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Subject:
Intersection of three cones
Category: Science > Math Asked by: kbo-ga List Price: $10.00 |
Posted:
24 Nov 2005 04:00 PST
Expires: 24 Dec 2005 04:00 PST Question ID: 597056 |
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There is no answer at this time. |
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Subject:
Re: Intersection of three cones
From: frde-ga on 24 Nov 2005 07:33 PST |
Surely, if they have the same gradient, and they are cones they must be congruent. Where they intersect (if at all) depends solely on where they are placed - if two cones intersect at all, they should have multiple points of intersection |
Subject:
Re: Intersection of three cones
From: frde-ga on 25 Nov 2005 02:10 PST |
Ah, I think I see what you are getting at. Three cones pushed into each other (let us say they are beams of light rather than solid) with 'intersection' defined as the surface layers touching each other. Because they are cones, the outside surfaces are moving away from each other as one looks at successive horizontal slices, so one can see that they can only intersect at one level. In that case one might as well forget about cones, and just use circles. I've just taken three different sized coins and tried placing them slightly on top of each other. One can easily get just one point of intersection, and there is a special case where one can get two points of intersection, subject to the diameters of the coins. Quite interesting, I think that if the cones are of different gradients one could possibly get two intersections at different levels - but visualizing that is slightly painful. |
Subject:
Re: Intersection of three cones
From: greyelf-ga on 25 Nov 2005 08:23 PST |
Using Mathematica and playing with the parameters I found a set that has four solutions for the double cones -- and hence two for the single cones that you are interested in. Here is what I used: Solve[ { x^2. + y^2 == (z - 1)^2, (x - 1)^2 + (y - 2)^2 == (z - 1)^2, (x - 4)^2 + (y - 3)^2 == 2(z - 1)^2 }, {x, y, z}] {{x -> -5.37298, y -> 3.93649, z -> -5.6607}, {x -> -5.37298, y -> 3.93649, z -> 7.6607}, {x -> 2.37298, y -> 0.0635083, z -> 3.37383}, {x -> 2.37298, y -> 0.0635083, z -> -1.37383}} |
Subject:
Re: Intersection of three cones
From: mathtalk-ga on 29 Nov 2005 07:55 PST |
The Question can be reformulated as concerning a common point of intersection of circles in a plane, whose centers are fixed and whose radii differ by fixed constants but are otherwise allowed to vary: R_1 = r_1 + x R_2 = r_2 + x R_3 = r_3 + x This assumes that it is the surface of the cones not their interiors which are being intersected. I believe that with this perspective it is obvious that no such common point of intersection is generally going to obtain, but with additional assumptions it can be shown. Consider the case of two circles. If one circle is within the other, then no point of intersection will obtain no matter how much their radii increase, since (as a corollary of the cones' "same gradient") the radii increase by equal increments. If two circles are mutually outside one another, then as they expand we will first obtain a single point of intersection (tangency) along the line connecting the centers of the two circles. Further expansion produces pairs of points of intersection, symmetric with respect to the line passing through both centers, which we hereafter term the "axis". The locus of points of intersection produce by the equal increase of radii is a line if the radii are equal (the bisector of the axis) or a parabola if the radii are unequal, symmetric with respect to the axis and opening in the direction of the circle of smaller radius. The problem of three unequal circles then amounts to one about the intersection of two parabolas, one generated by one pair of circles and another generated by a different pair. If two circles are allowed to be equal, the problem is simplified accordingly, but I'll not discuss it since the emphasis here is on cones with differing "heights". Two parabolas, both opening we may assume in the direction of the smallest of the three circles, will have two, three, or four points of intersection. Each such point represents a "solution" as a common intersection of three circles, or as posed in the original Question, a point common to the surfaces of three cones of parallel axes and equal slopes. regards, mathtalk-ga |
Subject:
Re: Intersection of three cones
From: mathtalk-ga on 29 Nov 2005 08:11 PST |
Oops! I erred in stating that the locus is a parabola... in fact, the locus of points whose distances to two fixed points have a constant difference is (one branch of) a hyperbola (which of course is analogous to defining an ellipse as the locus of points whose distances to two foci have a constant sum). However this method of analysis should prove effective. Given any three centers and the "initial radii" by passing a plane perpendicular to the cones' axes through the apex of the "highest cone, we obtain all the possible solutions by taking the points of intersection (if any) between two (branches of) hyperbolas. -- mathtalk-ga |
Subject:
Re: Intersection of three cones
From: mathtalk-ga on 30 Nov 2005 06:16 PST |
