[THE GREAT ONE]

hey people
can someone refresh my memory on how to trig out tangent points say for instance from .875 dia with 0.04radius off on a 30degree angle.
thank you ERIC

[gar]

060122-1426 EST USA

ERIC:

I think I understand the question in your title, but not your description.

I believe you want to machine along a straight line, then cut an arc, and come out of the arc with a straight line.

Draw the two straight lines so that they intersect at point C. The first line is called AC and the second line is BC. I shall define the points A and B as the other ends of the respective lines.

Assume the angle between AC and BC is less than 180 degree, call it theta, and that your arc will be less than 180 degree. phi will equal 1/2 * theta.

Draw a bisecting line between AC and BC and call this line DC.

Let value R equal the radius of your arc, and let R be the name of one radius of the arc, and R' another radius of the same arc.

For an arc to be tangent to a line its radius must be perpendicular to the line at the tangent point. Thus, R has to be perpendicular to AC and R' to BC.

Let the point of perpendicularity of R to AC be point E. Let the intersection of R with DC be point F.

Now we have a right triangle with legs EC and EF, and hypotenuse FC.

The the trig relations are CF = R / ( sin (1/2 * theta) ), and CE = R / ( tan (1/2 * theta ) ) .

Did I make any mistakes and is this what you want to know?

.

[THE GREAT ONE]

gar
thanks for the reply
now i need to see if your formula works the same
after some help from a friend
radius *cos(angle)=x 1/2 for start diameter
double for finish diameter
sine(radius)=z position

[gar]

060123-1527 EST USA

THE GREAT ONE:

The two equations you presented:

x = R * cos (angle)
z = R * sin (angle)

provide the polar to rectangular transformation of a circle centered at x = 0, and z = 0, in the X - Z plane. As defined here 0 angle lies along the positive X axis. The angle is positive for CCW rotation, assuming X is horizontal and positive to the right, and Z is positive in a CCW direction from positive X.

The inverse transform is:

R = sq-root ( x exp 2 + z exp 2 )
angle = arc tan ( z/x ) and when x = 0 the angle = 90 deg for positive z.

.