with m=ρ*A*Δx and k=E*A/Δx
But to solve this system with n equations, you either need to know F_in and F_out, or prescribe the motion of one of the nodes, like x_n = 0, which would lead to xpp_n = 0
As far as damping, usually, you employ proportional damping, with a damping matrix proportional to the stiffness matrix D = α*K multiplied by the vector of speeds.
Related
I'm really new to matlab and I need some help, I know java and python. Here's a code I want to understand
x(1) = 0
y(1) = 0
i = 1
x(i+1)=x(i)+vx*t+.5*a*t^2;
i=i+1;
end
I want to know what is happening here, "vx", "t" and "a" are variables tho
x(1) = 0 and y(1) = 0 is very similar to initializing a list / array in Python where x and y are the list variables and the first position is 1. MATLAB starts indexing at 1, and not 0 like in Java and Python. The similar syntax in Java or Python is: x[0] = 0; y[0] = 0. MATLAB uses round braces to index into an array / vector / list.
i = 1
x(i+1)=x(i)+vx*t+.5*a*t^2;
i=i+1;
This is pretty simple. i is some sort of loop variable or index variable... which you failed to show us that this code is probably part of a loop. The code just sets the next value of x or the second element in the array or list to x(i) + vx*t + 0.5*a*t^2. The * operator is multiplication and the ^ is the exponentiation operator. In Python, this is equivalent to saying x[i] + vx*t + 0.5*a*(t**2). Now the origin of this equation actually comes from calculating the displacement of a body mass using Newtonian physics - the Kinematic Equations actually. As such vx is the velocity of the body mass and a is the acceleration. t would be the time point you are considering. Specifically, the displacement can be calculated as follows:
Source: The Physics Classroom - Kinematic Equations
Look at the top left equation as that is what the statement is doing in the code. This only calculates the displacement at one point in time. Therefore, what you are doing is adding the displacement that this body mass encounters at some fixed point t a certain amount of times. x captures the total displacement overall from the beginning to end. The next statement after is just incrementing the loop counter. The thing about MATLAB is that you can dynamically extend the length of a list / array / vector whereas Python or Java will give you an out-of-bounds error. However, it's recommended you should pre-allocate the right amount of memory to use before using it for efficiency. See this informative post for more details: Efficient Array Preallocation in MATLAB.
I am new user of netlogo. I have a system of reactions (converted to Ordinary Differential Equations), which can be solved using Matlab. I want to develop the same model in netlogo (for comparison with matlab results). I have the confusion regarding time/tick because netlogo uses "ticks" for increment in time, whereas Matlab uses time in seconds. How to convert my matlab sec to number of ticks? Can anyone help me in writing the code. The model is :
A + B ---> C (with rate constant k1 = 1e-6)
2A+ C ---> D (with rate constant k2 = 3e-7)
A + E ---> F (with rate constant k3 = 2e-5)
Initial values are A = B = C = 500, D = E = F = 10
Initial time t=0 sec and final time t=6 sec
I have a general comment first, NetLogo is intended for agent-based modelling. ABM has multiple entities with different characteristics interacting in some way. ABM is not really an appropriate methodology for solving ODEs. If your goal is to simply build your model in something other than Matlab for comparison rather than specifically requiring NetLogo, I can recommend Vensim as more appropriate. Having said that, you can build the model you want in NetLogo, it is just very awkward.
NetLogo handles time discretely rather than continuously. You can have any number of ticks per second (I would suggest 10 and then final time is 60 ticks). You will need to convert your equations into a discrete form, so your rates would be something like k1-discrete = k1 / 10. You may have precision problems with very small numbers.
I am attempting to create a model whereby there is a line - represented as a 1D matrix populated with 1's - and points on the line are generated at random. Every time a point is chosen (A), it creates a 'zone of exclusion' (based on an exponential function) such that choosing another point nearby has a much lower probability of occurring.
Two main questions:
(1) What is the best way to generate an exponential such that I can multiply the numbers surrounding the chosen point to create the zone of exclusion? I know of exppdf however i'm not sure if this allows me to create an exponential which terminates at 1, as I need the zone of exclusion to end and the probability to return to 1 eventually.
