Simpler formula for central angle?

jeff hammond

so i hacked this formula together but i'm feeling like it's too convoluted.. it works fine but is there a simpler way to find a central angle (circle) when the input is radius(r) & chord length(c)?

2*(ACOS((SQRT((r*r)-(.25*(c*c))))/r))

that's a combination of this:

plus this for the d:

sqrt(radius*radius-(chord*chord*.25))

i feel like i should be able to use one of these (used for finding chord length):

except i can't seem to get the chord and theta on opposite sides of the = and have it work out right

(also,is there a better way to type r^2 [instead of (r*r)] that will work in a dc formula?

thanks

TIG

Use the 'Sine Rule':

diameter = chord_length / sin(center_angle_subtended_to_chord_ends)

and of course...

radius = diameter / 2

or

radius = chord_length / (2 * sin(center_angle_subtended_to_chord_ends))

rearranging formula...

chord_length = diameter * sin(center_angle_subtended_to_chord_ends)

or

chord_length = radius * 2 * sin(center_angle_subtended_to_chord_ends)

and to find the angle...

center_angle_subtended_to_chord_ends = asin(chord_length / diameter)

or

center_angle_subtended_to_chord_ends = asin(chord_length / (radius * 2))

Using your variable naming

angle=asin(c/(r*2))

jeff hammond

thanks Tig

(though i think that's for 1/2 the central angle.. but definitely what i was after.)

fwiw, here's a rough version of what i'm using it for..
it draws an arc with 3 different inputs -- radius, segment length, and # of segments

select the component then use the component options dialog

i don't necessarily see an immediate use for this because the arc segments are buried so deep.. i think when it i try to make a useful dc (such as that fillet example i posted the other day -- which i'm thinking is possible now), i'll just have it make guide points and i'll draw the arcs/lines over them..

TIG

It is the angle between the 2 radii at the 2 ends of the chord...

jeff hammond

@tig said:

It is the angle between the 2 radii at the 2 ends of the chord...

Right. After reading your post a few times, I realized that I was looking at the whole thing wrong (way over thinking it )
Your post made me realize there's a right triangle (or 2) in there. Right triangles I can deal with ok.