circle + ray
green_thumb_wannabe
with point (Rx, Ry) and angle Ra where (Rx,Ry) is on (not in) circle
C, and a positive radius Er. The circle is directional with Cr less
than 0 for clockwise. Find a circle E tangent to C and R with center
(Ex, Ey) with a radius matching that of the Er input. The resulting
circle E should flow from circle C to ray R. Solving for the tangent
point on the ray rather than the center of E is sufficient.
Narasimham
wrote:
It may help to solve for coincident roots ( equate quantity under
radical to zero). Or use three osculating circles result, where one
radius is infinity.
Philippe 92
It may surely help if it were clearer !
What is the radius of a ray ???
What is the sign of the radius to be clockwise ???
etc...
I didn't understand a word in that problem...
Regards.
--
Philippe C., mail : chephi...@free.fr
site : http://chephip.free.fr/ (recreational mathematics)
not_a_commie
The gist of the problem is this: how do I connect an intersecting arc-
line pair smoothly by the insertion of a second arc?
> What is the radius of a ray ???
The problem mentions no "radius of a ray". Er is the radius of the
circle tangent to the given ray and circle.
> What is the sign of the radius to be clockwise ???
A negative radius implies a clockwise circle and a positive radius
implies a counter-clockwise circle. The ray is directional by
definition.
Philippe 92
>> It may surely help if it were clearer !
> No doubt.
> line pair smoothly by the insertion of a second arc?
>
>> What is the radius of a ray ???
> The problem mentions no "radius of a ray". Er is the radius of the
> circle tangent to the given ray and circle.
OK, I misunderstood the 1st sentence...
>> What is the sign of the radius to be clockwise ???
> A negative radius implies a clockwise circle and a positive radius
> implies a counter-clockwise circle. The ray is directional by
> definition.
OK, its your convention.
If I understand correctly :
http://cjoint.com/?ewvMROItEZ
However the definition of 'angle Ra' is may be else, but
without changing a lot of things.
It seems also easier to get E rather than contact points :
E is intersect of a circle with radius Rc - Re (considering your
sign conventions) with a translated Ray in suitable direction.
A few trigonometry gives the result.
not_a_commie
correct, including the definition of Ra. A few questions: How do you
know which direction to translate the ray? How do you handle/detect
the degenerate case where the point E is inside the circle C? Or, as
is my hope, is it possible to ignore that case and have it fall out
correctly from the calculations of the external circle?
Philippe 92
> Thank you for looking at this problem. That picture is exactly
> correct, including the definition of Ra. A few questions:
> How do you know which direction to translate the ray?
see below
> How do you handle/detect the degenerate case where the point E is
> inside the circle C?
this case just can't happen : E is *allways* outside (C)
> Or, as is my hope, is it possible to ignore that case and have it
> fall out correctly from the calculations of the external circle?
So, how to know "which direction to translate the ray" and how
"correct case fall out from calculation".
I will discard the convention for "negative radius" and just
keep the problem simpler as :
Given a circle (C), centered in C with radius rC,
A point R on this circle and a ray (R) from R with angle Ra
with horizontal.
To connect the circle and the ray by a circle (E) with given
radius rE so that the path can be walked continuously starting
from (C) anticlockwise, then (E) clockwise, then on the ray,
in direction of the ray.
Without the orientation and ray constraints, the problem is
easy, and has at most 8 solutions.
With the given orientation, there is only one.
The problem is then to find (construct, calculate) directly
this correct solution, without the need to "choose" among the 8.
Consider what happens at contact point between (C) and (E).
We change direction from clockwise to anticlockwise, this just
implies that the circles touch externally.
Hence center E lies on a circle (C'), centered in C with radius
|rC| + |rE|.
(hence E is allways outside circle (C), as |rC| + |rE| > |rC|)
Now, at contact point between (E) and ray (R), anticlockwise
to ray direction just means that circle (E) lies "on the left"
of the ray, looking from R in direction of the ray.
Center E lies then on a parallel ray (R'), with the same
direction as (R), and starting at a point K with |RK| = |rE|,
and oriented angle ((R), RK) = +90°.
As K is inside circle (C'), except when the ray is tangent (in R)
to (C), this implies there is allways one and only solution :
Ray (R') allways intersects (C') in just one point E.
Now if you want calculations ...
Define K as :
x_K = x_R + rE*cos(Ra + 90°),
y_K = y_R + rE*sin(Ra + 90°)
Then the ray from K can be parametric defined as :
x = x_K + t*cos(Ra)
y = y_K + t*sin(Ra)
t > 0 (just ray and not line)
Intersecting this line (solving the quadratic in t)
with circle x^2 + y^2 = (|rC| + |rE|)^2 gives two solutions,
but the only one we want is for t > 0 (on the *ray* from K)
so in the quadratic, we choose the + sqrt.
Details let to reader...
I've put this on my (new) web site at
http://mathafou.free.fr/pbw_en/pb401.html
with a dynamic construction using JavaSketchpad (so patience with Java
on first load).
Don't try to navigate far in this web site : "design in progress".
(there are up to now only three topics !)
filmon mulugheta
10 degree
Dan in NY
I'm not sure what you mean but here are two relations.
sin(10) = tan(10) / Sqrt[1 +tan(10)*tan(10)]
tan(10) = sin(10) / Sqrt[1 -sin(10)*sin(10)]
Does this answer your question?
--
Dan in NY
(for email, exchange y with g in
dKlinkenbery at hvc dot rr dot com)