On 05/03/2011 12:59 PM, Tim Wescott wrote:
Consider two circles, of arbitrary diameter, and a point, all on a plane.

I want to inscribe an arc that is tangent to both circles, and which
passes through the point.

Anyone know a way to construct the arc? I'm not snickering in the
background here as I pose puzzles -- this is a drafting problem that I'm
running into quite a lot lately.

Well, I still haven't figured out a solution for the general problem
(and if there's a way to get QCad to do it I haven't found that either).
But after days of fumbling with math and compasses I found a solution
for two _equal_ sized circles and a point. It's so easy that it made me
feel stupid for a good hour. Kind of like spending a day
troubleshooting a TV for every problem known to man, then finding out
that the reason the power supply appeared to be shot was because it was
unplugged.

So, in spite of my embarrassment, I share the answer he

Inscribe your two circles, call their centers "A" and "B". Define your
point on the arc as "C". Draw line AB between the circle centers.

Now make line CD that's perpendicular to line AB (and passes through C).
Find the point on line CD that's one circle radius inward of point C
-- call this point E.

Now make an arc between points A, B, and E (QCad does have an "arc from
three points" function).

Now make an arc parallel to arc AEB, and larger by one circle radius.
It'll be tangent to the two circles, and pass through point E.

I'm still looking for a way to do this with circles of different
diameter -- hopefully it's almost as easy.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

In article ,
Tim Wescott wrote:

On 05/03/2011 12:59 PM, Tim Wescott wrote:
Consider two circles, of arbitrary diameter, and a point, all on a plane.

I want to inscribe an arc that is tangent to both circles, and which
passes through the point.

I assume that the arc is circular.


Anyone know a way to construct the arc? I'm not snickering in the
background here as I pose puzzles -- this is a drafting problem that I'm
running into quite a lot lately.

It's either easy or it's impossible. Most possible locations of Point X
(the center of the circular arc that's tangent to the two original
circles and passes through Point C) do not have a solution.

Turning this around, using algebra, you can derive the function giving
the space of locations for which a solution exists.

Does the arc need to be circular, or will a Bezier curve do?

Joe Gwinn





Well, I still haven't figured out a solution for the general problem
(and if there's a way to get QCad to do it I haven't found that either).
But after days of fumbling with math and compasses I found a solution
for two _equal_ sized circles and a point. It's so easy that it made me
feel stupid for a good hour. Kind of like spending a day
troubleshooting a TV for every problem known to man, then finding out
that the reason the power supply appeared to be shot was because it was
unplugged.

So, in spite of my embarrassment, I share the answer he

Inscribe your two circles, call their centers "A" and "B". Define your
point on the arc as "C". Draw line AB between the circle centers.

Now make line CD that's perpendicular to line AB (and passes through C).
Find the point on line CD that's one circle radius inward of point C
-- call this point E.

Now make an arc between points A, B, and E (QCad does have an "arc from
three points" function).

Now make an arc parallel to arc AEB, and larger by one circle radius.
It'll be tangent to the two circles, and pass through point E.

I'm still looking for a way to do this with circles of different
diameter -- hopefully it's almost as easy.

On Sun, 08 May 2011 10:28:21 -0700, Tim Wescott wrote:
On 05/03/2011 12:59 PM, Tim Wescott wrote:
Consider two circles, of arbitrary diameter, and a point, all on a
plane.

I want to inscribe an arc that is tangent to both circles, and which
passes through the point.

Anyone know a way to construct the arc? I'm not snickering in the
background here as I pose puzzles -- this is a drafting problem that
I'm running into quite a lot lately.

Well, I still haven't figured out a solution for the general problem
(and if there's a way to get QCad to do it I haven't found that either).
But after days of fumbling with math and compasses I found a solution
for two _equal_ sized circles and a point. [...]
Inscribe your two circles, call their centers "A" and "B". Define your
point on the arc as "C". Draw line AB between the circle centers.

Now make line CD that's perpendicular to line AB (and passes through C).
Find the point on line CD that's one circle radius inward of point C
-- call this point E.

Now make an arc between points A, B, and E (QCad does have an "arc from
three points" function).

Now make an arc parallel to arc AEB, and larger by one circle radius.
It'll be tangent to the two circles, and pass through point E.

Presumably you meant "pass through point C" at the end there. The larger
arc will go through C when point D bisects line AB, but not otherwise.

Note, your problem is the CCP form of Apollonius' Problem, as noted in (eg)
http://en.wikipedia.org/wiki/Special_cases_of_Apollonius%27_problem#Type_7:_Two _circles.2C_one_point_.28CCP.2C_generally_4_soluti ons.29
(About 80% thru http://en.wikipedia.org/wiki/Special_cases_of_Apollonius'_problem)

--
jiw

In article ,
James Waldby wrote:

On Sun, 08 May 2011 10:28:21 -0700, Tim Wescott wrote:
On 05/03/2011 12:59 PM, Tim Wescott wrote:
Consider two circles, of arbitrary diameter, and a point, all on a
plane.

I want to inscribe an arc that is tangent to both circles, and which
passes through the point.

Anyone know a way to construct the arc? I'm not snickering in the
background here as I pose puzzles -- this is a drafting problem that
I'm running into quite a lot lately.

