keeping the distance between objects
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Hello world,
Think of two linear trajectories. Hooks will be moving along those trajectories with "rope" tied between them - hook1 from traj1 with hook1 from traj2 with hook2 from traj1 with hook2 from traj2 and so forth. The two trajectories are in one plane with the center line, symmetrically at an angle around the center line and the hooks need to decelerate to 0 along the trajectory. To simplify, there'll be no queue forming at one end.
I know the timing of hooks because I know the speed of new rope coming in and the length of the rope between hooks. I have found equations for the trajectories, created 1x* matrizes for time and position according to the resolution I will be able to control the hooks at and decided the deceleration along the centerline should be linear for simplicity. What I'm now trying to do is output a lookup table of position (and speed) along either trajectory and the time for that position. As there's nothing dynamic this table will be looped through for each new hook.
I have thought of few ways to go about this, each looks equally beyond my ability right now so I'm hoping for your support.
1. Try to program "free particles" that will be bound by the boundary conditions a) the trajectory b) the distance to the one coming after. It seems to me this would incur great performance tribute.
2. Try to program a circle with the diameter of the distance between two hooks and let that decelerate linearly along the centerline. I don't know how to check the intersections with the trajectories and how to properly adapt the speed of either of the hooks along their trajectories.
3. Perhaps someone could do this with higher-than-I-could-come-up-with differential maths?
4. Perhaps I'm not seeing some easy relationship between the centerline and trajectoral deceleration that would save me all of the above work?
I do hope somebody can give me very easy to understand advice / instructions on which idea would be best and how to actually do it.
Thank you very much! Philipp
3 Comments
Image Analyst
on 22 Oct 2014
A diagram would be helpful so we could visualize this situation.
Image Analyst
on 23 Oct 2014
Accepted Answer
I assume:
1. The rope is fixed at Gx3 but free to move through all the other hooks.
2. Also you know the distance of each hook from the point A.
Since alpha and d(alpha)/dt determine the location and speed of all hooks, all you need is to compute d(alpha)/dt given vf - no?
So say (writing in "code" mode for clarity - not actually code):
dist(Gy1, A) = hy1
dist(Gy2, A) = hy2
dist(Gx2, A) = hx2
dist(Gx3, A) = hx3
The above are all known and constant.
Also say:
dist(Gy1, Gx1) = s11
dist(Gy1, Gx2) = s12
dist(Gx2, Gy2) = s22
dist(Gy2, Gx3) = s23
Then
d(s11)/dt + d(s12)/dt + d(s22)/dt + d(s23)/dt = vf % eq (1)
but:
s11 = hy1 = constant
s23 = hx3^2 + hy2^2 - 2*hx3*hy2*cos(2*alpha) % using cosine rule
s22 = hy2^2 + hx2^2 - 2*hy2*hx2*cos(2*aplha) % using cosine rule
s12 = hx2^2 + hy1^2 - 2*hx2*hy1*cos(2*alpha) % using cosine rule
Plug these into eq (1) and perform the differentiation. You get vf in terms of alpha and d(alpha)/dt. It is a first order differential equation:
4*C*sin(2*alpha)*d(alpha)/dt = vf
where:
C = (hx3*hy2 + hy2*hx2 + hx2*hy1).
You could use an ode solver to find the solution starting from some initial alpha0 and vf0 for time {0, h, 2h, 3h, .....}. Once you get the set of alpha angles, the locations of the hooks follow.
[Edited to make corrections]
18 Comments
The above approach will work for a fixed number of hooks at fixed locations, since every distance sij can be written in the form:
sij = hi^2 + hj^2 - 2*hi*hj*cos(2*alpha)
On the other hand if you are going to add hooks "on the fly" at arbitrary locations you'll have to think of some other (more computationally intensive) approach.
OK, I think I get the rough idea. But I still don't understand how the foil can fold unless the hooks bunch up near Gx3. How can the distance between Hooks remain constant. I don't have a mechanical/industrial engineering background, so some things that are obvious to you must be slipping past me. You may need to explain it as if you are describing it to a freshman student. For example what is a "twincat control".
SK
on 24 Oct 2014
OK, I get it now. That is just amazing. To think that machinery like that is probably used to manufacture commonplace items - we never think twice about. Anyway, I'd like to mull it over in my spare time - maybe I can come up with something in case you haven't already. Good luck with it anyway.