(2) How can I modify matrix values plus/minus a specific index (including that index)? I got as far as:
x(1:100) = 1; % Creates a 1D-matrix populated with 1's
p = randi([1 100],1,1);
x(p) =
But am not sure how to go about using the randomly generated number to alter values in the matrix.
Any help would be much appreciated,
Anna
Don't worry about exppdf, pick the width you want (how far away from the selected point does the probability return to 1?) and define some simple function that makes a small vector with zero in the middle and 1 at the edges. So here I'm just modifying a section of length 11 centred on p and doing nothing to the rest of x:
x(1:100)=1;
p = randi([1 100],1,1);
% following just scaled
somedist = (abs(-5:5).^2)/25;
% note - this will fail if p is at edges of data, but see below
x(p-5:p+5)=x(p-5:p+5).*somedist;
Then, instead of using randi to pick points you can use datasample which allows for giving weights. In this case your "data" is just the numbers 1:100. However, to make edges easier I'd suggest initialising with a "weight" vector which has zero padding - these sections of x will not be sampled from but stop you from having to make edge checks.
x = zeros([1 110]);
x(6:105)=1;
somedist = (abs(-5:5).^2)/25;
nsamples = 10;
for n = 1:nsamples
p = datasample(1:110,1,'Weights',x);
% if required store chosen p somewhere
x(p-5:p+5)=x(p-5:p+5).*somedist;
end
For an exponential exclusion zone you could do something like:
somedist = exp(abs(-5:5))/exp(5)-exp(0)/exp(5);
It doesn't quite return to 1 but fairly close. Here's the central region of x (ignoring the padding) after two separate runs:
I'm having trouble creating a random vector V in Matlab subject to the following set of constraints: (given parameters N,D, L, and theta)
The vector V must be N units long
The elements must have an average of theta
No 2 successive elements may differ by more than +/-10
D == sum(L*cosd(V-theta))
I'm having the most problems with the last one. Any ideas?
Edit
Solutions in other languages or equation form are equally acceptable. Matlab is just a convenient prototyping tool for me, but the final algorithm will be in java.
Edit
From the comments and initial answers I want to add some clarifications and initial thoughts.
I am not seeking a 'truly random' solution from any standard distribution. I want a pseudo randomly generated sequence of values that satisfy the constraints given a parameter set.
The system I'm trying to approximate is a chain of N links of link length L where the end of the chain is D away from the other end in the direction of theta.
My initial insight here is that theta can be removed from consideration until the end, since (2) in essence adds theta to every element of a 0 mean vector V (shifting the mean to theta) and (4) simply removes that mean again. So, if you can find a solution for theta=0, the problem is solved for all theta.
As requested, here is a reasonable range of parameters (not hard constraints, but typical values):
5<N<200
3<D<150
L==1
0 < theta < 360
I would start by creating a "valid" vector. That should be possible - say calculate it for every entry to have the same value.
Once you got that vector I would apply some transformations to "shuffle" it. "Rejection sampling" is the keyword - if the shuffle would violate one of your rules you just don't do it.
As transformations I come up with:
switch two entries
modify the value of one entry and modify a second one to keep the 4th condition (Theoretically you could just shuffle two till the condition is fulfilled - but the chance that happens is quite low)
But maybe you can find some more.
Do this reasonable often and you get a "valid" random vector. Theoretically you should be able to get all valid vectors - practically you could try to construct several "start" vectors so it won't take that long.
Here's a way of doing it. It is clear that not all combinations of theta, N, L and D are valid. It is also clear that you're trying to simulate random objects that are quite complex. You will probably have a hard time showing anything useful with respect to these vectors.
