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
<[email protected]> 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
[email protected] 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
[email protected] 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
<[email protected]> 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
<[email protected]> 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
[email protected] 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? Any help is greatly appreciated.
>>>> >> >
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