Tricky Math Problem
Tricky Math Problem
(OP)
Ok, looking for a little help with an analysis I'm doing for work. I can iterate to solve individually but my challenge here is that I'm trying to do it for a large population of random points that have been generated as part of a tolerance sensitivity analysis done in excel on a linkage mechanism.
Per the drawing attached I have coordinates relative to Origin 0 for points B and D and based on that everything else is known. What I would like to do is determine the resultant angle phi if I rotate O-B about point A such that the two arcs become tangent. I know that (Cx-Ex)^2+(Cy-Ey)^2=(2*r)^2 for tangency and I know my B1x^2+B1y^2=B2x^2+B2y^2. But is it possible to have excel solve? Tried using circular reference and turning on iteration but didn't appear to function like I was hoping.
I'm expecting I need something like a matlab to do this but I would like to do it for a large set of points so I can determine probability of manufacturing rejects based on that deflection. I'm hoping there is something obvious I'm missing, but my head is spinning at this point.
Thanks for any help.
Per the drawing attached I have coordinates relative to Origin 0 for points B and D and based on that everything else is known. What I would like to do is determine the resultant angle phi if I rotate O-B about point A such that the two arcs become tangent. I know that (Cx-Ex)^2+(Cy-Ey)^2=(2*r)^2 for tangency and I know my B1x^2+B1y^2=B2x^2+B2y^2. But is it possible to have excel solve? Tried using circular reference and turning on iteration but didn't appear to function like I was hoping.
I'm expecting I need something like a matlab to do this but I would like to do it for a large set of points so I can determine probability of manufacturing rejects based on that deflection. I'm hoping there is something obvious I'm missing, but my head is spinning at this point.
Thanks for any help.





RE: Tricky Math Problem
Alternatively, you could have a CAD jockey sketch the solution and get the point coordinates.
RE: Tricky Math Problem
one arc is (x-Ex)^2+(y-Ey)^2 = rE^2
the other is (x-Cx)^2+(y-Cy)^2 = rC^2
and (Cx-Ax)^2+(Cy-Ay)^2 = rA^2 (OB is rotating about A, C is fixed to OB)
but you know that C-E is the sum of the radii, rC + rE; the tangency is on a line joining the centers ...
i think i'd draw it in autocad
Quando Omni Flunkus Moritati
RE: Tricky Math Problem
RE: Tricky Math Problem
RE: Tricky Math Problem
Excel's "Solver" add-in can be used to solve non-linear simultaneous equations. Here's an approach for the case with three unknowns, readily extendable to larger or smaller problems.
» Convert your equations to the form
F(x,y,z) = 0
G(x,y,z) = 0
H(x,y,z) = 0
» Guesstimate a set of starting values for x, y & z. Place each of these guesstimates in a separate cell. (How close these need to be depends upon how non-linear your problem is. The closer the better.)
» Based on these starting values calculate the values of F, G & H in a separate cell for each. If you guesstimated correctly (which of course you didn't) the three cells should each be exactly zero.
» In a fourth cell calculate an "objective function" F²+G²+H².
» Fire up the Solver, and ask it to minimise the objective function by varying the (x,y,z) values.
» Bingo!
RE: Tricky Math Problem
-calculate the actual distance CE
-determine the change of this distance by the rotation: Δ=2r-CE
-the angle is Δ/CA
Otherwise (as in Occupant's sketch) you need to go through determining the initial and final angles of CE and CA.
prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads
RE: Tricky Math Problem
It can show how D and E rotate using CAD system.
RE: Tricky Math Problem
As I mentioned finding an individual solution is fairly easy, my challenge is that I have a large set of points I'm looking for solutions to use for statistical purposes.
Denial- I'll look into the solver, I'll have to get IT to install it for me as they seem wary of anyone installing anything on their own.
Thanks everyone.
RE: Tricky Math Problem
E rotates around "F", the other end of the line through D, yes?
for various angles phi, you want to calc the co-ord of E ?
you say "everything else is known" ... does that mean lengths OA (not that it matters), AB, BC, DF, DE, radius C, radius E ?
the co-ords of A and F ?
maybe solve as three known lengths AC,CE,EF = AF, the two variables are phi and the angle between AC and CE
now you have only two equations (x- and y- dim'ns) amd two unknowns.
Quando Omni Flunkus Moritati
RE: Tricky Math Problem
Make a triangle from A to B to D and back to A.
Presuming you know distance A-B.
Distance B to D at the point of tangency is the sum of the two radii.
Distance A to D is taken from the pythagorean theorem, assuming you know where Point D is relative to Point A.
You then know three sides and none of the angles of that triangle.
Using the Law of Cosines solves directly for Angle B-A-D.
The angle from horizontal down to D is the inverse tangent of (yA-YD)/(xD-XA)
And angle phi is the difference between those two angles.
RE: Tricky Math Problem
http://newtonexcelbach.wordpress.com/2009/07/25/us...
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/
RE: Tricky Math Problem
Calculate:
d=length of AC
δ=2r-length of CE (this is the elongation of CE)
sin β=sine of the angle of AC to CE (this is simple geometry: example)
Now the displacement of C is Δ=δ/sin β and the angle sought φ=Δ/d
This is still an approximation: it doesn't require anymore, like my previous one, β to be close to 90 deg, but still requires φ to be small (a few degrees).
prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads
RE: Tricky Math Problem
I'm assuming you know or can find X and Y coordinates for A and D and know the radius of both circles and the distance from A to B.
In that case, you know distance A-B and find distance A-D from the Pythagorean theorem.
Distance B to D is the sum of the two radii.
You then know 3 sides of the triangle A-B-D and find the angle B-A-D directly from the law of cosines, no iteration involved.
You find angle gamma from inverse tangent function, subtract that and angle B-A-D from 90 degrees to get phi.
Maybe I'm overlooking something that makes this harder and requires iteration, solvers, etc., but it looks pretty straightforward to me.
RE: Tricky Math Problem