Hello Alen, so nice of you to reply. I have already worked out the solution using sin, cos and tangent. I also used sympy to help solve the simultaneous equations and the solution was succinct. I used rational trig as it seemed interesting and I had thought that sympy might help to solve the problem in an even simpler form. As we have seen, that's not the case.
I'm now investigating the use of the parametric circle equation https://mathnow.wordpress.com/2009/11/06/a-rational-parameterization-of-the-unit-circle/. It seems that this model fits the problem space more closely. Again, thanks for the feedback and feel free to share anything else that comes to mind. On Mon, Aug 8, 2022 at 12:53 PM Alan Bromborsky <abrombo...@gmail.com> wrote: > It is not hard to write a closed form solution in terms of sine, cosine > and the lengths of Q1, Q2, and the distance of the pivot point from the > origin assuming the initial rotation angle is zero (makes it simpler). Is > there any reason to object to that solution letting sympy do the messy > algebra. I just takes a little vector algebra to set up the problem. > On 8/6/22 10:37 PM, Kevin Pauba wrote: > > Also, oscar, none of the radicals in the formulas are due to my getting > the lengths. All q* terms are quadrances ... no lengths are used in the > formulas. I should have said quadrance of DP (the square of the length of > the segment DP) instead of just segment DP. Once I have the values for q1, > q2, q_dp and s, I'll convert them to length and angles for human > consumption. > > On Saturday, August 6, 2022 at 2:26:36 PM UTC-7 Kevin Pauba wrote: > >> Oscar, I think I misstated something about rational trig that reinforces >> your statement that it avoids square roots. Maybe there is some >> manipulation of the equations (similar in spirit to using sympy.unrad()) >> that would keep them in a form that doesn't require the square roots. It >> would seem that sqrt would only be needed in the very last step when >> evaluating the quadrances when I want a final distance measure once solved >> ( d = sqrt(q1), for instance). >> >> On Saturday, August 6, 2022 at 10:35:38 AM UTC-7 Kevin Pauba wrote: >> >>> Rational trigonometry totally eschews the use of transcendental >>> functions like sine and cosine. The "Triple Spread Formula" relating the >>> speads to the quadrances, for example, is quadratic (as is the "Triple Quad >>> Formula") -- the use of square roots is unavoidable. >>> >>> After working the same problem using traditional circular functions >>> (sine & cosine), I was pleased to see a much simpler solution (that took >>> relatively little processing time). I am intrigued with rational >>> trigonometry but a bit disappointed that a solution is more elusive. >>> >>> On Saturday, August 6, 2022 at 9:14:19 AM UTC-7 Oscar wrote: >>> >>>> The equations you are attempting to solve lead to a very complicated >>>> Groebner basis that is slow to compute (I'm not sure how long it >>>> takes) and probably gives quite a complicated expression for the >>>> solution. >>>> >>>> It might be better if you can derive the equations without introducing >>>> radicals i.e. to work with the quadrances rather than square rooting >>>> them to get lengths. I'm not sure how you derived this but I imagine >>>> the intention of rational geometry is to avoid things like square >>>> roots. >>>> >>>> On Sat, 6 Aug 2022 at 16:22, Kevin Pauba <klp...@gmail.com> wrote: >>>> > >>>> > That might be another approach but this represents an applied >>>> geometry problem (mechanics). The segment DP (along with s) is an input >>>> measure that is easy to determine/specify. >>>> > >>>> > On Saturday, August 6, 2022 at 4:45:13 AM UTC-7 smi...@gmail.com >>>> wrote: >>>> >> >>>> >> >>>> >> Kevin, I took a look at some of the rational geometry links -- it >>>> makes me wonder if you might approach the relationship between Q1 and Q2 >>>> and the area of the triangle extending past the line instead of simply the >>>> length DP. >>>> >> >>>> >> /c >>>> >> On Friday, August 5, 2022 at 9:11:39 PM UTC-5 klp...