Tangent Angles

[David Bethel]

In plain Autolisp, I'm trying to find the angle A1 to form the tangent Lline T1 T2 ( of the green ARCs )

 

Known Values:

X Y R1

Max value for R1 is (min (* X 0.5) (* Y 0.5))

 

Simple Calculations:

P1 = (list 0 0) Preset

P2 = (list X Y)

C1 = (list R1 0)

C2 = (list (- X R1) Y)

 

Autocad's Fillet command moves C1 & C2

The fillet command would also crap out on full width 90 degree corners

Snapping tanget lines via autolisp is problematic at best

 

I would have thought after all these years that I had something like this and had seen a snippet.

 

The result is three entities (2) arcs (1) line ( well it's a beginning of an end result sort of )

 

Any ideas? TIA -David

[Hippe013]

I am just trying some thought here... If a line (lets call M1) was drawn from the known points c1 & c2 it's midpoint is the midpoint of your tangent line. Now you are left with two imaginary triangles. Using the sin law... 90 degrees is to half the distance of M1 as some angle (ang1) is to r1. Since you know the angle in the plane of M1 you can then add your angle (ang1) to that and now you have the angle in the plane of your tangent line. Your tangent line runs through the midpoint of M1. Now knowing the angle of your tangent line you then know the angle of C2 -> T2 and C1 -> T1. Does this make sense? If i had more time I would post an image of what I am trying to explain. Hope that this helps!

 

regards,

 

Hippe013

[CALCAD]

Hi Dave, now let's see if I can upload this

 

Hope this helps (hope this is correct!)

[Hippe013]

Always seeking the ever solvable triangle.

[David Bethel]

I had thought about the midpoint E but it didn't dawn on me to use calcad's C2-F as the right angle.

 

E = (list (* X 0.5) (* Y 0.5))

 

 

HMmmmmmmmmmmmmmmm........

 

Now that I think about it, how do you find point F

 

EF = (sqrt (- (* E E) (* R1 R1)))

 

That can give me A3

 

(angle EF C2 @0degree ) can make A2

 

OK I think I got it

 

Thanks! -David