From: israel@math.ubc.ca (Robert Israel)
Subject: Re: Linkage puzzle
Date: 29 Jan 1999 22:40:02 GMT
Newsgroups: sci.math
Keywords: When can this freely-moving linkage be constructed?
In article <78su9l$son$1@gannett.math.niu.edu>, rusin@vesuvius.math.niu.edu (Dave Rusin) writes:
|> Clive Tooth wrote:
|>
|> >Imagine three bars AB (length a), BC (length b) and CD (length c).
|> >The points A and D are fixed, and the bars have freely rotating joints
|> >at B and C.
|> >The bars are constrained to be in a plane.
|> >The distance AD is d.
|> >
|> >The bar AB can rotate freely through 2pi about A.
|> >Write down _all_ the inequalities which will ensure that no bar
|> >stretches, shrinks or breaks.
|> Inequalities in terms of which variables? You must have
|> | b - d | < (a+c)
|> but if this condition is met, there is at least one configuration. Are
|> you looking for constraints on the angles BAD and CDA or what?
I assume it's just inequalities on a, b, c, d.
The inequalities needed are
a + d <= b + c, |d - a| >= |b - c|.
First note that the bars BC and CD can accommodate any position of B
whose distance from D is between |b-c| and b+c. As AB rotates around A,
the distance from B to D varies between |d-a| and d+a.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
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From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: Linkage puzzle
Date: 30 Jan 1999 21:59:42 GMT
Newsgroups: sci.math
Clive Tooth wrote:
>Imagine three bars AB (length a), BC (length b) and CD (length c).
>The points A and D are fixed, and the bars have freely rotating joints
>at B and C.
>The bars are constrained to be in a plane.
>The distance AD is d.
>
>The bar AB can rotate freely through 2pi about A.
>Write down _all_ the inequalities which will ensure that no bar
>stretches, shrinks or breaks.
In this problem, I thought that we were to assume the linkage was
already in place, and the real question was to determine the range of
motion of the points B and C. This is a common problem in robotics,
and the Groebner Basis crowd has been all over this. Just to make sure
I understod the problem, I posted
> Inequalities in terms of which variables? You must have
> | b - d | < (a+c)
> but if this condition is met, there is at least one configuration. Are
> you looking for constraints on the angles BAD and CDA or what?
My constraint is necessary but not sufficient. To see why, note that
if the question is really about whether or not such a linkage exists,
then we are merely asking whether four line segments can be formed into
a quadrilateral in the specified order. Clearly the answer to this
question must be invariant under cyclic permutations of the four variables!
My condition may be written as requiring that neither b nor d be longer
than half the proposed perimeter, which is necessary but not sufficient
as similar conditions are required of a and c. (e.g. [a,b,c,d]=[10,1,1,1]
is not permitted, although it satisfies my condition).
Robert Israel noticed my error but his answer isn't
right either.
>I assume it's just inequalities on a, b, c, d.
>The inequalities needed are
> a + d <= b + c, |d - a| >= |b - c|.
These conditions may be sufficient but are certainly not necessary;
again it is clear that they lack the right symmetry, and they imply there
is no quadrilateral with sides of length [a,b,c,d]=[5,4,4,5], say, which
is clearly a contradiction.
I don't have a complete answer yet to the original problem, which may well
be very simple, but I have a complete method for testing proposed answers!
We had a similar thread a year ago in which some nameless person wanted
to know the constraints on the lengths of 6 line segments would form a
tetrahedron; see the "tetrahedral inequality" thread at
http://www.math.niu.edu/~rusin/known-math/index/52BXX.html
The upshot is that besides the various triangle inequalities, we must
have a non-negative volume.
In particular, we may describe the conditions on six lengths which may
be fitted together to form a planar quadrilateral and its diagonals.
Since in the present problem the diagonals are not given, I can present
the answer this way:
"Let A=a^2, B=b^2, C=c^2, D=d^2 be given. Then there is a planar
quadrilateral with lengths abcd (in that order) iff there exist two
other real numbers E, F satisfying
E > 0
F > 0
(E - B - C)^2 < 4BC
(E - A - D)^2 < 4AD
(F - A - B)^2 < 4AB
(F - C - D)^2 < 4CD
and
2AC(A+C) + 2BD(B+D) + 2EF(E+F) + (ABE+ADF+BCF+DEF) =
(A+B+C+D+E+F)(AC + BD + EF)
Assuming degenerate quadrilaterals are accepted, each inequality may be
allowed to be nonstrict."
If you take a,b,c,d as given, the equation may be scaled to give a
cubic EF(E+F) + EF + linear(E,F) = 0. So the problem is just to decide
whether the curve so described has any points inside a box (determined
as the intersection of 5 vertical and 5 horizontal half-planes, and
so possibly empty).