A bit more algebra and I found the "different heights, same gradient" problem can be reduced to intersecting a line and a (branch of a) hyperbola. For the sake of simplifying the algebra, I recommend translating the "highest" cone to the origin, so that the other two have apexes (apices?) below the xy-plane. (I'm picturing cones that open upward, since orginally the interest was in cones "above" the plane.) We can take (by scaling, rotation, reflection, and translation) these as the apexes: (0,0,0), (1,0,-p), (a,b,-q) with some restrictions on a,b,p,q: 0 < p < 1 (avoids having the first cone inside the second) 0 < q < SQRT(a^2 + b^2) (avoids first cone inside the third) The hyperbola formed by the intersection of first and second cone is then: (1-p^2)(x - 1/2)^2 - p^2 y^2 = (1/4)p^2 (1-p^2) A line may be derived using the equations of all three cones: (q - ap)x - bpy = (1/2)(q(1-p^2) - p(a^2 + b^2 - q^2)) A sufficient condition for a single point of intersection (between the line and the branch of the hyperbola that "contains" the origin) is that the slope of the line is between the slopes of the hyperbola's asymptotes. Note that if b = 0, then there may fail to exist any intersection. In effect the hyperbola branch formed by the first and second cones may nest inside or outside the hyperbola branch formed the the first and third cones. This could happen even if the second and third cones are not nested one within the other, so that they as a pair have non-empty intersection. If we assume b > 0, e.g. by reflection if necessary, then the slope condition alluded to is this: ap - b SQRT(1-p^2) < q < ap + b SQRT(1-p^2) So this gives a sufficient condition for there to be a unique point of common intersection between the three (upper) cones, taken in combination with our earlier assumptions (no cone nested within another). regards, mathtalk-ga |
Subject:
Re: Intersection of three cones
From: mathtalk-ga on 30 Nov 2005 18:40 PST |
kbo-ga asked, "[H]ow do I go from there to the coordinates of the intersection point(s)?" It's pretty easy, actually. The general data for your problem is a slope m for the cones together with the three apexes: (a_1,b_1,c_1), (a_2,b_2,c_2), (a_3,b_3,c_3) I'd "normalized" such data to unit slope (basically just scaling the z coordinate) and translating the highest apex down to z=0, followed by such affine transformation of the xy-plane to get these data: (0,0,0). (1,0,-p), (a,b,-q) If my yellow pad doesn't deceive me, then we need to solve these equations: (1-p^2)(x - 1/2)^2 - p^2 y^2 = (1/4)p^2 (1-p^2) (q - ap)x - bpy = (1/2)(q(1-p^2) - p(a^2 + b^2 - q^2)) For cases b=0, we use the second equation to solve for x: x = (1/2)(q(1-p^2) - p(a^2 + b^2 - q^2))/(q - ap) then plug that into the first equation to obtain y: y = +/- SQRT((p^-2 - 1)(x - 1/2)^2 - (1/4)(1 - p^2)) Conceivably this gives zero, one, or two real solutions depending on the sign of the value inside the square root for y. Check that any such solutions satisfy the condition: x <= (1 - p)/2 because that is required for the solution to lie on the branch of the hyperbola that "goes around" the origin (small cone/circle). If b nonzero, then: use the second equation to solve for y as a first degree polynomial in x, substitute the result for y in the first equation, and solve the resulting quadratic equation. Again check that any indicated solutions are on the right branch of the hyperbola (which is actually the "left" branch as we set things up), using the same inequality on x as above. Potentially one of the real solutions is on that branch and one is on the other branch, so it's possible your desire for a "unique" solution is met even when the quadratic has two real solutions. I'll try to find time tonight to work out an example. Feel free to post some sample data if you'd find that more convincing... regards, mathtalk-ga |
Subject:
Re: Intersection of three cones
From: mathtalk-ga on 08 Dec 2005 10:30 PST |
Hi, kbo-ga: I did make some progress, removing for example rotations from the method of solution as unnecessary. I'll have to take issue with the idea that "the gradient doesn't matter". Although the z-coordinate is not important to the solution you want, and the gradient m can be conveniently lumped into (for example) the Z data for any particular problem, it is still necessary to have at least this. Working backward from your solution, I've tried to determine the value of m that you had in mind. That is, by looking at the B point, taking the distance from your solution x = 375033, y = 760128 to (XB,YB) gives mz = 550849.214 or so. If your solution were consistent, then taking the distances from solution (x,y) to points (XA,YA) and (XC,YC) should give values that differ from mz by mZA and mZC, respectively: distance (x,y) to (XB,YB) minus distance (x,y) to (XA,YA) ----------------------------------------------------------- ~ 10.78 (ZA - ZB) distance (x,y) to (XB,YB) minus distance (x,y) to (XC,YC) ----------------------------------------------------------- ~ 0.4428 (ZC - ZB) Now the value for ZA = 46291 (as you labelled things earlier in the thread) is so much smaller than the other coordinates in the data, that I might suspect a missing leading digit there, but I wasn't able to find any "fix" that comes close to putting the solution right. * * * * * * * * * * * * * * * * * * * If you like, I can simply solve the problem taking m = 1 (or some other value you wish to specify), or perhaps you can check the results of your "boundary hunting program" to see if its input or output has gotten miscopied. regards, mathtalk-ga |