@gmail.com >>>> wrote: >>>> >>> >>>> >>> Hi Oscar, >>>> >>> >>>> >>> That very well may be (as you've noticed, I'm not well versed in >>>> math as I once was). This is an applied geometry problem (see attached >>>> diagram) where the triangle rotates about point A. I am solving for the >>>> relationships between q1 and q2, the perpendicular distance from P to the >>>> base of the triangle (segment DP) and the angular measure s using rational >>>> trigonometry (see https://stijnoomes.com/laws-of-rational-trigonometry/). >>>> I have a solution using sin, cos but I'm interested in this alternative >>>> method. >>>> >>> >>>> >>> equ1 in my working example solves for the segment DP given q1, q2 >>>> and s (and angular measure). The problem presented here is to determine q1 >>>> and q2 given two known segments and angular measures (q_dp1, s1) and >>>> (q_dp2, s2). >>>> >>> >>>> >>> I hope this makes sense and that the information might help you >>>> help me. >>>> >>> >>>> >>> Thanks for taking the time looking into it! >>>> >>> >>>> >>> On Friday, August 5, 2022 at 4:50:09 PM UTC-7 Oscar wrote: >>>> >>>> >>>> >>>> I just had a quick look and I think that maybe this has a positive >>>> >>>> dimensional solution set. >>>> >>>> >>>> >>>> On Fri, 5 Aug 2022 at 16:08, Kevin Pauba <klp...@gmail.com> >>>> wrote: >>>> >>>> > >>>> >>>> > Here's the minimal working example (except for it hanging on >>>> solve()): >>>> >>>> > >>>> >>>> > import sympy as sym >>>> >>>> > from sympy import sqrt >>>> >>>> > >>>> >>>> > q1, q2, s, s1, s2, q_dp1, q_dp2 = sym.symbols('Q_1, Q_2, s, s_1, >>>> s_2, Q_dp_1, Q_dp_2') >>>> >>>> > >>>> >>>> > equ1 = q1*s - q2*s + 2*q2 - 2*sqrt(q2*(q1*s - q2*s + q2 + >>>> 2*sqrt(q1*q2*s*(1 - s)))) + 2*sqrt(q1*q2*s*(1 - s)) >>>> >>>> > >>>> >>>> > eq1 = equ1.subs({ s: s1 }) - q_dp1 >>>> >>>> > print(f"{eq1} = 0") >>>> >>>> > >>>> >>>> > eq2 = equ1.subs({ s: s2 }) - q_dp2 >>>> >>>> > print(f"{eq2} = 0") >>>> >>>> > >>>> >>>> > soln = sym.solve([eq1, eq2], (q1, q2), simplify=False) >>>> >>>> > print(f"soln = {soln}") >>>> >>>> > >>>> >>>> > On Friday, August 5, 2022 at 7:47:07 AM UTC-7 Oscar wrote: >>>> >>>> >> >>>> >>>> >> You should be able to obtain a parametric Groebner basis to >>>> represent >>>> >>>> >> the solutions of this system. Whether that leads to an explicit >>>> >>>> >> solution in radicals is hard to say without trying. >>>> >>>> >> >>>> >>>> >> I would demonstrate how to do this but the code for putting >>>> together >>>> >>>> >> the equations is incomplete. >>>> >>>> >> >>>> >>>> >> On Fri, 5 Aug 2022 at 14:01, Chris Smith <smi...@gmail.com> >>>> wrote: >>>> >>>> >> > >>>> >>>> >> > If you remove the radicals >>>> (`sympy.solvers.solvers.unrad(eq1)`) and replace `Q1` and `Q2` with `x` and >>>> `y` and `Q_dp_1` with a and `s1` with `b` you will get an expression that >>>> is of degree 4 in every variable and can be split into a term with `a` and >>>> `b` and a term with only `b` -- both with `x` and `y`. >>>> >>>> >> > >>>> >>>> >> > u1 = a*(a**3 - 4*a**2*y*(2 - b) - 2*a*y**2*(-3*b**2 + 8*b - >>>> 8) - 4*b**3*x**3 + 2*b**2*(-x**2*(-3*a + 2*y*(b - 2)) + 2*y**3*(b - 2)) - >>>> 4*b*x*(a**2 + a*y*(b - 2) + y**2*(-b**2 - 8*b + 8))) + \ >>>> >>>> >> > b**2*(b*(x**2 + y**2) + 2*x*y*(b - 2))**2 >>>> >>>> >> > >>>> >>>> >> > Replace a,b with c,d (for `q_dp_2` and `s2`) to get `u2`. I >>>> can't imagine that solving a pair of quartics is going to give a nice >>>> solution. But solving this system with known values for `a` and `b` would >>>> be straightforward with `nsolve`. >>>> >>>> >> > >>>> >>>> >> > /c >>>> >>>> >> > >>>> >>>> >> > On Thursday, August 4, 2022 at 11:54:24 PM UTC-5 >>>> klp...@gmail.com wrote: >>>> >>>> >> >> >>>> >>>> >> >> And eq1=Q_1*s_1 - Q_2*s_1 + 2*Q_2 - Q_dp_1 - >>>> 2*sqrt(Q_2*(Q_1*s_1 - Q_2*s_1 + Q_2 + 2*sqrt(Q_1*Q_2*s_1*(1 - s_1)))) + >>>> 2*sqrt(Q_1*Q_2*s_1*(1 - s_1)) ... sorry, long day! >>>> >>>> >> >> >>>> >>>> >> >> On Thursday, August 4, 2022 at 9:39:53 PM UTC-7 Kevin Pauba >>>> wrote: >>>> >>>> >> >>> >>>> >>>> >> >>> Sorry, Jeremy. Good suggestion! >>>> >>>> >> >>> >>>> >>>> >> >>> s1, s2, q_dp1, q_dp2 = sym.symbols('s_1, s_2, Q_dp_1, >>>> Q_dp_2') >>>> >>>> >> >>> eq1 = equ1.subs({ s: s1 }) - q_dp1 >>>> >>>> >> >>> md( "$" + sym.latex(eq1) + " = 0$\n" ) >>>> >>>> >> >>> >>>> >>>> >> >>> eq2 = equ1.subs({ s: s2 }) - q_dp2 >>>> >>>> >> >>> md( "$" + sym.latex(eq2) + " = 0$\n" ) >>>> >>>> >> >>> >>>> >>>> >> >>> soln = sym.solve([eq1, eq2], q1, q2) >>>> >>>> >> >>> print(f"soln = {soln}") >>>> >>>> >> >>> >>>> >>>> >> >>> I'll set simplify to False and see how it goes ... >>>> >>>> >> >>> On Thursday, August 4, 2022 at 8:18:00 PM UTC-7 Kevin Pauba >>>> wrote: >>>> >>>> >> >>>> >>>> >>>> >> >>>> I've attached a portion of a jupyter notebook. I'm >>>> attempting to solve a simultaneous equation using sympy. The sym.solve() in >>>> the green input box doesn't return (well, I waited over night on my macbook >>>> pro). Might the solution be intractable? Is there another way to get a >>>> solution? 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