The set of points [a,b,c,d] which allow such E, F to be found will then
be a set in R^4 bounded by components of a certain algebraic variety.
Both Robert Israel and I were suggesting that the boundary would consist
of linear subspaces, which may indeed be true but right now I don't see
that that's necessary, and I haven't the time today to try to unravel
just what the boundary really is.
dave
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From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: Linkage puzzle
Date: 31 Jan 1999 03:00:22 GMT
Newsgroups: sci.math
I must apologize; I misread the problem several times (in different ways!)
The question, as re-emphasized by the original poster, is to determine the
combination of edge lengths which permit the construction of a quadrilateral
which allows one particular arm to _rotate completely_ while an adjacent
arm stays fixed.
(I see now that you can _construct_ the quadrilateral iff no side is longer
than the sum of the other three; and you can determine the _constraints
on rotation_ in general as I show below.)
Robert Israel posted a correct and succint answer. Let me corroborate it
with a longer answer amenable to more general problems.
The problem, slightly rephrased, is
>Imagine four bars AB (length a), BC (length b), CD (length c), DA (length d)
>The points A and D are fixed, and the bars rotate freely in a plane.
We are asked to find conditions on a,b,c,d to ensure that
>The bar AB can rotate freely through 2pi about A.
One approach is to parameterize the possible positions of B and C.
Fixing coordinates so that A=(0,0) and D = (d,0), we may write
B=(x1,y1) and C=(d-x2,-y2) where, for example,
x1=a*(1-t^2)/(1+t^2), y1=a*2*t/(1+t^2),
x2=c*(1-u^2)/(1+u^2), y2=c*2*u/(1+u^2).
As the bars move, we get a constraint relating t and u from substituting
these into (x1+x2-d)^2+(y1+y2)^2-b^2 . It's rather a messy formula so
I won't write it out, but because of the parameterizations I used, it's
just a long polynomial. Indeed, it's a quadratic in u, so for any choice of
point B we can determine t, then u, then C. More to the point, if
we want to know that _every_ point B on the left circle matches one or
more points C on the other circle, we just have to know that this
quadratic has a real root for every choice of t.
This is determined by requiring that the discriminant of the quadratic-in-u
be non-negative for all t. As it happens, that discriminant is (-4) times
the product of two simple factors,
2
(c + a - d + b) (-c + a - d - b) + (-c + a + d - b) (c + a + d + b) t ,
2
(-c + a - d + b) (c + a - d - b) + (c + a + d - b) (-c + a + d + b) t
so the condition on a,b,c,d which we seek is that these factors never
vanish as t ranges over the real line. (More generally, we observe that
in a linkage which does not permit full rotation, the limiting positions
are determined by those values of t making this discriminant vanish.)
(Actually the discriminant might not change sign as t moves past a
root of one of these factors. But the values of t^2 which cause the
two factors to vanish differ by a fraction with numerator abcd, i.e., they
are distinct for actual linkages. And neither quadratic has a multiple
root unless the quadrilateral is degenerate.)
So now we see that, assuming the linkage can be built at all, it allows
full motion at A iff both of these expressions are non-negative:
(c + a - d + b) (-c + a - d - b) (c + a + d + b) (-c + a + d - b),
(-c + a - d + b) (c + a - d - b) (c + a + d - b) (-c + a + d + b)
Now, in a real quadrilateral, each side is shorter than the sum of the
other three (and each of a,b,c,d > 0), so it is necessary and sufficient that
both of (b+c) - (a+d) and (a-d)^2 - (b-c)^2 be non-negative.
This is just what Robert Israel said:
>The inequalities needed are
> a + d <= b + c, |d - a| >= |b - c|.
Sorry to have been so obtuse.
dave
==============================================================================
From: padraig@celticweb.com
Subject: Deafening Silence re the challenge of Clive Tooth on Cassinian Ovals
Date: Sun, 31 Jan 1999 22:44:19 GMT
Newsgroups: sci.math
Hello Clive and other readers of this thread.
I found it profoundly rewarding examining your treatment of the Cassinian
Linkage.
Rather than append my thoughts on it onto the :-
http://www.angelfire.com/nm/cassinianoval/linkages.html
I've posted the additions to it to the following :-
http://www.angelfire.com/nm/cassinianoval/clive.html
>I have written a simple one already, quite hypnotic to watch.
What sort of one was it?
Thats an interesting page you have at
http://www.pisquaredoversix.force9.co.uk/
Thanks to you also TwoBirds for your helpful link to information on the
Peaucellier cell at:-
http://www.brockeng.com/mechanism/index.htm
Take the 2 X's from TwoBirds e-mail if you want to show your appreciation to
him/her XXrs.1@ix.netcom.com
Sincerely Padraig
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