Philipp
on 26 Oct 2014
Hi,
OK, so let us first ensure we are on the same page here. These are my notations:
By the point A on your diagram, I mean a fixed non-moving reference point (The hook Gx1 moves but when I refer to A, I mean the fixed point which serves as a transit point for the foil). One can think of the foil as sliding over a fixed bar A.
The hooks all "Grasp" the foil, i.e: there is no sliding through hooks (as I erroneously assumed in my earlier post). This means that the distances d(x3,y2), d(y2,x2), d(x2,y1) and d(y1,x1) are all the same and are equal to I_F.
Now let the hook Gx1 start moving away from A. After some time, the distance d(x1, A) becomes equal to I_F. At this time, a new hook (call it Hyy) appears at A on the upper side and grasps the foil. The combination of the three hooks Gy1, Gx1 and Hyy then moves preserving the distances between them until the distance d(yy, A) becomes equal to I_F. Again a new hook (call it Hxx) appears at A on the lower side and grasps the foil. The combination of the four hooks then continues to move preserving distance ... and so on. Of course the number of simultaneous hooks that are moving depends on the angle alpha - If alpha is small, the number of simultaneous hooks will be more. If alpha is larger then there may only be a few simultaneously moving hooks - because old hooks will merge on the left before new hooks appear on the right.
OK so far?
So we see that there is one leading hook (just call it hook0). In theory, the leading hook can move at any speed along the conveyor. The speed of the following hook (hook1 - on the opposite side of hook0) has a relationship with the speed of hook zero that is determined by the distance constraint.
Similarly the speed of hook2 (which is on the opposite side of hook 1) has a relationship with the speed of hook1 that is determined by exactly the same kind of distance constraint.
So as you had mentioned in your first post, there is essentially one relationship to figure out - that between the speed of a hook and the speed of the hook that is following it (on the opposite side). And this relationship is given to us, surprise, surprise, by the cosine rule:
y1^2 + x1^2 - 2*x1*y1*cos(alpha) = L^2
where L is the length of I_F.
The above actually expresses the relationship between the distance of Gx1 from A (which I've denoted by x1) and the distance of Gy1 from A (which I've denoted by y1). A triangle is formed with sides x1, y1 and I_F. Use the cosine rule on this triangle to get the above relationship. Moreover any two consecutive hooks (on opposite sides) satisfy this relationship.
We can solve the equation above for x1 in terms of y1 and alpha. SO at every location of a hook, the location of the opposite following hook is known.
The speed of the leading hook can be arbitrary - unless you want to set it using some other condition on how exactly the hooks move around the conveyor.
{edited later for corrections]
By the way, I haven't looked at your code yet, but perhaps the above would help.
As a "by the way", the diagram is somewhat misleading. I now see that the foil should be drawn like a ray of light reflecting in-between two angled mirrors. Does that place a constraint on the angle alpha??
Note also that in that in my first post there were three equations like:
s23 = hx3^2 + hy2^2 - 2*hx3*hy2*cos(2*alpha) % using cosine rule
The LHS should be squared, so these should actually be:
s23^2 = hx3^2 + hy2^2 - 2*hx3*hy2*cos(2*alpha) % using cosine rule
and things modified accordingly. But anyway it is irrelevant now as these were based on a complete misunderstanding.
take the linear deceleration of the first hook and generate a lookup table of time vs x0. I then loop through the cosine rule for those x0 to generate a lookup table time vs y0.
Right.
Do you think I will need to use the cosine rule on the y0 table to find the x1 table or shouldn't it be the same as the y0 just triggered appropriately later?
You need to use the cosine rule iteratively / recursively until the (smaller) solution becomes less than L. For example, given the alpha of round 17 degrees that you have, it may be only around four or five iterations /recursions max.
You said in your comment that any two following hooks would satisfy the formula which I agree with yet it's not clear to me if they would follow the exact same profile as the second hook or would somehow need to slow down earlier than the second one.
If you solve for x1 in terms of y1 (or vice versa) you will see that x1 and y1 are in a constant ratio (R say) that depends only on alpha and L. So in fact the acceleration profiles will also be related to each other by the same ratio. So whatever profile you have for the leading hook gets multiplied by R, R^2, R^3, ... etc for successive hooks.
So in fact, we realize now that it is actually much simpler than we originally thought (I'm not feeling bad about that and I recommend that you don't either :-) ). In fact you probably do not even need to write a single loop (get familiar with using the .* and ./ operators and vector operations in general in Matlab).
As to what profile to use for the first hook, why not just take constant deceleration. The distance profile (x0) would then be:
x0 = u*t - 0.5*a*t^2
where a is the constant deceleration and u is the initial speed with:
t = [0, 0.001, 0.002, ...]
