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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