直线与圆弧间拐角倒圆算法求解：给定参数与输出要求

Alright, let's walk through how to implement this fillet (corner rounding) algorithm step by step. This is a typical problem in CAD, CNC path planning, or geometric computing, so I'll break it down into actionable, easy-to-follow steps.

First, let's formalize all given inputs to avoid confusion:

• Points: Start point P₀, corner point P, end point P₂
• Radii: Radius R₂ of the arc P-P₂, fillet radius R (the radius of the rounding arc we need to create)
• Target outputs: Tangent point C₀ (between line P₀-P and fillet arc), tangent point C₂ (between original arc P-P₂ and fillet arc), fillet arc center O

Before we can find the fillet center, we need to get O₂—the center of the arc P-P₂:

1. Compute vector PP₂ = P₂ - P (subtract corresponding x/y coordinates)
2. Normalize PP₂ to get unit vector u = PP₂ / |PP₂| (where |PP₂| is the length of PP₂)
3. Find the two perpendicular unit vectors to u: (-u.y, u.x) and (u.y, -u.x)
4. Choose the correct perpendicular direction based on the arc's winding (clockwise or counter-clockwise; this should be known from your input context)
5. Compute O₂ = P + R₂ * correct_perpendicular_vector

The fillet center O(x,y) must satisfy two key constraints:

Constraint 1: Distance from O to line P₀-P equals R #

First, write the line P₀-P in general form Ax + By + C = 0:

• A = P.y - P₀.y
• B = P₀.x - P.x
• C = P.x * P₀.y - P₀.x * P.y

The distance from O(x,y) to this line is exactly R. Use the distance formula, and drop the absolute value by choosing the correct sign (based on which side of the line the fillet should be on—this is the "outside" of the corner formed by P₀-P and arc P-P₂):

(Ax + By + C) / sqrt(A² + B²) = R

Constraint 2: Distance from O to O₂ equals R₂ + R #

Since the fillet arc is externally tangent to the original arc P-P₂, the distance between their centers is the sum of their radii. This gives us a circle equation:

(x - O₂.x)² + (y - O₂.y)² = (R₂ + R)²

Solve the system of equations #

Now you have a system of one linear equation and one quadratic equation. Solve for (x,y)—you'll get two possible solutions. Pick the one that lies on the correct side of the corner (the side where you want the fillet to be).

The tangent point C₀ is the foot of the perpendicular from O to the line P₀-P:

1. Compute the unit normal vector of line P₀-P that points towards the line (this is the opposite direction of the sign we used in Step 2.1)
2. C₀ = O - R * unit_normal_vector
3. Verify that C₀ lies on the line segment (or extended line) P₀-P—it should be positioned between P₀ and P (or on the extension towards the corner)

For two tangent circles, the tangent point lies on the line connecting their centers. So:

1. Compute vector O₂O = O - O₂
2. Normalize O₂O to get unit vector w = O₂O / |O₂O|
3. C₂ = O₂ + R₂ * w (this points from O₂ to the tangent point on the original arc)
   • Alternatively, C₂ = O - R * w will give the same point, since O₂O has length R₂ + R

• Ensure O is on the correct side of the corner (not inside the angle formed by P₀-P and the arc P-P₂)
• Confirm C₀ is on the line P₀-P and C₂ is on the arc P-P₂
• Handle edge cases: If R is too large (larger than the maximum possible fillet radius for the corner), the system of equations will have no real solutions—add error handling for this scenario

内容的提问来源于stack exchange，提问作者UsrDS