The series you're trying to simulate seems similar to the Wiener process. So I started with that, you can start with anything that is random yet reasonable. I then use that as a starting point for an optimization that tries to satisfy 2,3 and 4. The closer your initial value to a valid vector (satisfying all your conditions) the better the convergence.
function series = generate_series(D, L, N,theta)
s(1) = theta;
for i=2:N,
s(i) = s(i-1) + randn(1,1);
end
f = #(x)objective(x,D,L,N,theta)
q = optimset('Display','iter','TolFun',1e-10,'MaxFunEvals',Inf,'MaxIter',Inf)
[sf,val] = fminunc(f,s,q);
val
series = sf;
function value= objective(s,D,L,N,theta)
a = abs(mean(s)-theta);
b = abs(D-sum(L*cos(s-theta)));
c = 0;
for i=2:N,
u =abs(s(i)-s(i-1)) ;
if u>10,
c = c + u;
end
end
value = a^2 + b^2+ c^2;
It seems like you're trying to simulate something very complex/strange (a path of a given curvature?), see questions by other commenters. Still you will have to use your domain knowledge to connect D and L with a reasonable mu and sigma for the Wiener to act as initialization.
So based on your new requirements, it seems like what you're actually looking for is an ordered list of random angles, with a maximum change in angle of 10 degrees (which I first convert to radians), such that the distance and direction from start to end and link length and number of links are specified?
Simulate an initial guess. It will not hold with the D and theta constraints (i.e. specified D and specified theta)
angles = zeros(N, 1)
for link = 2:N
angles (link) = theta(link - 1) + (rand() - 0.5)*(10*pi/180)
end
Use genetic algorithm (or another optimization) to adjust the angles based on the following cost function:
dx = sum(L*cos(angle));
dy = sum(L*sin(angle));
D = sqrt(dx^2 + dy^2);
theta = atan2(dy/dx);
the cost is now just the difference between the vector given by my D and theta above and the vector given by the specified D and theta (i.e. the inputs).
You will still have to enforce the max change of 10 degrees rule, perhaps that should just make the cost function enormous if it is violated? Perhaps there is a cleaner way to specify sequence constraints in optimization algorithms (I don't know how).
I feel like if you can find the right optimization with the right parameters this should be able to simulate your problem.
You don't give us a lot of detail to work with, so I'll assume the following:
random numbers are to be drawn from [-127+theta +127-theta]
all random numbers will be drawn from a uniform distribution
all random numbers will be of type int8
Then, for the first 3 requirements, you can use this:
N = 1e4;
theta = 40;
diffVal = 10;
g = #() randi([intmin('int8')+theta intmax('int8')-theta], 'int8') + theta;
V = [g(); zeros(N-1,1, 'int8')];
for ii = 2:N
V(ii) = g();
while abs(V(ii)-V(ii-1)) >= diffVal
V(ii) = g();
end
end
inline the anonymous function for more speed.
Now, the last requirement,
D == sum(L*cos(V-theta))
is a bit of a strange one...cos(V-theta) is a specific way to re-scale the data to the [-1 +1] interval, which the multiplication with L will then scale to [-L +L]. On first sight, you'd expect the sum to average out to 0.
However, the expected value of cos(x) when x is a random variable from a uniform distribution in [0 2*pi] is 2/pi (see here for example). Ignoring for the moment the fact that our limits are different from [0 2*pi], the expected value of sum(L*cos(V-theta)) would simply reduce to the constant value of 2*N*L/pi.
How you can force this to equal some other constant D is beyond me...can you perhaps elaborate on that a bit more?
Can we use Dijkstra's algorithm with negative weights?
STOP! Before you think "lol nub you can just endlessly hop between two points and get an infinitely cheap path", I'm more thinking of one-way paths.
An application for this would be a mountainous terrain with points on it. Obviously going from high to low doesn't take energy, in fact, it generates energy (thus a negative path weight)! But going back again just wouldn't work that way, unless you are Chuck Norris.
I was thinking of incrementing the weight of all points until they are non-negative, but I'm not sure whether that will work.
As long as the graph does not contain a negative cycle (a directed cycle whose edge weights have a negative sum), it will have a shortest path between any two points, but Dijkstra's algorithm is not designed to find them. The best-known algorithm for finding single-source shortest paths in a directed graph with negative edge weights is the Bellman-Ford algorithm. This comes at a cost, however: Bellman-Ford requires O(|V|·|E|) time, while Dijkstra's requires O(|E| + |V|log|V|) time, which is asymptotically faster for both sparse graphs (where E is O(|V|)) and dense graphs (where E is O(|V|^2)).