Subject:
Re: Intersection of three cones
From: mathtalk-ga on 11 Dec 2005 19:39 PST |
Hi, kbo-ga: Starting over to try and simplify the presentation as well as the conclusions. You gave these sample coordinates for the cones' apexes: X Y Z ------ ------ ------ A 329137 736281 46291 B 408263 210282 0 C 729192 602876 368889 You didn't specify the gradient m, which I'd represent in the equations for the cone in this way: (x - XA)² + (y - YA)² = m²(z - ZA)² [1] (x - XB)² + (y - YB)² = m²(z - ZB)² [2] (x - XC)² + (y - YC)² = m²(z - ZC)² [3] because the square term makes it easy to combine the gradient m with the z-coordinates, effectively reducing to the case m = 1 (after Z* is replaced by m times Z*). Simply for the sake of illustration, I'll assume m = 1 was the gradient in this example. We will omit any scaling or rotation, because although it helps by eliminating a couple of the free parameters, it is a relatively big price to pay for that small amount of simplification. On the other hand, subtracting off the x,y,z coordinates of one apex from the other two is easy (both to "do" and "undo" at the end of the solution procedure). Since you already have the lowest z-coordinate at zero, I'll keep that the same by subtracting the "B" coordinates from the the other two apexes: X Y Z ------ ------ ------ A -79126 525999 46291 B 0 0 0 C 320929 392594 368889 [Bear in mind we will add x = 408263, y = 210282 back at the end.] Consequently we can always simply equations [1]-[3] into: (x - XA)² + (y - YA)² = (z - ZA)² [1] x² + y² = z² [2] (x - XC)² + (y - YC)² = (z - ZC)² [3] Notice that m is gone, and [2] has become reduced. Subtracting each of [1] and [3] from [2] gives a pair of first degree polynomial equations in x,y,z: 2XA x + 2YA y = 2ZA z + (XA² + YA² - ZA²) 2XC x + 2YC y = 2ZC z + (XC² + YC² - ZC²) Divide both sides (of each equation) by 2 and use the substitutions: RA = (1/2)(XA² + YA² - ZA²) RC = (1/2)(XC² + YC² - ZC²) and we have a system of two linear equations compactly expressed in matrix form: / XA YA \ / x \ / ZA*z + RA \ | | | | = | | \ XC YC / \ y / \ ZC*z + RC / This system can be solved for x,y in terms of z and constant data if and only if the matrix of coefficients on the left- hand side is nonsingular, ie. the determinant: D = XA*YC - XC*YA is nonzero. The geometric interpretation of this is that the line through points (XA,YA) and (XC,YC) does not pass through the origin, ie. not through (XB,YB) in reference to our original choice of input coordinates. We defer for now the discussion of the "degenerate" cases where these are three colinear points, in part because for the data at hand: -79126*392594 - 320929*525999 = -199872725915 is clearly nonzero, and in part because I suspect your "physical application" precludes this possibility. Recall that a 2x2 matrix inverse can be expressed quite simply in terms of its entries and its determinant: / XA YA \ -1 1 / YC -YA \ | | = --- | | \ XC YC / D \ -XC XA / Therefore, after multiplying both sides of the matrix equation above by the inverse so expressed: / x \ 1 / YC -YA \ / ZA*z + RA \ | | = --- | | | | \ y / D \ -XC XA / \ ZC*z + RC / we have essentially solved for x,y (dependent on z): YC(ZA*z + RA) - YA(ZC*z + RC) x = ----------------------------- D YC*ZA - YA*ZC YC*RA - YA*RC = ------------- z + ------------- D D -XC(ZA*z + RA) + XA(ZC*z + RC) y = ------------------------------ D -XC*ZA + XA*ZC -XC*RA + XA*RC = -------------- z + -------------- D D These first degree expressions may then be substituted respectively for x and y in equation [2] above, giving a polynomial of (at most) degree 2 to solve for z. Let's define: p = (YC*ZA - YA*ZC)/D d = (YC*RA - YA*RC)/D q = (-XC*ZA + XA*ZC)/D b = (-XC*RA + XA*RC)/D so that: x = pz + d and y = qz + b . The substitution into [2] then reduces as follows: (pz + d)² + (qz + b)² = z² (p² + q² - 1)z² + 2(pd + qb)z + (d² + b²) = 0 A quick check may be made on roots provided by applying the quadratic formula, to see if the value of z derived is above each of the Z* input values (thus intersection among the "upper" cones is indicated). Evaluation of the various intermediate expressions in the analysis above can be performed in a convenient spreadsheet: RA = 140396507598 RC = 60523188778 p = 0.87986830 d = -116492.58108278 q = 0.22036441 b = 249390.04756139 p² + q² - 1 = -0.17727130 2(pd + qb) = -95082.87790347 d² + b² = 75765917270.00 Solving this quadratic found two roots: z = -974813.2131 , 438443.9282 where clearly only the latter could correspond to an intersection of the upper cones. For z = 438443.9282: x = pz + d = 269280.3342 y = qz + d = 346007.4852 Finally, once the (XB,YB) coordinates are added back: x = 677543.3342 y = 556289.4852 gives the point of intersection in the original terms. I used the spreadsheet to verify that all three cones' equations are satisfied by these final coordinates, at least to within rounding errors. regards, mathtalk-ga |
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