Thanks, Michele.
With your earlier post, I noticed the difference in the solution and I was
able to modify the code to work as I needed -- I figured I just didn't
describe it well enough. I had everything working perfectly on Friday.
Thanks again for your assistance.
On Sunday, February 26, 2017 at 12:15:32 PM UTC-6, Michele Zaffalon wrote:
>
> Hi Kevin,
> I mixed up the coordinates and the solution I gave you is wrong. To make
> it up to you, here is the complete (and this time I hope correct) solution.
>
> from sympy import init_session
> init_session()
> y, x0 = symbols("y, x0", real=True)
> # I define d positive, but the equation for the circle is
> # (x-x0)^2 + (y+d)^2 = r^2
> d, m, r = symbols("d, m, r", positive=True)
>
> #method I
> # write explicit equation for circle
> x = x0 - sqrt(r**2 - (y + d)**2)
> # find the point on circle for which slope dx/dy = 1/m
> # is the same as line x = 1/m * y
> y_v = solve(diff(x,y) - 1/m, y)[0]
> # find x coordinate of tangent
> x_v = y_v / m
> # from x = x0 - sqrt(r**2 - (y + d)**2)
> # find center of circle
> x0_v = simplify(x_v + sqrt(r**2 - (y_v + d)**2))
>
> # check
> x = symbols("x", real=True)
> y = -d + sqrt(r**2 - (x - x0_v)**2)
> # the derivative at point x_v should be m
> diff(y, x).subs(x, x_v)
>
>
>
>
> On Thursday, February 23, 2017 at 1:40:21 PM UTC+1, Kevin Pauba wrote:
>>
>> Wow, thanks Michele!
>>
>> Having two solutions will help me to understand sympy even better. I
>> really, really appreciate your replies and I'll be plugging in this code
>> interactively line-by-line along with the docs to see how every argument
>> and line contributes to the solution.
>>
>> On Thursday, February 23, 2017 at 12:49:58 AM UTC-6, Michele Zaffalon
>> wrote:
>>>
>>> Another quicker way is to impose that the circle slope is the same as
>>> the line.
>>> The first expression is the explicit equation for the top part of the
>>> circle. Then I impose that the slope is that of the line and I find the
>>> tangent `x` position.
>>>
>>> y = y0 + sqrt(r**2 - (x-d)**2)
>>> solve(diff(y,x) - m, x)
>>>
>>>
>>> On Thursday, February 23, 2017 at 7:40:17 AM UTC+1, Michele Zaffalon
>>> wrote:
>>>>
>>>> This is how I would do analytically: the equation for the circle is
>>>> (x-d)^2 + (y-y0)^2 = r^2, a circle of radius `r`, centered at `(d,y0)`.
>>>> You
>>>> impose that `y` that belongs to the circle, belongs to the line as well.
>>>> To
>>>> do this, you substitute the equation for the line in that for the circle
>>>> and you obtain `expr` below (y -> mx).
>>>> If you solve this second order equation, you find two solutions because
>>>> in general, the line crosses the circle in two points (or never). You want
>>>> these two points to be coincident because this is an alternative
>>>> definition
>>>> of tangent and you solve for `y0`. There are again two solutions, one with
>>>> the circle to the right of the line, one to the left. You should pick the
>>>> correct one: on my computer, this is the second solution.
>>>> If you now substitute this into any of `sols`, you will find your
>>>> tangent `x` point.
>>>> Does this help?
>>>>
>>>> x, y, d, y0 = symbols("x, y, d, y0", real=True)
>>>> m, r = symbols("m, r", positive=True)
>>>> expr = (x+d)**2 + (m*x-y0)**2 - r**2
>>>> sols = solve(expr, x)
>>>> y0_sols = solve(sols[0] - sols[1], y0)
>>>> simplify(sols[1].subs(y0, y0_sol[1]))
>>>>
>>>>
>>>>
>>>> On Thursday, February 23, 2017 at 6:54:12 AM UTC+1, Kevin Pauba wrote:
>>>>>
>>>>> I'm trying to wrap my brain around sympy and am looking for a little
>>>>> help.
>>>>>
>>>>> Let's say I have a line 'y=m*x' (y-intercept is zero). Let's also say
>>>>> I have a circle of radius 'r' whose center is constrained to be along a
>>>>> horizontal line a specific distance 'd' below the x-axis.
>>>>> Given m, r and d, how might I code sympy to determine the circle's
>>>>> x-value where the given line is tangent to the circle (i'm only
>>>>> interested
>>>>> in the tangent on the left side of the circle). I also would like to
>>>>> determine the (x,y) of the tangent point.
>>>>>
>>>>> Feel free to nudge me in the right direction if you don't have the
>>>>> time to offer a more complete solution.
>>>>>
>>>>> For those interested, my interest is related to the design a cam
>>>>> profile that uses a roller-type follower.
>>>>>
>>>>
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