Well, I still haven't figured out a solution for the general problem
(and if there's a way to get QCad to do it I haven't found that either).
But after days of fumbling with math and compasses I found a solution
for two _equal_ sized circles and a point. [...]
Inscribe your two circles, call their centers "A" and "B". Define your
point on the arc as "C". Draw line AB between the circle centers.

Now make line CD that's perpendicular to line AB (and passes through C).
Find the point on line CD that's one circle radius inward of point C
-- call this point E.

Now make an arc between points A, B, and E (QCad does have an "arc from
three points" function).

Now make an arc parallel to arc AEB, and larger by one circle radius.
It'll be tangent to the two circles, and pass through point E.

Presumably you meant "pass through point C" at the end there. The larger
arc will go through C when point D bisects line AB, but not otherwise.

Note, your problem is the CCP form of Apollonius' Problem, as noted in (eg)
http://en.wikipedia.org/wiki/Special...blem#Type_7:_T
wo_circles.2C_one_point_.28CCP.2C_generally_4_solu tions.29
(About 80% thru
http://en.wikipedia.org/wiki/Special_cases_of_Apollonius'_problem)

Cute. It figures that the ancients solved the problem, with only
straightedge and compass. It was a competitive sport back then.

Joe Gwinn

On 05/08/2011 01:01 PM, Joseph Gwinn wrote:
In ,
Tim wrote:

On 05/03/2011 12:59 PM, Tim Wescott wrote:
Consider two circles, of arbitrary diameter, and a point, all on a plane.

I want to inscribe an arc that is tangent to both circles, and which
passes through the point.

I assume that the arc is circular.


Anyone know a way to construct the arc? I'm not snickering in the
background here as I pose puzzles -- this is a drafting problem that I'm
running into quite a lot lately.

It's either easy or it's impossible. Most possible locations of Point X
(the center of the circular arc that's tangent to the two original
circles and passes through Point C) do not have a solution.

Turning this around, using algebra, you can derive the function giving
the space of locations for which a solution exists.

Does the arc need to be circular, or will a Bezier curve do?

It's for a CAD program that handles ellipses badly, and Bezier's even worse.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

On Mon, 09 May 2011 08:05:47 -0400, Joseph Gwinn wrote:
In article ,
James Waldby wrote:
On Sun, 08 May 2011 10:28:21 -0700, Tim Wescott wrote:
On 05/03/2011 12:59 PM, Tim Wescott wrote:
Consider two circles, of arbitrary diameter, and a point, all on a
plane.

I want to inscribe an arc that is tangent to both circles, and which
passes through the point.

Anyone know a way to construct the arc? I'm not snickering in the
background here as I pose puzzles -- this is a drafting problem that
I'm running into quite a lot lately.

Well, I still haven't figured out a solution for the general problem
(and if there's a way to get QCad to do it I haven't found that
either).
But after days of fumbling with math and compasses I found a
solution
for two _equal_ sized circles and a point.
[snip description of not-quite-a-solution]
Now make an arc parallel to arc AEB, and larger by one circle radius.
It'll be tangent to the two circles, and pass through point E.

Presumably you meant "pass through point C" at the end there. The
larger arc will go through C when point D bisects line AB, but not
otherwise.

Note, your problem is the CCP form of Apollonius' Problem, as noted in
[section 7 of]
http://en.wikipedia.org/wiki/Special_cases_of_Apollonius'_problem)

Cute. It figures that the ancients solved the problem, with only
straightedge and compass. It was a competitive sport back then.

I'm not certain which of the 10 combinations of 3 circles / lines /
points problems were solved geometrically in antiquity, but all of
the cases have been solved both algebraically and geometrically in
modern times. I've written a program for two of four solutions for
the CCP case, and it produces output as shown below when given input
shown below. Note, each input line contains Circle A [center &
radius], Circle B, Point C, and inner/outer circle flag (+1 or -1
to select outer or inner solution). In output, circle i gives the
center and radius of arc through C, tangent to A,B.
--------input----------
0 0 1 4 0 1 1 2 +1
0 0 1 6 6 1 2 4 +1
0 0 1 6 6 1 2 4 -1
1 2 3 11 5 5 6 -4 +1
1 2 3 11 5 5 6 -4 -1
--------output----------
a: 0.000000 0.000000 1.000000
b: 4.000000 0.000000 1.000000
c: 1.000000 2.000000
i: 2.000000 -1.154701 3.309401

a: 0.000000 0.000000 1.000000
b: 6.000000 6.000000 1.000000
c: 2.000000 4.000000
i: 18.217514 -12.217514 22.935029

a: 0.000000 0.000000 1.000000
b: 6.000000 6.000000 1.000000
c: 2.000000 4.000000
i: 4.782486 1.217514 3.935029

a: 1.000000 2.000000 3.000000
b: 11.000000 5.000000 5.000000
c: 6.000000 -4.000000
i: 6.501130 5.515071 9.528259

a: 1.000000 2.000000 3.000000
b: 11.000000 5.000000 5.000000
c: 6.000000 -4.000000
i: 6.004348 -1.108697 2.891306

ATM (traveling) I don't have a CAD program installed to
view those to see if they look ok

--
jiw