You will have to choose the constants u and a so that x0 becomes equal to I_V at some time (T say) of your choice and the speed u - a*t becomes 0 at that t = T. The distance profiles of successive hooks are then related by the the constant multiplier R.
More generally you could write a function that takes an arbitrary distance profile for the leading hook, and returns the distance profiles of all the successive hooks. Note that, in general each successive hook will have a shorter distance profile than the previous one due to the cutoff condition, so the function is not entirely trivial.
The speed and acceleration profiles can then be found (approximately) by taking first and second differences, respectively, of the distance profiles.
How about I take a linearly decelerating point on the center-line say w_-1 ...
Do you mean to take the projection of the point onto the conveyor. In that case, a linearly decelerating point on the centre-line would translate into a linearly decelerating point on the conveyor, wouldn't it?
You could also start with acceleration profiles or speed profiles (instead of distance profiles). They also are related by the same ratio R. It would probably be a good Matlab exercise to write a general function that takes either a distance profile or speed profile or acceleration profile, along with some initial conditions and produces all the remaining profiles correctly with correct cutoff lengths.
Forget the remark about light rays :-P. The thing to note is that the shape of the foil forms a set isosceles triangles on the conveyor.
OK, I had a quick read through your post. I'll read it in more detail later. But a couple of points:
1. You are right about there being no constant factor. That was a mistake on my part. We have
y1^2 + x1^2 - 2*x1*y1*cos(2*angle) = L_F^2
I divided throughout by y1^2 and forgot to divide the RHS - which is how I obtained a constant ratio between x1 and y1. So you are right, it should be:
x1 = y1*cos(2*angle) - sqrt(L_F - y1^2sin^2(2*angle))
( using 1 - cos^2(2*angle) = sin^2(2*angle) )
2. You are eventually looking for a numerical set of profiles - right? So suppose that you are given an arbitrary initial profile (say distance profile - although it could be a velocity or acceleration profile also with initial conditions). Then:
a) Find the profile for the next hook. You have already figured that out in (1) above.
b) In the profile found in (a), remove all the distances that are less than L_F. (That is a single line in Matlab)
c) Go back to step (a).
My point is that if you try to start with a profile given by a complicated expression, integrate it mathematically, find the succeeding profiles, all in one step, the likelihood of error is high. So first get (a), (b) and (c) coded. Then you can worry about what exact initial profile you want. So basically you will have a convenient function:
function following = get_hook_profiles(leading, type)
The 'type" argument will be a one of the strings: 'dist', 'vel' or 'acc'. The 'leading' argument is the profile of the leading hook. The return result 'following' can perhaps be a cell-array of vector profiles, with following{i} being the profile for the i-th hook.
Also, before I read your post in detail, am I to infer from your post that the distance/velocity/acceleration profiles of all the hooks (including the leading one) is determined (constrained) by a few scalar parameters. I had thought that, since the hooks were free to move along the conveyor, that the speed of the leading hook could be more or less arbitrary. But of course, in any given span of time, the number of hooks leaving on the left side must equal the number of hooks coming in on the right - so is that the condition that constrains the speed of the leading hook? In other words - why should the profile of the leading hook matter at all? Whatever its speed, the overall job of folding the foil still gets accomplished right?
Philipp
on 29 Oct 2014
SK
on 29 Oct 2014
I really hope I'm not annoying you with my thickness.
Not at all. It is good to see someone who follows through with his own ideas rather than blindly accepting somebody else's suggestion.
You probably just made a mistake in the computation. Try this. I get the leading hook and 2 following hooks for your distance profile. All the code below can be executed (in order) in Matlab.
angle = 17.698*pi/180;
L_F = 2*500*sin(angle);
From you graph, I see that the deceleration a is:
a = 500/40.5;
Initial speed is:
u = 9*a;
Take vector of times:
t = 0 : 0.01 : 9;
leading distance profile (hook 0):
y0 = u*t - 0.5*a*t.^2;
plot(t', y0);
Next hook (hook 1):
x0 = y0*cos(2*angle) - sqrt(L_F^2 - (y0.^2)*sin(2*angle)^2);
Remove distances that are less than LF (don't forget to modify the time vector):
t1 = t(x0 >= L_F);
x0 = x0(x0 >= L_F);
figure;
plot(t1', x0);
Next hook (hook 2)
y1 = x0*cos(2*angle) - sqrt(L_F^2 - (x0.^2)*sin(2*angle)^2);
figure;
plot(t1', y1);
Cant go any further, since all elements in y1 are less than L_F. SO there is a leading hook and two following hooks for this particular angle (17.698 degrees)
Now of course when y0 reaches its endpoint, x0 becomes the leading hook and its distance profile has to be extended upto the endpoint. Presumably you would give it the same distance profile that you gave y0, and then compute the following two hooks. The process goes on.