In your example of a mountainous terrain (necessarily a directed graph, since going up and down an incline have different weights) there is no possibility of a negative cycle, since this would imply leaving a point and then returning to it with a net energy gain - which could be used to create a perpetual motion machine.
Increasing all the weights by a constant value so that they are non-negative will not work. To see this, consider the graph where there are two paths from A to B, one traversing a single edge of length 2, and one traversing edges of length 1, 1, and -2. The second path is shorter, but if you increase all edge weights by 2, the first path now has length 4, and the second path has length 6, reversing the shortest paths. This tactic will only work if all possible paths between the two points use the same number of edges.
If you read the proof of optimality, one of the assumptions made is that all the weights are non-negative. So, no. As Bart recommends, use Bellman-Ford if there are no negative cycles in your graph.
You have to understand that a negative edge isn't just a negative number --- it implies a reduction in the cost of the path. If you add a negative edge to your path, you have reduced the cost of the path --- if you increment the weights so that this edge is now non-negative, it does not have that reducing property anymore and thus this is a different graph.
I encourage you to read the proof of optimality --- there you will see that the assumption that adding an edge to an existing path can only increase (or not affect) the cost of the path is critical.
You can use Dijkstra's on a negative weighted graph but you first have to find the proper offset for each Vertex. That is essentially what Johnson's algorithm does. But that would be overkill since Johnson's uses Bellman-Ford to find the weight offset(s). Johnson's is designed to all shortest paths between pairs of Vertices.
http://en.wikipedia.org/wiki/Johnson%27s_algorithm
There is actually an algorithm which uses Dijkstra's algorithm in a negative path environment; it does so by removing all the negative edges and rebalancing the graph first. This algorithm is called 'Johnson's Algorithm'.
The way it works is by adding a new node (lets say Q) which has 0 cost to traverse to every other node in the graph. It then runs Bellman-Ford on the graph from point Q, getting a cost for each node with respect to Q which we will call q[x], which will either be 0 or a negative number (as it used one of the negative paths).
E.g. a -> -3 -> b, therefore if we add a node Q which has 0 cost to all of these nodes, then q[a] = 0, q[b] = -3.
We then rebalance out the edges using the formula: weight + q[source] - q[destination], so the new weight of a->b is -3 + 0 - (-3) = 0. We do this for all other edges in the graph, then remove Q and its outgoing edges and voila! We now have a rebalanced graph with no negative edges to which we can run dijkstra's on!
The running time is O(nm) [bellman-ford] + n x O(m log n) [n Dijkstra's] + O(n^2) [weight computation] = O (nm log n) time
More info: http://joonki-jeong.blogspot.co.uk/2013/01/johnsons-algorithm.html
Actually I think it'll work to modify the edge weights. Not with an offset but with a factor. Assume instead of measuring the distance you are measuring the time required from point A to B.
weight = time = distance / velocity
You could even adapt velocity depending on the slope to use the physical one if your task is for real mountains and car/bike.
Yes, you could do that with adding one step at the end i.e.
If v ∈ Q, Then Decrease-Key(Q, v, v.d)
Else Insert(Q, v) and S = S \ {v}.
An expression tree is a binary tree in which all leaves are operands (constants or variables), and the non-leaf nodes are binary operators (+, -, /, *, ^). Implement this tree to model polynomials with the basic methods of the tree including the following:
A function that calculates the first derivative of a polynomial.
Evaluate a polynomial for a given value of x.
[20] Use the following rules for the derivative: Derivative(constant) = 0 Derivative(x) = 1 Derivative(P(x) + Q(y)) = Derivative(P(x)) + Derivative(Q(y)) Derivative(P(x) - Q(y)) = Derivative(P(x)) - Derivative(Q(y)) Derivative(P(x) * Q(y)) = P(x)*Derivative(Q(y)) + Q(x)*Derivative(P(x)) Derivative(P(x) / Q(y)) = P(x)*Derivative(Q(y)) - Q(x)*Derivative(P(x)) Derivative(P(x) ^ Q(y)) = Q(y) * (P(x) ^(Q(y) - 1)) * Derivative(Q(y))