The only thing is that both x0 and y1 come to a stop when y0 reaches the endpoint (look at the graphs) so you have to "restart" x0. So when you look at the profiles over the full time, they are going to be jerky - although they still accomplish the job of folding the foil. If you feel this is ugly, you need to choose the initial profile in such a way that the transition from the constrained (ie: following) profile to the leading profile is smooth. You get what I mean? You would probably have to choose an initial profile whose speed is non-zero right up to the endpoint (ie: it just stops suddenly at the endpoint)
To do these computations in a loop, you'll have to decide on how you are going to store each profile and also make few other programming decisions.
Anyway, you have really worked hard over this, so I don't want to burden you with any further advice. :-)
More Answers (2)
with this cutoff function you wrote, am I not retaining something like
5 s 304 mm
5.01 s 305 mm
etc.
when I really want the whole remainder x0>=L_F shifted to begin at the origin?
It depends on how you want to represent the final data. I think you need to think about this carefully, ie:
What does the final data representation look like?
But before considering this question, there is another issue:
The problem of choosing the leading profile is under-determined. We would probably want to use some sort of continuity / periodicity condition to choose the leading profile. Ideally we want periodicity - after all it is a machine that repeats things over and over.
could I not let my seeding y0 continue beyond L_V so as to avoid having to restart x0?
Yes you could and you would get correct results - in the sense that the job will still get done. But you would have to keep on "seeding" y0 arbitrarily to keep going. Since this is after all a machine, we expect some sort of periodicity and you won't get that by arbitrarily seeding y0.
Now that I would be getting a single profile (I suppose without having tried it that x0 and y1 are the same) that can be repeated by all hooks, ...
This is the key. In a sense we want all the profiles to merge smoothly into one another. Moreover the profile of one hook should look the same as the profile of any other hook - only shifted in time. So that should give us a condition to fix the leading profile shape, ie we want:
x(t+tau) = x(t)*cos(2*angle) + sqrt(L_F^2 - (x(t).^2)*sin(2*angle)^2) -- (1)
which can alternately be written in its original form (before we solved for the root):
x(t+tau)^2 + x(t)^2 - 2*x(t)*x(t+tau)*cos(2*angle) = L_F^2 ----- (1a)
What is the time tau? It is the time for which:
x(tau) = L_F ----- (2)
because that's when the next hook comes online (after the leading hook has moved L_F from the zero position). Note that we don't know this time yet because we don't yet know the profile. We have to simultaneously solve for both the time tau and the profile x(t) from the two equations (1) and (2) above.
How does one solve this? I'm not sure - I've never really come across a "delay" equation like the above, but it certainly looks like it should be solvable. Note that simple ansatz's like polynomials in t wont work since you will get more equations than unknowns. For example - 2nd degree polynomial (constant acceleration) would be of the form x(t) = at + bt^2. There is no constant term since x(0) = 0 (equivalent to x(tau) = L_F) . If we plug this in to eq (1a), we get 5 equations in three unknowns (a, b and tau).
Thus we need a more sophisticated ansatz, or we need to solve (1) numerically.
Can you think of an ansatz? How about x(t) = A(1 - e^(-bt)). What if we plug this into (1a) and try to equate coefficients (we will have e^(2bt) terms, e^(bt) terms and constant terms (so three equations) and three unknowns A, b and tau. Could you try it?
[edited for corrections]
SK
on 31 Oct 2014
0 votes
I don't think the problem has an answer.
There is one very obvious fact that has been embarrassingly missed. It is not possible to keep all the foil lengths constant simultaneously. At the very least the length of the foil from Gy2 to Gx3 has to change. So somehow the problem statement has to be modified.
Where do you get this problem from?
3 Comments
Philipp
on 2 Nov 2014
Philipp
on 2 Nov 2014
I'll get back to you with a drawing. It really is quite easy to see if you draw a bunch of isosceles triangles starting from the stopping point Gx3.
However, I think we should probably continue this via private e-mail, as this thread may have outlasted its welcome on this site. We are taking up MathWorks bandwidth on a problem not directly related to the Matlab product. I wonder if you can see my e-mail address. If you do, send me an identifying e-mail. Let me know.
Regards.
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